Introduction to
Chemical Processes
Principles, Analysis, Synthesis
Second Edition
Regina M. Murphy
University of Wisconsin, Madison
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INTRODUCTION TO CHEMICAL PROCESSES: PRINCIPLES, ANALYSIS, SYNTHESIS,
SECOND EDITION
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Library of Congress Cataloging-in-Publication Data
Names: Murphy, Regina M., author.
Title: Introduction to chemical processes : principles, analysis, synthesis
/ Regina M. Murphy, University of Wisconsin, Madison.
Description: Second edition. | New York, NY : McGraw Hill Education, [2022]
| Includes index.
Identifiers: LCCN 2021047252 (print) | LCCN 2021047253 (ebook) | ISBN
9781259883972 | ISBN 9781260791372 (spiral bound) | ISBN 9781264474134
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mheducation.com/highered
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Dedication
To my wonderful family
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Contents
Preface
List of Nomenclature (Typical Units)
List of Important Equations
CHAPTER 1
1.1
1.2
1.3
Converting the Earth’s Resources into
Useful Products
Introduction
Raw Materials
Balanced Chemical Reaction Equations
Example 1.1
1.3.1
1.4
Balanced Chemical Reaction Equation:
Nitric Acid Synthesis
Example 1.2
Balanced Chemical Reaction Equations:
Adipic Acid Synthesis
Using Matrices to Balance Chemical Reactions
Example 1.3
Balancing Chemical Equations with
Matrix Math: Adipic Acid Synthesis
Generation-Consumption Analysis
Example 1.4
1.4.1
1.5
Generation-Consumption Analysis:
Ammonia Synthesis
Example 1.5
Generation-Consumption Analysis:
The Solvay Process
Using Matrices in Generation-Consumption Analysis
Example 1.6
Generation-Consumption Analysis
Using Matrix Math: Nitric Acid Synthesis
xvi
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1
2
3
5
7
8
10
12
12
15
16
19
20
A First Look at Material Balances and Process Economics
22
1.5.1
1.5.2
23
25
25
1.5.3
1.5.4
Mass, Moles, and Molar Mass
Atom Economy
Example 1.7
Atom Economy: LeBlanc versus Solvay
Example 1.8
Atom Economy: Improved Synthesis
of 4-ADPA
Process Economy
Example 1.9
Process Economy: The Solvay Process
Process Capacities and Product Values
Case Study: Six-Carbon Chemistry
Summary
ChemiStory: Changing Salt into Soap
26
29
30
31
32
41
42
iv
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Contents
Quick Quiz Answers
References and Recommended Readings
Chapter 1 Problems
CHAPTER 2
Process Flows: Variables, Diagrams,
Balances
44
44
45
61
2.1
2.2
Introduction
Process Variables
62
63
2.3
Chemical Process Flow Sheets
72
2.4
Process Flow Calculations
81
2.4.6
87
88
2.5
2.2.1
2.2.2
2.2.3
2.2.4
2.2.5
2.3.1
2.3.2
2.3.3
2.3.4
2.4.1
2.4.2
2.4.3
2.4.4
2.4.5
A Brief Review of Dimensions and Units
Mass, Moles, and Composition
Temperature and Pressure
Volume, Density, and Concentration
Flowrates
Input-Output Flow Diagrams
Block Flow Diagrams
Process Flow Diagrams (PFD)
Modes of Process Operation
Systems, Streams, and Specifications
Material Balance Equation
Components
Generation, Consumption, Accumulation
A Systematic Procedure for Process Flow
Calculations
Helpful Hints for Process Flow Calculations
A Plethora of Problems
Example
Example
Example
Example
Example
2.1
2.2
2.3
2.4
2.5
Example 2.6
2.6
Mixers: Battery Acid Production
Reactors: Ammonia Synthesis
Separators: Fruit Juice Concentration
Splitter: Fruit Juice Processing
Separation with Accumulation:
Air Drying
Reaction with Accumulation: Light
from a Chip
Process Flow Calculations with Multiple Process Units
Example 2.7
2.6.1
2.6.2
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Multiple Process Units: Toxin
Accumulation
Example 2.8
Multiple Process Units: Soap
Manufacture
Synthesizing Block Flow Diagrams
The Art of Approximating
63
65
67
69
71
73
74
75
78
81
83
85
85
89
90
92
95
97
100
103
105
106
108
112
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vi
Contents
Case Study: Biological Routes to Specialty Chemicals
Summary
ChemiStory: Guano and the Guns of August
Quick Quiz Answers
References and Recommended Readings
Chapter 2 Problems
CHAPTER 3
3.1
3.2
Mathematical Analysis of Material Balance
Equations and Process Flow Sheets
Introduction
The Material Balance Equation—Again
3.2.1
3.2.2
3.2.3
3.2.4
3.2.5
3.3
Linear Models of Process Flow Sheets
3.3.1
3.3.2
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Stream Variables
System Variables
The Differential Material Balance Equation: Molar Units
Example 3.1
Decomposition Reactions
Example 3.2
Differential Mole Balances: Manufacture
of Urea
Example 3.3
Differential Mole Balances: Urea
Manufacture from Cheaper Reactants
Example 3.4
Total Mole Differential Balance: Urea
Manufacture from Cheaper Reactants
The Differential Material Balance Equation: Mass Units
Example 3.5
Differential Mass Balance: Sugar
Dissolution
Example 3.6
Differential Mass Balance: Glucose
Consumption in a Fermentor
The Integral Material Balance Equation
Example 3.7
Integral Equation: Blending and Shipping
Example 3.8
Integral Equation with Unsteady Flow:
Jammin’ with Cherries
System Performance Specifications and Linear
Models of Process Units
Example 3.9
Linear Models of Mixers: Sweet Mix
Example 3.10 Linear Model of a Splitter: Sweet Split
Example 3.11 Linear Model of a Reactor: GlucoseFructose Isomerization
Example 3.12 Linear Model of Separators: Sweet
Solutions
Process Topology
Example 3.13 Multiple Process Units and Recycle:
Taking an old Plant out of Mothballs
115
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121
125
125
126
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159
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Contents
3.4
Degree of Freedom Analysis
3.4.1
3.4.2
194
Degree of Freedom Analysis for Single Process Units
Example 3.14 DOF Analysis: Fruit Juice Processing
Example 3.15 DOF Analysis: Air Drying
Example 3.16 DOF Analysis: Ammonia Synthesis
Example 3.17 DOF Analysis: Battery Acid Production
Degree of Freedom Analysis for Block Flow Diagrams
with Multiple Process Units
Example 3.18 DOF Analysis: Adipic Acid Production
194
196
197
198
199
200
201
Case Study: Manufacture of Nylon-6,6
Summary
ChemiStory: Of Toothbrushes and Hosiery
Quick Quiz Answers
References & Recommended Reading
Chapter 3 Problems
CHAPTER 4
4.1
Synthesis and Analysis of Reactor
Flow Sheets
4.1.1
4.1.2
4.1.3
Industrially Important Chemical Reactions
Heuristics for Selecting Chemical Reactions
A Brief Review: Generation-Consumption Analysis
and Atom Economy
Example 4.1
Generation-Consumption and Atom
Economy: Improved Synthesis of Ibuprofen
Reactor Design Variables
Reactor Material Balance Equations
4.2.1
4.2.2
4.2.3
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231
Introduction
4.1.4
4.2
204
214
216
218
219
219
Reactors with Known Reaction Stoichiometry
Example 4.2
Continuous-Flow Steady-State Reactor with
Known Reaction Stoichiometry: Sustainable
Synthesis of Acetic Acid
Example 4.3
Continuous-Flow Steady-State Reactor with
Multiple Chemical Reactions: Combustion of
Natural Gas
Example 4.4
Batch Reactor with Known Reaction
Stoichiometry: Ibuprofen Synthesis
Example 4.5
Semibatch Reactor with Known Reaction
Stoichiometry: Ibuprofen Synthesis
Independent Chemical Reactions
Example 4.6
Independent Chemical Reactions
Reactors with Unknown Reaction Stoichiometry
Example 4.7
Material Balance Equation with Elements:
Combustion of Natural Gas
232
232
234
234
235
237
239
239
240
243
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249
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Contents
Example 4.8
Example 4.9
Example 4.10
4.3
Stream Composition and System Performance
Specifications for Reactors
4.3.1
4.3.2
4.3.3
4.3.4
4.4
Mass Rates of Reaction: Microbial
Degradation of Soil Contaminants
Integral Equation with Unsteady Flow
and Chemical Reaction: Controlled
Drug Release
Differential Equation with Unsteady
Flow and Chemical Reaction: Glucose
Utilization in a Fermentor
Stream Composition Specification: Excess
and Limiting Reactants
Example 4.11 Excess Reactants: A Badly
Maintained Furnace
System Performance Specifications
System Performance Specification:
Fractional Conversion
Example 4.12 Fractional Conversion: Ammonia
Synthesis
Example 4.13 Effect of Conversion on Reactor Flow:
Ammonia Synthesis
System Performance Specifications: Selectivity
and Yield
Example 4.14 Selectivity and Yield Definitions:
Acetaldehyde Synthesis
Example 4.15 Using Selectivity in Process Flow
Calculations: Acetaldehyde Synthesis
Fractional Conversion and Its Effect on Reactor
Flowsheet Synthesis
Example 4.16
Effect of Conversion on Reactor Flows:
Ammonia Synthesis
4.4.1Fractional Conversion and Recycle
Example 4.17 Low Conversion and Recycle:
Ammonia Synthesis
4.4.2Fractional Conversion, Recycle, and Purge
Example 4.18 Recycle with Purge: Ammonia
Synthesis
Case Study: Evolution of a Greener Process
Summary
ChemiStory: Quit Bugging Me!
Quick Quiz Answers
References and Recommended Readings
Chapter 4 Problems
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257
260
264
266
268
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284
285
288
296
298
301
301
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Contents
CHAPTER 5
5.1
Why Reactors Aren’t Perfect: Reaction
Equilibrium and Reaction Kinetics
Introduction
5.1.1
5.1.2
5.1.3
5.1.4
5.1.5
5.1.6
5.2
The Chemical Reaction Equilibrium Constant Ka
Example 5.1
Deriving Equations for Ka: Three Cases
Gibbs Energy of Reaction
Calculating Ka
Example 5.2
Calculating Ka: Ethyl Acetate Synthesis
Equilibrium Considerations in Reaction Pathway Selection
Example 5.3
Chemical Equilibrium Considerations in
Selection of Reaction Pathway: Safer
Routes to Dimethyl Carbonate
Chemical Reaction Equilibrium and Conversion
Example 5.4
Reactor Performance and Ka:
Ammonia Synthesis
Example 5.5
Equilibrium Conversion as a Function
of T and P: Ammonia Synthesis
Chemical Reaction Equilibrium, Selectivity, and Yield
Example 5.6
Multiple Chemical Equilibria and
Reactor T: NOx Formation.
Example 5.7
Multiple Chemical Equilibria and
Selectivity: DEE from Waste Ethanol
Chemical Reaction Kinetics and Reactor Performance
5.2.1
5.2.2
Irreversible First-Order Reaction in a Stirred-Tank
Batch Reactor
Example 5.8
Reaction Kinetics and Reactor
Performance: Vegetable Processing
Growth Kinetics in a Stirred-Tank Continuous-Flow Reactor
Example 5.9
Growth Kinetics in a Stirred-Tank
Continuous-Flow Reactor: Microbial
Degradation of Toxins in Wastewater
Case Study: Hydrogen and Methanol
Summary
Quick Quiz Answers
References and Recommended Readings
Chapter 5 Problems
CHAPTER 6
6.1
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Selection of Separation Technologies and
Synthesis of Separation Flow Sheets
Introduction
6.1.1
Physical Property Differences: The Basis for
All Separations
ix
321
322
322
324
325
328
330
331
331
334
335
337
338
340
343
345
348
349
351
352
353
361
362
362
362
375
376
376
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Contents
Example 6.1
6.1.2
6.2
Classification of Separation Technologies
6.2.1
6.2.2
6.2.3
6.2.4
6.2.5
6.3
392
6.3.2
394
395
6.3.1
Continuous-Flow Steady-State Separators
Example 6.7
Continuous-Flow Steady State Separators:
CO2 Removal from Flue Gas
Batch Separators
Example 6.8
Batch Separators: Caffeine from
Coffee Beans
Semibatch Separators
Example 6.9
Semibatch Mechanical Separation:
Filtration of Beer Solids
Example 6.10 Rate-Based Separation: Membranes
for Kidney Dialysis
Stream Composition and System Performance
Specifications for Separators
Example 6.12
Example 6.13
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379
Mechanical Separations
379
Example 6.2
Mechanical Separations: Matching the
Problem with the Technology
380
Rate-Based Separations
381
Example 6.3
Rate-Based Separations: Fresh
Water from the Sea
383
Equilibrium-Based Separations
384
Heuristics for Selecting Separation Technologies
387
Example 6.4
Selection of Separation Technology:
Separating Benzene from Toluene
388
Example 6.5
Selection of Separation Technology: Removing
Viruses from Engineered Antibodies
388
Heuristics for Sequencing Separations
390
Example 6.6
Sequencing of Separation Technologies:
Aromatics and Acid
390
Example 6.11
6.5
376
377
Separator Material Balance Equations
6.3.3
6.4
Physical Property Differences:
Separating Salt from Sugar
Mixtures and Phases
396
397
398
399
401
Defining Separator Performance
Specifications: Separating Benzene
from Toluene
406
Purity and Recovery Specifications in
Process Flow Calculations: Separating
Benzene and Toluene
407
Fractional Recovery in Rate-Based Separations:
Membranes for Kidney Dialysis
409
Recycling in Separation Flow Sheets
Example 6.14
392
Separation with Recycle: Separating Sugar
Isomers
411
413
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Contents
6.6
Entrainment: Incomplete Mechanical Separation
415
Case Study: Recovering Proteins From Fermentation Broths
Summary
ChemiStory: How Sweet It Is
Quick Quiz Answers
References and Recommended Readings
Chapter 6 Problems
418
422
423
426
427
427
Equilibrium-Based Separation Technologies
445
Example 6.15
CHAPTER 7
Accounting for Entrainment: Coffee Making 416
7.1
Introduction
446
7.2
Multicomponent Phase Equilibrium and the
Equilibrium Stage Concept
449
Equilibrium-Based Separation Technologies with
Energy-Separating Agents
453
7.1.1
7.1.2
7.2.1
7.2.2
7.3
7.3.1
7.3.2
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Phases: A Brief Review
Single-Component Phase Equilibrium
The Gibbs Phase Rule
The Equilibrium Stage
446
447
450
451
Liquid-Solid Equilibrium and Crystallization
454
Example 7.1
Process Flow Calculations with LiquidSolid Equilibrium Data: Potassium
Nitrate Crystallization
458
Example 7.2
Entrainment Effects in Equilibrium-Based
Separations: Separation of Benzene and
Naphthalene by Crystallization
459
Vapor-Liquid Equilibrium and Associated
Separation Technologies
462
Example 7.3
Using Raoult’s Law: Dew Point and
Bubble Point Temperatures of
Hexane-Heptane Mixtures
466
Example 7.4
Process Flow Calculations with Raoult’s Law:
Dehumidification of Air by Condensation
467
Example 7.5
Vapor-Liquid Separations with Raoult’s
Law: Equilibrium Flash of a Hexane/
Heptane Mixture
469
Example 7.6
Vapor-Liquid Separations with Nonideal
Solutions: Equilibrium Flash Separation of
Ethanol-Water Mixture
470
Example 7.7
The Power of Multistaging: Distillation
versus Equilibrium Flash for
Hexane/Heptane Separation
473
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Contents
7.4
Equilibrium-Based Separation Technologies with
Material-Separating Agents
475
7.4.2
480
481
7.4.1
7.4.3
7.4.4
Gas-Liquid Equilibrium and Absorption
Example 7.8
Process Flow Calculations Using
Gas-Liquid Equilibrium Data: Cleaning
up Dirty Air by Absorption
Solid-Fluid Equilibrium and Adsorption
Example 7.9
Process Flow Calculations Using
Adsorption Isotherms: Monoclonal
Antibody Purification
Liquid-Liquid Phase Equilibrium and Solvent Extraction
Example 7.10 Process Flow Calculations Using
Liquid-Liquid Distribution Coefficients:
Cleanup of Wastewater Stream by
Solvent Extraction
Example 7.11 Process Flow Calculations Using
Triangular Phase Diagrams: Separating
Acetic Acid from Water
Multistaged Separations Using Material
Separating Agents
Example 7.12 The Power of Multistaging: Recovery
of Acetic Acid from Wastewater
Case Study: Scrubbing Sour Gas
Summary
Quick Quiz Answers
References and Recommended Readings
Chapter 7 Problems
CHAPTER 8
8.1
Process Energy Calculations and
Synthesis of Safe and Efficient
Energy Flow Sheets
Introduction
8.1.1
8.1.2
8.1.3
8.1.4
8.2
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Energy Sources
Energy Distribution: Electricity, Heating Fluids,
and Cooling Fluids
Energy Transfer Equipment
A Brief Review of Energy-Related Dimensions
and Units
The Energy Balance Equation
8.2.1
8.2.2
The Energy Balance Equation
System Energy and Energy Flows
477
483
485
489
491
492
494
496
503
504
504
504
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524
524
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Contents
8.2.3
8.2.4
8.2.5
8.3
Kinetic and Potential Energy and the Energy
Balance Equation
8.3.1
8.3.2
8.4
Two Forms of Energy: Kinetic and Potential
Example 8.1
Kinetic and Potential Energy:
Toddler Troubles
Example 8.2
Change in Potential Energy:
Snow Melt
Example 8.3
Change in Kinetic Energy of a
Stream: Thomas Edison
or Rube Goldberg?
Process Energy Calculations with Kinetic and
Potential Energy
Example 8.4
Potential Energy into Work:
Water over the Dam
Internal Energy and Enthalpy and the Energy
Balance Equation: Pressure, Temperature,
and Phase Effects
8.4.1
8.4.2
8.4.3
8.4.4
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Heat and Work
The Energy Balance Equation—Again
Process Energy Calculations
Using Tables and Graphs to Find ​​Û ​​ and ​​Ĥ ​​
Example 8.5
Using Steam Tables to Find H
​​ ̂ ​​:
Several Cases
Example 8.6
Using Steam Tables: Pumping
Water, Compressing Steam
Example 8.7
Comparing Kinetic, Potential,
and Internal Energy: Frequent Flyer
Using Model Equations to Find ​​Û ​​ and ​​Ĥ ​​
Example 8.8
Enthalpy Calculations:
Enthalpy of Vaporization of
Water at High Pressure
Minisummary
Process Energy Calculations: Pressure, Temperature,
and Phase Effects
Example 8.9
Integral Energy Balance with a
Closed System: Unplugging the
Frozen Pipes
Example 8.10 Differential Energy Balance:
Heat Exchanger
Example 8.11 Simultaneous Energy and Material
Balances: Mel and Dan’s
Lemonade Stand
Example 8.12 Energy Balance with Equilibrium Flash:
Separation of Hexane and Heptane
xiii
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536
538
541
541
542
543
544
545
545
546
547
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561
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Contents
Example 8.13
8.5
Unsteady-State Heat Loss: Cooling
a Batch of Sterilized Broth
Internal Energy and Enthalpy and the Energy
Balance Equation: Composition Effects
8.5.1
8.5.2
Finding Effect of Composition on ​​U ​​̂ and ​​H ​​̂
Example 8.14 Enthalpy Calculations: Enthalpy
of Reaction at High Temperature
Example 8.15 Using Enthalpy-Composition Graphs:
Ammonia-Water Mixtures
Process Energy Calculations: Composition Effects
Example 8.16 Temperature Change with Dissolution:
Caustic Tank Safety
Example 8.17 Energy Balance with Chemical Reaction:
Adiabatic Flame Temperature
Example 8.18 Energy Balances with Multiple Reactions:
Synthesis of Acetaldehyde
Case Study: Energy Management in a Chemical Reactor
Summary
ChemiStory: Get the Lead Out!
Quick Quiz Answers
References and Recommended Readings
Chapter 8 Problems
CHAPTER 9
9.1
9.2
A Process Energy Sampler
Introduction
Work and the Engineering
Bernoulli Equation
Example 9.1
9.3
Heat Exchangers and the Synthesis of Heat
Exchange Networks
Example 9.2
9.4
Heat Exchanger Sizing: Steam
Heating of Methanol Vapor
Energy Conversion Processes
Example 9.3
Example 9.4
Example 9.5
mur83973_fm_i-xxxiv.indd 14
The Engineering Bernoulli Equation:
Sizing a Pump
Converting Reaction Energy to Heat:
Furnace Efficiency
Converting Reaction Energy to Work:
Heat Engine Analysis
Converting Reaction Energy to Work:
Hydrogen Fuel Cells
569
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572
577
579
581
581
583
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599
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614
614
615
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618
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622
626
630
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Contents
9.5
Chemical Energy and Chemical Safety: Explosions
633
Chapter 9 Problems
639
Example 9.6
Estimating Explosive Potential:
Trinitrotoluene
Mathematical Methods
APPENDIX B Physical Properties
Glossary
Index
APPENDIX A
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651
675
709
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Preface
Introduction to Chemical Processes: Principles, Analysis, Synthesis is intended
for use in an introductory one-semester or two-quarter course for students in
chemical engineering and related disciplines. The text assumes that the students
have had one semester of college-level general chemistry and one or two semesters of college-level calculus. Although student understanding of the material will
be deeper with greater background in linear algebra or organic chemistry, the text
is organized so that this background is not required for successful completion.
Course Trends
Introductory chemical engineering courses traditionally focus on chemical process calculations. Material and energy balances are taught, a few concepts in
thermodynamics are introduced and miscellaneous information on units, dimensions, and curve fitting are included. By the end of the semester most students,
given a well-defined problem, can set up and solve material and energy balance
equations, but they do not have a good understanding of how these calculations
are related to actually designing chemical processes to make products.
Several years ago the chemical engineering faculty at UW—Madison
decided to redesign our introductory course. Our goals were twofold: (1) to give
the students a better flavor of how chemical processes convert raw materials to
useful products and (2) to provide the students with an appreciation for the ways
in which chemical engineers make decisions and balance constraints to come
up with new processes and products. At the end of the semester, we wanted
students to be able, with a minimum amount of information, to synthesize a
chemical process flowsheet that would approximate real industrial processes.
This includes selection of appropriate separation technology, determination of
reasonable operating conditions, optimization of key process variables, integration of energy needs, and calculation of material and energy flows. This becomes
possible at the introductory level through use of limiting cases, idealizations,
approximations, and heuristics. We also wished to integrate concepts in sustainable resource utilization, process safety, environmental protection, and economics at the earliest levels of engineering education, so that these principles become
naturally embedded in a student’s problem-solving practices.
The modern approach equips students with the tools necessary for thinking
about the creative strategies of chemical process synthesis and greatly enhances
students’ understanding of the connection between the chemistry and the process.
xvi
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Preface
xvii
It provides the students a framework for much of the rest of the curriculum:
Students are more motivated to struggle through the rigor and abstraction of
engineering science courses in thermodynamics, transport, and kinetics, because
the connection between fundamental concepts and practical engineering problem solving has been made. Senior process design courses revisit the same
terrain but at a more sophisticated level. Students learn that the principles of
chemical processes, and the strategies of process synthesis and analysis, can
be advantageously applied to an enormous diversity of problems, from intracellular trafficking of a drug to accumulation of pollutants in the ecosystem. The
ready availability of easy-to-use computational tools means that students in an
introductory course can tackle challenging and complex problems.
Organization
Many times, students decide to major in chemical engineering because they
like chemistry and math, and are interested in practical applications. In designing this text, we have tried to keep this motivation in mind. We start right off
the bat, in Chapter 1, providing a link to freshman chemistry courses. We show
how simple stoichiometric concepts are used to make informed choices about
raw materials and reaction pathways. Students should understand that engineering is not simply about doing calculations, but about using calculations wisely
to make good choices. The idea of combining calculations, data and heuristics
to make choices is a central theme throughout the text.
Chapter 2 introduces the simple but powerful idea of process flow sheeting
as the chemical engineer’s means to communicate ideas about raw materials,
reaction chemistry, processing steps, and products. Here students learn the 10 Easy
Steps for process flow calculations, and are introduced, in a very conceptual
manner, to system variables, system and stream specifications, and material
balances. Many example problems, drawn from a wide diversity of applications, are worked out in detail.
In Chapter 3 we revisit material balance equations, reaction stoichiometry,
and process flow sheeting, but with a more rigorous and mathematical
approach. Throughout, the text retains this spiral organization, in which we
first reinforce concepts introduced in earlier chapters, and then expand and
deepen student understanding of these concepts. In this chapter, material balance equations are derived from conservation-of-mass principles, using a notation that students will see in more advanced classes, and we do not shy away
from transient processes. Students learn degree-of-freedom analysis as an
essential tool for organizing information, identifying constraints, and developing logical problem-solving strategies.
Chapters 4 and 5 delve in greater depth into chemical reactions and reactors. In Chapter 4, students receive additional practice in applying material
balance equations to reacting systems. Quantitative measures of reactor performance are introduced, and students learn how performance specifications
mur83973_fm_i-xxxiv.indd 17
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xviii
Preface
influence reactor material balance calculations. Descriptive information on the
major kinds of industrially important reactions is provided, and heuristics for
synthesizing reactor flow sheets are discussed. Chapter 5 introduces key concepts in chemical reaction thermodynamics and chemical kinetics, and students
learn how to integrate reaction thermodynamic or kinetic constraints with
material balances to select reactor operating conditions for better performance.
Chapters 6 and 7 focus on separators. In Chapter 6, the major separation
technologies are described, and heuristics for selecting an appropriate separation method and for sequencing multiple separation steps are provided.
Quantitative measures of separator performance are introduced, and students
learn how these performance specifications are used in separator material balance calculations. Chapter 7 delves more deeply into equilibrium-based separations. Students gain considerable experience in using physical property data,
graphs and equations to obtain phase equilibrium information. They learn how
phase equilibrium constraints are coupled with material balance equations to
design and analyze common separation units, and learn how to select process
operating conditions to improve separator performance.
In Chapter 8, the energy balance equation is derived, and students learn
the 12 Easy Steps for solving process energy calculations. Concepts such as
work and heat are introduced. Students learn how to calculate changes in
kinetic or potential energy, and how to find internal energy or enthalpy from
equations, charts, and graphs. Plenty of worked-out example problems illustrate
how to apply thermodynamic information and the energy balance equation to
solve important problems. Chapter 9 is a “sampler” of more complex applications of the basic concepts taught in Chapter 8.
Chapters 1–4 and Chapter 6 provide an excellent introduction to material
balances and chemical processes for instruction in a one-quarter course or for
those wishing a more leisurely approach. Instructors who do not want to introduce reaction thermodynamics or phase equilibria can omit Chapters 5 and 7.
Chapters 8 and 9 can be omitted if energy balances are taught in thermodynamic classes. For students with less mathematical background, all linear
­algebra sections as well as the unsteady-state problems can be skipped.
Changes from the first edition. The author kept the general flavor and
approach of the first edition. The major changes include:
1. Simple matrix manipulations (for example, for balancing chemical reactions) were integrated into the text when the topic was introduced, rather
than relegated to a separate section. Students have greater access and familiarity with programs such as MatLab or Python, or calculators that can
easily solve matrix equations.
2. The Degree-of-Freedom analysis was moved from Chapter 2 to Chapter 3.
This provides students with more practice with solving problems and
builds their intuition before the introduction of a systematic means to count
equations and variables. This also allows for the introduction of the extent
of reaction concept before DOF analysis, which then makes the method of
counting reaction variables clearer.
mur83973_fm_i-xxxiv.indd 18
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Preface
xix
3. The development of the differential material balance equation in Chapter 3
was substantially reorganized. Notation for the integral material balance
equation was simplified. The more advanced material on linear models of
flowsheets was deleted, as the author found she never taught that material.
4. The old Chapter 4 was split into two chapters, with one focused on reactor
performance and practice with the material balance equations in reacting
systems, and the second covering more advanced materials on reaction thermodynamics and kinetics. Whereas the old Chapter 4 was rather hefty, the
new Chapters 4 and 5 are more similar in scope and size to Chapters 1–3.
5. Similarly, the old Chapter 5 was split into two, with the new Chapter 6
focused on descriptive information on separators and practice on applying
the material balance equations to separation flowsheets, and the new
Chapter 7 covering more advanced material on phase equilibria and
­equilibrium-based separations. The later material was also reorganized so
that the introduction of a specific type of phase equilibrium was followed
immediately by application of that equilibrium data to a separation problem.
6. Additionally, the old Chapter 6 was split into two, for the same reason of
making more equal-sized “bites.” The new Chapter 8 provides an introduction to energy balances. The development of the energy balance equation
was reorganized and the order of some material was rearranged. Now
students learn how to find changes in enthalpy as temperature, pressure,
and phase change, and then immediately apply this new knowledge to
energy balance problems, before moving on to study enthalpy changes due
to reaction or mixing. Chapter 9 includes several more advanced topics
that illustrate applications of energy balances.
7. Some end-of-chapter problems were omitted, and many new problems were
added. For the “Warm-up” and “Drills and Skills” problems, the problems
were linked explicitly to the corresponding section in the text.
Features of the Text
The text is written to encourage students to:
∙ Link to chemistry. The text provides a clear link to freshman chemistry
courses. Students will remain more interested in the processes and get a
better flavor of what chemical processes do if they understand how chemistry relates to processing.
∙ Synthesize chemical processes. The text treats process calculations as a
means to an end: the design of safe, reliable, environmentally sound, and
economical chemical processes. The author’s approach gives students a
good understanding of how these calculations inform choices that must be
made in designing chemical processes to make desired products.
∙ Develop solid problem-solving strategies. Developing good problemsolving strategies is an important outcome of this introductory course.
Readers will find a systematic approach to deriving equations and accounting
mur83973_fm_i-xxxiv.indd 19
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xx
Preface
for specifications. A novel feature of this text is the use of heuristics,
introducing beginning students to the notion that practicing engineers rely
not just on calculations but also on collected experiences.
∙ Invent and analyze. The text integrates the best of the “process synthesis” philosophy with modern approaches, problems, and techniques.
Students learn that principles of process synthesis are gainfully applied to
problems in biotechnology, medicine, materials science, and environmental
protection.
∙ Let pedagogy lead. The text is heavily laden with pedagogy, tools to
guide the reader and enhance the subject matter. A few of the pedagogical
elements in this text include Helpful Hints , Quick Quiz, ChemiStory, and
Case Study sections. For a complete overview of the pedagogical elements
see the Guided Tour section.
∙ Explore software. This text is not directly tied to one software program,
allowing students to use software as a common tool to solve problems. An
appendix illustrates the use of Excel to find fit data to equations or to find
roots of equations, and the use of MATLAB to solve matrix equations.
mur83973_fm_i-xxxiv.indd 20
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About the Author
Regina Murphy received her S.B. in Chemical Engineering in 1978 from MIT,
then took a job at Chevron’s Richmond Refinery to learn about real engineering.
During her 5 years at Chevron she wore several hats, all of them hard. She
returned to MIT in 1983 where she obtained her Ph.D. under the guidance of
Clark Colton and Martin Yarmush. In 1989 she joined the faculty in the
Department of Chemical Engineering at the University of Wisconsin—Madison,
where she has happily stayed for her entire academic career. Her research
interests are in protein misfolding and aggregation, especially related to
­
­neurodegenerative disease. She has taught several courses throughout the undergraduate curriculum, from an introductory course for freshmen to senior design.
She is the recipient of multiple teaching awards including the Chancellor’s
Distinguished Teaching Award, the highest campus-level recognition for contributions to education. She served as department chair from 2018–2021, where
she hired several new faculty and initiated a major project to renovate instructional and research laboratory space. She lives in an old Victorian house on
Lake Monona in Madison with her husband Mark Etzel, also a professor at UW.
Their twin sons are both proud graduates of UW Madison.
xxi
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Acknowledgements
Eray S. Aydil
University of California, Santa
Barbara
Chelsey D. Baertsch
Purdue University
Paul Blowers
University of Arizona
Paul C. Chan
University of Missouri—Columbia
Wayne R. Curtis
The Pennsylvania State University
Janet De Grazia
University of Colorado at Boulder
Jeffrey Derby
The University of Minnesota
Douglas Lloyd
The University of Texas at Austin
Teng Ma
Florida State University
Michael E. Mackay
Michigan State University
Susan Montgomery
University of Michigan
James F. Rathman
Ohio State University
James T. Richardson
University of Houston
Richard L. Rowley
Brigham Young University
Gregory L. Griffin
Louisiana State University
Michael S. Strano
University of Illinois at
Urbana-Champaign
Sarah W. Harcum
Clemson University
Eric Thorgerson
Northeastern University
Joseph Kisutcza
New Jersey Institute of Technology
Timothy M. Wick
Georgia Institute of Technology
Dana E. Knox
New Jersey Institute of Technology
Lale Yurttas
Texas A&M University
The author would like to acknowledge the many people who have contributed in
various ways to this project. In particular:
Dale Rudd, who co-authored Process Synthesis and provided wise counsel
about dealing with publishers.
Michael Mohr, a warm and witty instructor who provided my first introduction to chemical engineering.
Thatcher Root, who cheerfully taught for many semesters from early drafts
of the first edition of this text, and provided not only many end-of-chapter problems but also practical insights and moral support.
xxii
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Acknowledgements
xxiii
Harvey Spangler, for his sponsorship of the named chair that provided funds
to make completion of the first edition feasible.
Many students at UW, who were enthusiastic participants in this experiment.
Many teachers at Lapham and Marquette Elementary Schools, who gave
me a fresh perspective on learning and teaching, who taught me that learning
is risky and that the best teachers provide an environment where students are
unafraid to take big leaps into the unknown.
Kevin and Nick, who grew from toddlers to teenagers during the writing
of the first edition, and who inspired at least one cartoon figure, one quick quiz
answer and one example problem, and who typed some of the tables in App. B
(for a fee). Despite now being fully independent, they are still willing to shoot
hoops with their mom.
Mark, for his contributions of problems and ideas, but mostly for his unwavering support and love over many, many years.
mur83973_fm_i-xxxiv.indd 23
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Section 7.2
Helpful Hint
For a pure component at liquid-vapor
or solid-vapor equilibrium, P = P sat.
Multicomponent Phase Equilibrium and the Equilibrium Stage Concept
449
P-T diagrams of the type shown in Fig. 7.1 are available in handbooks (see
References at the end of this chapter) for many common substances. Sometimes
the information is shown in tabular form instead, which is more convenient if
precision is required.
There are several useful model equations available to describe Psat as a function of T. One of the simplest and most widely used is the Antoine equation.
Guided Tour
B
log10 P sat = A − _
T+C
Eq. (7.1)
where A, B, and C are empirically-determined constants for a specific material.
Refer to App. B for Antoine constants for many common chemicals.
Tools That Reinforce Concepts
Illustration: The saturation pressure of water, according to Antoine’s equa-
Quick Quiz 7.4
From the Antoine
equation, what’s the
saturation pressure for
H2O at 100°C?
Intion
This
Chapter in App. B, is
coefficients
Words to Learn
sat
1750.286
_______________
log10 P (mmHg)
=provides
8.10765a −
from 0°C to 60°C
An
section
brief
introduction
100 In This
Chapter Chapter
2 Process Flows:
Variables,
Diagrams, Balances
T(°C)
+ 235.0 of the subject matter and a
bulleted list of questions that are addressed in each chapter. A list of Words to Learn is
1668.21
also outlined
atsatthe
beginning
of each chapter.
These elements
help
the$ reader
to focus
log10 P
(mmHg)
= 7.96681
− _______________
from
60°C
to 150°C
3 lb solids
of juice
by +
the228.0
little Fruitys = S4 × _______ = _________
T(°C)
on the fundamental points Value
as they
readconsumed
each
chapter.
lb solids
day
$5.42
$16.26
to
150°C.
Compare
We use this equation to calculate P sat versus T from 0°C
× _______
= ______
lb solids CHAPTER
day
ONE
the plot to Fig. 7.1; notice that we have just constructed the
vapor-liquid
coex(At nearly $6000/year,
istence curve for this temperature
range! the little Fruitys can drink water!)
Quick Quizzes
1
10. Check.
if the solids
content
of
Converting
the
Earth’s
The Quick Quizzes are sprinkled within theStep
chapters
and One way to check the answer is to see
the juice consumed by the little Fruitys is 12
wt%, sinceinto
we Useful
did not
Resources
are intended to test student understanding of the 10
topics
4
use that information in our calculations yet.
Products
covered in each chapter. Answers to the quizzes are proS
3
4
_______
______
vided at the end of each chapter.
= 0.125 (close enough)
=
S4 + W4
P sat, mmHg
1000
Helpful Hints
Helpful Hint
Balance the element that appears
in the fewest
compounds first.
3 + 21In This Chapter
We begin our study of chemical process synthesis. Chemical reactions are at
NoticeSection
that in 1.3
this problem
we carried
moreso digits
the heart of along
chemical processes,
we start by than
reviewingsignificant
reaction stoichiomBalanced
Chemical
Reaction
Equations
7
etry and balanced chemical equations. We show how chemical reactions convert raw is
materials
to desired
products. idea,
We learn how
calculate the
quantities
in our intermediate calculations. This
often
a good
to toavoid
roundof raw material required to produce a desired product, the quantities and identities
of
waste
products,
and
the
process
economics
involved
with
choosing
rates carrying just significant digits.
100off errors. Try recomputing all flow
raw materials and chemical reactions.
are some of of
the questions
we addressin
in this
chapter: 4?
canWhat
be do you find to be the solidsHerecontent
the juice
stream
Helpful Hints sections
whereinthe
is taken over all I compounds. In our example, the elefound
thesummation
margins sprinkled
ment
Cl
appears
in
two
compounds, SiCl4 (εCl,SiCl4 = 4) and HCl (εCl,HCl = 1),
throughout the text.So
Helpful
Hints
10
far, we’ve
always set the accumulation term equal to zero. This happens either
and
Eq.
(1.1)
for
the
element
is simplyand steady state, or batch over a fixed time
are designed to helpwhen
students
with Cl
the process
is continuous
Words to Learn
interval
with
all
materials
added to the system
at the beginning and all materidifficult points.
_1
∙ What raw materials are used most frequently in chemical processes?
∙ How can chemical reactions be combined into pathways to efficiently convert raw materials to products?
∙ How much raw material is consumed? What byproducts are generated?
∙ What are simple measures for comparing the economic and environmental
impact of different raw materials or chemical reaction pathways?
εalsCl,SiCl
ν 1 from
+ the
εCl,HCl
νHCl
) + let’s
1(+2)
removed
system
by =
the4(−
end. 2Now,
turn=
to 0two problems where
4 SiCl
4
Watch for these words as you read Chapter 1.
Chemical process synthesis
150
0
100
Balanced chemical reaction equations
material accumulates
in the50system during
the process.
Stoichiometric
coefficients
analysis
In addition to Cl, there are two other Temperature,
elements, SiGeneration-consumption
H, so
there are two more
°Cand
Atom
economy
similar equations:
Example 2.5
Basis
Scale factor
Process economy
Separation with Accumulation: Air Drying
εSi,SiCl4 νSiCl4 + εSi,Si νSi = 1(−1) + 1(+1) = 0
Air is used throughout a process plant to move control valves (special valves that
Multicomponent
Phase Equilibrium and
regulate
air is humid, it needs to be dried before being used. To proεH,H flow).
νH +Ifεthe
H,HCl νHCl = 2(−1) + 1(+2) = 0
dry air for instrument use, filtered and compressed humid room air at 83°F
theduce
Equilibrium
Stage
Concept
and 1.1
atm pressure,
containing
mol%
O (as vapor), is pumped
through a
Eq. (1.1) is very
useful
because
we can1.5use
it toH systematically
find unknown
7.2
1
2
2
mur83973_ch01_001-060.indd 1
3
2
07/10/21 5:35 PM
tank at a flow rate of 100 ft /min. The tank is filled with 60 lbs of alumina (Al2O3)
Everything in the
previous section
applies for
systems
containofone
compostoichiometric
coefficients.
For example,
suppose
thethat
reaction
interest
is
pellets. The water vapor in the air adsorbs (sticks) onto the pellets. Dry instrument
nent. But,ofif methane
we air,
arecontaining
interested
separations
problems,
then
we
have
at
least
two
oxidation
(CH4)just
toin
CO
and
water.
Written
in
an
unbalanced
form
2
0.06 mol%
H2O, exits from the tank. The maximum amount of
water
that can
adsorb
to the alumina
pellets is 0.22
H2O per
lb alumina. How
components!
This
means
that
we have
to understand
a bitlbabout
multicomponent
the
reaction is:
long can
tank bebehavior
operated before
the alumina pellets need
to be replaced?
phase equilibrium.
Thethephase
of multicomponent
systems
is described
CH4 + O2 → CO2 + H2O
Solution
Examples
Steps 1 and 2.(C,Draw
diagram,
choosefour
a system.
The system (CH
is the4separator—
There are three elements,
H, aand
O), and
compounds
, O2, CO2,
Quick Quiz 1.2
thethe
tank
containing
the
aluminato
pellets.
there areidea
three
equations
four
unknown
H2O),thesoconceptual
Over 100 worked examplesand
indicate
problem
is involving
designed
illustrate
as wellstoichiometric
as the specific
=
Why did wechosen.
set νCH Classical
application
and modernKnowing
topics are used
in the
example
problems.
each
compound
and applying Eq. (1.1) to each
coefficients.
εhi for
−1 and not νCH = 1?
Alumina pellets
element, we derive:
4
4
mur83973_ch07_445-522.indd 449
xxiv
Humid air
with adsorbed
C: 1νCH4 + 0νO2 + 1νCO2 + 0νHwater
=0
2O
Dry air
10/11/21 1:25 PM
H: 4νCH4 + 0νO2 + 0νCO2 + 2νH2O = 0
Quick Quiz 1.3
Instead of setting
mur83973_fm_i-xxxiv.indd
24
O: 0νCH4 + 2νO2 + 2νCO2 + 1νH2O = 0
Since there are four unknowns but only three equations, there are many possible
23/11/21 7:47 PM
32
Chapter 1 Converting the Earth’s Resources into Useful Products
Six-Carbon Chemistry
In this case study, we illustrate how the concepts introduced in Chap. 1 are
used to make decisions about raw materials, products, and reaction pathways,
by looking in some depth at specific processes of importance in the organic
chemicals business. These processes are linked by their connection to 6-carbon
compounds. We’ll look at two questions:
41
Summary
1. Benzene is a 6-carbon compound purified from petroleum. Suppose we
have available 15,000 kg/day benzene. What are some useful 6-carbon
products we might make from benzene?
2. Could we replace benzene with a raw material from a renewable resource
to make the same 6-carbon products?
Remember that these calculations are minimum waste generation; we have
not accounted for any inefficiencies in the process. The processes using benzene produce less waste than those using glucose. A lot of the carbon in glucose
ends up as CO2 rather than as product (as we already saw in the atom economy
H
calculations). Why? One reason is this: in fermentation, glucose conversion to
H
H
C
CO2 produces energy for bacterial survival and growth. For a fairer compariC
son, we
should see examples
if energy needsillustrate
for the benzene
are met byof
burnCase Studies are provided
atCthe end of most chapters. These
in-depth
theprocesses
application
key
C
C
ing fuels and thus producing CO2. If so, then the waste calculations must
concepts from that chapter
problems. Case studies
integrate
analysisas and
and boost student
Cto modern
H
H
consider
energy requirements
well assynthesis,
raw material requirements.
Case Studies
confidence in their ability to
H tackle complex problems and issues.
Figure 1.5 Three different representations of the structure of benzene, C6H6, one of the
most important raw materials in the synthetic organic chemicals industry.
Simple organic compounds like benzene serve as raw materials in the pro-
126
216
duction of the plastics,
detergents, pharmaceuticals, and fibers that are ubiquiEnd-of-Chapter
Summaries
tous in modern societies. Think, for example, of nylon. Nylon was first sold
Summary
The Summary sections appear at the end of each chapter and provide an overview of the key definitions and
equations from that chapter.
∙ Chemical processes convert raw materials into useful products. In the
initial stages of chemical process synthesis, we choose raw materials to
make a specific product, or products to make from a specific raw material.
We choose a chemical reaction pathway for converting the chosen raw
materials into desired products. These choices all have profound consequences on the technical and economic feasibility of the process.
commercially in 1940, in the early days of World War II. The fiber rapidly
became an indispensable element in the war effort, as it was used for parachutes,
tents, ropes, airplane tire cords, and other military essentials. Perhaps nylon’s
greatest commercial success was in women’s hosiery, as nylon stockings
replaced the silk stockings formerly supplied by the Japanese.
There are several kinds of nylon, of which one of the most important is
called nylon 6,6. Nylon 6,6 is a polymer—a very large macromolecule containing many small repeating units linked by covalent bonds. Nylon 6,6 contains
Chapter 2 Process Flows: Variables, Diagrams, Balances
Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
O
OH
mathematics, and some physical and chemical data. The bulk of the handO
book is concerned with chemical engineering principles and methods; the
O
ChemiStory:
Of
Toothbrushes
and
Hosiery
book includes many sketches of chemical process equipment. Perry’s is
OH
published by McGraw Hill, New York.
The
Roaring
20s
aEncyclopedia
wild,
excitingoftime
in U.S.
history—a is
time
of
Benzene,
C6Hwas
Cyclohexane,
CChemical
O boot- Adipic acid, C6H10O4
Cyclohexanone,
C6aH10
3. The
Kirk-Othmer
Technology
multivolume
6
6H12
leg booze
and speakeasies,
rising on
skirts
and rising
The
DuPontThe
compendium
of information
chemicals
anda fortunes.
chemical
processes.
Figure
1.6
Benzene
is converted
to adipic
acid through
series
of chemical
reactions
family
was
one
of
several
fabulously
wealthy
families
of
the
time.
The involving intermediates
coverage
is and
truly
encyclopedic, and includes data on process economics,
cyclohexane
cyclohexanone.
DuPont
Company
started
as
a
gunpowder
manufacturer,
and
had
grown
to It
market size, physical and chemical properties, and process technology.
become
the
major
supplier
of
explosives
to
the
Allied
forces
in
World
War
I. are
is published by Wiley, New York. Two other books in the same vein
With the end of WWI and the beginning of the peacetime economic expanShreve’s Chemical Process Industries, McGraw Hill, New York, and the
sion, the company wisely moved from explosives to consumer goods.
McGraw Hill Encyclopedia of Science and Technology, McGraw Hill,
DuPont illustrated its new consumer focus through their famous motto:
New York.
“Better Things for Better Living through Chemistry.” Using their expertise
4. The Knovel Engineering and Scientific Online Database is a comprehenin cellulose and nitrocellulose chemistry, the company developed and sold
sive
of searchable
Perry’s Handbook
and CRC
mur83973_ch01_001-060.indd
32
a host
of source
new
consumer
products:information.
cellophane packaging,
rayon stockings,
Handbook
are a few
the many
lacquers
for painting
cars of
bright
colors.authoritative references that are searchable
from
the
Knovel
database.
Cellulose and nitrocellulose are plant-derived polymers, although at the
∙ Balanced chemical equations are needed to begin process calculations.
Chemical equations are balanced if
ChemiStories
∑ εhi νi = 0
all i
for all elements, where εhi is the number of atoms of element h in molecule
ChemiStories
describe historical events in the lives of the people
i, and νi is the stoichiometric coefficient for compound i; νi is negawho contributed
the chemical
industry
and for
itscompounds
products.
tive for to
compounds
that are reactants
and positive
that The
are
stories bringproducts.
to life the chemical products we take for granted,
∙ A generation-consumption analysis is a systematic way to analyze chemillustrate the
humanity of the heroes of chemical technology,
ical reaction pathways involving I compounds and K reactions. To complete
generation-consumption
analysis: forces drive scientific and
demonstratea that
social and political
Write balanced
equations
for all Kthat
reactions.
engineering (1)progress,
andchemical
caution
readers
technological
(2) List all I compounds (reactants and products).
breakthroughs
sometimes
have
unwanted
adverse
effects.
(3) For each reaction k, write the stoichiometric coefficient
νik associated
true nature. Debates raged among European
chemists: Were polymers true molecules, albeit very large, or were they
Chapter 2 Problems
aggregates of small molecules held together by some as-yet-unknown non(a) force?
WaterVirtually
is pumped
intowell-respected
a large tank. System:
Component:
water
covalent
every
chemist tank.
believed
the latter—
(b) Water
is pumped
intoofa alarge
tank with
that ahad
been preloaded
Warm-Ups
they could
not fathom
the idea
molecule
molecular
weight ofwith
sugar
crystals.
The
sugar
dissolves,
and
a
sugar
solution
is
pumped
100,000,
Section
2.2any more than they could imagine “an elephant. . . 1500 ft long
outhigh.”
of theAt
tank.
System: tank,
and 300 ft
a conference
heldComponent:
in Europe sugar.
in 1926, Hermann
(c) Same
(b) except
water
P2.1
You
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100-mL
volumetric
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then tare
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the component:
Nobelflask
Prizeon
ina Chemistry)
virtually
(d)
and
airg.(aThen
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of oxygen
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are(C
pumped
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it reads
0.00
you
add anhydrous
fructose
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as heEthylene
that
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true
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a reactor
at the
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ofreads
the
ethylene
major
sugarnamed
inoperating
fruit)
into
flask
until
theone
balance
15.90
g.reacts
You
organic
chemist
Wallace
Carothers
was
of the
small
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with
theStaudinger.
oxygen
to form
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reactor.
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fill the
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up to ethylene
the 100 mL
line.System:
The balance
reads
105.97 g.
who agreed
with
ethylene.
Calculate
the wt%born
fructose
andhad
the an
mol%
fructose start;
of thehesolution.
Wallace
Carothers,
in 1896,
inauspicious
attended
(e) Same
as (d)
except
component:
ethylene
oxide
his father’s
secretarial
and
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typing
and penmanship.
laterbeen
P2.2
Soybean
meal
isschool
a product
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soybeans
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would
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at the
University
of
(f ) Same The
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component:
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wt% protein
along
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1928 1234.
Carothers
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am,
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Your chemical
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taught processing
inInRoom
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and indigestible
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Charles
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Between
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of
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be recovered
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protein
isolate,”
can
mur83973_ch01_001-060.indd
believed
that
corporations
should
have
funda42
students
enter
the
classroom
and
3
leave.
If
the
classroom
is
the
be spun, mixed, and shaped into soy “bacon,” “burgers,” or other meat
mental
research
groups
for is
the“Input,”
prestige
system and For
students
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component,
what
substitutes.
100 lb
soybean
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about
how
many “Output,”
lbthey
of soy
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the
company.
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“Accumulation”?
protein
isolate “Consumption,”
can be made? If and
soybean
meal sells for $375/metric ton,
time.
DuPont
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per lbat ofthesoy
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ested (molar
in Carothers
Carothers
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P2.3 1000 grams of polystyrene
mass = because
20,800 g/gmol)
is dissolved
interested
polymers.
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Drills and
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mass =Carothers
104 g/gmol).
Calculate
in 4000
grams of styrene
(C8H8,inmolar
prove
that
Staudinger
was
right
about
the
the mole percent of polystyrene in the mixture.
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molecular
polymers,
andonDuPont
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P2.4 Air is approximately 79
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be the
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to pressure
make
is
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Calculate
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is mixed
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(H2)vessel
Wallace
Carothers.
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he mass
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Photo Researchers/
vessel using
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in units
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). What
is the
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the mole
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hexane
(C6H14the
reaction
between
an
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Sciencehydrogen,
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Images/
psig, bar, kPa,
and and
atm.cyclohexane
(b) The nitrogen
completely with
benzene,
in the reacts
mixture?
Alamy Stock Photo
to produce an ester. He reasoned that if both the
hydrogen to make ammonia
(NH3). If the temperature is still 298 K
and the ammonia is a gas, what is the change in pressure in the vessel
(in bar)?
P2.38 Your company needs on-site storage for 45,000 lb ammonia (NH3).
What is the diameter (in ft) of a spherical vessel needed to store the
ammonia (a) as a gas at 80°F and 5 atm, (b) as a liquid at 80°F and
12 atm, or (c) as a liquid at −30°F and 1 atm? The density of liquid
mur83973_ch03_155-230.indd 216
ammonia is 42.6 lb/ft3 at −30°F and 37.5 lb/ft3 at 80°F. Gas density can
mur83973_ch02_061-154.indd 126
be calculated from the ideal gas law. Which temperature and pressure
would you choose, and why?
P2.39 Seawater contains about 5 grams gold per trillion grams water. About
how much seawater would you need to recover 1.0 ounce gold? If there
are about 3.32 × 108 cubic miles of seawater on the planet, and the
density of seawater is 1.05 g/cm3, how much gold (tons) is dissolved in
the ocean?
P2.40 The following table lists data from the EPA for production and recycling
of plastics in the United States. The annual global production of plastics
is estimated at 78 million metric tons. Assuming that the table includes
the major plastics, calculate (a) the total metric tons of plastics produced
mur83973_fm_i-xxxiv.indd 25in the United States, (b) the percentage of global production that is in
130
Chapter 2
was known
about
their
Process time
Flows:little
Variables,
Diagrams,
Balances
07/10/21 5:36 PM
with each compound i in a column. There will be K columns, one for
each reaction.
(4) For all compounds i that should have zero net generation or consumption, zero net generation or consumption of compound i, find χk
such that
Homework Problems
Homework Problems are broken into four categories:
∑ χk νik = 0
all k
-Warm-Ups: Short-answer questions that cover basic
­definitions and straightforward calculations. Minimal
­proficiency.
41
-Drills and Skills: Drills and Skills problems cover the
fundamental skills and concepts learned in that chapter.
Average proficiency.
23/11/21 4:27 PM
-Scrimmage: Scrimmage problems require application of
more than one skill or concept and may involve material
from multiple (previous) chapters. Creativity is needed and
some problems require students to make judicious decisions
in the absence of complete information.
-Game Day: Game Day problems are best suited for use as
group projects and can be used to promote teamwork and
improve communication skills.
21/10/21 5:11 PM
23/11/21 6:31 PM
xxv
13/12/21 9:53 AM
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mur83973_fm_i-xxxiv.indd 27
23/11/21 7:47 PM
List of Nomenclature
(Typical Units)
ai
activity of compound i (dimensionless)
Cp
heat capacity at constant pressure, (J/gmol °C or J/g K)
Cv
heat capacity at constant volume, (J/gmol °C or J/g K)
Ek
kinetic energy (kJ)
Ep
potential energy (kJ)
fCi
fractional conversion of reactant i (dimensionless)
fRij
fractional recovery of component i in stream j (dimensionless)
fSj
fractional split to stream j (dimensionless)
g
acceleration due to gravity (m/s2)
ΔĜ°f
standard molar Gibbs energy of formation (kJ/gmol)
ΔĜr
molar Gibbs energy of reaction (kJ/gmol)
ΔĜ°r
standard molar Gibbs energy of reaction (kJ/gmol)
h
height above a reference plane (m)
Hi
Henry’s law constant (atm)
H
enthalpy (kJ)
Ḣ
enthalpy flow (kJ/s)
Ĥ
molar or specific enthalpy (kJ/gmol or kJ/g)
ΔĤ°c
standard enthalpy of combustion (kJ/gmol)
ΔĤ°f
standard molar enthalpy of formation (kJ/gmol)
ΔĤm
molar or specific enthalpy of melting (kJ/gmol or kJ/g)
ΔĤmix
molar or specific enthalpy of mixing (kJ/gmol or kJ/g)
ΔĤr
molar enthalpy of reaction (kJ/gmol)
ΔĤ°r
standard molar enthalpy (kJ/gmol)
ΔĤsoln
molar or specific enthalpy of solution (kJ/gmol or kJ/g)
xxviii
mur83973_fm_i-xxxiv.indd 28
11/12/21 1:55 PM
xxix
List of Nomenclature (Typical Units)
ΔĤv
molar or specific enthalpy of vaporization (kJ/gmol or kJ/g)
Ka
chemical reaction equilibrium constant
M
molar mass (g/gmol)
msys
mass in system (g)
ṁ
mass flow rate (g/s)
nsys
moles in system (gmol)
ṅ
molar flow rate (gmol/s)
P
pressure (atm, N/m2, bar)
pi
partial pressure of compound i (atm, N/m2, bar)
Pisat
saturation pressure of compound i (atm, N/m2, bar)
Q
heat (kJ)
Q̇
rate of heat transfer (kJ/s)
R
ideal gas constant (J/gmol-K, bar cm3/gmol K)
Ṙik
mass rate of reaction of compound i in reaction k (g/s)
ṙik
molar rate of reaction of compound i in reaction k (gmol/s)
SA→P
mur83973_fm_i-xxxiv.indd 29
selectivity for conversion of reactant A to product P
­(dimensionless)
t
time (s)
T
temperature (°C, K)
Tb
normal boiling point temperature (°C, K)
Tm
normal melting point temperature (°C, K)
U
internal energy (kJ)
Û
molar or specific internal energy (kJ/gmol or kJ/g)
v
velocity (m/s)
V
volume (m3)
V̂
molar or specific volume (m3/gmol, m3/kg)
wi
weight fraction of i (dimensionless)
W
work (kJ)
Ws
shaft work (kJ)
Ẇ
rate of work transfer (kJ/s, kW, hp)
11/12/21 1:55 PM
xxx
List of Nomenclature (Typical Units)
Ẇs
rate of shaft work transfer (kJ/s, kW, hp)
xi
mole fraction of i, typically in the liquid phase (dimensionless)
xis
mole fraction of i in the solid phase (dimensionless)
yi
mole fraction of i in the vapor phase (dimensionless)
yA→P
fractional yield for conversion of reactant A to product P
(dimensionless)
zi
mole fraction of i, typically when phase is undefined
(dimensionless)
Subscripts
f
final
h
element
i
compound or component
j
stream
k
reaction
sys
system
0
initial
Greek Letters
mur83973_fm_i-xxxiv.indd 30
αAB
separation factor for components A and B (dimensionless)
εhi
number of atoms of element h in compound i
νik
stoichiometric coefficient of compound i in reaction k
ρ
density (kg/m3 or gmol/m3)
ξ
extent of reaction (gmol)
ξ̇
extent of reaction (gmol/s)
χk
multiplying factor for reaction k
η
efficiency (dimensionless)
11/12/21 1:55 PM
List of Important Equations
Material Balance Equations
Differential form:
Total mass:
d​m​ sys​​
_
​
​ = ​​  ∑ ​​​​ ​​ṁ ​​  j​​ − ​​  ∑ ​​​​ ​​ṁ ​​  j​​
dt
all j​​in​
all j​​out​
Mass of i:
d​m​ i,sys​​
_
​
​ = ​​  ∑ ​​​​ ​​ṁ ​​  ij​​ − ​​  ∑ ​​​​ ​​ṁ ​​  ij​​ + ​​  ∑ ​​​​ ​ν​ ik​​ ​Mi​  ​​ ​​ξ ​​ ̇ k​​
dt
all k
all j​​in​
all j​​out​
Total moles:
d​n​ sys​​
_
​
​ = ​​  ∑ ​​​​ ​​ṅ ​​  j​​ − ​​  ∑ ​​​​ ​​ṅ ​​  j​​ + ​  ∑ ​​​ ​  ∑ ​​​ ​ν​ ik​​ ​​ξ ​​ ̇ k​​
dt
all j​​in​
all j​​out​
all k all i
Moles of i:
d​n​ i,sys​​
_
​
​ = ​​  ∑ ​​​​ ​​ṅ ​​  ij​​ − ​​  ∑ ​​​​ ​​ṅ ​​  ij​​ + ​  ∑ ​​​ ​ν​ ik​​ ​​ξ ​​ ̇ k​​
dt
all j​​in​
all j​​out​
all k
Integral form
Total mass:
​m​ sys, f​​ − m
​ ​ sys,0​​ = ​​  ∑ ​​​​ ​m​ j​​ − ​​  ∑ ​​​​ ​m​ j​​
all j​​in​
all j​​out​
Mass of i:
​m​ i,sys, f​​ − ​m​ i,sys,0​​ = ​​  ∑ ​​​​ ​m​ ij​​ − ​​  ∑ ​​​​ ​m​ ij​​ + ​  ∑ ​​​ ​Mi​  ​​ ​νi​  k​​ ​ξk​  ​​ all j​​in​
all j​​out​
all k
Total moles:
​n​ sys, f​​ − n​ ​ sys,0​​ = ​​  ∑ ​​​​ ​n​ j​​ − ​​  ∑ ​​​​ ​n​ j​​ + ​  ∑ ​​​ ​  ∑ ​​​ ​ν​ ik​​ ​ξk​  ​​
all j​​in​
all j​​out​
all k all i
xxxi
mur83973_fm_i-xxxiv.indd 31
23/11/21 7:47 PM
xxxii
List of Important Equations
Moles of i:
​n​ i,sys, f​​ − n​ ​ i,sys,0​​ = ​​  ∑ ​​​​ ​n​ ij​​ − ​​  ∑ ​​​​ ​n​ ij​​ + ​  ∑ ​​​ ​ν​ ik​​ ​ξk​  ​​
all j​​in​
all j​​out​
all k
System Performance Specifications
Splitter
Fractional split:
​​ṅ ​​  j​​
moles leaving in stream j _
_____________________
​fS​  j​​ = ​
​ = ​   ​
n
​​
̇ ​​  in​​
moles fed to splitter
Reactor
Fractional conversion:
− ​  ∑ ​​​ ​ν​ ik​​ ​​ξ ​​ ̇ k​​
moles of i consumed by reaction ______________
___________________________
​fC​  i​​ = ​
​ = ​  all k  ​
​​ṅ ​​  i,in​​
moles of i fed to reactor
Selectivity:
​  ∑ ​​​ ​ν​ Pk​​ ​​ξ ​​ ̇ k​​
moles
of
reactant
A
converted
to
product
P
ν
​
​
​​
all k
A1 _____________
​sA​  →P​​ = ​  ___________________________________
​ = ​ _
​νP​  1 ​ ​ ​​ ​  ∑ ​​​ ​ν​  ​​ ​​ξ ​​ ̇  ​​​
moles of reactant A consumed
Ak k
all k
Yield:
moles of reactant A converted to desired product P
_________________________________________
​
yA​  →P​​ = ​
​
moles of reactant A fed
​  ∑ ​​​ ​ν​ Pk​​ ​​ξ ​​ ̇ k​​
​_
ν
​  1​​ _____________
A
= − ​  ​ν​   ​ ​ ​​ all k  ​
​​ṅ ​​  A,in​​
P1
Separator
Fractional recovery:
​​n ​​i̇ j​​
moles of i leaving in stream j _
​fR​  ij​​ = ​  ________________________
​ = ​   ​
​​n ​​i̇ ,in​​
moles of i fed to separator
Separation factor:
​z​  ​​ ​z​ B2​​ _
​​ṅ ​​  ​​ ​​ṅ ​​  B2​​
​α​ AB​​ = _
​  ​z​ A1 ​ ​ ​​ _
​ = ​  A1 ​ ​ _
​
​​ṅ ​​  A2​​ ​​ṅ ​​  B1​​
A2 ​z​  B1​​
mur83973_fm_i-xxxiv.indd 32
23/11/21 7:47 PM
xxxiii
List of Important Equations
Chemical Reaction Equilibrium
​Ka​  ​​ = ​  ∏ ​​​ ​a​ i​ν​  ​ i​​​
all i
where, to a first approximation,
​y​ i​​ P
​a​ i​​ = _
​
​ for a gas
1 atm
ai = xi for a liquid
ai = 1 for a solid
−Δ​Ĝ ​r° _______
Δ​Ĥ ​°r ____
ln ​Ka​  ,T​​ = _____________
​
​  + ​   ​ ​​
​  1 ​− __
​  1 ​ ​​
R [ 298 T ]
298R
where Δ​Ĝ ​r° = ∑ ​ν​ i​​ Δ​Ĝ ​i°, f and Δ​Ĥ ​r° = ∑ ​ν​ i​​ Δ​Ĥ ​i°, f.
Phase Equilibrium
Raoult’s law:
​P​  sat
​  ​
​y​ i​​ = _
​  i ​ ​x​ i​​
P
Henry’s law:
​H​  ​​
​y​ i​​ = _
​  i ​ ​x​ i​​
P
Energy Balance Equations
Differential form:
d(​Ek​  ,sys​​ + E
​ p​  ,sys​​ + ​Us​  ys​​) ​
​ ________________
dt
= ​​  ∑ ​​​​ ​​ṁ ​​  j​​ (​​Ê ​​  kj​​ + E
​​ ̂ ​​  ​pj​ ​ ​​​ + ​​Ĥ ​​  j​​) − ​​  ∑ ​​​​ ​​ṁ ​​  j​​ (​​Ê ​​  ​kj​ ​ ​​​ + ​​Ê ​​  ​pj​ ​ ​​​ + ​​Ĥ ​​  j​​ ) + ∑
​ ​ ​​ ​​Q̇ ​​  j​​ + ​∑​ ​​ ​​Ẇ ​​  sj​​
all j​​in​
all j​​out​
j
j
Integral form:
​(E
​ k​  ,sys​​ + ​Ep​  ,sys​​ + ​Us​  ys​​)​  f​​ − ​(E
​ k​  ,sys​​ + ​Ep​  ,sys​​ + U
​ s​  ys​​)​  0​​
̂
̂
̂
̂
̂
̂
​ ​ ​​ ​Qj​  ​​ + ∑
​ ​ ​​ ​Ws​  j​​
= ​​  ∑ ​​​​ ​m​ j​​ (​​E ​​  kj​​ + ​​E ​​  ​pj​ ​ ​​​ + ​​H ​​  j​​) − ​​  ∑ ​​​​ ​m​ j​​ (​​E ​​  ​kj​ ​ ​​​ + ​​E ​​  ​pj​ ​ ​​​ + ​​H ​​  j​​ ) + ∑
j
j
all j​​in​
mur83973_fm_i-xxxiv.indd 33
all j​​out​
23/11/21 7:47 PM
CHAPTER ONE
1
Converting the Earth’s
Resources into Useful
Products
In This Chapter
We begin our study of chemical process synthesis. Chemical reactions are at
the heart of chemical processes, so we start by reviewing reaction stoichiometry and balanced chemical equations. We show how chemical reactions convert raw materials to desired products. We learn how to calculate the quantities
of raw material required to produce a desired product, the quantities and identities of waste products, and the process economics involved with choosing
raw materials and chemical reactions.
Here are some of the questions we address in this chapter:
∙ What raw materials are used most frequently in chemical processes?
∙ How can chemical reactions be combined into pathways to efficiently convert raw materials to products?
∙ How much raw material is consumed? What byproducts are generated?
∙ What are simple measures for comparing the economic and environmental
impact of different raw materials or chemical reaction pathways?
Words to Learn
Watch for these words as you read Chapter 1.
Chemical process synthesis
Balanced chemical reaction equations
Stoichiometric coefficients
Generation-consumption analysis
Atom economy
Basis
Scale factor
Process economy
1
mur83973_ch01_001-060.indd 1
07/10/21 5:35 PM
2
Chapter 1 Converting the Earth’s Resources into Useful Products
1.1
Introduction
Chemical processes convert raw materials into needed products by changing
the chemical and/or physical properties of the materials (Fig. 1.1). Why do
humans synthesize, design, build, and operate chemical processes?
To make a product that has a specific desired function. Many children
bring lunch to school every day. Wouldn’t it be great to have a lightweight,
safe, easy-open packaging material for carrying juice or milk? Aseptically
packaged drink boxes fulfill these product requirements and have replaced
heavy, bulky thermoses in the nation’s lunch bags. But, although throwaway
products are convenient, they carry with them waste disposal concerns.
To convert waste materials into useful products. It takes about 10 pounds
of milk to make 1 pound of cheese. The other 9 pounds end up as whey. Whey
used to be simply a waste product, dumped in nearby waterways or sprayed on
farmers’ fields. Now processes have been developed that recover the useful
components of whey. For example, the protein lactoferrin is purified from whey
and used in infant formula to improve iron uptake. Whey sugars serve as a
feedstock for production of biodegradable polymers.
To improve the performance of a natural material. Vincristine is a vinca
alkaloid present in minute quantities in the periwinkle plant. Concentrated and
purified, vincristine has proved to be a powerful drug for treating leukemia and
lymphomas. Its success has led to synthesis in the laboratory of structurally related
compounds, any of which might serve as effective medicines to treat cancers.
Si
Chemical process
Figure 1.1 Chemical processes convert raw materials into desired products. In synthesizing
chemical processes, we choose appropriate raw materials, then select chemical reactions and
physical operations to change the properties of the raw materials to those of the desired
products. We aim to design a chemical process that is safe to operate, that uses raw materials efficiently and economically, that reliably produces the desired products, and that has
minimal environmental impact.
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Section 1.2 Raw Materials
To convert material into energy. Huge quantities of energy are used every
day to heat or cool our homes, power our motor vehicles, and cook our food.
Much of this energy is derived from combustion of fossil fuels—natural gas,
oil, or coal. In this process, the raw material reacts with oxygen to form carbon
dioxide and water. It is the energy released by the reaction, not the reaction
products, that is useful.
An enormous breadth of industries—paper, foods, plastics, fibers, glass,
electronic materials, fuels, pharmaceuticals, to name a few—depend on chemical processes. The art and science of chemical process synthesis is in choosing
appropriate raw materials and chemical reaction pathways, and in developing
an efficient, economical, reliable, and safe chemical process. Articulation of
product requirements must be made before process development can begin;
thus, product engineering and process engineering are inextricably linked. The
quality and availability of raw materials, economic forecasts, product safety and
reliability, marketing concerns, patents, and proprietary technology all influence
process design.
1.2
Raw Materials
Ultimately, we derive all of our raw materials from the earth. The fundamental
raw materials are air, water, minerals, fossil fuels, and agricultural products.
Air. Plentiful, readily available, and cheap, air serves as the source of
oxygen and nitrogen in many chemical processes. Oxygen is used widely for
oxidation reactions, the most important of which is the burning of fuels to
generate heat and electricity. Discovery of a method to convert atmospheric
nitrogen to liquid ammonia spawned the agricultural fertilizer industry, with
enormous repercussions for production of sufficient food to feed the growing
world population.
Petroleum
exploration and
production
Crude oil
Petroleum
refinery
Benzene
Commodity
chemicals
Plastic
fabricators
Polycarbonate
Plastics and
polymers
Phenol
Specialty
chemicals
Airplane windows
Eyeglasses
Baby bottles
Bisphenol A
Bicycle helmets
Figure 1.2 Many companies and processes are needed to convert a raw material such as crude oil to products such
as bicycle helmets. Companies and municipalities are trying to close the loop, by recovering consumer products at
the end of their useful life and reprocessing the materials into new products.
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Chapter 1 Converting the Earth’s Resources into Useful Products
Sodium chloride
Limestone
Chlorine
Caustic soda
Soda ash
Sodium
bicarbonate
Pulp and paper
Solvents
Plastics
Pesticides
Antifreeze
Refrigerants
Soap
Dyes
Fibers
Paper
Drugs
Rubber
Soap
Glass
Drugs
Paper
Water softening
Ceramics
Baking soda
Baking powder
Carbonated
beverages
Fire
extinguishers
Figure 1.3 Important commodity chemicals as well as common household products are
made from sodium chloride and limestone, as part of the chlor-alkali industry. Adapted from
Chemical Process Industries, 4th ed. by R. N. Shreve and J. A. Brink, 1977.
Water. Water is used as a reactant in many chemical processes and serves
an important role as a solvent. This is especially true for the biotechnology
industries—old (e.g., beer making), middle-aged (antibiotic production by fermentation), or new (antibody production from genetically engineered cells).
Water may eventually serve as a source of hydrogen, a clean-burning fuel.
Fossil fuels. Natural gas, crude oil, and coal are all hydrocarbon materials
produced by the decay of once-living things. Besides providing us with heat,
light, and electricity, fossil fuels serve as the raw material for the synthesis of
carbon-based products like polymers for plastic soft drink bottles and contact
lenses, fibers for clothing and furnishings, medicines, and pesticides (Fig. 1.2).
Minerals. Minerals are solid inorganic elements or compounds. One important mineral is salt (sodium chloride), which, besides its use as a preservative
and a flavoring, serves as the raw material for the enormous chlor-alkali industry
(Fig. 1.3). Minerals are the feedstocks for the inorganic chemicals industries,
which produce silicon chips for computers and aluminum for bicycles.
Agricultural and forest products. Living plants are carbon-based, but they
also contain a significant quantity of fixed oxygen and (sometimes) nitrogen.
Our food, of course, is produced from these raw materials. Other products derived
from agricultural raw materials include paper, natural fibers such as wool or
cotton, natural rubber, and medicines. There is an increasing interest in using
agricultural materials (also called biomass) as raw materials for production of
carbon-containing chemicals, thus reducing our reliance on non-renewable fossil fuels. For example, DuPont and partners developed processes in which cornderived glucose is fermented, using engineered bacteria, to make 1,3-propanediol.
The 1,3-propanediol is purified and then reacted to form a polymer called 3GT,
which is spun into fibers and woven into a fabric (Fig. 1.4).
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Section 1.3 Balanced Chemical Reaction Equations
1,3 Propanedol
Fiber and
polymer
Corn or other
renewable
sugar source
Fermentation
with engineered
bacteria
Polymerization and
purification
Figure 1.4 New processes to make chemical products from renewable resources are being
developed, like this process to synthesize fiber from corn.
Some chemical processors start with raw materials and make intermediates
that are sold to industrial partners, which then will further process those intermediates into consumer products. For example (Fig. 1.2):
∙
∙
∙
∙
∙
∙
An oil company extracts crude oil from underground reservoirs.
A petroleum refining company processes the oil to recover benzene.
A commodity chemicals company reacts the benzene to phenol.
A fine chemicals company converts phenol to bisphenol A.
A plastics company polymerizes bisphenol A to polycarbonate.
Fabricators use polycarbonate to make airplane windows, bullet-proof
glass, eyeglasses, baby bottles, compact discs, and football helmets.
∙ Consumers purchase eyeglasses, baby bottles, and compact discs, use them,
and then discard them to the landfill or recycling bin.
∙ Recyclers reprocess discarded materials into new products.
1.3
Balanced Chemical Reaction Equations
At the heart of most chemical processes lies one or more chemical reactions.
If A and B are reactants that undergo a chemical reaction to form products C
and D, we write:
​A + B → C + D​
As an example, in making electronics-grade silicon, silicon tetrachloride (SiCl4)
reacts with hydrogen to make pure silicon and hydrogen chloride:
​SiC​l​4​​ + ​H2​ ​​→ Si + HCl​
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Chapter 1 Converting the Earth’s Resources into Useful Products
The arrow indicates the direction of reaction, from reactants to products. This
reaction as written is not balanced.
If we are interested in showing not only the identity but also the quantity
of compounds taking part in a chemical reaction, we write a balanced chemical
reaction equation. A chemical equation is balanced if the number of atoms of
each element on the left-hand side of the equation equals the number of atoms
of that element on the right-hand side. To emphasize that the reaction is balanced, we can replace the arrow with an equals sign. For example, the reaction
of silicon tetrachloride with hydrogen to make silicon and hydrogen chloride
is balanced if we write
​SiC​l4​ ​​ + 2​H2​ ​​= Si + 4HCl​
Because the coefficients are relative rather than absolute quantities, it is also
true that
​2SiC​l4​ ​​ + 4​H2​ ​​= 2Si + 8HCl​
or
​​ _12 ​ ​SiCl​4​​ + ​H2​ ​​ = _​ 21 ​ Si + 2HCl​
because the coefficients do not have to be integers.
Since the reaction is written as an equation, we can collect all compounds
on the right-hand side and write:
​0 = − ​ _12 ​ SiC​l4​ ​ − ​H2​ ​ + _​ 12 ​ Si + 2HCl​
Now let’s define stoichiometric coefficients νi for each chemical compound i, and specify that νi is negative for compounds that are reactants and
positive for compounds that are products. For example, in the above reaction,
Quick Quiz 1.1
What is the numerical
value of ν​
​​ ​H2​ ​​?​
​ν​S​  iCl​4​​​​ = −​ _12 ​​
​
and
​​ν​ HCl​​ = +2​
We define
​εh​  i​​≡ number of atoms of the element h in molecule i​.
​
For example, ​​ε​ Cl,SiC​l​4​​​= 4​, because there are 4 Cl atoms in each molecule
of SiCl4. A chemical equation is stoichiometrically balanced with respect to
the hth element if and only if
​
​∑​ ​​ ​εh​  i​​ ​νi​  ​​ = 0​
i
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Eq. (1.1)
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Section 1.3 Balanced Chemical Reaction Equations
Helpful Hint
Balance the element that appears
in the fewest
compounds first.
7
where the summation is taken over all compounds. In our example, the element
Cl appears in two compounds, SiCl4 (​​εC​  l,​SiCl​4​​​​= 4​) and HCl (​​ε​ Cl,HCl​​ = 1​), and
Eq. (1.1) for the element Cl is simply
​ε​C​  l,SiCl​4​​ ​​​ ν​S​  iCl​4​​​​ + ​εC​  l,HCl​​ ​νH​  Cl​​ = 4(− ​ _12 ​ ) + 1(+2) = 0​
​
In addition to Cl, there are two other elements, Si and H, so there are two more
similar equations:
​
εS​  i,SiC​l4​ ​​​ ​νS​  iC​l4​ ​​​ + ​εS​  i,Si​​ ​νS​  i​​ = 1(−​​ _12 ​​) + 1(+​​ _12 ​​) = 0
​
εH​  ,​H2​ ​​ ​​ν​H​  2​ ​​​ + ​εH​  ,HCl​​ ​νH​  Cl​​= 2(−1) + 1(+2) = 0
Eq. (1.1) is very useful because we can use it to systematically find unknown
stoichiometric coefficients. For example, suppose the reaction of interest is
oxidation of methane (CH4) to CO2 and water. Written in an unbalanced form
the reaction is:
CH4 + O2 → CO2 + H2O
Quick Quiz 1.2
Why did we set ν​ ​​ ​CH​4​​​​ =
−1​and not ν​ ​​ ​CH​4​​​​ = 1​?
There are three elements, (C, H, and O), and four compounds (CH4, O2, CO2,
and H2O), so there are three equations involving four unknown stoichiometric
coefficients. Knowing εhi for each compound and applying Eq. (1.1) to each
element, we derive:
C:​1​ν​C​  H​4​​​​ + 0​ν​O​  2​ ​​​​ + 1​ν​C​  O​2​​​​ + 0​ν​H​  2​ ​​O​​ = 0​
H:​4​ν​C​  H​4​​​​ + 0​ν​O​  2​ ​​​​ + 0​ν​C​  O​2​​​​ + 2​ν​H​  2​ ​​O​​ = 0​
O:​0​ν​C​  H​4​​​​ + 2​ν​O​  2​ ​​​​ + 2​ν​C​  O​2​​​​ + 1​ν​H​  2​ ​​O​​ = 0​
Quick Quiz 1.3
Instead of setting
​​ν​C​  H​4​​​​ = −1​, suppose
we had chosen to set ​​
ν​ ​O2​ ​​​​ = −1​. What would
be the balanced chemical reaction equation?
Example 1.1
Since there are four unknowns but only three equations, there are many possible
solutions. To proceed, we arbitrarily set one of the stoichiometric coefficients.
For example, we can pick ​​ν​ ​CH​4​​​​​as the basis and set ν​ ​​ ​CH​4​​​​= −1​. There are now
only three unknowns, and we solve to find ν​ ​​ ​O2​ ​​​​= −2, ν​ ​C​  O​2​​​​= 1, ν​ ​H​  2​ ​​O​​ = 2​. The
balanced chemical reaction equation is
(1)CH4 + 2O2 → (1)CO2 + 2H2O
Balanced Chemical Reaction Equation: Nitric Acid Synthesis
Nitric acid (HNO3) is an important industrial acid used, among other things, in the
manufacture of nylon. In one of the reactions for making nitric acid, ammonia
(NH3) and oxygen (O2) react to form NO and H2O. Write the balanced chemical
equation.
Solution
We’ll write the unbalanced chemical reaction as
​​NH​3​​ + ​O2​ ​​→ NO + ​H2​ ​​O​
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Chapter 1 Converting the Earth’s Resources into Useful Products
There are three elements and four compounds, so there are three equations in four
unknowns:
N:​1​ν​N​  H​3​​​​ + 1​νN​  O​​ = 0​
H:​3​ν​N​  H​3​​​​ + 2​ν​H​  2​ ​​O​​ = 0​
O:​2​ν​O​  2​ ​​​​ + 1​νN​  O​​ + 1​ν​H​  2​ ​​O​​ = 0​
We’ll choose to set ​​ν​ ​NH​3​​​​= −1​. Starting with the N balance, we solve to get
​​νN​  O​​= 1​. From the H balance, ​​ν​ ​H2​ ​​O​​ = ​ _32 ​​. Finally, from the O balance, ν​ ​​ ​O2​ ​​​​ = − ​ _54 ​​.
The balanced equation is:
​​NH​3​​ + _​  54 ​ ​O2​ ​​→ NO + _​ 32 ​ ​H2​ ​​O​
[Try choosing to set a different stoichiometric coefficient. Do you get the same
balanced equation?]
Example 1.2
Balanced Chemical Reaction Equations: Adipic Acid Synthesis
Adipic acid is an intermediate used in the manufacture of nylon. (We’ll discuss
this process in greater detail later in this chapter.) Several chemical reaction steps
are involved in synthesis of adipic acid:
Reaction 1.
Reaction 2.
Cyclohexane (C6H12) reacts with oxygen (O2) to produce cyclohexanol (C6H12O).
Cyclohexane (C6H12) reacts with oxygen (O2) to produce cyclohexanone (C6H10O).
Water (H2O) is a byproduct of one of these reactions.
Reaction 3.
Reaction 4.
Cyclohexanol reacts with nitric acid to produce adipic acid
(C6H10O4).
Cyclohexanone reacts with nitric acid to produce adipic acid
(C6H10O4).
NO and H2O are byproducts of both Reactions 3 and 4.
Write the four balanced chemical equations corresponding to these four reactions.
Solution
Reaction 1.
From the problem statement, water may be a byproduct of this
reaction. Let’s assume it is, and see what happens. The unbalanced
chemical reaction is
​​C​6​​​H1​ 2​​ + ​O2​ ​​ → ​C6​ ​​​H1​ 2​​​O + H​2​​O​
There are three elements and four compounds, so we can set one
stoichiometric coefficient arbitrarily. C appears in two compounds,
O and H in three each, so we apply Eq. (1.1) to the element that
appears in the fewest number of compounds:
C: ​6​ν​C​  ​6​​​H1​ 2​​​​ + 6​ν​C​  ​6​​​H1​ 2​​O​​ = 0​
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Section 1.3 Balanced Chemical Reaction Equations
9
If we choose to set one of these two stoichiometric coefficients, we
can solve for the other immediately. Let’s set ν​ ​​ ​C​6​​​H1​ 2​​O​​ = +1​. Then
Helpful Hint
If one of the
elements appears
in only two
compounds, set
the stoichiometric
coefficient of one of
those compounds
to a fixed value.
​​ν​C​  6​ ​​​H1​ 2​​​​ = −1​
We then move on to the other two elements:
H:​12​ν​C​  ​6​​​H1​ 2​​​​ + 12​ν​C​  6​ ​​​H1​ 2​​O​​ + 2​ν​H​  2​ ​​O​​ =​
​12(−1) + 12(1) + 2 ​ν​H​  2​ ​​O​​ = 0​
O:​2​ν​O​  2​ ​​​​ + ​ν​C​  ​6​​​H1​ 2​​O​​ + ​ν​H​  2​ ​​O​​ = 2​ν​O​  2​ ​​​​ + 1 + ν​ ​H​  2​ ​​O​​ = 0​
These are easily solved to yield
​​ν​H​  2​ ​​O​​ = 0​
​​ν​O​  2​ ​​​​ = − ​ _12 ​​
The balanced chemical equation is
​​C6​ ​​​H1​ 2​​ + _​  12 ​ ​O2​ ​​ → ​C6​ ​​​H1​ 2​​O​
Reaction 2.
Finding the stoichiometric coefficients led us to the conclusion that
water is not a byproduct of Reaction 1 after all.
The unbalanced reaction of cyclohexane with oxygen to produce
cyclohexanone, with water as a possible byproduct, is
​​C​6​​​H1​ 2​​​ + O​2​​ → ​C6​ ​​​H1​ 0​​​O + H​2​​O​
Proceeding in a manner similar to that used for Reaction 1, we
write three equations:
C:​6​ν​C​  ​6​​​H1​ 2​​​​ + 6​ν​C​  6​ ​​​H1​ 0​​O​​ = 0​
H:​12​ν​C​  ​6​​​H​12​​​​ + 10​ν​C​  6​ ​​​H1​ 0​​O​​ + 2​ν​H​  2​ ​​O​​ = 0​
O:​2​ν​O​  2​ ​​​​ + ν​ ​C​  6​ ​​​H1​ 0​​O​​ + ​ν​H​  2​ ​​O​​ = 0​
We arbitrarily set ν​ ​​ ​C​6​​​H​12​​​​= −1​and solve the equations in order to
find the other three stoichiometric coefficients. The balanced chemical equation is
​​C6​ ​​​H1​ 2​​ + ​O2​ ​​ → ​C6​ ​​​H1​ 0​​O + ​H2​ ​​O​
Reaction 3.
In the third chemical reaction, cyclohexanol (C6H12O) and nitric
acid (HNO3) react to make adipic acid (C6H10O4), with nitric oxide
(NO) and water (H2O) as byproducts. The unbalanced reaction is
​​C6​ ​​​H1​ 2​​O + ​HNO​3​​ → ​C6​ ​​​H1​ 0​​​O4​ ​​+ NO + ​H2​ ​​O​
There are four elements and five compounds:
C:​6​ν​C​  ​6​​​H1​ 2​​O​​ + 6​ν​C​  6​ ​​​H1​ 0​​​O4​ ​​​​ = 0​
H:​12​ν​C​  ​6​​​H1​ 2​​O​​ + ​ν​H​  NO​3​​​​ + 10​ν​C​  6​ ​​​H1​ 0​​​O4​ ​​​​ + 2​ν​H​  2​ ​​O​​ = 0​
O:​​ν​ ​C​6​​​H1​ 2​​O​​ + 3​ν​H​  NO​3​​​​ + 4​ν​C​  6​ ​​​H1​ 0​​​O4​ ​​​​ + ν​ N​  O​​ + ​ν​H​  2​ ​​O​​ = 0​
N:​​ν​ ​HNO​3​​​​ + ​νN​  O​​ = 0​
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Chapter 1 Converting the Earth’s Resources into Useful Products
Starting with either the C or the N balance is a good choice. Let’s
set ​​ν​C​  ​6​​​H​12​​O​​= −1​. We immediately solve the C balance to find
​​ν​C​  ​6​​​H​12​​​O​4​​​​= 1​. Substituting these values into the remaining three
equations yields
H:​−12 + ​ν​H​  NO​3​​​​+ 10 + 2​ν​H​  2​ ​​O​​ = 0​
O:​−1 + 3​ν​H​  NO​3​​​​ + 4 + ν​ N​  O​​ + ν​ ​H​  2​ ​​O​​ = 0​
N:​​ν​ ​HNO​3​​​​ + ​νN​  O​​ = 0​
We can’t immediately solve any of the remaining equations. To
solve “by hand,” we
1. Subtract the N balance from the O balance to eliminate νNO:
​3 + 2​ν​H​  NO​3​​​​ + ​ν​H​  2​ ​​O​​ = 0​
2. Subtract the H balance from 2× this equation to eliminate ν​ ​​ ​H2​ ​​O​​​:
​8 + 3​ν​H​  NO​3​​​​ = 0​
3. Solve for ​​ν​ ​HNO​3​​​​​and work backwards to find the other stoichiometric coefficients:
​ν​ ​H​  NO​3​​​​ = − ​ _83 ​​
Quick Quiz 1.4
In the balanced
chemical equation
for Reaction 3 of
Example 1.2, noninteger coefficients appear.
Rewrite the equation,
using only integer
coefficients.
ν​​ N​  O​​ = _​ 83 ​​
​ν​ ​H​  2​ ​​O​​ = + ​ _73 ​​
The balanced chemical equation is:
Reaction 4.
​​C6​ ​​ ​H1​ 2​​ O + _​ 83 ​ ​HNO​3​​ → ​C6​ ​​ ​H1​ 0​​ ​O4​ ​​ + _​ 83 ​ NO + _​ 73 ​ ​H2​ ​​O​
The balanced chemical equation is (details are left for you)
​​C​6​​ ​H1​ 0​​ O + 2 ​HNO​3​​ → ​C6​ ​​ ​H1​ 0​​ ​O4​ ​​+ 2NO + ​H2​ ​​O​
1.3.1
Using Matrices to Balance Chemical Reactions
Recall that we balance chemical reactions by invoking the element balance
equation:
​​∑​ ​​​εh​  i​​ ​νi​  ​​= 0 ​
Eq. (1.1)
i
where ​​εh​  i​​= ​the number of atoms of element h in molecule i and ​​ν​ i​​ =​ the
(unknown) stoichiometric coefficients. If there are H elements, then H equations
must be solved simultaneously to find the unknown ν​ ​​i​​​. This is a system of
linear equations, and it is straightforward to use matrices to set up and solve
these equations.
Suppose we are interested in the reaction of NH3 and O2 to NO and H2O,
which is the topic of Example 1.1. There are 3 elements and 4 compounds, so
3 equations must be solved simultaneously to find the 4 unknown stoichiometric
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Section 1.3 Balanced Chemical Reaction Equations
11
coefficients. These 3 equations are given in Example 1.1. We can write this
set of equations in matrix form Ax = b:
⎢ ⎥
⎡​νN​  ​H3​ ​​​​⎤
0
1
0 ​ν​  ​​
0
N: 1
​O2​ ​​
​​H:​ ​​
3​ ​
0​
0​
2​ ​​ ​  ​​​ ​= ​ ​  0 ​​  ​​
[
] ​νN​  O​​
[0]
O: 0
2
1
1 ⎣​ν​  ⎦​​
​H2​ ​​O
Notice that each column in matrix A represents the chemical formula for one
compound! For example, column 1 represents N1H3O0 (NH3). In other words,
we can write this matrix from the known chemical formulas for each compound,
without bothering to derive the element balance equation! The vector x simply
contains the stoichiometric coefficients for each of the 4 compounds.
Because there are four variables but only three equations, the matrix A is
not square. Furthermore, the vector b is full of zeros. This system of equations
has an infinite number of possible solutions. To find one solution, we arbitrarily
specify one of the stoichiometric coefficients of the reactants to equal −1. Let’s
pick ν​​ N​  ​H3​ ​​​​ = −1​. We’ll call NH3 our basis compound. We then plug this value
into the element balance equations and simplify so only terms involving the
unknowns are on the left-hand side:
N:​​ν​ NO​​ = 1​
H:​2 ​ν​H​  2​ ​​O​​ = 3​
O:​2 ​ν​O​  2​ ​​​​ + ν​ N​  O​​ + ν​ ​H​  2​ ​​O​​ = 0​
(Of course it would be easy to solve this set of equations, but we continue on
for illustration purposes.) We write this new set of three equations in three
unknowns in matrix form:
​ν​O​  2​ ​​​​
N: 0
1
0
1
ν
​
​
​​
​​H:​ ​​ 0​ ​  0​  2​ ​​ ​  NO​​​ ​ = ​ ​  3 ​​  ​​
[
][ ] [ 0 ]
O: 2
1
1 ​ν​H​  2​ ​​O​​
Notice three things. First, we now have a 3 × 3, or square matrix. This is a
necessary (but not sufficient) condition for finding a unique solution. Second,
matrix A can be written down by inspection: Each column is simply the chemical formula of the compounds for which the stoichiometric coefficient is not
known. Third, vector b can be written down by inspection: it is simply the
chemical formula for the compound (NH3) chosen as the basis! The solution
is straightforward; you can solve by using matrix functions on a calculator or
by using equation-solving software.
Let’s recap how we use matrices to balance a chemical equation involving
I compounds and H elements:
1. List the elements involved (e.g., C, H, O, N).
2. Choose one of the reactants to serve as a basis. Set its stoichiometric coefficient equal to −1.
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Chapter 1 Converting the Earth’s Resources into Useful Products
3. List the chemical composition of all other compounds except the basis
compound in a column in a matrix A. Be sure to list the elements in the order
chosen in step 1, and do not forget the zeros. The matrix will have H rows
(corresponding to the H elements) and I − 1 columns (I compounds − 1
basis compound).
4. List the unknown stoichiometric coefficients in vector x. The vector will
have I − 1 entries. Be sure to list the coefficients in the same order as the
compounds were entered into the matrix.
5. List the chemical composition of the basis compound in vector b. Be sure
to list the elements in the order chosen in step 1, and do not forget the zeros.
6. Find the solution to the matrix equation, using a calculator or an equationsolving program.
Example 1.3
Balancing Chemical Equations with Matrix Math: Adipic Acid Synthesis
Cyclohexanol (C6H12O) and nitric acid (HNO3) react to make adipic acid (C6H10O4),
with nitrogen oxide (NO) and water (H2O) as byproducts. Find the stoichiometric
coefficients using a matrix equation.
Solution
We solved this already in Example 1.2 (Reaction 3), but this time we will solve using
the matrix method. We select C6H12O as the basis compound, set ν​ ​​ ​C​6​​​H1​ 2​​O​​ = −1​, and
proceed to write by inspection:
6
0
0⎤ ⎡ ​νH​  N​O3​ ​​​​ ⎤ ⎡ 6 ⎤
C: ⎡0
H: 1
10
0
2 ​ν​C​  6​ ​​​H1​ 0​​​O4​ ​​​​
12
​​  ​​ ​
​ ​
​  ​
​
​ ​ ​ ​
​ ​
​ ​ ​= ​ ​   ​​  ​  ​​
​νN​  O​​
O: 3
4
1
1
1
⎣
⎣
⎦
ν
​
​
​​
⎣
⎦
​H2​ ​​O
N: 1
0⎦
0
1
0
⎢
The solution is
⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎡ ν​ H​  N​O3​ ​​​​ ⎤ ⎡−8/3⎤
​ν​C​  6​ ​​​H1​ 0​​​O4​ ​​​​
1
​ ​
​ ​ ​​
​​ ​
​ ​ ​= ​ ​ ​  ​
​νN​  O​​
8/3
⎣ ​ν​H​  2​ ​​O​​ ⎦ ⎣ 7/3 ⎦
(Compare to the solution given in Example 1.2.)
1.4
Generation-Consumption Analysis
Choosing the raw materials and writing balanced chemical reaction equations are
early steps in chemical process synthesis. Many chemical processes require that
we combine multiple chemical reactions together, in order to convert available
raw materials to the desired products. This is done, in order to make the most
product out of the least (and least expensive) raw material. We also want to avoid
making waste byproducts, especially if those materials are toxic or hazardous.
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13
Section 1.4 Generation-Consumption Analysis
Generation-consumption analysis is a systematic method for synthesizing
reaction pathways involving multiple chemical reactions with these goals in
mind. This analysis allows us to calculate the moles of raw materials consumed
in generating a given quantity of product, and the moles of byproducts generated per mole of product.
As an example, suppose a company has developed a fermentation process
that produces isobutanol (C4H10O) from glucose, which was derived sustainably
from agricultural waste products. The balanced reaction is:
​​C​6​​H1​ 2​​O6​ ​ → ​C4​ ​​H1​ 0​O + 2C​O2​ ​ + ​H2​ ​O​
(R1)
The company’s goal is to use isobutanol as a “platform” chemical—a chemical
that serves as an intermediate toward making other chemicals from renewable
resources that are currently made from fossil fuels. One idea is to make xylene
(C8H10), a chemical that is blended into high-octane gasoline, used to make plastic bottles, and used as a solvent for waxes (even ear wax!). Xylene is produced
from isobutanol in two steps. First isobutanol is dehydrated to isobutene (C4H8)
​​C​4​​H1​ 0​O → ​C4​ ​​H8​ ​ + ​H2​ ​O​
(R2)
And then isobutene reacts to make xylene, with hydrogen as a byproduct
​2​C​4​​H8​ ​ → ​C8​ ​​H1​ 0​ + 3​H2​ ​​
(R3)
Let’s focus first on just reactions (R2) and (R3). In reaction (R2), one mole
of C4H8 is generated for every mole of isobutanol consumed. In reaction (R3),
one mole of xylene is generated for every two moles of isobutene consumed.
Isobutanol is the desired reactant and xylene is the desired product: we want
no net generation or consumption of isobutene. So we need two moles of isobutene generated in (R2). This can easily be achieved by simply multiplying
(R2) by 2!
​2​C​4​​H1​ 0​O → 2​C4​ ​​H8​ ​+ 2​H2​ ​O​
2 ​×​ (R2)
These concepts can be expressed mathematically. We define
​​ν​ ik​​≡ stoichiometric coefficient of compound i in reaction k.​
For example, ν​ ​​ ​H2​ ​,3​​= +3.​We define
​​χ​ k​​≡ multiplying factor for reaction k.​
For example, we used χ​ ​​ 2​​= 2​to ensure no net generation or consumption of
isobutene when combining (R2) and (R3).
The net generation or consumption of a compound i from a system of
reactions is then
​
νi​  ,net​​ = ​  ∑ ​​​ ​νi​  k​​ ​χk​  ​​
all k
Eq. (1.2)
For a compound that is an overall product (net generated) of a reaction
pathway, ν​​ i​  ,net​​ > 0.​
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Chapter 1 Converting the Earth’s Resources into Useful Products
For a compound that is an overall reactant (net consumed) of a reaction
pathway, ν​​ i​  ,net​​ < 0.​
For a compound that serves as an intermediate, with no net generation or
consumption,
​​ν​ i,net​​ = 0.​
Eq. (1.3)
This equality is used to find the correct multiplying factors when combining a
system of chemical reactions into a pathway. In the example of making xylene
from glucose ((R1) through (R3)), both isobutanol and isobutene serve as intermediates and therefore ideally have no net generation or consumption. Applying
Eqs. (1.2) and (1.3) to these two intermediates yields a system of two equations
involving three multiplying factors:
​​ν​ isobutanol,net​​ = ​νi​  sobutanol,1​​ ​χ1​  ​​ + ​νi​  sobutanol,2​​ ​χ2​  ​​ + ​νi​  sobutanol,3​​ ​χ3​  ​​ = 0​
​​ν​ isobutene,net​​ = ​νi​  sobutene,1​​ ​χ1​  ​​ + ​νi​  sobutene,2​​ ​χ2​  ​​ + ​νi​  sobutene,3​​ ​χ3​  ​​ = 0​
Inserting the known stoichiometric coefficients yields
​​ν​ isobutanol,net​​ =​ ​(+1) ​χ1​  ​​ + (−1) ​χ2​  ​​ + (0) ​χ3​  ​​ = 0​
​​ν​ isobutene,net​​ =​ ​(0) ​χ1​  ​​ + (+1) ​χ2​  ​​ + (−2) ​χ3​  ​​ = 0​
Since there is one more unknown than equation, we cannot yet solve for χ​ ​​k​​​.
To proceed, we pick one of the multiplying factors and set it arbitrarily to a
specific value. Let’s pick χ​ ​​ 3​​= 1​. Then the system of equations becomes
​(+1) ​χ1​  ​​ + (−1) ​χ2​  ​​ = 0​
(+1) ​χ2​  ​​ = 2
This system of equations can be solved to find: χ​ ​​ 1​​ = 2​ and χ​​ 2​  ​​ = 2.​
We can generalize this method as follows. Suppose we have K reactions
involving I compounds. To complete a generation-consumption analysis, we
1. Write balanced chemical equations for all K reactions.
2. Make a (vertical) list of all I compounds (reactants and products) in a column.
3. For each reaction k, write νik associated with each compound in a column.
There will be I rows, one for each compound, and K columns, one for each
reaction.
4. Adjust the net generation or consumption of compounds by finding multiplying factors χk. If we wish to have zero net generation or consumption
of compound i, we find χk such that
​
νi​  ,net​​ = ​  ∑ ​​​ ​χk​  ​​ ​νi​  k​​= 0
all k
Eq. (1.3)
5. Calculate the net generation or consumption of all compounds using
Eq. (1.2) and the χk that you found in step 4.
Why can we do step 4? Because stoichiometric coefficients give relative
quantities, or ratios, of reactants and products, not absolute quantities. If we
multiply a balanced chemical equation by a common factor, the equation is still
balanced.
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Section 1.4 Generation-Consumption Analysis
15
Why should we do step 4? There are many reasons; for example, we may
want to avoid net consumption of an expensive compound, or net generation
of a toxic or hazardous byproduct.
Note: it is not always possible to find χk that will satisfy Eq. (1.3). In that
case, we may need to search for additional or different chemical reactions to
achieve our goals.
Generation-consumption analysis is illustrated in the next two examples.
Example 1.4
Generation-Consumption Analysis: Ammonia Synthesis
Ammonia is one of the largest-tonnage chemicals produced today. Ammonia synthesis proceeds by reacting steam (H2O) with methane (CH4) to make carbon monoxide and hydrogen. Then CO and water react to make CO2 and more H2. Finally
nitrogen (N2) and hydrogen combine to produce ammonia, NH3.
How can we combine these reactions so there is no net generation or consumption of CO or H2?
Solution
We start with balanced chemical equations:
​C​H​4​​ + ​H2​ ​​O → CO + 3​H2​ ​​​
(R1)
​CO + ​H​2​​O → C​O2​ ​​ + ​H2​ ​​​
(R2)
​​N​2​​ + 3​H2​ ​​ → 2N​H3​ ​​​
(R3)
Let’s look at the generation-consumption table, using these balanced chemical equations as written, without yet considering multiplying factors.
Compound
R1
νi1
R2
νi2
R3
νi3
CH4
−1 −1
H2O
−1
−1 −2
CO
+1
−1
H2
+3
+1
−3
Net
νi,net
0
+1
CO2+1 +1
N2 −1
−1
NH3 +2
+2
This solution does satisfy the constraint that net CO = 0, but doesn’t satisfy the
requirement that net H2 = 0. Let’s write Eq. (1.3) for these two intermediates:
​CO: ​χ​ 1​​ − ​χ2​  ​​ = 0​
​H2:
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3​
χ1​  ​​ + ​χ2​  ​​ − 3​χ3​  ​​ = 0​
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Chapter 1 Converting the Earth’s Resources into Useful Products
All we need to do is find a set of values for ( χ1, χ2, χ3) that satisfies these two equations. Since there are two equations but three variables, there is more than one
valid solution. Because there is more than one valid solution, we can pick any
number greater than zero for the value of one of the multiplying factors, and then
solve for the other two. Let’s pick
χ1 = 1
Then,
​​χ2​  ​​= 1, ​χ3​  ​​ = _​ 43 ​​
By multiplying all entries in the (R1), (R2), and (R3) columns by these values for
χ1, χ2, and χ3, respectively, we get the result we want:
Compound
χ1νi1
χ2 νi2
χ3 νi3
νi,net
CH4
−1 −1
Quick Quiz 1.5
H2O
−1
−1 −2
In Example 1.4, what’s
the net reaction if you
set χ2 = 3 and solve for
χ1 and χ3?
CO
+1
−1 0
H2
+3
+1
Would it be possible to
combine the set of reactions in Example 1.4
such that there is no
net generation or consumption of CO2?
N2 −4/3
−4/3
NH3 +8/3
+8/3
Example 1.5
−4 0
CO2+1 +1
The net reaction for ammonia synthesis is read from the last column:
​C​H4​ ​​ + 2​H2​ ​​O + _​ 43 ​ ​N2​ ​​ → C​O2​ ​​ + _​  83 ​ N​H3​ ​​​
Generation-Consumption Analysis: The Solvay Process
Limestone (CaCO3) decomposes to lime (CaO), and lime reacts with water to form
“milk of lime,” Ca(OH)2
​​CaCO​3​​→ CaO + ​CO​2​​​
(R1)
​CaO + ​H2​ ​​O → Ca​(OH)​2​​​
(R2)
Milk of lime reacts with ammonium chloride to make ammonia and calcium chloride:
​​Ca(OH)​2​​ + 2​NH​4​​Cl → 2​NH​3​​ + ​CaCl​2​​ + 2​H2​ ​​O​
(R3)
Ammonia dissolved in water makes ammonium hydroxide, which reacts with CO2
to make ammonium carbonate and then ammonium bicarbonate:
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​​NH​3​​ + ​H2​ ​​O → ​NH​4​​OH​
(R4)
​2​NH​4​​OH + ​CO​2​​ → ​(​NH​4​​)​2​​​CO​3​​ + ​H2​ ​​O​
(R5)
​​(​NH​4​​)2​ ​​​CO​3​​ + ​CO​2​​ + ​H2​ ​​O → 2​NH​4​​​HCO​3​​​
(R6)
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17
Section 1.4 Generation-Consumption Analysis
Ammonium bicarbonate reacts with sodium chloride to produce sodium bicarbonate
and generate more ammonium chloride:
​​NH​4​​​HCO​3​​+ NaCl → ​NH​4​​Cl + ​NaHCO​3​​​
(R7)
Finally, sodium bicarbonate (NaHCO3, common baking soda) decomposes to the
desired product, sodium carbonate, releasing carbon dioxide and water as byproducts:
​2NaHC​O3​ ​​ → N​a2​ ​​C​O3​ ​​ + C​O2​ ​​ + ​H2​ ​​O​
(R8)
Can we use these reactions to come up with a process for making sodium carbonate
from limestone and salt that makes efficient use of raw materials?
Solution
These 8 reactions involve 14 different compounds. Let’s use the generationconsumption analysis in Table 1.1a to evaluate this set of chemical reactions,
without yet considering any multiplying factors.
Table 1.1a
Generation-Consumption Analysis of the Solvay
Process (first try)
Compound
R1
νi1
CaCO3
−1
CaO
+1
CO2
+1−1
R2 R3
νi2 νi3
R4
νi4
R5
νi5
R6
νi6
R7
νi7
R8
Net
νi8 νi,net = ∑ νik
−1
−1
+1
−1
0
−1 +1
0
−1 +1
+1
H2O−1
+2
Ca(OH)2
+1
−1
0
NH4Cl
−2
+1
−1
NH3
+2
+1
−1
NH4OH
+1
−2
(NH4)2CO3+1
−1
NH4HCO3+2
−1
−1
0
+1
NaCl
−1
−1
NaHCO3
+1
−1
−2
CaCl2
+1
+1
Na2CO3
+1
+1
We are using 1 mole CaCO3 and 1 mole NaCl to make 1 mole Na2CO3 (the
desired product), but we are also consuming or generating a lot of other chemicals.
Could we synthesize a reaction pathway with no net consumption of any raw materials other than CaCO3 and NaCl, and no net generation or consumption of any of the
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Chapter 1 Converting the Earth’s Resources into Useful Products
ammonia-containing compounds? In other words, can we find multiplying factors χk
such that the entry in the Net column equals zero for NH4Cl, NH3, NH4OH, (NH4)2CO3,
NH4HCO3, and NaHCO3? Applying Eq. (1.3) to these six compounds gives
NH4Cl:​−2​χ3​  ​​ + χ​ 7​  ​​ = 0​
NH3:​2​χ3​  ​​ − ​χ4​  ​​ = 0​
NH4OH:​​χ​ 4​​ − 2​χ5​  ​​ = 0​
(NH4)2CO3:​​χ​ 5​​ − ​χ6​  ​​ = 0​
NH4HCO3:​​2χ​ 6​​ − ​χ7​  ​​ = 0​
NaHCO3:​​χ​ 7​​ − 2​χ8​  ​​ = 0​
One solution that satisfies all these constraints is
​​χ3​  ​​​ = ​​χ5​  ​​​ = χ​​ 6​  ​​​ = ​​χ8​  ​​​ = 1
​​χ4​  ​​​ = ​​χ7​  ​​​ = 2
Let’s see what happens if we multiply the stoichiometric coefficients for reactions
(R4) and (R7) by 2 and multiply all other reactions by 1 (Table 1.1b):
Table 1.1b
Generation-Consumption Analysis of the Solvay
Process (second try)*
Compound R1 R2 R3 R4 R5 R6 R7 R8
Net
νi1 νi2 νi3 χ4 νi4 νi5 νi6 χ7 νi7 νi8 νi,net = ∑ χk νik
CaCO3
−1
CaO
+1
CO2
+1−1
−1
+1
0
−1+1
0
−1 +1
0
H2O −1
+2
Ca(OH)2
+1
−1
−2
−1
0
NH4Cl−2+2
0
NH3+2
0
−2
NH4OH+2
−2
(NH4)2CO3+1
−1
NH4HCO3+2
0
0
−2
0
NaCl−2
−2
NaHCO3+2
−2
0
CaCl2+1
+1
Na2CO3+1
+1
*Changes are shown in bold.
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Section 1.4 Generation-Consumption Analysis
19
Perfect! All of the ammonia-containing compounds are now strictly intermediates, with no net generation or consumption. Furthermore, there is no net consumption of NaHCO3. Remarkably, the net effect of this pathway of 8 chemical reactions
involving 14 compounds is simply (from the last column of Table 1.1b):
​CaC​O3​ ​​+ 2NaCl → N​a2​ ​​C​O3​ ​​ + CaC​l2​ ​​​
1.4.1
Using Matrices in Generation-Consumption Analysis
Matrix math can be used to find the correct values of xk in Eq. (1.2) and (1.3).
These methods are particularly useful for systems of large numbers of reactions, because the matrix equation can be developed by inspection. Matrix
methods are even useful in cases where it is not obvious whether the reactions
can be combined in such a way that a compound can serve as an intermediate
or must be a byproduct! Our goal is to find an equation Ax = b, where A is
a matrix containing the stoichiometric coefficients of all compounds that have
net-zero generation/consumption, x is the vector containing the (unknown) multiplying factors, and b is a vector containing the stoichiometric coefficients for
one reaction chosen as the basis reaction. Then we solve for x and complete
the generation-consumption analysis. Here is the procedure to follow:
1. List all the compounds that appear in any reaction. To write the matrix A,
list the stoichiometric coefficients for each reaction in a column, in the
order of the compounds in your list. A will have I rows (one for each
compound) and K columns (one for each reaction). There should be at least
as many compounds as there are reactions.
2. Scan the rows of A. Cross out any rows that have only a single nonzero
entry. These rows correspond to compounds that appear in only a single
reaction in the reaction system. Such compounds cannot have net-zero
generation/consumption and so cannot be intermediates: They must be
either a reactant or a product.
3. If there is one fewer row than column in matrix A, go to step 4. If not,
scan the remaining rows of A and identify any compounds that are acceptable
as raw materials and/or products. Such compounds may be “acceptable”
because they are nontoxic, or because they are cheap raw materials or
valuable byproducts. Continue crossing out rows of acceptable compounds
until there is one fewer row than column in A (equivalently, there is one
fewer compound than reaction).
4. Choose one of the reactions (one of the columns) to serve as a basis reaction. Let b = a column vector containing the negative of the stoichiometric
coefficients of the basis reaction. Delete that column from matrix A.
5. Check that you now have a square coefficient matrix A with an equal
number of rows and columns, a variable vector x that is the unknown
multiplying factors, and a vector b that describes your basis reaction. Solve
for x. Use the solution to complete the generation-consumption analysis.
The procedure sounds more complicated than it is. Example 1.6 illustrates the idea.
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Chapter 1 Converting the Earth’s Resources into Useful Products
Example 1.6
Generation-Consumption Analysis Using Matrix Math:
Nitric Acid Synthesis
We want to develop a reaction pathway to make nitric acid (HNO3) from readily
available and cheap raw materials. We think some combination of the following
reactions might be useful:
​​O​2​ + 2C​H4​ ​→ 2CO + 4​H2​ ​​
(R1)
​CO + ​H​2​O → C​O2​ ​ + ​H2​ ​​
(R2)
​​N​2​ + 3​H2​ ​ → 2N​H3​ ​​
(R3)
​4N​H​3​ + 5​O2​ ​→ 4NO + 6​H2​ ​O​
(R4)
​2NO + ​O2​ ​ → 2N​O2​ ​​
(R5)
​3N​O​2​ + ​H​2​O → 2HN​O​3​ + NO​
(R6)
Use matrix methods to combine these reactions into a pathway to make nitric acid.
Preferably, we’d like to use inexpensive and readily available raw materials like
water, methane (CH4), and oxygen and nitrogen from air, and we want to avoid
any net generation of toxic or environmentally damaging compounds such as NO,
NO2, NH3, and CO.
Solution
1. We list the compounds involved and immediately write down the matrix of
stoichiometric coefficients from the balanced chemical reactions:
R1
R2
R3
R4
R5
R6
O2
−1
0
0
−5
−1
0
0
0
−1
0
0
0
CH4
−2
0
0
0
0
0
H2O
0
−1
0
6
0
−1
CO
2
−1
0
0
0
0
CO2
N2
0
1
0
0
0
0
H2
4
1
−3
0
0
0
NH3
0
0
2
−4
0
0
NO
0
0
0
4
−2
1
NO2
0
0
0
0
2
−3
0
0
0
0
0
2
HNO3
2. Next we scan the list and eliminate any rows (compounds) with just one entry.
This includes necessary reactants N2 and CH4 and the desired product HNO3.
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Section 1.4 Generation-Consumption Analysis
21
We also observe that CO2 must be a product of this reaction pathway, because
it appears in only one reaction. Our matrix becomes:
R1
O2
H2O
R2
R3
R4
R5
−1
0
0
−5
−1
0
0
−1
0
6
0
−1
2
−1
0
0
0
0
4
1
−3
0
0
0
0
0
2
−4
0
0
0
0
0
4
−2
1
0
0
0
0
2
−3
CO
H2
NH3
NO
NO2
R6
3. We have seven compounds but only six reactions; according to our procedure
we need to have one fewer compound than reaction. We look for two materials
that are acceptable raw materials or byproducts. O2 and H2O fit the bill. We
eliminate them from consideration. The remaining five compounds can all be
net-zero compounds! The matrix becomes:
CO
H2
⎢
R1 R2 R3 R4
⎡2 −1
0
0
4
1 −3
0
⎥
R5 R6
0
0⎤
0
0
0​
2​  −4​
0​
0​ ​​
​​NH​ ​3​ ​​ ​​ 0​
​
NO
0
0
0
4 −2
1
NO2
⎣0
0
0
0
2 −3⎦
4. We arbitrarily choose one of the reactions to serve as the basis reaction—let’s
choose (R1). We create the b vector by multiplying the column corresponding
to (R1) by −1, and then we delete that column from A. The x vector is simply
the listing of the multiplying factors for the remaining reactions. We end up with
⎢
⎡−1
0
0
0
⎥⎢ ⎥ ⎢ ⎥
0⎤ ⎡ χ2⎤
0 χ3
⎡−2⎤
1 −3
0
0
−4
χ
0
2
−4
0
0
​​ ​
​
​
​
​
​ ​​ ​​ ​ 4​ ​​ = ​​ ​ 0​​  ​​
0
0
4 −2
1 χ5
0
0
0
2 −3⎦ ⎣ χ6⎦ ⎣ 0⎦
⎣ 0
5. The columns in the matrix A correspond to the five remaining reactions,
(R2) through (R6). The rows in the matrix correspond to the stoichiometric
coefficients of the remaining compounds: CO, H2, NH3, NO, and NO2. These
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Chapter 1 Converting the Earth’s Resources into Useful Products
are the compounds where we want to have no net generation or consumption.
We solve, by calculator or by computer, and find the multiplying factors:
⎢⎥
⎡2⎤
2
x = ​​ 1​ ​​  ​​
3
⎣2⎦
Finally, we multiply the stoichiometric coefficients ν​ ​​​i ​k̇ ​​​ by the corresponding
multiplying factor x​ ​​ k​​​to complete the generation-consumption table:
Compound
R1
R2
R3
R4
O2
−1−5
R5
R6
Net
−3 −9
N2−2 −2
CH4
−2−2
H2O−2+6−2
CO
+2
+2
−2 0
CO2
+2+2
H2
+4
+2
−6
NH3+4
0
−4 0
NO +4
−6
+2
0
NO2
+6
−6
0
HNO3
+4
+4
The net overall reaction is
​9​O2​ ​​ + 2​N2​ ​​ + 2C​H4​ ​​ → 2​H2​ ​​O + 2C​O2​ ​​ + 4HN​O3​ ​​​
1.5
A First Look at Material Balances
and Process Economics
In this section, we’ll examine how to use the results from a generation-consumption
analysis to calculate the mass of raw materials needed to produce a specified
mass of product, the mass of byproducts produced per mass of desired product,
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Section 1.5 A First Look at Material Balances and Process Economics
23
and the cost of raw materials per mass of desired product. These are simple but
essential calculations in the early stages of chemical process synthesis, as we
evaluate alternative choices of raw materials and chemical reaction pathways.
1.5.1
Mass, Moles, and Molar Mass
Let’s briefly review a few definitions.
Atomic mass is expressed in terms of atomic mass units (amu). One amu is
equal to one-twelfth the mass of a carbon 12C atom, or 1.66053873 × 10−27 kg.
The atomic mass reported in periodic tables is the compositional average mass
of that element, averaged over the distribution of isotopes in nature. The atomic
mass of carbon = 12.011 amu (taking into account 12C, 13C, and 14C isotopes),
while the relative atomic mass (dimensionless) of 12C = 12. Refer to Appendix B
for a listing of atomic mass and number of the elements.
Molecular mass is the sum of the atomic masses of all the atoms in a
molecule. Molar mass is the mass in grams of one mole (6.02214199 × 1023)
of atoms or molecules. The molar mass is numerically equivalent to molecular
mass but has units of [grams/gram-mole], abbreviated as [g/gmol].
Illustration: Glucose (C6H12O6) contains 6 carbons (atomic mass of 12.011 amu),
12 hydrogens (atomic mass of 1.0079 amu), and 6 oxygens (atomic mass of
15.9994 amu). The molecular mass of glucose is 6(12.011) + 12(1.0079) +
6(15.9994) = 180.157 amu. The molar mass of glucose is 180.157 g/gmol.
To convert from moles to mass, multiply the total moles by the molar mass.
To convert from mass to moles, divide the total mass by the molar mass.
180.157 g glucose
Illustration: ​100.0 gmol glucose​ ​  _______________
​​ = 18,020 g glucose​
( gmol glucose )
gmol glucose
_______________
​100.0 g glucose​ ​
​ ​ = 0.5551 gmol glucose​
( 180.157 g glucose )
For calculations that do not warrant a high level of accuracy, it is common
practice to use approximate values for atomic mass. It is worth memorizing the
following:
H=1
C = 12
N = 14
O = 16
S = 32
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Chapter 1 Converting the Earth’s Resources into Useful Products
Illustration: The molar mass of glucose (C6H12O6) is 6(12) + 12(1) + 6(16) =
180 g/gmol.
180 g glucose
​100 gmol glucose​ ​  ____________
​ ​ = 18,000 g glucose​
( gmol glucose )
gmol glucose
____________
​100 g glucose​ ​
​ ​ = 0.5​5̅ ​g glucose​
( 180 g glucose )
You will encounter many different systems of units throughout your career.
Although SI units (kilograms, meters, seconds, K) are used in most scientific
venues, many industries still use the British system (pounds, feet, seconds, °F).
To convert from one unit of mass to another, use the following conversion
factors:
​1 lb = 453.59 g = 16 oz = 0.45359 kg = 5 × 1​0​−4​ ton​
​1 kg = 1000 g = 35.274 oz = 2.2046 lb = 1​0​−3​metric ton​
​1 ton = 907,180 g = 2000 lb = 907.18 kg = 0.90718 metric ton​
​1 metric ton = 1​0​6​g = 2204.6 lb = 1000 kg = 1.1023 ton​
Quick Quiz 1.6
How many gmol of
ethanol (C2H5OH) are
contained in 104 grams
of ethanol?
How many grams of
ethanol are contained in
104 gmol of ethanol?
How many kg of
ethanol are in 104 lb?
How many kgmol of
ethanol are in 104 lb?
lb ​ = 10,150 lb​
_______
Illustration: ​5.075 tons​(​  2000
​
ton )
907.18 kg
35.274 oz
​5.075 ton​ ​  _________
​
​​ ​  _________
​
​ = 162,400 oz​
( ton )(
)
kg
Since many of our calculations will give mass in units of lb, kg, tons—units
other than grams—it is useful to define molar mass in different units. The
molar mass may be written as [lb/lbmol], [kg/kgmol], [ton/tonmol], or any
other convenient units. The numerical value of the molar mass of a compound
in any of these units is identical. The conversion factors given for mass units
can be used to convert from one molar unit to another.
​1 lbmol = 453.59 gmol = 16 oz.mol = 0.45359 kgmol = 5 × 1​0​−4​ tonmol​
​1 kgmol = 1000 gmol = 35.274 oz.mol = 2.2046 lbmol = 1​0​−3​metric tonmol​
1 tonmol = 907,180 gmol = 2000 lbmol = 907.18 kgmol
= 0.90718 metric tonmol
1 metric tonmol = 1​0​6​gmol = 2204.6 lbmol = 1000 kgmol = 1.1023 tonmol
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Section 1.5 A First Look at Material Balances and Process Economics
Illustration: The molar mass of glucose (C6H12O6) is 180.157 g/gmol; it is also
180.157 lb/lbmole, 180.157 kg/kgmol, or 180.157 ton/tonmol.
The mass of 104.2 kgmol of glucose is
180.157 kg glucose
104.2 kgmol glucose​​ ​  ________________
​​​ = 18,770 kg glucose
( kgmol glucose )
1.5.2
Atom Economy
Atom economy gives a rapid and simple measure of the efficiency of a reaction pathway in converting reactants to products:
Helpful Hint
The sum in the
denominator
includes only the
compounds with
net consumption.
mass of desired product
____________________
​Fractional atom economy = ​
​​
total mass of reactants
A mathematical expression for atom economy is
​νP​ ​​MP​ ​
Fractional atom economy = ​ __________________
​
− ∑ ​νi​​​ M​i​
Eq. (1.4)
all reactants
where νP is the stoichiometric coefficient and MP is the molar mass for the
desired product P, while νi is the stoichiometric coefficient and Mi is the molar
mass of reactant i. In pathways where multiple reactions are combined, the
stoichiometric coefficients in Eq. (1.4) are the net coefficients.
Calculating the fractional atom economy for a reaction pathway is straightforward once a generation-consumption analysis has been completed. Notice that
the atom economy tells you the best you could ever do, given the chosen reaction
pathway. A real process will never achieve quite as good utilization of raw materials as the calculated atom economy.
All else being equal, reaction pathways with high fractional atom economy
are preferable; these should have fewer waste products and, by making good
use of the raw materials, should be more cost-efficient.
Example 1.7
Atom Economy: LeBlanc versus Solvay
The LeBlanc process was an old way to make sodium carbonate. The net reaction is
​2NaCl + ​H2​ ​S​O4​ ​+ 2C + CaC​O3​ ​ → N​a2​ ​C​O3​ ​+ 2HCl + 2C​O2​ ​+ CaS​
Compare the atom economy of the LeBlanc process to that of the Solvay process
from Example 1.5.
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Chapter 1 Converting the Earth’s Resources into Useful Products
Solution
We list the stoichiometric coefficient νi and the molar mass Mi of all reactants and
the desired product, Na2CO3, in table form. For the LeBlanc process:
Compound
νi
Mi
νi Mi
NaCl
−2
58.5
−117
H2SO4
−1
98
−98
C
−2
12
−24
CaCO3
−1
100
−100
Na2CO3
+1
106
+106
We then calculate the fractional atom economy, using Eq. (1.4):
​ν​p​ ​M​p​
106
_________
​
​ = _______________________________
​
​= 0.31
− ∑ ​νi​​ ​Mi​​ −[(−117) + (−98) + (−24) + (−100)]
all reactants
The net reaction of the Solvay process (Example 1.5) is
​2NaCl + CaC​O3​ ​ → N​a2​ ​C​O3​ ​ + CaC​l2​ ​​
From the stoichiometric coefficients and the molar masses, we calculate
​νp​ ​ ​Mp​ ​
106
​​ _________ ​​ = _________________
​​
​​ = 0.49
− ∑ ​ν​i​ ​M​i​ −[(−117) + (−100)]
all reactants
The Solvay process makes much better use of its raw materials.
Example 1.8
Atom Economy: Improved Synthesis of 4-ADPA
4-ADPA (4-aminodiphenylamine, C6H5NHC6H4NH2) is used to make compounds
that reduce degradation of rubber tires. The traditional process required four
reactions: chlorination of benzene to chlorobenzene, reaction with nitric acid to
make PNCB (p-nitrochlorobenzene), reaction of PNCB with formaniline to make
4-NDPA, and hydrogenation of 4-NDPA to 4-ADPA. The balanced chemical
equations are
mur83973_ch01_001-060.indd 26
​​C​6​​​​H6​ ​​ + C​​l2​ ​​ → ​​C6​ ​​​​H5​ ​​Cl + HCl
(R1)
​​C​6​​​​H5​ ​​Cl + HN​​O3​ ​​ → ​​C6​ ​​​​H4​ ​​ClN​​O2​ ​​ + ​​H2​ ​​O
(R2)
​​C​6​​​​H4​ ​​ClN​​O2​ ​​ + ​​C6​ ​​​​H5​ ​​NHCHO + 0.5​​K​2​​C​​O3​ ​​ →
​​C​6​​​​H5​ ​​NH​​C6​ ​​​​H4​ ​​N​​O2​ ​​+ KCl + CO + 0.5C​​O​2​​ + 0.5​​H2​ ​​O
(R3)
​​C​6​​​​H5​ ​​NH​​C6​ ​​​​H4​ ​​N​​O2​ ​​ + 3​​H2​ ​​ → ​​C6​ ​​​​H5​ ​​NH​​C6​ ​​​​H4​ ​​N​​H2​ ​​ + 2​​H2​ ​​O
(R4)
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27
Section 1.5 A First Look at Material Balances and Process Economics
In the early 1990s, a new process was developed and commercialized. The new
process requires only two reaction steps, starting with nitrobenzene and aniline:
​​C​6​​H5​ ​N​O2​ ​ + ​C6​ ​​H5​ ​N​H2​ ​ → ​C6​ ​​H5​ ​NH​C6​ ​​H4​ ​NO + ​H2​ ​O​
(R1)
​​C​6​​H5​ ​NH​C6​ ​​H4​ ​NO + 2​H2​ ​ → ​C6​ ​​H5​ ​NH​C6​ ​​H4​ ​N​H2​ ​ + ​H2​ ​O​
(R2)
What is the difference in atom economy between the traditional and the new
processes?
Solution
First let’s complete a generation-consumption analysis of the traditional scheme:
Compound
νi1
C6H6
−1
−1
Cl2
−1
−1
C6H5Cl
+1
HCl
+1
+1
νi2
νi3
νi4
νi,net
−1 0
HNO3 −1−1
C6H4ClNO2 +1
H2O +1
−1 0
+0.5
+2
+3.5
C6H5NHCHO−1
−1
K2CO3−0.5−0.5
C6H5NHC6H4NO2 +1
−1
0
KCl+1
+1
CO+1
+1
CO2+0.5+0.5
H2
−3
−3
C6H5NHC6H4NH2
+1
+1
Now let’s calculate the atom economy, using the stoichiometric coefficients from
the “net” (last) column. We need to consider only the reactants (negative stoichiometric coefficient) and the desired product (4-ADPA) in the atom economy
calculation.
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Chapter 1 Converting the Earth’s Resources into Useful Products
Compound
νi
Mi
νi Mi
C6H6
−1
78
−78
Cl2
−1
71
−71
HNO3
−1
63
−63
C6H5NHCHO
−1
121
−121
K2CO3
−0.5
138
−69
H2
−3
2
−6
C6H5NHC6H4NH2
+1
184
+184
​ν​ ​ ​M​ ​
184
_________
__________________________________________
​​  P P  ​​ =
​​
​​ = 0.45
− ∑ ​νi​​ ​Mi​​ −[(−78) + (−71) + (−63) + (−121) + (−69) + (−6)]
all reactants
Now let’s complete a generation-consumption analysis of the new scheme:
Compound
νi1
C6H5NO2
−1−1
C6H5NH2
−1−1
C6H5NHC6H4NO
+1
H2O
+1+1
νi2
−1
νi,net
0
H2−2
−2
C6H5NHC6H4NH2+1
+1
and let’s calculate the atom economy of the new scheme:
mur83973_ch01_001-060.indd 28
Compound
νi
Mi
νi Mi
C6H5NO2
−1
123
−123
C6H5NH2
−1
93
−93
H2
−2
2
−4
C6H5NHC6H4NH2
+1
184
+184
23/11/21 4:15 PM
Section 1.5 A First Look at Material Balances and Process Economics
​ν​ ​ ​M​ ​
184
_________
____________________
​​  P P  ​​ =
​​
​​ = 0.84
− ∑ ​νi​​ ​Mi​​ −[(−123) + (−93) + (−4)]
Quick Quiz 1.7
Refer to Example 1.4.
What is the fractional
atom economy for
ammonia synthesis?
29
all reactants
Converting from the traditional to the new process increases the atom economy
from 0.45 to 0.84. This is a remarkable achievement, which was recognized by a
Presidential Green Chemistry Challenge Award.
1.5.3
Process Economy
Once we have chosen raw materials, desired products, and the reaction pathway,
and we have completed the generation-consumption analysis, we can calculate
the process economy. We ask: what is the total quantity of raw material required
for a given amount of product? what is the cost of the raw materials, the value
of the product, and the net profit (or loss)?
We start with the results from the generation-consumption analysis, which
gives the relative molar quantities of raw materials consumed and products
generated. Then we need to do a few simple steps:
1. Convert moles to mass.
2. Scale up or scale down.
3. Convert mass to money.
Let’s discuss each one of these steps in turn.
1. Convert moles to mass. To convert moles to mass, we simply multiply the
stoichiometric coefficient νi by its molar mass Mi. This calculation gives
a relative mass quantity, since the stoichiometric coefficients give a relative
rather than absolute molar quantity.
2. Scale up or scale down. The desired production rate provides a basis for all
subsequent calculations. A basis is a quantity or flow rate that indicates the
size of the process. Either a raw material or a product can serve as the basis
compound. We scale up from a relative mass quantity to the basis quantity
by using a scale factor, which is simply:
basis
Scale factor = __________________
​​
​​
relative mass quantity
The quantity of any raw material or product is calculated by simply multiplying the relative quantity of the raw material or product by the scale factor.
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Chapter 1 Converting the Earth’s Resources into Useful Products
3. Convert mass to money. We calculate the raw material costs and the product
value by multiplying the quantity of each compound by its unit price.
To evaluate the overall process economy, we sum up the cost of all
raw materials and the value of all products. For a process to be profitable,
this sum must be greater than zero.
Example 1.9
Process Economy: The Solvay Process
The Solvay process (Example 1.5) consumes limestone (CaCO3) and salt (NaCl)
to produce soda ash (Na2CO3), with calcium chloride (CaCl2) as a byproduct. If
we wish to produce 1000 tons soda ash/day, what are the required feed rates of
limestone and salt?
Suppose current prices for bulk quantities are $87/ton for CaCO3, $95/ton for
NaCl, $105/ton for Na2CO3, and $250/ton for CaCl2. Does the Solvay process make
economic sense if the byproduct CaCl2 cannot be sold? How does the economic
picture change if there is a market for the byproduct?
Solution
From Table 1.1b of Example 1.5, we see that there is net consumption or generation of 4 compounds: NaCl, CaCO3, Na2CO3, and CaCl2. These compounds, along
with their stoichiometric coefficients from their net reaction, are listed in the first
two columns of Table 1.2.
Table 1.2
Raw Material Requirements and Process Economics for 1000 tons/day
Soda Ash Production
$/day,
tons/day,
tons/day ×
Compound
νi
Mi
νi Mi
νi Mi × SF
$/ton
$/ton
NaCl
−2
CaCO3
−1
Na2CO3
CaCl2
58.5
−117
−1104
95
−105,000
100
−100
−943
87
−82,000
+1
106
+106
1000
105
+105,000
+1
111
+111
1047
250
+262,000
Sum (w/o CaCl2)−82,000
Sum (w/ CaCl2)
0
0+180,000
*A negative number indicates a material consumed, or a cost. A positive number indicates a material generated, or income. SF = 1000/106.
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Section 1.5 A First Look at Material Balances and Process Economics
Helpful Hint
The sum of the
mass of all materials
consumed should
equal the mass
of all materials
generated.
31
To convert moles to mass, we multiply the stoichiometric coefficients νi by
the molar mass Mi to get the relative mass (column 4). Then, we scale up. The
relative mass of Na2CO3 is 106, and the desired production rate is 1000 tons/day
of Na2CO3, so the scale factor SF = 1000/106. We multiply the numbers in column
4 by the scale factor to get the tons/day consumed or generated for all compounds.
Finally, we convert mass to money. The cost of raw materials and the selling price
of products, per ton, are listed in column 6. By multiplying tons/day by $/ton, we
get $/day.
The mass of raw materials consumed should equal the mass of products made.
We check this by summing up all the numbers in the tons/day column (pay attention to the sign of each number). It should sum to zero.
Considering just raw materials costs, if we were unable to sell the calcium
chloride, there would be a net loss of $82,000/day! This does not include energy
costs, labor costs, or capital equipment costs—all of which contribute substantially to the overall process economics. If there is a market for calcium chloride,
then we make a profit of $180,000/day of Na2CO3. In fact, at these prices, we
might consider the Solvay process as a way to make CaCl2, with soda ash as a
byproduct!
Quick Quiz 1.8
In Example 1.9 we
evaluated a process for
making 1000 tons/day
sodium carbonate,
which had a value of
$105/ton. Use Table 1.3
to categorize this process as “commodity”
or “specialty.”
Table 1.3
1.5.4
Process Capacities and Product Values
Chemical process facilities vary enormously in scale; some are small enough
to fit in your hand while others occupy several city blocks. Chemical products
vary enormously in value; some are bought with the spare change in your
pocket while others are more precious than gold. Table 1.3 gives some useful
order-of-magnitude numbers regarding the scale of chemical processes and the
value of chemical products.
Typical Plant Capacities, Product Values, and Waste Generation for
Chemical Processes
Typical plant
Typical product
Chemical category
capacity, lb/year
value, $/lb
Typical waste
generation,
lb waste/lb product
Petroleum
1 billion–100 billion
0.1
0.1
Bulk (commodity)
10 million–1 billion
0.1–2
<1–5
Fine (specialty)
100 thousand–10 million
2–10
2–50
Pharmaceuticals
1 thousand–100 thousand
10–infinity
10–100
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Chapter 1 Converting the Earth’s Resources into Useful Products
Six-Carbon Chemistry
In this case study, we illustrate how the concepts introduced in Chap. 1 are
used to make decisions about raw materials, products, and reaction pathways,
by looking in some depth at specific processes of importance in the organic
chemicals business. These processes are linked by their connection to 6-carbon
compounds. We’ll look at two questions:
1. Benzene is a 6-carbon compound purified from petroleum. Suppose we
have available 15,000 kg/day benzene. What are some useful 6-carbon
products we might make from benzene?
2. Could we replace benzene with a raw material from a renewable resource
to make the same 6-carbon products?
H
H
C
C
H
C
C
C
C
H
H
H
Figure 1.5 Three different representations of the structure of benzene, C6H6, one of the
most important raw materials in the synthetic organic chemicals industry.
Simple organic compounds like benzene serve as raw materials in the production of the plastics, detergents, pharmaceuticals, and fibers that are ubiquitous in modern societies. Think, for example, of nylon. Nylon was first sold
commercially in 1940, in the early days of World War II. The fiber rapidly
became an indispensable element in the war effort, as it was used for parachutes,
tents, ropes, airplane tire cords, and other military essentials. Perhaps nylon’s
greatest commercial success was in women’s hosiery, as nylon stockings
replaced the silk stockings formerly supplied by the Japanese.
There are several kinds of nylon, of which one of the most important is
called nylon 6,6. Nylon 6,6 is a polymer—a very large macromolecule containing many small repeating units linked by covalent bonds. Nylon 6,6 contains
O
OH
O
O
OH
Benzene, C6H6
Cyclohexane, C6H12
Cyclohexanone, C6H10O
Adipic acid, C6H10O4
Figure 1.6 Benzene is converted to adipic acid through a series of chemical reactions involving intermediates
cyclohexane and cyclohexanone.
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Section 1.5 A First Look at Material Balances and Process Economics
33
two repeating units, both of which are 6-carbon compounds: hexamethylene
diamine and adipic acid. We will focus our attention on the manufacture of
adipic acid from benzene. The structures of adipic acid and of important intermediates are shown; notice that the 6-carbon structure is conserved.
Reaction 1.
Benzene is hydrogenated to cyclohexane:
​​C​6​​H6​ ​ + 3​H2​ ​ → ​C6​ ​​H1​ 2​​
Reaction 2.
Cyclohexane is partially oxidized with oxygen, producing
cyclohexanone (C6H10O) and water:
​​C​6​​H1​ 2​ + ​O2​ ​ → ​C6​ ​​H1​ 0​O + ​H2​ ​O​
Reaction 3.
(R1)
(R2)
Cyclohexanone is oxidized with nitric acid to make adipic acid:
​​C​6​​H1​ 0​O + 2HN​O3​ ​ → ​C6​ ​​​H1​ 0​O4​ ​+ 2NO + ​H2​ ​O​
(R3)
The generation-consumption analysis is shown in Table 1.4. There is zero net
generation/consumption of the intermediates cyclohexane and cyclohexanone,
so no further adjustments are needed.
H2
O2
O
H2O
HNO3
OH
O
NO, H2O
O
OH
Figure 1.7 Reaction pathway from benzene to adipic acid, showing other raw materials and
byproducts.
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Chapter 1 Converting the Earth’s Resources into Useful Products
Table 1.4
Generation-Consumption Analysis of Benzene-to-Adipic
Acid Process
Compound
νi1
C6H6
−1
−1
H2
−3
−3
C6H12
+1
νi2
νi3
νi,net
−1
0
O2−1 −1
C6H10O+1
−1
0
HNO3−2
−2
C6H10O4+1
+1
NO+2
+2
H2O+1
+2
+1
The net reaction is:
​​C​6​​H6​ ​ + 3​H2​ ​ + ​O2​ ​ + 2HN​O3​ ​ → ​C6​ ​​H1​ 0​​O4​ ​+ 2NO + 2​H2​ ​O​
We consume one mole of benzene, 3 moles of hydrogen, 1 mole of oxygen,
and 2 moles of nitric acid to produce one mole of adipic acid. There are two
waste products: nitric oxide, which is released to the atmosphere, and water,
which goes down the drain (via a water treatment system, of course!). Release
of nitrogen oxide compounds is an environmental concern, but so far no commercial process has been developed that avoids the nitric acid oxidation that
leads to generation of nitrogen oxides.
Adipic acid, of which about 85 percent is used to make nylon 6,6, is one
possible value-added product to make from benzene. Are there other options?
One idea is catechol, an important feedstock for fine-chemical production.
Catechol is used to make pharmaceuticals like L-Dopa (used to treat Parkinson’s
OH
OH
OH
OH
OH
OH
Benzene,
C6H6
Phenol,
C6H5OH
Catechol,
C6H6O2
Resorcinol,
C6H6O2
OH
Hydroquinone,
C6H6O2
Figure 1.8 Dihydroxybenzenes and their precursors, benzene + phenol.
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35
Section 1.5 A First Look at Material Balances and Process Economics
disease) and flavorings like vanillin. Catechol is one of three isomers of dihydroxybenzene C6H6O2; the other two, hydroquinone (p-dihydroxybenzene) and
resorcinol (m-hydroxybenzene) are also industrially important chemicals.
(Isomers have identical molecular formulas, but the atoms are arranged in different geometries.) From the structure of catechol, it is easy to see why benzene
makes sense as a raw material.
Let’s look at the reaction pathway from benzene to catechol.
Reaction 1:
Benzene and propylene (C3H6) combine to make isopropylbenzene (C9H12, also called cumene):
​​C​6​​H6​ ​ + ​C3​ ​​H6​ ​ → ​C9​ ​​H1​ 2​​
Reaction 2:
(R1)
Cumene reacts with oxygen to give the unstable intermediate
cumene hydroperoxide (C9H12O2):
​​C​9​​H1​ 2​ + ​O2​ ​ → ​C9​ ​​H1​ 2​​O2​ ​​
Reaction 3:
(R2)
Cumene hydroperoxide breaks down into phenol (C6H6O) and
the byproduct acetone (C3H6O):
​​C​9​​H1​ 2​​O2​ ​ → ​C6​ ​​H6​ ​O + ​C3​ ​​H6​ ​O​
Reaction 4:
(R3)
Phenol reacts with hydrogen peroxide (HOOH), a strong oxidizing agent, to produce catechol:
​​C​6​​H6​ ​O + ​H2​ ​​O2​ ​ → o-​​C6​ ​H6​ ​​O2​ ​ + ​H2​ ​O​
(R4)
The generation-consumption analysis is shown in Table 1.5.
Table 1.5
Generation-Consumption Analysis for Benzene-toCatechol Process
Compound
νi1
C6H6
−1 −1
C3H6
−1 −1
C9H12
+1
νi2
νi3
νi4
−1
νi,net
0
O2−1 −1
C9H12O2 +1
−1
C6H6O +1
−1
0
0
C3H6O+1 +1
mur83973_ch01_001-060.indd 35
H2O2−1
−1
o-C6H6O2+1
+1
H2O+1
+1
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36
Chapter 1 Converting the Earth’s Resources into Useful Products
The net result is:
​​C​6​​H6​ ​ + ​C3​ ​​H6​ ​ + ​O2​ ​ + ​H2​ ​​O2​ ​ → o-​C6​ ​​H6​ ​​O2​ ​ + ​H2​ ​O + ​C3​ ​​H6​ ​O​
Overall, we’ve consumed 1 mole of benzene, 1 mole of propylene, and two
different oxygen sources, O2 and H2O2, to make 1 mole of catechol. Unlike
the adipic acid case, we’ve produced a byproduct that is valuable: acetone is
a useful solvent and feedstock for synthesis of other organic chemicals.
We’ve identified two useful products we might make from benzene. How
do the two processes compare on atom economy? Considering only the cost of
the raw material and the value of the products, what is the best course of
action? Assume benzene is valued at $0.41/kg. (The price of benzene changes
dramatically with changes in crude oil prices.)
Option 1:
Sell the benzene. Selling 15,000 kg/day benzene at this price
generates
15,000 kg benzene __________
$0.41
$6150
________________
​​
​ × ​
​ = ​ ______
​​
day
kg benzene
day
Option 2:
Make adipic acid. The generation-consumption analysis for this
option was shown in Table 1.4. We obtain pricing information
from ICIS Chemical Business or other sources and complete the
analysis in tabular form, as shown below. The basis for the
calculation is 15,000 kg/day benzene consumed.
The fractional atom economy is
​ν​ ​ ​M​ ​
146
_________
​​  P P  ​​ = ______________
​​
​​ = 0.60
− ∑ ​ν​i​ ​M​i​ (78 + 6 + 32 + 126)
all reactants
Table 1.6
Process Economy for Benzene-to-Adipic Acid Process
kg/day
Compound
νi1
Mi
νi1 Mi (SF = 192.3)* $/kg
$/day
C6H6
−1
78
−78
−15,000
0.41
−6,150
H2
−3
2
−6
−1,154
0.2
−230
O2
−1
32
−32
−6,154
~0
0
HNO3
−2
63
−126
−24,230
0.40
−9,700
C6H10O4
+1
146
+146
+28,076
1.54
+43,200
NO
+2
30
+60
+11,538
~0
0
H2O
+2
18
+36
+6,923
~0
0
Sum
0
0+27,100
*The scale factor SF is (15,000 kg benzene/day)/(78 g benzene) = 192.3.
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37
Section 1.5 A First Look at Material Balances and Process Economics
The process economics are attractive: we could make a tidy $27,000/day, a
considerable increase over the value of the benzene itself. Of course, we’ve
neglected the cost of building and operating the facility, and we’ve assumed that
the price of adipic acid will remain stable despite the increase in worldwide plant
capacity that would occur if such a plant were built. This very preliminary assessment simply tells us that it is worth considering this process in greater detail.
Option 3:
Make catechol. Now, consider the possibility of producing catechol from benzene (Table 1.7).
If we consider both acetone and catechol as useful products, the atom economy
is very high at 0.90. The net profit is a whopping $89,300/day!
Let’s take a step back and consider the raw material, benzene, which is a
widely used reactant. Why benzene? Benzene is derived from crude oil and is
plentiful and relatively cheap; decades of research and development in the
petroleum industry have made it that way. We know how to recover crude oil
from the ground, how to purify benzene from crude oil, and how to use all the
other components of crude oil for numerous functions.
So what’s the problem? First, petroleum is a nonrenewable resource.
Second, benzene is carcinogenic. Third, it is volatile, so some of it ends up in
the air and contributes to smog. Additionally with the benzene-to-adipic acid
process, nitrogen oxides are produced, which may contribute to ozone depletion
and the greenhouse effect.
Is there another raw material that might substitute for benzene? What other
6-carbon compounds are readily available, perhaps from renewable resources?
Glucose (C6H12O6) is one such compound. It’s nontoxic and is produced from
renewable resources like corn. Compare the structure of glucose to those of
Table 1.7
Process Economy of Benzene-to-Catechol Process
kg/day
Compound
νi
Mi
νi
(SF = 192.3)
$/kg
$/day
C6H6
−1
78
−78
−15,000
0.41
−6,150
C3H6
−1
42
−42
−8,077
0.26
−2,100
O2
−1
32
−32
−6,154
~0
0
H2O2
−1
34
−34
−6,538
1.49
−9,740
C3H6O
+1
58
+58
+11,154
0.86
+9,600
o-C6H6O2
+1
110
+110
+21,153
4.62
+97,700
H2O
+1
18
+18
+3,462
~0
0
Sum
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0+89,300
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Chapter 1 Converting the Earth’s Resources into Useful Products
adipic acid and catechol: glucose is chemically more similar to these two products
than is benzene.
O
OH
H
HO HO
O
O
HO
HO
OH
H
HO
H
OH
OH
OH
HO
OH
OH
OH
Glucose, C6H12O6
O
Adipic acid, C6H10O4
Catechol, C6H6O2
Figure 1.9 Linear and cyclic structures of glucose, compared to adipic acid and catechol.
Not all hydrogens are shown.
Is glucose a suitable substitute for benzene as a raw material in adipic acid
and catechol production? The first challenge is to identify reaction pathways
that convert glucose to the desired products. Unfortunately, glucose does not
have the same chemical reactivity as benzene. It cannot withstand the high
pressures and temperatures, frequently used with benzene chemistry, without
degrading. On the other hand, glucose is a very useful feedstock for microorganisms like yeast and bacteria (not to mention humans!). Bacteria and yeast
consume glucose for energy, maintenance, growth, and reproduction. With
modern genetic engineering methods, microorganisms can often be tricked into
converting some of the glucose into products that are useful for humans.
The bacteria E. coli has been genetically engineered in a research laboratory
to convert glucose to muconic acid (C6H6O4). E. coli needs _​​ 73 ​​mole of glucose
17
and ​​ __
​​moles of oxygen to produce 1 mole of muconic acid; carbon dioxide and
2
water are the byproducts:
​​ _73 ​ ​C​6​​H1​ 2​​O6​ ​ + __
​ 172 ​ ​O2​ ​ → ​C6​ ​​H6​ ​​O4​ ​ + 8C​O2​ ​ + 11​H2​ ​O​
(R1)
Muconic acid can then be hydrogenated to adipic acid in a more conventional
chemical reactor:
​​C​6​​H6​ ​​O4​ ​ + 2​H2​ ​ → ​C6​ ​​H1​ 0​​O4​ ​​
(R2)
The generation-consumption analysis for conversion of glucose to adipic acid
is shown in Table 1.8.
E. coli has also been engineered to convert glucose directly to catechol.
Bacterial conversion of glucose to catechol requires 2 ​​ _13 ​​moles of glucose plus
oxygen to produce 1 mole of catechol, with carbon dioxide and water as
byproducts:
​​ _73 ​ ​C6​ ​​H1​ 2​​O6​ ​ + __
​ 152 ​ ​O2​ ​ → ​C6​ ​​H6​ ​​O2​ ​ + 8C​O2​ ​ + 11​H2​ ​O​
mur83973_ch01_001-060.indd 38
(R1)
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39
Section 1.5 A First Look at Material Balances and Process Economics
Table 1.8
Generation-Consumption Analysis of Glucose-Adipic
Acid Process
Compound
νi1
νi2
νi,net
C6H12O6
−7/3−7/3
O2
−8.5−8.5
C6H6O4
+1
−1
0
CO2
+8+8
H2O
+11+11
H2−2
−2
C6H10O4+1
+1
How do the atom and process economies compare for glucose versus
benzene as a raw material? The comparison must be based on the same rate
of production of desired products: 21,150 kg catechol/day or 28,100 kg adipic
acid/day. (Rates were rounded off to reflect level of accuracy of these calculations.) The price of glucose fluctuates somewhat with crop prices, purity, and
location. Let’s use a price of $0.60/kg glucose. We’ll assume that oxygen is
free, and that carbon dioxide and water have no value.
Table 1.9 shows that glucose is clearly not a good choice as a raw material
for adipic acid production. The fractional atom economy is only 0.21, because
so much of the carbon is consumed to make CO2 (to produce the energy for
bacterial survival and growth). The process loses money.
Table 1.9
Process Economy of Glucose-to-Adipic Acid Process
kg/day
Compound
νi
Mi
νi Mi (SF = 192.5)* $/kg
$/day
C6H12O6
−7/3
180
−420
−80,850
0.60
O2
−8.5
32
−272
−52,360
~0
0
CO2
+8
44
+352
+67,760
~0
0
H2O
+11
18
+198
+38,120
~0
0
H2
−2
2
−4
−770
0.2
−150
C6H10O4
+1
146
+146
+28,100
Sum
1.54
−48,500
+43,300
0 −5,400
*The scale factor SF = 28,100/198 = 192.5.
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Chapter 1 Converting the Earth’s Resources into Useful Products
Table 1.10
Process Economy of Glucose-to-Catechol Process
Mi
Compound
νi
g/g-mol νi Mi
kg/day
(SF = 192.3) $/kg
$/day
C6H12O6
−7/3
180
−420
−80,770
0.60
−48,500
O2
−7.5
32
−240
−46,150
~0
0
CO2
+8
44
+352
+67,690
~0
0
H2O
+11
18
+198
+38,080
~0
0
+1
110
+110
+21,150
4.62
+97,700
o-C6H6O2
Sum
0+49,200
Table 1.10 shows that the glucose-to-catechol process is poor in atom
economy (0.17), but is profitable because of the high value of the catechol
product. Still, glucose is not competitive with benzene if just raw material costs
are considered. Other considerations (such as environmental impact, reliability
of raw material source, patent protection, energy costs, cost of equipment,
safety, technical feasibility, and projected changes in raw material costs) may
swing a decision toward the more expensive raw material. In catechol manufacture, for example, a significant amount of the isomer hydroquinone is made
as a byproduct when benzene is used as the raw material, but not when glucose
is used. If it is expensive to separate the hydroquinone from the catechol, the
glucose process becomes more economically competitive.
In comparing different processes, besides considering the costs of raw
materials and the value of the products, we need to consider waste production.
Production of wastes means that some of our valuable raw materials, for which
we’ve paid good money, have been converted into things we didn’t want. At
best, “waste” products are valuable byproducts. At worst, if the waste products
are toxic, costly disposal is required. Let’s compare waste generation for four
processes: benzene to adipic acid, benzene to catechol, glucose to adipic acid,
and glucose to catechol (Table 1.11).
Table 1.11
Waste Generation from Four Processes
Raw
material Product
kg/day
kg/day
Identity of kg/day Identity of kg waste per
product byproduct byproduct wastes
wastes
kg product
Benzene
Adipic acid
28,100
Benzene
Catechol
21,150
Glucose
Adipic acid
28,100
Glucose
Catechol
21,150
mur83973_ch01_001-060.indd 40
0
18,500
NO, H2O
0.66
3,460
H2O
0.16
0
105,880
CO2, H2O
3.77
0
105,770
CO2, H2O
5.00
11,150
acetone
07/10/21 5:37 PM
Summary
41
Remember that these calculations are minimum waste generation; we have
not accounted for any inefficiencies in the process. The processes using benzene produce less waste than those using glucose. A lot of the carbon in glucose
ends up as CO2 rather than as product (as we already saw in the atom economy
calculations). Why? One reason is this: in fermentation, glucose conversion to
CO2 produces energy for bacterial survival and growth. For a fairer comparison, we should see if energy needs for the benzene processes are met by burning fuels and thus producing CO2. If so, then the waste calculations must
consider energy requirements as well as raw material requirements.
Summary
∙ Chemical processes convert raw materials into useful products. In the
initial stages of chemical process synthesis, we choose raw materials to
make a specific product, or products to make from a specific raw material.
We choose a chemical reaction pathway for converting the chosen raw
materials into desired products. These choices all have profound consequences on the technical and economic feasibility of the process.
∙ Balanced chemical equations are needed to begin process calculations.
Chemical equations are balanced if
​  ∑ ​​​​ε​hi​​ ν​i​= 0
all i
for all elements, where εhi is the number of atoms of element h in m
­ olecule
i, and νi is the stoichiometric coefficient for compound i; νi is negative for compounds that are reactants and positive for compounds that are
products.
∙ A generation-consumption analysis is a systematic way to analyze chemical reaction pathways involving I compounds and K reactions. To complete
a generation-consumption analysis:
(1) Write balanced chemical equations for all K reactions.
(2) List all I compounds (reactants and products).
(3) For each reaction k, write the stoichiometric coefficient νik associated
with each compound i in a column. There will be K columns, one for
each reaction.
(4) For all compounds i that should have zero net generation or consumption, zero net generation or consumption of compound i, find χk
such that
​  ∑ ​​​​χ​k​​ ν​ik​= 0
all k
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42
Chapter 1 Converting the Earth’s Resources into Useful Products
(5) Calculate
​νi​  ,net​​ =​  ∑ ​​​ ​νi​  k​​ ​χk​  ​​
all k
This sum is the net generation or consumption of compound i.
Compounds that have negative sums are raw materials, indicating net
consumption. Compounds that have positive sums are products, indicating net generation. Compounds that have “zero” sums are intermediates, indicating no net generation or consumption.
∙ A basis is a quantity or flow rate that indicates the size of a process. A scale
factor provides an easy way to scale a process up or down to the desired
basis. A scale factor is a ratio of the basis quantity to a relative quantity.
∙ Atom economy is a simple indicator of the efficiency of utilization of raw
materials in a given reaction pathway
​ν​ ​ ​M​ ​
Fractional atom economy = _________
​​  P P  ​​
− ∑ ​νi​​ ​Mi​​
all reactants
∙ A simple measure of the process economy is made by starting from the
generation-consumption analysis, scaling up or down to a desired production rate, and then computing the difference between the product values
and the raw material costs.
ChemiStory: Changing Salt into Soap
Soap is made by combining fats or oils from animals or plants with an
alkaline material. Today caustic soda (sodium hydroxide, NaOH) is the
alkali used for making soap, but in the past sodium carbonate (Na2CO3),
potassium hydroxide (KOH), and potassium carbonate (K2CO3) were common choices.
In the 1700s in Europe, soap was a luxury reserved for the wealthy. But technology
to make cheaper cotton clothing was rapidly
developing. Cotton needed to be cleaned
before it could be dyed and sold, so demand
for soap for textile manufacture increased
dramatically. (Use of soap for personal
hygiene was still uncommon.) At the same
time, glass and paper factories were expanding; both products required sodium carbonate
for their manufacture. Growth of the textile,
Library of Congress Prints &
glass, and soap industries led to a huge
Photographs Division
[LC-USF34-T01-034006-D]
demand for new sources of alkali.
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43
Summary
King Louis XVI of France issued a proclamation offering an award—
the equivalent of half a million dollars—to the person who invented a process to turn common table salt (NaCl) into “washing soda,” better known
now as sodium carbonate or soda ash. Why was sodium carbonate so valuable to the French King? At that time, alkali for French factories was
imported from two sources: Spanish and Irish peasants who harvested and
burned seaweed and recovered the ashes, and New England settlers who
burned brush to clear land and to make “potash” (a mix of mainly KOH,
NaOH, K2CO3, and Na2CO3). A few problems arose. First, there wasn’t
enough seaweed to meet the increasing demand. Second, France supported
the American War of Independence, which led Britain to block potash
exports from New England to France. French industry was threatened
because of the loss of access to raw materials.
Luckily for the King, chemistry was quite fashionable at this time. The
Duke of Orleans, the wealthiest man in France and an outspoken critic of
the French absolute monarchy, established a large chemistry research lab on
his palace grounds. One of the talented but poor young men who benefited
from the Duke’s patronage was Nicolas LeBlanc. LeBlanc knew that common salt (NaCl) was very stable, but it could be converted to more reactive
sodium sulfate by treating it with sulfuric acid. It took him 5 more years to
stumble upon the idea of reacting sodium sulfate with limestone to make
sodium carbonate. In 1789, LeBlanc planned to collect his prize. Unfortunately
for him, in 1789 the French Revolution happened. LeBlanc never received
his award from King Louis. The king was even more unfortunate—he lost
his head.
Despite these setbacks, with financial help from the Duke, LeBlanc built
his first factory in 1791. The next few years were not easy, however. In
1793, the Duke of Orleans was arrested and executed. Then, LeBlanc’s factory was seized by the government, he lost his job, and his daughter died.
At the age of 63, depressed and broke, LeBlanc killed himself with a gunshot
to the head.
The story doesn’t end here, however. The LeBlanc process survived and
thrived in the next decades; demand for washing soda exploded, and his
process was the only reliable and economical way to make it on a large
scale. However, the LeBlanc process was an environmental disaster. Acid
gas spewed out over the landscape, wasting forests, destroying farmland,
and poisoning workers. HCl gas combined with the waste sulfur solids to
make hydrogen sulfide, a deadly and smelly gas. In 1863, disgusted Britons
passed the Alkali Act, one of the earliest pieces of legislation dealing with
chemical pollution. Tall chimneys were installed to disperse the gases over
a broader area. Facilities were built next to the soda ash plant to recover
and reuse the waste products.
In the mid-1860s, Ernest Solvay figured out a way to exploit a different
series of chemical reactions. Solvay used the same raw materials, limestone
(continued)
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Chapter 1 Converting the Earth’s Resources into Useful Products
and salt, to make the same product, soda ash.
However, the reaction pathway was different, no
acid was consumed or generated, and a useful
byproduct, CaCl2, was made. Furthermore, whereas
the LeBlanc process operated in batch mode, the
Solvay process was continuous. One might argue
that Ernest Solvay was to large-scale chemical manufacturing what Henry Ford was to large-scale automaking. The Solvay process completely displaced
the LeBlanc process by about 1915. Ernest Solvay
made millions and gave it all away to charity.
Today in the United States, it is cheaper to
Historic Collection /
mine soda ash, mainly in Wyoming, than to make
Alamy Stock Photo
it from salt and limestone. In countries where there
are no natural sources of soda ash, the Solvay process is still used. In the
1890s, the availability of cheap hydroelectric power led to the development
of an electrolytic process to convert NaCl to produce NaOH—the alkali used
today to make soap.
Quick Quiz Answers
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
​ν​​H2​ ​​​ = −1.
Because methane is a reactant, not a product.
0.5CH4 + O2 → 0.5CO2 + H2O.
3C6H12O + 8HNO3 → 3C6H10O4 + 8NO + 7H2O.
3CH4 + 6H2O + 4N2 → 3CO2 + 8NH3. No, because CO2 is in only
one reaction.
2.26 gmol; 4790 g; 47.2 kg; 1.024 kgmol.
0.51.
At 730 million lb/yr and $0.05/lb, clearly commodity.
References and Recommended Readings
1. ICIS Chemical Business is a weekly periodical that contains current
pricing information on commodity chemicals as well as stories about
the chemical business. Available online through subscription and at many
university libraries.
2. Chemical and Engineering News is a highly readable weekly news magazine published by the American Chemical Society. Research news, business developments, and policy issues of interest to chemists and chemical
engineers are covered. Regular features include salary surveys, data on performance of specific companies in various industry sectors, and information
mur83973_ch01_001-060.indd 44
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Chapter 1 Problems
45
on chemical production. Available online, at most university libraries, and
to all members of the American Chemical Society.
3. Chemical Engineering Progress is a news magazine published by the
American Institute of Chemical Engineers. Technology and business developments, practical solutions to engineering problems, and career advice of
interest to chemical engineers are covered. Available at many university
libraries and to all members of the American Institute of Chemical Engineers.
4. The Kirk-Othmer Encyclopedia of Chemical Technology is a multi-volume
treasure trove of information about chemical products and processes.
5. Prometheans in the Lab: Chemistry and the Making of the Modern World,
by Sharon Bertsch McGrayne, is a balanced and engaging study of the
people and places behind important chemical advances. The book served as
a major reference for some of the ChemiStories. Published in 2001 by
McGraw Hill.
Chapter 1 Problems
Warm-Ups
Section 1.2
P1.1 Consider the following common household products: paper bag, plastic
soda bottle, wine glass. For each product, list the raw material (minerals,
fossil fuels, agricultural materials) that is used to make the product.
P1.2 List five major sources of raw materials. Which of these raw materials
might you choose if you needed a source of: (a) N2 (b) Fe (c) C (d) H?
Section 1.3
P1.3 Balance the following reaction for synthesis of acrylonitrile from propylene, ammonia, and oxygen:
​​C3​ ​​H6​ ​ + N​H3​ ​ + ​O2​ ​ → ​​C3​ ​H3​ ​N (acrylonitrile) + ​H2​ ​O​
P1.4 Urea [(NH2)2CO] is a solid that is mixed into lawn fertilizer. When applied
to the ground, urea spontaneously decomposes to ammonia (NH3) to
provide nitrogen to the soil and CO2. Write down the balanced chemical reaction, noticing that air and water are available and could serve
as reactants or byproducts.
P1.5 Urea (NH2CONH2) and ethylene glycol (C2H6O2) react to ethylene carbonate C3H4O3 and ammonia (NH3). Find the stoichiometric coefficients.
P1.6 Carborundum (silicon carbide, SiC) is one of the hardest materials in
the world, and is used as an industrial abrasive for grinding wheels.
It is made by reacting SiO2 (from sand) and coke (C). CO is the only
byproduct. What is the balanced chemical reaction?
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Chapter 1 Converting the Earth’s Resources into Useful Products
P1.7 You are interested in the combustion of iso-octane (C8H18), a major
component of gasoline. Complete oxidation with O2 leads to CO2 and
H2O as the only products of the reaction. If ​​ν​​C8​ ​​H1​ 8​​​= −1, what is ​​ν​​O2​ ​​​?​​
ν​C​O2​ ​​​? ν​​ ​H​ 2​ ​O​​?
P1.8 Gallium nitride can be used to make light-emitting diodes (LEDs). GaN
is made by reacting trimethyl gallium [(CH3)3Ga] with ammonia (NH3).
Methane (CH4) is a byproduct. Write down the 4 element balance equations. Then choose νGaN = 1 and solve for the remaining stoichiometric
coefficients.
P1.9 “Milk of lime” (Ca(OH)2) reacts with ammonium chloride (NH4Cl) to
make calcium chloride (CaCl2) with ammonia (NH3) and water (H2O)
as byproducts. How many element balance equations can you write?
How many element balance equations are needed to balance this reaction?
P1.10 Propylene (C3H6), ammonia (NH3), and oxygen (O2) react to make acrylonitrile (C3H3N), with water as the byproduct. Determine the stoichiometric coefficients for all other compounds, if ν​
​​ ​C​3​H​6​​ = −2.
P1.11 Nitroglycerin [C3H5(NO3)3] explosively decomposes to CO2, H2O, N2,
and O2. Write the balanced chemical equation.
P1.12 Platinum is an important metal used in catalytic converters in automobiles, as well as in catalysts in industrial reactors. It is made by the
decomposition of (NH4)2PtCl6, with NH4Cl, N2, and HCl as byproducts.
Write down the element balance equations. Then set νHCl = 6 and solve
for all the other stoichiometric coefficients.
Section 1.4
P1.13 Sulfuric acid is made in three steps: (R1) Sulfur S is combusted with
O2 to make SO2, (R2) SO2 is further oxidized with O2 to make SO3,
and (R3) SO3 is dissolved into water to make H2SO4. Write down the
three balanced chemical reactions, and then show that if χ1 = χ2 =
χ3 = 1, there will be no net generation or consumption of SO2 or SO3.
P1.14 Is the value for ​∑ ​χk​  ​​ ​νi​  k​​​positive, negative, or zero for (a) a reactant,
(b) an intermediate, (c) the desired product, and (d) a byproduct? Explain
what the subscript k means, and why χ​ ​​ k​​​has only one subscript but ν​ ​​ ik​​​
has two subscripts.
Section 1.5
P1.15 Urea [(NH2)2CO] is source of nitrogen for fertilizers. Calculate (a) the
number of grams of urea per gmol and (b) the number of g of N per
100 g of urea.
P1.16 If 1 gmol of N2 and 3 gmol of H2 are consumed to make 2 gmol of
NH3, how many pounds of N2 and H2 are required to make 1 billion lb
of NH3?
P1.17 I have 4.4 lbmol H2O. What is the mass of H2O in units of lb, kg, and
metric tons?
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Chapter 1 Problems
47
P1.18 One gallon of water weighs about 8.35 lb. Calculate the lbmol and the
gmol water in one gallon.
P1.19 Amino acids have the general formula NH2CH(R)COOH, where R
stands for a side chain connected to the alpha carbon. There are 20
naturally occurring amino acids, where each has a different side chain.
Calculate the molar mass of the amino acid methionine, where R is
CH2CH2SCH3 (the dash indicates the carbon that is attached to the
main chain).
P1.20 Polyethylene terephthalate (PET) is a common polymer that is used to
manufacture products such as plastic beverage bottles. The polymer has
the molecular formula CH3[C10H8O4]nOH, where n indicates the number
of the repeating units of the monomer building block. What is the molar
mass of PET if n = 50?
P1.21 About 245 million tons of sulfuric acid (H2SO4) are manufactured every
year worldwide. Calculate the annual production of H2SO4 in kg,
pounds, grams, and gmole. Then estimate the kg H2SO4 per person by
dividing by the world’s population.
P1.22 Sucrose (table sugar) has a molecular weight of 342 g/gmol. What is
sucrose’ molecular weight in lb/gmol? In lb/lbmol?
P1.23 An elemental analysis of yeast shows that the yeast contains 50 wt% C,
6.94 wt% H, 9.72 wt% N, and 33.33 wt% O. If the yeast is modeled as
though it were a single chemical compound CHxOyNz, find the values
of x, y, and z.
P1.24 About 40 billion lb of ammonia (NH3) is produced every year. What is
the annual production of NH3 in lbmol? In tons? In tonmoles?
P1.25 How many kg CO2 are generated when 1 kg of CH4 is burned?
P1.26 In Example 1.4, three reactions were combined to produce ammonia
from methane, water, and nitrogen. Carbon dioxide is a byproduct of the
overall reaction. How many grams of CO2 are generated per g of NH3?
P1.27 If ammonia costs $0.0045 per gmol, what is its cost in $ per metric ton?
P1.28 If chlorine (Cl2) costs $0.45/kg, what is the cost of 1 gmol Cl2?
P1.29 3 gmoles CH4, 6 gmoles H2O, and 4 gmoles N2 are required to make
8 gmoles NH3. What is the fractional atom economy for ammonia
synthesis?
P1.30 A new route to diesel fuel additive from renewable resources is the
gas-phase dehydration of ethanol (C2H5OH) to diethyl ether (C2H5)2O.
Write the stoichiometrically balanced reaction. Calculate the fractional
atom economy.
P1.31 You are designing a process to make cyclohexane (C6H12) from benzene
(C6H6) and hydrogen (H2). The rate of cyclohexane production is 84,000
lb/day. How much hydrogen (lb/day) will your process need? Suppose
benzene sells for $0.15/lb, hydrogen for $0.09/lb, and cyclohexane for
$0.18/lb. What is your net earnings ($/day)?
P1.32 In Example 1.8, we calculated the fractional atom economy of producing
4-ADPA by conventional versus new reaction pathway. If 300 million lb
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Chapter 1 Converting the Earth’s Resources into Useful Products
of 4-ADPA are produced per year, compare the waste production (lb/year)
of using the conventional versus the new reaction pathway.
P1.33 List three reasons why waste generation per pound of product might
be higher for pharmaceuticals like ibuprofen compared to commodity
products such as urea.
P1.34 Write the definition of the following terms: basis, scale factor, fractional
atom economy. Then describe, with words and/or sketches, how you
would explain the concept to a 10-year-old.
P1.35 A 12-oz. bottle of water sells for $1.75. What is the product value in $/lb?
Based on its price, is this product classified as commodity, specialty, or
pharmaceutical?
Drills and Skills
P1.36 Antacids like Alka-Seltzer® contain aspirin (C9H8O4), sodium bicarbonate (NaHCO3), and citric acid (C3H5O(COOH)3). What stoichiometrically balanced reaction between the last two compounds produces the
“fizz” when the tablet dissolves in water?
P1.37 One cause of upset stomach is excess stomach acid (HCl). You are in
charge of designing a product that neutralizes excess acid by reacting
with HCl and producing salt (along with some CO2 and water). If you
wanted to get the most neutralizing activity per gram of product, would
you choose sodium bicarbonate (NaHCO3), calcium carbonate (CaCO3),
or magnesium carbonate (MgCO3)?
P1.38 Urea [(NH2)2CO] is mixed into lawn fertilizer and sold to home gardeners. When applied to the ground, urea spontaneously decomposes to
ammonia (the active ingredient) and carbon dioxide. Write the balanced
chemical reaction. Calculate the fractional atom economy of the reaction.
P1.39 Cyclohexanone (C6H10O) reacts with nitric acid (HNO3) to produce
adipic acid (C6H10O4). NO is a byproduct. Use matrix methods to find
the correct stoichiometric coefficients and determine whether H2O is
also a byproduct.
P1.40 Consider the reactions of ammonia (NH3) and oxygen (O2) to form nitric
oxide (NO), nitrous oxide (N2O, sometimes called laughing gas), or
nitrogen dioxide (NO2, a brown air pollutant). Water is a byproduct. Use
matrix methods to find the correct stoichiometric coefficients for all
three reactions. (Once you set up the problem it is easy to solve for each
of the cases.)
P1.41 Hexane (C6H14) is purified from crude oil, while glucose (C6H12O6) is
made from agricultural products such as corn. For either six-carbon
compound, complete combustion with oxygen produces carbon dioxide
and water. How many grams of CO2 are produced per gram of hexane
combusted? Per gram of glucose?
P1.42 (a)Methane (CH4) is the main compound in natural gas. How many kg
of CO2 are produced when 1 kg of methane is burned?
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Chapter 1 Problems
P1.43
P1.44
P1.45
P1.46
49
(b) Gasoline is a complex liquid mixture of hydrocarbons, but let’s
assume that the average molecule in gasoline is iso-octane (C8H18).
How many kg of CO2 are produced when 1 kg of iso-octane is burned?
(c) Coal is a solid complex mixture of hydrocarbons. Let’s assume that
the average molecular composition is C23H15O. How many kg of
CO2 are produced when 1 kg of coal is burned?
Comment on any trends you notice.
Glycerol (C3H8O3) is a byproduct of biodiesel manufacture, and researchers are coming up with new uses for this chemical. One idea is to react
glycerol with water to make CO2 and H2, and then use the H2 in fuel
cells to generate electricity. Calculate the pounds of H2 and of CO2 produced per pound of glycerol consumed. What do you think of this idea?
Freon-12 is a chlorofluorocarbon (CF2Cl2) that has been widely used in
refrigerators and air conditioners, but when it leaks into the atmosphere
it can deplete ozone in the upper atmosphere. Its manufacture was
banned, but there are still stores of Freon-12 in older equipment. The
chemical is extremely stable and difficult to destroy. One possible
method is to react CF2Cl2 with sodium oxalate (Na2C2O4), which produces three solids: sodium fluoride (NaF), sodium chloride (NaCl),
and coke (C), as well as carbon dioxide. Write the stoichiometrically
balanced reaction between Freon-12 and sodium oxalate. Calculate the
grams of sodium oxalate required, and the grams of solid products produced, per gram of Freon-12 destroyed.
Magnetic nanoparticles are of interest for medical imaging and drug
delivery. In one experiment, 1.52 mmol Fe(CO)5 was mixed with 1.28 g
oleic acid to form an iron oleate complex. Then 0.34 g trimethylaminoxide [(CH3)3NO] is added. Precipitation yields highly pure crystalline
nanoparticles of γ-Fe2O3. What is the fractional atom economy of this
scheme?
Nitrates in well water are problematic in rural regions due to runoff
from fertilized farms. If ingested, the nitrates can cause “blue baby”
syndrome. Some bacteria can remove nitrates from water if methanol is
added: the reaction can be written as
​​HNO​3​​​ + CH​3​​OH → ​C3​ ​​​H7​ ​​​NO​2​​​ + CO​2​​​ + H​2​​O​
The reaction as written is not balanced. Find the correct stoichiometric coefficients. Calculate the grams of methanol required to reduce
the nitrate content of 10 liters of well water from 64 mg/L to 10 mg/L.
P1.47 Amino acids have the general formula NH2CH(R)COOH, where R is a
group connected to the first carbon. There are 20 different amino acids
in proteins, each of which has a different R group. For example:
​ lycine: R = H​
G
​Methionine: R = C​H2​ ​C​H2​ ​SC​H3​ ​​
​Tryptophan: R = C​H2​ ​(​C4​ ​N​H2​ ​)(​C4​ ​​H4​ ​)​
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Chapter 1 Converting the Earth’s Resources into Useful Products
A protein is a linear polymer of a mix of amino acids, where the carboxy
terminus of one amino acid is linked to the amino terminus of the next
to form an amide, with release of water:
​​NH​2​​CH​(R)​1​​COOH + ​NH​2​​CH(​R)​2​​COOH →
​NH​2​​CH(​R)​1​​CONHCH(​R)​2​​COOH + ​H2​ ​​O​
P1.48
P1.49
P1.50
P1.51
This same reaction happens over and over again as the amino acids are
linked together.
First calculate the molar mass of glycine, methionine, and tryptophan.
Then determine the molar mass of a protein that contains 16 glycines,
3 tryptophans, and 8 methionines. (Note—this would be a very unusual
protein! Hint: How many amide bonds are made?)
“Biocementation” is a new idea to prevent leakage of toxic materials
from contaminated soils. Basically, special microorganisms are spread
on the soil along with urea [(NH2)2CO] and calcium chloride (CaCl2).
The microorganisms process the chemicals and deposit calcium carbonate (CaCO3), which then binds with the soil and hardens in place.
Ammonium chloride is a byproduct. Determine the balanced chemical
reaction, keeping in mind that water could be either a reactant or
byproduct. Also calculate the pounds of urea and calcium chloride you
need to add in order to make 1 pound of calcium carbonate.
Polyethylene terephthalate (PET) is a polymer used for plastic food packaging and soda bottles. Recycling plastics is an important method for
reducing solid waste, but direct reuse of polymers is technically challenging. One option is to “unzip” polymers into their constituent monomer
chemicals, then reuse them as reactants to make new polymers. For example, PET can be unzipped using a reaction called methanolysis. In methanolysis, PET (CH3[C10H8O4]nOH, where n = number of monomer repeat
units) reacts with methanol (CH3OH) to make dimethyl terephthalate
(DMT, C10H10O4), with ethylene glycol (C2H6O2) as a byproduct. DMT
can then be reused to make more PET. Suppose n = 50. How many moles
of methanol are needed per mole of PET to completely break it down
into DMT? How many grams of methanol are required per g of PET, and
how many grams of DMT are produced per gram of PET?
Sulfuric acid is made by reacting S with O2 to make SO3 and then mixing SO3 with H2O to make H2SO4. Calculate the tons of S, O2, and H2O
needed to supply the worldwide annual demand of 245 million tons
sulfuric acid.
High-purity silicon for the chips in your laptop or smart phone is made
from sand (SiO2) and coke (C) using three reactions:
(R1) ​​SiO​2​​+ C → Si + CO​
(R2) ​​Si + Cl​2​​ → ​SiCl​4​​​
(R3) ​​SiCl​4​​​ + H​2​​→ Si + HCl​
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Chapter 1 Problems
P1.52
P1.53
P1.54
P1.55
P1.56
mur83973_ch01_001-060.indd 51
51
The reactions as written are not balanced. First balance the reactions.
Then calculate the quantity of reactants (grams) required to produce
100 grams of high-purity Si. Also calculate the quantity of each byproduct.
When making French bread dough, a small amount of yeast is added to
a mix of flour and water. The yeast serve as a catalyst—consuming
glucose (C6H12O6) from starch, and producing CO2, which causes the
dough to rise. In addition, yeast consume glucose and ammonia (NH3)
from protein in the flour and generate more yeast, which can be modeled
as a “pseudo” compound using the formula CH1.66O0.6N0.166. A commercial bakery was experimenting with adjusting their dough recipe. In
one experiment, 3.9 g CO2 was generated for every gram yeast generated. Write two balanced chemical reactions, one in which glucose is
consumed to make CO2, and the other in which glucose and ammonia
are consumed to make yeast. O2 and H2O can be either reactants or
products in either reaction. Then calculate the fraction of glucose that
was consumed to make CO2, and the fraction that was used to make
more yeast.
The reducing agent sodium borohydride (NaBH4) is a classic way to
hydrogenate a chemical using water. With the advent of modern catalysts, hydrogen can be used directly instead. Compare the hydrogenation
of methyl phenyl ketone (acetophenone) C6H5COCH3 to 1-phenylethanol
(C6H5CH(OH)CH3)—an intermediate in the production of various pharmaceuticals. To do this, first determine the balanced chemical reactions
for (a) the conventional reaction of NaBH4 and water with acetophenone
to 1-phenylethanol, with NaB(OH)4 as the byproduct, and (b) the catalyzed
reaction of hydrogen with acetophenone to 1-phenylethanol. Then calculate and compare the fractional atom economy of the conventional
versus the catalytic approach.
1-phenylethanol (C6H5CH(OH)CH3) can be converted to the carboxylic
acid (C6H5CH(CH3)COOH) using Mg, HCl, and CO2. The byproducts of
this reaction are a salt and water. Write down the balanced chemical reaction and calculate the fractional atom economy, then compare this conventional approach to a newer reaction that uses carbon monoxide (CO).
Baking soda (sodium bicarbonate, NaHCO3) is used in baking cookies,
quick breads, and cakes. When baking soda is heated or mixed with
acidic foods, the CO2 helps the baked goods to rise. Sodium carbonate
(Na2CO3) and one other compound are the other products when baking
soda decomposes. Determine the other compound, and write the stoichiometrically balanced reaction. Calculate the grams of CO2 produced
per gram baking soda.
A chemist might carry out the benchtop oxidation of 1-phenylethanol
(C6H5CH(OH)CH3) to acetophenone (C6H5COCH3) by adding sulfuric
acid (H2SO4) and chromium trioxide (CrO3). This reaction produces
chromium sulfate (Cr2(SO4)3) as a byproduct. CrO3 is a suspected carcinogen, and if this reaction were carried out at a manufacturing scale,
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Chapter 1 Converting the Earth’s Resources into Useful Products
a large amount of the chromium sulfate waste would be produced. An
alternative technology, that is more feasible on a larger industrial scale,
uses hydrogen peroxide (H2O2) as the oxidizing agent. Compare the two
methods by deriving the balanced chemical reactions (be sure to include
any additional byproducts), calculating the fractional atom economy, and
then finding the pounds of byproducts generated per pound of acetophenone generated for the two approaches.
P1.57 The proteins in our food contain C, H, O, and N. When we digest food
proteins, some of the nitrogen is recirculated and used to build muscle
proteins, while some end up as ammonium ions. ​​NH​ 4+​  ​​ is extremely
toxic, so our bodies work hard to get rid of it. Here is a simplified
summary of the detoxification reactions that happen in our cells:
​​NH​4​​​HCO​3​​​ + C​5​​​H1​ 2​​​O2​ ​​​N2​ ​​ (ornithine) → ​C6​ ​​​H1​ 3​​​O3​ ​​​N3​ ​​ (citrulline)​
​​C6​ ​​​H1​ 3​​​O3​ ​​​N3​ ​​​ + C​4​​​H7​ ​​​O4​ ​​N (aspartic acid) → ​C1​ 0​​​H1​ 8​​​O6​ ​​​N4​ ​​ (arginosuccinate)​
​​C1​ 0​​​H1​ 8​​​O6​ ​​​N4​ ​​ → ​C4​ ​​​H4​ ​​​O4​ ​​ (fumarate) ​+ C​6​​​H1​ 4​​​O2​ ​​​N4​ ​​ (arginine)​
​​C6​ ​​​H1​ 4​​​O2​ ​​​N4​ ​​ (arginine) → ​CH​4​​​ON​2​​ (urea) ​+ C​5​​​H1​ 2​​​O2​ ​​​N2​ ​​ (ornithine)​
As usual in biological systems, water is ubiquitous and can be a reactant
or a product. Balance the reactions, adding water where needed. Use a
generation-consumption analysis to determine the overall reaction. Where
does the excess ammonia end up? Is there net generation or consumption
of water?
P1.58 Soaps are the sodium salts of fatty acids, derived from natural products
such as animal fat. In a typical soap-making process, glycerol stearate
is contacted with hot water to produce stearic acid and glycerol. The
balanced reaction is
​​​(​C​17​​​H3​ 5​​COO)​​3​​​C3​ ​​​H5​ ​​​ + 3H​2​​O → ​3C​17​​​H3​ 5​​​COOH + C​3​​​H5​ (​​​​ OH)​​3​​​
After the glycerol is removed, stearic acid is neutralized with sodium
hydroxide (NaOH) to produce sodium stearate soap:
​​C1​ 7​​​H3​ 5​​COOH + NaOH → ​C1​ 7​​​H3​ 5​​​COONa + H​2​​O​
If glycerol stearate sells for $2.50/kg, NaOH costs $1.25/kg and glycerol
can be sold as a byproduct for $2.20/kg, what is the lower bound on
the sales price of a kg of sodium stearate soap? How does that compare
to the selling price at your local store?
P1.59 Estimate the size of the market (lb/year) in the United States for gasoline, polystyrene, and aspirin. Do this by simply estimating the typical
annual consumption of yourself and your family and friends, and then
multiplying by 330 million people. Compare to the typical plant values
listed in Table 1.3.
P1.60 With your kitchen faucet on full throttle, you collect water for exactly
ten seconds. You weigh the water on a kitchen scale and find that it
weighs 1441 grams. If your kitchen faucet was a chemical plant, would
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Chapter 1 Problems
53
this rate be most similar to the annual production rate of a refinery, a
commodity chemical plant, a specialty chemical plant, or a pharmaceutical plant (compare to Table 1.3)?
P1.61 You are conducting a preliminary economic analysis of a proposed new
process. The required raw materials are Compound A (27,000 lb/h,
$0.14/lb) and Compound B (8100 lb/h, $1.01/lb). The product is
Compound C (32,000 lb/h) that will sell for $0.86/lb. There is also a
byproduct of negligible value. The process will operate 350 days/year,
leaving 2 weeks for a plant turnaround for maintenance. If you were
writing a memo to management summarizing your analysis, what would
you report for the expected (a) annual production rate, (b) annual raw
material cost, (c) annual product sales, and (d) annual profit? Think
about appropriate units and appropriate number of digits in your numerical
answers. (e) What category of plant capacity does this process fall
under? (See Table 1.3.)
Scrimmage
P1.62 Citral (C10H16O) is extracted from lemongrass and is blended into consumer products such as perfumes, soaps, and soft drinks, imparting a
lemon-lime fragrance. Synthetic citral can be made from butene (C4H8),
formaldehyde (CH2O), and oxygen by combining three reactions:
​​C4​ ​​​H8​ ​​​ + CH​2​​O → ​C5​ ​​​H1​ 0​​O​
​​C5​ ​​​H1​ 0​​​O + O​2​​ → ​C5​ ​​​H8​ ​​O​
​​C5​ ​​​H1​ 0​​​O + C​5​​​H8​ ​​O → ​C1​ 0​​​H1​ 6​​O​
Water may participate in any of these reactions, either as a reactant
or as a product. Determine the stoichiometrically balanced reactions,
complete a generation-consumption analysis so that there is no net consumption of any compounds other than butene, formaldehyde, and oxygen. Calculate the kg of reactants consumed, and the kg of byproducts
generated, per kg of citral produced.
P1.63 One route to acetic acid (CH3COOH) involves the dehydration of methanol to dimethyl ether (CH3)2O, then the subsequent carbonylation
(reaction with CO and water) of dimethyl ether to make acetic acid. CO
is made from CO2 and H2 by the water—gas shift reaction, with H2O
as a byproduct. Write the three balanced chemical reactions. Calculate
the overall atom economy for converting the raw materials to acetic
acid. If you make 10,000 metric tons of acetic acid per year, how much
methanol will you consume?
P1.64 Isobutanol can be produced by fermentation and has been proposed as
an excellent renewable feedstock to make a variety of chemical products
that are currently synthesized from fossil fuels. Your company has discovered a new catalyst that converts isobutanol to isobutene:
​C4​ ​​​H1​ 0​​O → ​C4​ ​​​H8​ ​​​  + H​2​​O​
​
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Chapter 1 Converting the Earth’s Resources into Useful Products
Isobutene then reacts to make xylenes (C8H10), with hydrogen as a
byproduct. Xylenes are used as octane boosters in gasoline, as industrial solvent, or to make various plastics and polymers. First calculate
the kg of xylene produced per 1000 kg of isobutanol. If bio-based
isobutanol costs $950/ton and xylene sells for $1500/ton, is this a viable business?
P1.65 Nitric acid (HNO3) is an important commodity chemical that is made
in a three-step process: (R1) ammonia (NH3) reacts with oxygen (O2)
to form nitric oxide (NO) and water; (R2) NO reacts further with O2 to
make nitrogen dioxide (NO2); (R3) NO2 is bubbled through water to
produce HNO3 and NO. Write out the three stoichiometrically balanced
reactions. Use generation-consumption analysis to synthesize a reaction
pathway to make nitric acid from ammonia and oxygen with no net generation or consumption of NO or NO2. Then calculate the flows (kg/day)
of reactants and byproducts for a plant that makes 126,000 kg nitric
acid per day. Estimate the profit ($/year), assuming the plant operates
350 days per year. Use the following prices: ammonia: $0.14/kg; oxygen:
$0.033/kg; water: $0.012/kg; nitric acid: $0.26/kg.
P1.66 Hexamethylenediamine (HMD, H2N(CH2)6NH2), is one of the two reactants used to make nylon-6,6. Connie Chemist has proposed two alternative reaction pathways for making HMD. Eddie Engineer has to decide
which one to use for a process making 116,000 lb HMD/day.
Reaction pathway 1: React butadiene (C4H6) with hydrogen cyanide
(HCN) to make adiponitrile (NC(CH2)4CN). Then react adiponitrile
with hydrogen (H2) to produce HMD.
Reaction pathway 2: React acrylonitrile (CH2CHCN) with hydrogen to make adiponitrile. Then react adiponitrile with hydrogen to
make HMD.
Which pathway should Eddie Engineer recommend?
Compound
Formula
Cost, $/lb
Butadiene
C4H6
0.25
Hydrogen cyanide
HCN
1.13
Hydrogen
H2
0.12
Acrylonitrile
CH2CHCN
0.78
Adiponitrile
NC(CH2)4CN
??
Hexamethylenediamine
H2N(CH2)6NH2
??
P1.67 Polycarbonates are strong, lightweight, impact-resistant, and transparent
polymers used to make products like football helmets, eyeglass lenses,
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Chapter 1 Problems
55
and airplane windows. The synthesis of polycarbonates involves the
reaction of phenol with acetone to make bisphenol A:
​​C6​ ​​​H5​ ​​​OH + CH​3​​​COCH​3​​ →​ C​15​​​H1​ 6​​​O2​ ​​​
and then bisphenol A reacts with phosgene and sodium hydroxide to
produce the desired polymer, along with sodium chloride:
​​C1​ 5​​​H1​ 6​​​O2​ ​​​ + COCl​2​​+ NaOH → ​[OC​6​​​H4​ ​​C(​​ C​H3​ ​​)​​2​​​C​6​​​H4​ ​​​OCO]​n​​ + NaCl​
where n indicates that the chemical unit repeats itself n times in the polymer
chain. Assume n = 50.
Phosgene is made by steam reforming of methane to make CO, then
reaction of CO with chlorine gas:
​​CH​4​​​ + H​2​​O → ​CO + H​2​​​
​​Cl​2​​+ CO → ​COCl​2​​​
Besides the reactants and products shown, there might be other simple
compounds such as CO2 or H2O generated or consumed as well. Write
the balanced chemical reactions. Complete a generation-consumption
analysis to develop a reaction network with no net generation or consumption of CO or phosgene, because these compounds are extremely
toxic. Determine the kg of reactants consumed per kg of polycarbonate
generated.
P1.68 Since phosgene is so toxic, you are interested in developing a greener
process for synthesis of polycarbonates. The process that you discovered
requires dimethylcarbonate (DMC, C3H6O3), so you are now searching
for environmentally friendly and economical means of producing DMC.
The following set of reactions is of interest:
Syngas (mixture of CO and H2) production from methane (CH4) and
steam (H2O):
​C​H​4​ + ​H2​ ​O → CO + 3​H2​ ​​
(R1)
Methanol (CH3OH) production from CO and H2:
​CO + 2​H2​ ​ → C​H3​ ​OH​
(R2)
Dimethyl carbonate production from methanol, CO, and O2:
​2C​H​3​OH + CO + 0.5​O2​ ​ → ​C3​ ​​H6​ ​​O3​ ​ + ​H2​ ​O​
(R3)
Synthesize a reaction pathway to make DMC by combining these three
reactions, with no net generation or consumption of CO or CH3OH.
Calculate the fractional atom economy of the overall reaction. Calculate
the flows of reactants and byproducts for a plant that produces 1800 kg
DMC/day. Estimate the profit ($/year) assuming that the plant operates
350 days per year and the following prices per kg: methane: $0.22;
hydrogen: $0.88; methanol: $0.38; oxygen: $0.12; DMC: $2.11.
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Chapter 1 Converting the Earth’s Resources into Useful Products
P1.69 Acrylonitrile is the basic building block of synthetic rubbers and orlon
fibers. Your company is interested in building a new plant to satisfy
increased demand for acrylonitrile, and your supervisor has asked you
to consider three different reactions:
(a)​​C​2​​​H2​ ​​+ HCN → ​C3​​​​H3​ ​​N​
(b)​​C​3​​​H6​ ​​​ + NH​3​​​ + 1.5O​2​​ → ​C3​ ​​​H3​ ​​​N + 3H​2​​O​
(c)​​C​2​​​H4​ ​​O + HCN → ​C3​​​​H3​ ​​​N + H​2​​O​
Calculate the fractional atom economy for each of these three different
choices. Analyze the process economics of each, using the following
prices: $4.92/kg ethylene; $1.58/kg hydrogen cyanide; $0.2/kg ammonia;
$0.04/kg oxygen; $1.48/kg ethylene oxide; $2.97/kg acrylonitrile.
Considering economics, environmental and safety considerations, which
reaction would you choose? Write a brief memo to your supervisor, summarizing the results of your analysis and explaining your recommendation. Append documentation of supporting calculations to the memo.
P1.70 Urea (CON2H4), used as a fertilizer, can be manufactured from methane
(CH4), water and nitrogen in a reaction pathway involving four reactions.
​​CH​4​​ + ​H2​ ​​O → CO + 3​H2​ ​​​
(R1)
​CO + ​H​2​​O → ​CO​2​​ + ​H2​ ​​​
(R2)
_​  1 ​​ N​ ​​
​​H​2​​ + 3
2
_​  2 ​​ NH​ ​​​
→3
3
​​NH​3​​ + _​  12 ​​ CO​2​​ → _​ 12 ​​ CON​2​​​H4​ ​​ + _​  12 ​​ H​2​​O​
(R3)
(R4)
Synthesize a reaction pathway to make urea from methane, water, and
nitrogen by combining these four reactions, with no net generation or
consumption of CO, CO2 or NH3. Calculate the fractional atom economy
of the overall reaction pathway. If a process makes 120,000 kg/day urea,
what are the flows (kg/day) of reactants and byproducts? What is the
net profit ($/day), if methane costs $0.11/kg, nitrogen costs $0.011/kg,
hydrogen costs $0.18/kg, water costs $0.012/kg, and urea sells for
$0.36/kg?
P1.71 When heated together, limestone (CaCO3) and charcoal (C) react to
form calcium carbide (Ca2C) and carbon dioxide (CO2).
​2CaC​O​3​+ 2C → C​a2​ ​C + 3C​O2​ ​​
(R1)
If water is then dripped over the calcium carbide, it makes acetylene
(C2H2), hydrogen (H2), and calcium hydroxide (Ca(OH)2).
​2C​a​2​C + 8​H2​ ​O → ​C2​ ​​H2​ ​ + 3​H2​ ​ + 4Ca(OH​)2​ ​​
(R2)
Acetylene is a useful chemical compound, and we’d like to see if a
process to make acetylene from limestone and charcoal is feasible.
Combine the two reactions into an overall reaction with no net generation or consumption of Ca2C. Calculate the fractional atom economy of
the overall reaction. If limestone costs $0.03/lb, charcoal costs $0.02/lb,
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Chapter 1 Problems
57
water is essentially free, and acetylene sells for $0.52/lb, is the process economically feasible? What if the hydrogen can also be sold at
$0.40/lb?
P1.72 Chalcocite (Cu2S) is one form of copper deposits found in mines. To
recover metallic Cu (for pennies, wiring, and other products), chalcocite
is leached in an acid solution of ferric sulfate to form cupric sulfate.
Cupric sulfate is then reduced with iron to produce metallic copper and
iron sulfate. The iron sulfate then is reacted with sulfuric acid and
hydrogen peroxide to recover ferric sulfate. The main reactions are:
(R1) ​​Cu​2​​​S + Fe​2​​​​(​SO​4​​)​​3​​→ ​CuS + CuSO​4​​​ + 2FeSO​4​​​
(R2) ​​CuS + Fe​2​​​​(​SO​4​​)​​3​​ → ​CuSO​4​​​ + 2FeSO​4​​​ + ​​S
(R3) ​​CuSO​4​​+ Fe → ​Cu + FeSO​4​​​
(R4) ​​2FeSO​4​​​ + H​2​​​SO​4​​​ + H​2​​​O2​ ​​ → ​Fe​2​​​​(​SO​4​​)​​3​​​ + H​2​​O​
Can the above reactions be combined so there is no net generation or
consumption of CuS, Fe2(SO4)3, CuSO4, or FeSO4? If yes, determine
the χ values and the overall net reaction. If not, explain why, and derive
an overall net reaction that sets some of these compounds to zero net
generation or consumption. Then calculate the grams of Cu2S and other
reactants required to produce 1 g of metallic Cu.
P1.73 1,4-butanediol (C4H10O2) is used in the synthesis of polyurethanes,
which are used to make tires, adhesives, seals and gaskets, foam seating,
and many other products. Conventionally 1,4-butanediol is made using
chemicals derived from fossil fuels. A biological route is proposed using
fermentation of glucose (C6H12O6) to succinic acid (C4H8O4), with CO2
and H2O as the major byproducts. This is followed by hydrogenation
of succinic acid to 1,4-butanediol. H2 required for the hydrogenation
step is made by steam reforming of methane:
​​CH​4​​​ + H​2​​O → ​CO + 3H​2​​​
The water–gas shift reaction is used to consume CO made by steam
reforming:
​​CO + H​2​​O → ​CO​2​​​ + H​2​​​
Write down the stoichiometrically balanced fermentation and hydrogenation reactions. Oxygen, water, and CO2 can be other reactants or
products in these reactions. Then combine those two reactions with
steam reforming and water–gas shift reactions to come up with a reaction pathway that has no net generation or consumption of CO, H2, or
succinic acid.
The glucose comes from a renewable resource, but the methane
used in steam reforming comes from non-renewable natural gas. For
every 1000 kg of 1,4-butanediol made, how many kg of glucose
and how many kg of methane are used? How many kg of CO2 are
produced?
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Chapter 1 Converting the Earth’s Resources into Useful Products
P1.74 In the case study, we described the chemical reactions for making adipic acid, using benzene as the raw material, with oxygen and nitric acid
as oxidizing agents.
Your company is a large supplier of adipic acid, but there are concerns about the safety and environmental impact of the conventional
process. You have learned about two alternative reaction pathways to
make adipic acid, both of which look promising in laboratory tests:
(a) Cyclohexene (C6H10) reacts with hydrogen peroxide (H2O2) to adipic acid over a sodium tungstate catalyst in aqueous solution, with
water as a byproduct. Cyclohexene is priced at $0.20/kg, and 35 wt%
hydrogen peroxide solution (in water) is $0.25/lb.
(b) N-hexane (C6H14) reacts with oxygen over a proprietary catalyst.
One-third of the n-hexane reacts to adipic acid, while the rest is
oxidized to different unwanted products. N-hexane is priced at
$0.33/kg and oxygen is essentially free.
Analyze the process economics of these two pathways, and compare to
the conventional process described in the case study. Assume an adipic
acid production rate of 28,100 kg/day and a sales price of $1.54/kg.
Considering also any safety or environmental concerns, write a brief
(1 page) memo to your supervisor summarizing your analysis and recommending a course of action. Append documentation of your supporting
calculations to your memo.
Game Day
P1.75 Polyvinyl chloride (PVC) is produced by the catalytic polymerization
of vinyl chloride and is used extensively to make products like plastic
pipe and film. Your assignment is to design a process for making vinyl
chloride (C2H3Cl). A brief survey of the synthetic chemistry literature
unearths the following reactions involving vinyl chloride or similar
molecules:
C2H2 + HCl → C2H3Cl
C2H4 + Cl2 → C2H4Cl2
C2H4Cl2 → C2H3Cl + HCl
2 HCl + 1/2 O2 + C2H4 → C2H4Cl2 + H2O
C2H4Cl2 + NaOH → C2H3Cl + H2O + NaCl
Come up with several different reaction pathways for the production of
vinyl chloride by mixing and matching these five reactions. Using the
prices given below, analyze which of your pathways look most promising. Consider atom economy and process economy. Since you work for
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Chapter 1 Problems
59
a PVC manufacturer, management is not interested in setting up a business to sell HCl or C2H4Cl2.
Compound
Price, $/lb
Ethylene (C2H4)
0.27
Dichloroethane (C2H4Cl2)
0.17
Acetylene (C2H2)
1.22
Chlorine (Cl2)
0.10
Hydrogen chloride (HCl)
0.72
Sodium hydroxide (NaOH)
1.13
Vinyl chloride (C2H3Cl)
0.22
P1.76 Round-up® is a popular biodegradable (and somewhat controversial)
nonselective herbicide. A key intermediate in the synthesis of Round-up
is DSIDA (disodium iminodiacetate, C4H5NO4Na2). DSIDA is made by
mixing ammonia (NH3), formaldehyde (CH2O), and hydrogen cyanide
(HCN) under acidic conditions to make IDAN (iminodiacetonitrile).
(This is called a Strecker synthesis, and similar reactions are used to
synthesize amino acids, the building blocks of proteins.) Then an aqueous solution of sodium hydroxide (NaOH) is mixed with IDAN to
make DSIDA, with NH3 as a byproduct.
H
N
C
C
H
N
C
H
NaO
C
H
H
Iminodiacetonitrile (IDAN), C4H5N3
C
O
C
H
O
H
H
C
C
H
H
C
H
N
ONa
H
H
Disodium iminodiacetate (DSIDA), C4H5O4NNa2
H
OH
C
C
N
H
H
N
C
OH
H
H
H
Diethanolamine, C4H11O2N
There are several problems with this reaction pathway. HCN is highly
toxic. The reaction is exothermic (gives off heat), and without tight
control of the reactor a runaway reaction situation can develop. Because
of some side reactions (not considered here), the process generates 1 lb
cyanide-contaminated waste per 7 lb DSIDA produced.
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Chapter 1 Converting the Earth’s Resources into Useful Products
Your job is to investigate an alternative, safer approach, that uses
DEA, diethanolamine. Over a copper catalyst in a NaOH solution, DEA
reacts to form DSIDA. The structure of DEA is shown above.
First, figure out the two stoichiometrically balanced reactions (with
byproducts) for production of DSIDA from ammonia, formaldehyde,
hydrogen cyanide, and sodium hydroxide. Develop a generationconsumption table for the overall process. Determine the cost per day,
in raw materials, for production of 1770 lb/day DSIDA.
Next, figure out the reaction stoichiometry for synthesis of DSIDA
from DEA. What is the byproduct? Compare the cost of producing
DSIDA for this process to the cost for the conventional HCN process.
Is the safer process economically competitive?
DEA is synthesized by oxidation of ethylene (C2H4) with oxygen
to ethylene oxide (C2H4O), then reaction of C2H4O with NH3. Assume
that air is used as the source of O2 and it is free. Write the stoichiometrically balanced reactions, complete the production-consumption
analysis, and develop the input–output table. From a raw materials cost
point of view, would it be a good idea to make the DEA inhouse
instead of purchasing it?
Assume the following values for raw materials used in this process:
mur83973_ch01_001-060.indd 60
Compound
Price
Formaldehyde (37 wt% in water)
0.15/lb solution
Ammonia
$182/metric ton
Hydrogen cyanide
$1.92/kg
Sodium hydroxide
$0.40/lb
Diethanolamine
$1.60/kg
Ethylene
$0.48/lb
Hydrogen
~0
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2
CHAPTER TWO
Process Flows: Variables,
Diagrams, Balances
In This Chapter
We take the next step along the path toward designing a chemical process. We
describe the key process variables used to characterize a chemical process. We
illustrate how to translate a generation-consumption analysis into an inputoutput flow diagram. We introduce two important visual representations of
chemical processes: the block flow diagram and the process flow diagram. We
briefly describe the major processing units in a block flow diagram, and we
begin our study of material balance calculations.
Here are some of the questions we address in this chapter:
∙ What are the important process variables?
∙ What are the four major processing units common to most chemical
processes?
∙ What are three different ways to diagram a process flowsheet?
∙ How do I begin to synthesize a process flow sheet?
∙ How are process flow rates calculated?
∙ What kind of information is useful for completing process flow calculations?
Words to Learn
Watch for these words as you read Chapter 2.
Process flow sheet
Process variables
Process streams
Process units: Mixers, Reactors,
Separators, Splitters
Input-output diagrams
Block flow diagrams
Process flow diagrams
Batch or continuous-flow
Steady-state or transient
Material balance equations
Basis
System
Components
Stream composition specification
System performance specification
61
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Chapter 2 Process Flows: Variables, Diagrams, Balances
2.1
Introduction
In order to manufacture products, we need to design, build, and operate a chemical process plant. The chemical process plant is a physical facility in which the
raw materials undergo chemical and physical changes in order to make the
desired products (Fig. 2.1). Chemical process plants come in all shapes and
sizes, but share many common features.
The flow of materials through a chemical process plant is shown visually
on process flow sheets. In general, a chemical process plant carries out some
or all of the following functions (Fig. 2.2):
∙ Feed preparation. Bring raw materials to the correct composition or
physical state.
∙ Reaction. Provide conditions to allow desired chemical reactions to take
place under control.
Figure 2.1 Chemical process plants come in many shapes and sizes. Left: clean diesel plant at the Danube Refinery
in Hungary. Holly Curry/McGraw Hill. Center: part of a pharmaceutical plant. Monty Rakusen/Cultura/Image Source.
Right: cheese-making facility. Monty Rakusen/Cultura/Getty Images.
Raw
materials
Reactants
Feed
preparation
Chemical
reactors
Reaction
products
Separation
units
Desired
products
Product
formulation/
storage
Final products
to customers
Waste
products
Environmental
control facilities
Discharge to
environment
Figure 2.2 The flow of raw materials to desired products is illustrated on process flow sheets.
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Section 2.2 Process Variables
63
∙ Separation and purification. Separate desired products from raw materials,
byproducts, and wastes.
∙ Environmental control. Handle wastes for safe reuse or disposal.
∙ Product formulation. Mix, formulate, package, and store the final product.
To carry out these functions, the chemical plant has a diverse array of
equipment including: mixers, reactors, columns, driers, pumps, heat exchangers, piping, and instrumentation.
In this chapter, we identify the key process variables needed to describe
chemical processes, and we describe three kinds of flow sheets used to illustrate
chemical processes. We learn heuristics for choosing processing units and connecting them together in a flow sheet. We learn how to complete simple process
flow calculations, using material balance equations and process specifications.
2.2
Process Variables
Process flow sheets incorporate quantitative information about process variables. The process variables that we are interested in now are: moles, mass,
composition, concentration, pressure, temperature, volume, density, and flow
rate. Before we proceed, we will briefly review what you need to know about
dimensions and units.
2.2.1
Quick Quiz 2.1
What is the derived
dimension of volume
and of density in terms
of the base dimensions?
A Brief Review of Dimensions and Units
A dimension is a fundamental quantity, a property of a physical entity. There
are only a few base dimensions; the ones we concern ourselves with in this
book are: mass m, length L, time t, thermodynamic temperature T, and amount
of substance n. Quantities such as area, volume, density, and pressure are all
derived from these base units. For example, area has the derived dimension of
L2 and pressure has the derived dimension of m/Lt2.
A unit is a specific magnitude of a dimension, either base or derived. You
have probably used the Système Internationale (SI) system (kilograms, meters,
seconds) in your science classes, but you will encounter many different units
in the chemical process industry. For example:
Base dimension
Units
Mass m
Kilogram, gram, ounce, ton, metric ton, . . .
Length L
Meter, foot, centimeter, mile, light-year, . . .
Time t
Day, hour, second, year, semester, . . .
Temperature T
Degree Celsius, degree Fahrenheit, Kelvin, degree Rankine, . . .
Amount of substance n
Gram-mole, kilogram-mole, ton-mole, . . .
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Chapter 2 Process Flows: Variables, Diagrams, Balances
There are all kinds of crazy units. One of my personal favorites is the EFOB—
equivalent fuel oil barrel—which is not a measure of volume but of energy!
There’s not much you can do about unit profusion—except to work hard to
avoid unit confusion.
Helpful hint: Whenever you write a number, write down the associated
units. A quantity is meaningless without units. If an equation is dimensionally
consistent, all the terms in the equation are of the same dimension. Check that
the units on both sides of any equals, plus, or minus sign are the same.
Illustration: When you go to the store and buy groceries, the cashier doesn’t
say “that’ll be 52 and 47,” she says “that’ll be 52 dollars and 47 cents.”
Helpful Hint
Whenever you do a
calculation, check
for dimensional
consistency.
Helpful Hint
Whenever you convert between different units, keep track
by crossing out in
both numerator and
denominator.
Illustration: Every Monday you buy two gallons of milk and five boxes of
cookies. Since you have to carry your purchases home, you’d like to know
how many pounds of food you’ve bought. Can we just add together gallons of
milk plus boxes of cookies to get pounds of food?
“Gallons” has dimension of volume, or [L3]. “Boxes” has dimension of amount
of substance, or [N]. “Pounds” has dimension of mass, or [m]. The dimensions
are not consistent. We want every term to have dimension of mass. To convert
volume [L3] to mass [m], we need to multiply by [m]/[L3], or density. Dimensional
analysis has guided us to determine that we need to multiply the volume of milk
by the density of milk (8.3 lb/gallon). To convert amount (boxes, N) to mass m,
we need to know the mass per unit box. We check the cookie package for information and find that there are 2.0 lb of cookies per box. Now we can solve:
​​ 2 gal milk × _____
​ 8.3 lb
​​​ + ​​ 5 boxes of cookies × _____
​ 2.0 lb ​)​​
(
gal ) (
box
= (16.6 lb milk) + (10.0 lb cookies) = 26.6 lb food
Illustration: If one gallon of milk costs 11 U.S. quarters, you can trade in
one euro for $1.20, and you can trade in 107 yen per euro, how much will you
spend, in yen, for a gallon of milk?
107 yen __________
245 yen
11 ​U.S. quarters​ __________
​$​ 0.25
1 ​euro​ ​​ × ​​ _______
______________
​​
​​ × ​​
​​ × ​​ ______
​euro​ ​​ = ​​  gallon milk ​​
​U.S. quarter​ ​$​ 1.20
gal milk
Helpful Hint
Choose units so
that numbers are
neither too large
nor too small.
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Illustration: Don’t report 0.004167 kg/s. Choose instead 4.167 g/s or 0.25 kg/min
or 15 kg/h. Don’t report 15,000,000,000,000 mg/h. Choose instead 15,000 kg/h
or 15 metric tons/h.
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Section 2.2 Process Variables
2.2.2
65
Mass, Moles, and Composition
In Sec. 1.5.1, we reviewed mass and mole units and unit conversions. Recall
that molar mass is the mass in grams of 1 mole (6.02214 × 1023) of atoms or
molecules, and has units of [g/gmol]. For convenience, the molar mass may be
written as [lb/lbmol], [kg/kgmol], [ton/tonmol], or any other similar units with
dimension [m/n]. The numerical value of the molar mass of a compound in
any of these units is identical. To convert from moles to mass, multiply the
moles by the molar mass. To convert from mass to moles, divide the mass by
the molar mass. In equation form, if ni is the moles of compound i, mi is the
mass of compound i, and Mi is its molar mass, then
​​m​i​​ = ​​n​i​​ ​​M​i​​
Eq. (2.1)
We will sometimes be interested in the total moles of an element in a given
number of moles or given mass of a compound. If nhi is the moles of element
h present in compound i, then
​m​​
​​n​hi​​ = ε​​ ​hi​​ ​​n​i​​ = ​​ε​hi​​ ___
​​  i  ​​
​M​i​
Eq. (2.2)
where εhi is the moles of element h per mole of compound i.
Illustration: 12 gmol glucose (C6H12O6) sits in a beaker. We’d like to know
how many grams of glucose and how many gram-moles of carbon (C) are in
the beaker.
180 g glucose
____________
​​m​glucose​​ = ​​n​glucose​​ ​​M​glucose​​= 12 gmol glucose × ​​
​​= 2160 g glucose
gmol glucose
6 gmol C
​​n​C in glucose​​ = ​​ε​C in glucose​​ ​​n​glucose​​ = ___________
​​     ​​× 12 gmol glucose = 72 gmol C
gmol glucose
We often deal with mixtures of compounds. If m is the total mass and n
is the total moles, then
m = ​​∑​ ​​​ ​​m​i​​
Eq. (2.3a)
n = ​​∑​ ​​​ ​​n​i​​
Eq. (2.3b)
i
i
where the summation sign indicates that the sum is taken over all compounds.
The composition of a mixture is defined as the quantities of compounds
in that mixture relative to one another. Composition can be reported on a mass
basis (fraction or percent) or a mole basis (fraction or percent), or composition
can be expressed as a mass or molar ratio.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Unless otherwise stated, we use mass and weight interchangeably and we
indicate mass (or weight) fraction by wi and mole fraction by zi.
​m​​
​m​i​
Mass fraction of i = _________
​​  mass of i  ​​ = ​​w​i​​ = _____
​​  i  ​​ = ​​ ___
​​
total mass
​∑​ ​​ ​m​i​ m
i
Mass percent of i = _________
​​  mass of i  ​​× 100% = ​​w​i​​ × 100%
total mass
​n​​
​ni​​
Mole fraction of i = __________
​​  moles of i  ​​ = ​​z​i​​ = ____
​​  i  ​​ = ​​ __
​​
total moles
​∑​ ​​​n​i​ n
i
Mole percent of i = __________
​​  moles of i  ​​× 100% = ​​z​i​​ × 100%
total moles
Illustration: 12 g glucose (C6H12O6) and 3 g salt (NaCl) are dissolved in 85 g
water (H2O).
12 g glucose
mass fraction glucose = ​​ ______________________________
​​ = 0.12
12 g glucose + 3 g salt + 85 g water
12 g glucose
mass percent glucose = ​​ ______________________________
​​ × 100%
12 g glucose + 3 g salt + 85 g water
= 12 mass%
(Note: do not add a percent sign to a mass fraction! The mass fraction glucose
is NOT 0.12%.)
Mass and mole fractions are dimensionless. Notice that the mass or mole fractions in a stream must sum to 1:
​​∑​ ​​​ ​​w​i​​ = 1
i
​​∑​ ​​​ ​​z​i​​ = 1
i
Illustration: 12 g glucose (C6H12O6) and 3 g salt (NaCl) are dissolved in 85 g
water (H2O).
12 g glucose
mass fraction glucose = w​
​​ glucose​​ = ​​ _____________________________
​​ = 0.12
12 g glucose + 3 g salt + 85 g water
3 g salt
______________________________
mass fraction salt = w​
​​ salt​​ =
​​      ​​ = 0.03
12 g glucose + 3 g salt + 85 g water
mass fraction water = w​
​​ water​​= 1 − 0.12 − 0.03 = 0.85
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67
Section 2.2 Process Variables
Converting between mass fraction and mole fraction requires knowledge of the
molar mass of all the species in the mixture:
​z​​ ​M​​
​​w​i​​ = ______
​  i i  ​
​∑​ ​​ ​zi​​ ​Mi​​
i
​wi​​∕​Mi​​
​​z​i​​ = _________
​​
​​
​∑​ ​​ (​wi​​∕M
​ i​​)
i
Illustration: 12 g of glucose (C6H12O6) and 3 g sodium chloride (NaCl) are
dissolved in 85 g of water. Molar masses are 180, 58.5, and 18 g/gmol,
respectively.
12 g glucose
​
​w​glucose​​ = ​ ​_______________________________
​​ = 0.12
12 g glucose + 3 g NaCl + 85 g water
mass percent glucose = 0.12 × 100% = 12 mass% (12 wt%)
12 g glucose∕180 g∕gmol
​
​z​glucose​​ = ​​ _______________________________________________________________
​​ = 0.014
12 g glucose∕180 g∕gmol + 3 g NaCl∕58.5 g∕gmol + 85 g water∕18 g∕gmol
mole percent glucose = 0.014 × 100% = 1.4 mol%
Quick Quiz 2.2
A beaker contains
15 g glucose and
85 g water. Does
the beaker contain
15 wt% glucose or
0.15 wt% glucose?
Illustration: 12 gmol glucose (C6H12O6) and 3 gmol sodium chloride (NaCl)
are dissolved in 85 gmol water.
12 gmol glucose
​
​z​glucose​​= ​ ​________________________________________
​​= 0.12
12 gmol glucose + 3 gmol NaCl + 85 gmol water
mole percent glucose = 0.12 × 100% = 12 mol%
12 gmol glucose (180 g∕gmol)
​w​glucose​= ​ ​___________________________________________________
​​ = 0.56
12 gmole glucose (180 g∕gmol) + 3 g NaCl (58.5 g∕gmol) + 85 g water (18 g∕gmol)
mass percent glucose = 0.56 × 100% = 56 mass% (56 wt%)
Since the composition of a mixture describes the quantities of compounds relative to one another, composition can also be reported as a mole or mass ratio.
Illustration: 12 g glucose (C6H12O6) and 3 g salt (NaCl) are dissolved in 85 g
water (H2O). This mixture has a mass ratio of 4 g glucose/g salt.
2.2.3
Temperature and Pressure
Temperature T is a base dimension. Kelvin (K) and Rankine (°R) scales are
both absolute scales (no negative temperatures). Celsius (°C) and Fahrenheit (°F)
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Chapter 2 Process Flows: Variables, Diagrams, Balances
are displaced from Kelvin and Rankine scales by a constant number. To convert
among various temperature scales, use the equations below.
T(K) = T(°C) + 273.15
T(°R) = T(°F) + 459.67
Quick Quiz 2.3
T(°R) = 1.8T(K)
If a system is at
0 K, what is the
temperature in °R?
T(°F) = 1.8T(°C) + 32
T(°C) = _​​ 59 ​​ [T(°F) − 32]
Illustration:
25°C = 77°F = 298.15 K = 536.67°R
Pressure P has dimension of [m/Lt2]. You might never know this from looking
at the plethora of units used. Use the conversion factors below to convert from
one system of pressure units to another.
1 bar = 0.1 MPa = 100 kPa = 105 Pa = 105 N/m2 = 106 dyn/cm2 = 750.062 mm Hg
(at 0°C) = 33.4553 ft H2O (at 4°C) = 14.50377 lbf /in2 (psi) = 0.9869233 atm
1 atm = 1.01325 × 105 Pa = 101.325 kPa = 1.01325 bar = 0.101325 MPa =
760 mm Hg (at 0°C) = 33.89854 ft H2O (at 4°C) = 14.69595 lbf /in2 (psi)
Illustration:
1 atm
760 mm Hg
5.075 bar ​​(__________
​
​)​​ ​​(__________
​
​
​​ = 3807 mm Hg
1 atm )
1.01325 bar
Quick Quiz 2.4
You use a pressure
gauge to measure that
the pressure in your
bicycle tire is 0 mm Hg.
Is your tire under
vacuum?
A pressure gauge open to the atmosphere reads 0, not 1, so gauge pressure =
(absolute pressure) − (atmospheric pressure). Gauge pressure is indicated by a
“g” after the pressure unit, for example, “psig” (pounds of force per square
inch gauge). When you pump up a bicycle tire, the pressure you read is the
gauge pressure. If you have a bicycle pump with a pressure gauge, try reading
the pressure with the pump disconnected from the tire.
Illustration: If a pressure gauge reads 4.06 barg, the absolute pressure is
4.06 + 1.01 = 5.07 bar.
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Section 2.2 Process Variables
2.2.4
Volume, Density, and Concentration
Volume V has dimension of [L3]. Use the conversion factors below to convert
between different units of volume.
1 cm3 = 1 mL = 0.001 L = 0.033814 fl. oz. (U.S.) = 0.06102374 in3
= 2.6418 × 10−4 gallons (U.S.) = 3.532 × 10−5 ft3
1 liter (L) = 1000 cm3 = 1 dm3 = 0.001 m3 = 61.02374 in3 = 0.03531467 ft3
= 33.814 fl. oz. (U.S.) = 2.11376 pints (U.S. liquid)
= 1.056688 qt (U.S. liquid) = 0.26417205 gallons (U.S.)
= 0.21997 gallons (U.K.)
1 ft3 = 28316.847 cm3 = 28.316847 L = 0.028316847 m3 = 1728 in3
= 7.480519 gallons (U.S.) = 0.803564 bushels (U.S. dry) = 0.037037 yd3
1 barrel (oil) = 158.987 L = 42 gallons (U.S.) = 1.333 barrels (U.S. liquid)
= 5.614583 ft3 = 0.15899 m3
Illustration:
7.480519 gal
___________
5.075 ​​ft​3​​​​ ​
​
​​ = 37.96 gal (U.S.)
(
)
​ft​3​
The volume per mass is called the specific volume and has dimension of
[L3/m]. The volume per mole is called the molar volume and has dimension
of [L3/n]. We will use ​​V̂ ​ to denote either specific or molar volume; you will
know which is meant by looking at the units.
Density ρ is mass or moles per unit volume. Specific density is the
inverse of the specific volume and has dimension of [m/L3]. Molar density is
the inverse of the molar density and has dimension of [n/L3]. We will use ρ to
denote either specific or molar density.
To find the volume of a material from its mass, divide the mass by the
density, or multiply the mass by the specific volume.
̂
V = __
​​  m
ρ ​​ = m​​V ​​
Similarly, volume is calculated from moles by dividing moles by molar density.
Illustration: Liquid water at 4°C has a specific density ρ of 62.43 lb/ft3.
Its specific volume ​V
​ ̂ ​is (1/62.43 lb/ft3) = 0.160 ft3/lb. 234 lb of water occupy
a volume of (234 lb)/(62.43 lb/ft3) = 3.75 ft3. The molar mass of water is
18 lb/lbmol. The molar density of liquid water at 4°C is 62.43 lb/ft3/
18 lb/lbmol = 3.47 lbmol/ft3.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Specific gravity is the ratio of the specific density of a substance to the
specific density of water at 4°C. (Sometimes temperatures other than 4°C are
used as reference. If so, this will be indicated.) Specific gravity is dimensionless. The density of solids is nearly independent of temperature and pressure.
The density of liquids is nearly independent of pressure but somewhat dependent on temperature.
Use the conversion factors below to convert between different units of density.
1 g/cm3 = 1000 kg/m3 = 1 kg/L = 62.42796 lb/ft3
= 8.345404 lb/gal (U.S.) = 0.0361279 lb/in3
Illustration: The specific gravity of liquid benzene is reported as 0.876520/4,
indicating that this is the specific gravity of benzene at 20°C relative to liquid
water at 4°C. Liquid water at 4°C has a density of 1.00 g/cm3 = 62.43 lb/ft3.
The density of benzene at 20°C is therefore 0.8765 g/cm3 or 54.72 lb/ft3.
The density of gases is strongly dependent on both pressure and temperature. For most gases and gas mixtures at moderate temperatures and pressures,
the molar density can be calculated from the ideal gas law
n  ​​ = ​​ ___
P  ​​
ρ = __
​​  1  ​​ = ​​ __
V
RT
̂
​V ​
Eq. (2.4)
where T is the absolute temperature and R is the ideal gas constant.
R = 83.144 bar cm3/gmol K = 82.057 atm cm3/gmol K = 62.361 mmHg L/gmol K
= 1.314 atm ft3/lbmol K = 0.083144 bar L/gmol K = 0.082057 atm L/gmol K
= 555.0 mm Hg ft3/lbmol °R = 10.73 psi ft3/lbmol °R = 0.7302 atm ft3/lbmol °R
The ideal gas law is not a law at all. Rather, the “law” is a model equation,
which relates one physical property of a material (in this case, molar density
of a gas) to process variables (in this case, temperature and pressure). In this
text we will assume that gases obey the ideal gas law, because this simple
model equation is sufficiently accurate for our purposes. See App. B for other
model equations for more accurately estimating the molar density of a gas from
its pressure and temperature.
Specific volumes and densities of gases are often reported at standard
temperature and pressure (STP), 0°C and 1 atm pressure.
Helpful Hint
The ideal gas law
does not apply to
liquids and solids!
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Illustration: A gas at 100°C and 3.50 atm pressure has a specific density of
gmol
3.50 atm
______________________________
ρ =
​​
​​ = 0.114 ​​ _____
​​
L
0.082057 L atm∕gmol K (373.15 K)
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Section 2.2 Process Variables
71
At STP, the specific density is
gmol
1 atm
ρ = ______________________________
​​
​​ = 0.0446 ​​ _____
​​
L
0.082057 L atm∕gmol K (273.15 K)
Quick Quiz 2.5
Johnnie Genius says
that pure H2O at 77°F
and 1 atm has a molar
density of 0.0019
ft3/lbmol. He says he
calculated this from
the ideal gas law. Is
Johnnie right?
The specific molar volume of an ideal gas at STP is 22.414 L/gmol or 359 ft3/lbmol.
Concentration gives the mass (or moles) of a solute per volume of solution. It has dimension of [m]/[L3] or [n]/[L3]. Concentration has the same
dimension as density, but the meaning is distinctly different. The density of a
liquid solution is the mass (or moles) of solution per volume of solution.
2.2.5
Flowrates
Flow rates have dimension of [m/t] (mass flow rate), [n/t] (molar flow rate) or
[L3/t] (volumetric flow rate). Units for mass, molar, and volumetric flow rates
are interconverted in the same way as those for mass, moles, and volumes.
Figure 2.3 Taking a process from the lab into full-scale commercial production requires
more than just purchase of bigger beakers!
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Illustration: The mass flow rate of a process stream is 115.0 lb
oxygen/min. The molar mass of O2 is 31.9988 g/gmol, and the gas behaves
the ideal gas law, so ​V
​ ̂ ​ = 22.414 L/gmol at STP. The volumetric flow rate in
3
cm /s at STP is:
115.0 lb
453.59237 g
1 gmol
22.414 L
1 min
608.97 L
​​ _______
​
​​​​​ ___________
​
​
​​​​ _________
​
​ ​​​​ ________
​
​​​​​ _____
​
​ ​​ = ________
​​
s ​​
(
)
( min )(
)
(
)
(
)
31.9988 g
lb
gmol
60 s
In process flow rates, time may be given in units of seconds, minutes, hours,
days, or years. Commodity chemical process plants operate 24 hours per day,
7 days per week. Typically they operate year-round except for a short shutdown
period of 1–2 weeks for cleaning, maintenance and upgrades. Thus, a typical
operating year is approximately 350 days. Hours and days of operation of
specialty-chemical process plants are more variable.
Illustration: A process plant produces 125 lb/h of vanilla extract. Its annual
production of vanilla extract is:
350 operating days
125 ​​
lb × ​​ ____
24 h ​​ × ​​  ________________
​​ ______
year ​​= 1,050,000 lb/yr or 1.05 million lb/yr.
h
day
2.3
Chemical Process Flow Sheets
Process flow sheets are compact and precise diagrams that present a large
amount of technical information about chemical processes. They are the language that chemical process engineers use, and you should become fluent at
translating from words to flow sheet and back again.
The three types of chemical process flow sheets that we will use are listed
in Table 2.1. In the following sections, we first describe the key features of
each type of flow sheet, then illustrate each, using ammonia synthesis as an
example. Generation-consumption analysis of the reaction pathway from methane and air to ammonia was the subject of Example 1.4. For more on the long
and checkered history of this simple and ubiquitous chemical, read the
ChemiStory at the end of this chapter.
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Section 2.3 Chemical Process Flow Sheets
Table 2.1
73
Three Types of Chemical Process Flow Sheets
Diagram
Information
Input-output flow diagram
Raw materials
Reaction stoichiometry
Products
Block flow diagram
Everything above, plus
Material balances
Major process units
Process unit performance specifications
Process flow diagram (PFD)
Everything above, plus
Energy balances
Process conditions (T and P)
Major process equipment specifications
2.3.1
Input-Output Flow Diagrams
An input-output diagram (Fig. 2.4) is the simplest process flow sheet. It has
the following features:
∙ A single block represents all of the physical and chemical operations in
the process.
∙ Lines with arrows represent material moving into and out of the process.
∙ Raw materials enter on left.
∙ Products leave on right.
∙ Raw material and product flow rates (or quantities) may be shown.
Figure 2.5 shows an input-output diagram for ammonia synthesis, using a
basis of production of 1000 metric tons ammonia/day. Quantities of raw materials and products were calculated by using the methods described in Chap. 1.
An input-output diagram is the simplest process flow sheet we can imagine. Despite its simplicity, the input-output diagram might still trigger a few
thoughts. For example, in developing the input-output flow diagram for ammonia synthesis, we might consider such questions as: What is the source of
Raw materials
Process
Products
Figure 2.4 A generic input-output diagram shows the raw materials consumed and the
products (and byproducts) generated.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Quick Quiz 2.6
In Figure 2.5, does the
total mass of raw
materials into the
process equal the total
mass of products out?
350 tons CH4/day
795 tons H2O/day
Ammonia synthesis
process
825 tons N2/day
1000 tons NH3/day
970 tons CO2/day
Figure 2.5 This input-output diagram for ammonia synthesis was developed by scaling
up the generation-consumption analysis of Example 1.4.
nitrogen? Should we use air? If so, what should we do with the oxygen? Asking
and answering these questions constitute vital steps in the journey from a balanced chemical equation to a functioning chemical process.
2.3.2
Block Flow Diagrams
Block flow diagrams represent the next step up in complexity and detail. We
will use block flow diagrams extensively in this text. These diagrams have the
following features:
∙ Chemical processes are represented as a group of connected blocks, or
process units.
∙ Lines with arrows connect the blocks and represent process streams:
inputs (material moving into each process unit) and outputs (material
moving out of each unit).
∙ Raw materials typically enter on left.
∙ Products typically leave on right.
∙ Quantities or flow rates of inputs and outputs may be indicated, either
directly on the diagram or in an accompanying table.
There are only four kinds of process units that are included in a block flow
diagram: Mixers, Reactors, Splitters, and Separators (Fig. 2.6). Each process
unit represents a specific process function.
Mixers. Mixers combine two or more inputs into a single output.
Reactors. The input streams contain reactants. One or more chemical
reactions take place inside a reactor. The output streams contain reaction
products as well as unconsumed reactants. In the simplest case, there is
one input and one output stream.
Splitters. Splitters split a single input into two or more outputs without
changing the stream composition. If the input is a mixture of two or
more components, then the output streams of a splitter all have the same
composition as the input.
Separators. An input stream is separated into two or more outputs, where
the outputs have different compositions from each other and from the
input. In the simplest case, there is one input and two output streams.
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Section 2.3 Chemical Process Flow Sheets
75
Butter
C12H22O11
Eggs
Flour
NaCl
Mixer
Raw chocolate chip
cookie dough
Reactor
Baked chocolate chip cookies
NaHCO3
Chocolate chips
Raw chocolate chip
cookie dough
4 dozen cookies
75% chocolate chip cookies
25% peanut butter cookies
4 dozen cookies
75% chocolate chip cookies
25% peanut butter cookies
3 dozen cookies
75% chocolate chip cookies
25% peanut butter cookies
Splitter
1 dozen cookies
75% chocolate chip cookies
25% peanut butter cookies
3 dozen cookies
97.2% chocolate chip cookies
2.8% peanut butter cookies
Separator
1 dozen cookies
8.3% chocolate chip cookies
91.7% peanut butter cookies
Figure 2.6 Examples of the four kinds of process units included on block flow diagrams.
These four kinds of process units are connected in myriad ways in a block
flow diagram. Output streams from one process unit become input streams
to other units. A simplified block flow diagram for the ammonia synthesis
process is shown in Figure 2.7. Read the description carefully and compare
with the flow diagram.
2.3.3
Process Flow Diagrams (PFD)
Block flow diagrams are useful sketches that show the major process units and
the materials flowing between the units, but they are far from realistic pictures
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Air
Natural gas
Mixer
Steam
Reactor
Mixer
H2
O2
N2
CH4
CO
H2O
CO2
Separator
Mixer
Separator
Reactor
Reactor
H2
N2
NH3
H2
N2
CO2
H2O
H2
N2
Separator
H2
N2
Water
Ammonia
Figure 2.7 Simplified block flow diagram for synthesis of ammonia from natural gas. Natural gas, which is mostly
methane, and steam are mixed and then reacted to CO and H2. Air is added to facilitate further oxidation of methane
and to supply nitrogen. A water-gas shift reactor converts CO and H2O to CO2 and more H2. Excess water and CO2
are removed in separators, then the H2/N2 mix is sent to the ammonia synthesis reactor. Ammonia is separated from
unreacted gases, which are recycled back to the reactor.
Quick Quiz 2.7
What four compounds
are in the output from
the first reactor in
Figure 2.7?
Quick Quiz 2.8
Why is the last process
unit in Figure 2.7 a
separator and not a
splitter?
mur83973_ch02_061-154.indd 76
of a functional chemical process. The process units in a block flow diagram
might correspond one-to-one with a specific piece of equipment, or to several
pieces of equipment. Alternatively, one piece of equipment might accomplish
multiple process functions. Process Flow Diagrams (PFDs) are a step up in
complexity and information content compared to block flow diagrams. Most
PFDs have these features:
∙ Chemical processes are represented as a connected group of process equipment.
∙ All major pieces of process equipment are drawn representationally.
Reactor type and separation methods have been chosen.
∙ Equipment used to move material around (e.g., pumps, compressors, conveyer
belts) and to heat or cool material (e.g., heat exchangers, furnaces) is included.
∙ Each piece of equipment is assigned a number and given a descriptive name.
∙ Major utilities (steam, cooling water, etc.) are shown.
∙ Lines with arrows connect the process equipment and represent inputs and
outputs.
∙ Process streams are numbered.
∙ Generally, materials enter from the left and leave from the right.
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Section 2.3 Chemical Process Flow Sheets
Cooling
To process steam
High-temperature
shift converter
Air
Condensate
stripper
Heat
recovery
Air compressor
Natural
gas load
Steam
Feed gas
compressor
Sulfur
removal
Process
steam
Secondary
reformer
Heat
recovery
Primary
reformer
Low-temperature
shift converter
Heat
recovery
Condensate to
boiler feedwater
system
Cooling
Heat
recovery
Methanator
Heat
recovery
CO2
absorber
CO2
product
Syn gas
cooling
Cooling
Drier
Cooling
Refrigeration
compressor
Cooling
Cooling
Synthesis gas
compressor
Refrigeration
drums
Ammonia
product
CO2
stripper
Cooling
Heat
recovery
Refrigeration
exchanger
Purge
gas to
fuel
C
Hydraulic
turbine
Cooling
water
Heat
recovery
Start-up
heater
Ammonia
converter
Figure 2.8 Simplified Process Flow Diagram, for an ammonia synthesis facility. Adapted from a figure in KirkOthmer Encyclopedia of Chemical Technology.
∙ Generally, gas streams are at the top, liquid streams are in the middle, solid
streams are toward the bottom.
∙ The flow rate or quantity, composition, temperature, pressure, and/or phase
of process streams are indicated, usually in an accompanying table.
Figure 2.8 illustrates a Process Flow Diagram for an ammonia plant. Some
typical process equipment representational icons are shown in Figure 2.9.
Chemical processes are brought from idea to reality by methodically moving from the simplest to the most complex diagrams. The engineering cost
to produce each diagram increases dramatically as we move down the list in
Table 2.1, because so much more information is required. Therefore, at each
stage, the process is re-evaluated for economics and feasibility. For example,
initial screening of process economics considers only the difference between
product value and raw material costs; input-output diagrams are sufficient at
this stage. A clearer picture of the required process operations as well as alternative process schemes emerges after generation of block flow diagrams. A
more accurate economic analysis requires information about capital costs (cost
of purchasing land, equipment, buildings, etc.) and operating costs (material,
energy, and labor costs). Preliminary estimates of capital and operating costs
require completion of a PFD. Even more detailed drawings are needed for
construction and operation of the process.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Mixer
Reactor
Mixer
Packed-bed reactor
Flash drum
Cyclone
Tray column
Blower
Centrifugal pump
Conveyor
Turbine
Blower
Centrifugal pump
Conveyor
Turbine
Heat exchanger
Furnace
Kettle-type reboiler
Heat exchanger
Furnace
Kettle-type reboiler
Evaporator
Figure 2.9 A selection of process equipment icons used in process flow diagrams. Many other icons are used in
addition to the ones shown. The level of detail and accuracy in representation is variable.
2.3.4
Modes of Process Operation
We categorize process operations on the basis of their dependence on time.
∙ Steady-state processes are time-independent. Process variables do not
change with time.
∙ Transient or unsteady-state processes are time-dependent. One or more
process variable changes with time.
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Section 2.3 Chemical Process Flow Sheets
79
Illustration: At 9 a.m., the input to a reactor is 20% glucose and 80% water,
with a flow rate of 20 g/s and a temperature of 25°C, while the output from
the reactor is 10% glucose, 10% fructose, and 80% water, at a temperature of
50°C. At 9 p.m., the input to a reactor is 20% glucose and 80% water, with a
flow rate of 20 g/s and a temperature of 25°C, while the output from the reactor is 10% glucose, 10% fructose, and 80% water, at a temperature of 50°C.
This is a steady-state process.
We also categorize process operations by how input and output streams
are handled (Fig. 2.10).
∙ In batch processes, input streams enter the process unit all at once, and at
some later time output streams are removed from the process unit all at once.
Input and output streams are quantified in dimensions of mass or moles.
∙ In continuous-flow processes, input streams continuously flow into the
process unit and output streams continuously flow out of the process unit.
Input and output streams are quantified in dimensions of mass per unit time
or moles per unit time.
∙ Semibatch processes are some combination of batch and continuous. For
example, input streams might be added all at once while output streams
are removed continuously.
Quick Quiz 2.9
Consider your digestive system as a block
flow diagram. Identify
the type of processing
unit (mixer, splitter,
reactor, separator) of
(a) mouth, (b) stomach,
(c) intestine. Does
your digestive system
operate in batch, continuous, or semibatch
mode? Steady state or
unsteady state?
mur83973_ch02_061-154.indd 79
Any of the process units—Mixers, Reactors, Separators, or Splitters—
can be operated as either batch, continuous, or semi-batch. Any can be operated as either steady state or unsteady-state. It is possible to have mixed
modes of operation in a single process. For example, a fermentor in which
microorganisms convert sugars into amino acids could operate in batch mode
for 48 h. Once the fermentation is complete, the broth is removed from
the fermentor and stored in a tank. The tank contents are fed continuously
to a separator.
Batch processes are common in the pharmaceutical, specialty polymer, and
personal care product industries. In these industries, the annual production rate
is often low (50 to 500 tons/yr or less), and the same equipment can be used
over and over again for different products. Home cooking is an example of
batch processing. Continuous-flow processes are common in the petroleum and
industrial chemicals businesses, where annual production rates are large (1000
to 5000 tons/yr or more at a single manufacturing site). In continuous-flow
processes, equipment is dedicated to a single purpose and may be in operation
24 hours a day, 7 days a week.
Batch and semibatch processes by their nature are unsteady state.
Continuous-flow processes are almost always operated under steady-state conditions, except during start-up and shutdown, when they are unsteady state.
Product quality is more consistent, and operating costs are generally lower, with
steady-state operation.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Batch operation
A
C
B
8.00 A.M.
9.00 A.M.
11.00 A.M.
10.00 A.M.
tf
0 < t < tf
t=0
Continuous steady-state operation
A
A
A
B
B
A
C
C
8.00 A.M.
C
B
B
10.00 A.M.
9.00 A.M.
C
11.00 A.M.
Semibatch operation
A
B
8.00 A.M.
t=0
B
C
10.00 A.M.
9.00 A.M.
0 < t < tf
11.00 A.M.
tf
Figure 2.10 Examples of different modes of process operation for a reactor. In all three cases, compounds A and
B are mixed and reacted to C, but the way in which inputs and outputs are handled is different, as is the time dependence of the process. In batch operation, reactants A and B are added all at once to the empty tank at 8 AM. Then
for several hours, A and B are consumed while C is generated. At 11 AM, the tank is emptied. In continuous-flow
steady-state operation, A and B are constantly pumped into the reactor, a reaction occurs in the reactor as they flow
through, and product C is constantly removed. In this example of semibatch operation, reactant A is loaded into the
empty tank all at once at 8 AM. Then reactant B is slowly and continuously added to the reactor over several hours;
during this time, A and B react to produce C. At 11 AM, the tank is emptied.
All three kinds of flowsheets—input-output diagram, block flow diagram,
and PFD—are used in batch, semibatch, or continuous mode, and for either
steady or unsteady state.
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Section 2.4 Process Flow Calculations
81
Process Flow Calculations
2.4
In Sec. 2.2 you reviewed units and dimensions of important process variables
and in Sec. 2.3 you learned about the various kinds of process flow sheets and
modes of process operation. In this section, we will put together what we know
about generation-consumption analysis, process flow sheets, and process variables, so that we can complete preliminary process flow calculations. We focus
on calculations for input-output diagrams and block flow diagrams. These calculations are needed to synthesize process flow sheets and to evaluate alternative
processing schemes.
There are several important definitions required to complete process flow
calculations. We will briefly describe each in turn.
2.4.1
Systems, Streams, and Specifications
Systems
A system is a specified volume with well-defined boundaries. Within
these boundaries we define what material is inside the system, what material is
outside the system, and what material is crossing the system boundaries. On a
block flow diagram, a system might correspond to a process unit. Or we might
draw a boundary around several process units and group them into a single
system. (See Fig. 2.11.) A system variable describes a change in a quantity
inside the system.
Streams
Streams are inputs to and outputs from the system. A stream variable
describes the quantity or flow rate of a material in a stream.
Specifications There are three kinds of specifications of importance to pro-
cess flow calculations:
∙ basis
∙ stream composition specifications
∙ system performance specifications
A basis is a flow rate or quantity that indicates the size of a process. Most
often, the basis is the quantity or flow rate of either a raw material entering
the process or a desired product leaving the process. However, the flow rate or
quantity of any stream on the flow diagram could serve as the basis.
Illustration:
∙ 1000 tons of ammonia are produced every day.
∙ The fresh juice flow rate into the evaporator is 260 lb/h.
∙ The batch fermentor is initially charged with 5000 mL broth.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Stream
Stream
Mixer
Stream
Stream
Mixer
Reactor
Separator
Splitter
Stream
Stream
Stream composition
specification
90% A
10% B
System
Stream
System performance
specification
60% of A is
converted to C
by chemical reaction
Figure 2.11 Examples of streams, systems and specifications. Top: A mixer is defined
as the system (shaded). Middle: Several process units are grouped together in a single­
system (shaded). Only the streams shown with heavy lines are inputs or outputs to the
system. Bottom: Stream composition specifications describe a single stream, while system
performance specifications describe the chemical and/or physical changes occurring inside
the system.
Stream composition specifications provide information about the composition
of a process stream. This information could be in the form of mass or mole
percent, mass or mole fraction, mass or mole ratio, or concentration. A stream
composition specification provides information about the quantities of components in one single stream.
Illustration:
∙ Glucose is fed to a process as a 10 wt% glucose solution in water.
∙ A customer requires that a hydrogen product must be at least 99.9 mol% pure.
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Section 2.4 Process Flow Calculations
83
∙ Environmental regulations state that a wastewater stream must contain no
more than 1 g acetic acid per 1000 kg water.
System performance specifications describe quantitatively the extent to which
chemical and/or physical changes have occurred inside the system. A system
performance specification provides information about the relationship between
two different streams that enter and/or leave the system, or between a stream
and a system variable. With a mixer, one might describe, for example, the
relative quantities of two input streams. A system performance specification on
a reactor might describe the relationship between the amount of a reactant fed
and the amount consumed by reaction. For a splitter, the specification could
be the fraction of the input that is sent to one of the output streams, whereas
for a separator, the quantity of a component in the feed that is recovered in
one of the product streams might be specified.
Illustration:
∙
∙
∙
∙
The nitrogen feed rate to the mixer is three times that of the hydrogen feed.
65% of the nitrogen fed to a reactor is converted to ammonia
Two-thirds of a fruit juice stream fed to a splitter is sent to a bottling plant.
98% of the fat in milk processed in a centrifuge is recovered in the fluid
skimmed off the top.
2.4.2
Material Balance Equation
Consider the cartoon in Fig. 2.12. The sketch shows a large lake with fish.
We’ll call the lake our system. The lake contains material—water, fish, perhaps some plants—and has defined boundaries—the surface in contact with
the air, the surface in contact with the earth, and the points at which streams
and rivers enter and exit the lake. Material enters and leaves the system via
streams crossing its boundaries: Water and fish enter the lake through a mountain stream, and both water and fish leave the lake through the river. Water
enters the lake when it rains, and water leaves by evaporation when the sun
shines. If a hungry bear comes along, more fish may leave the lake. Inside
the lake, fish generate baby fish by reproduction, and some small fish are
consumed by bigger fish.
Suppose we want to know whether the number of fish in the lake is increasing or decreasing. The number of fish in the lake increases because fish swim
into the lake from the mountain stream and because fish reproduce. The number of fish in the lake decreases because fish swim out of the lake into the
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Figure 2.12 The lake is the system, with system boundary indicated by dashed line. Some fish are in the system,
and within the system fish might be generated (by reproduction) and consumed (by being eaten). Fish swimming in
the streams enter and leave the system.
river and because some fish are eaten by other fish. The net change in the
number of fish in the lake is the sum of all these factors:
Number of fish entering the lake from the stream
− the number of fish leaving the lake in the river
+ the number of fish born (generated) in the lake
− the number of fish eaten (consumed) in the lake
= change in number of fish in lake
This is a material balance on the number of fish in the lake. There are a number of different kinds of material balances we could write. We could write a
balance on each species of fish. We could write a balance on water, with terms
accounting for water entering the lake from rain, or leaving the lake by evaporation. We could write balances on nitrogen, phosphates, or oxygen. In all cases
we would consider the same items—material entering or leaving the lake through
its boundary, and material generated or consumed inside the lake. Notice that
this balance does not tell us the total number of fish in the lake, only the change
in the number, and that the change in the number of fish in the lake is due to
inputs and outputs (streams) as well as processes happening inside the lake
(system). Notice too that we could alternatively have written a balance on the
mass of fish rather than on the number of fish. The balance on mass would
be different from the balance on number. For example, when a big fish eats a
little fish, the number of fish changes but the total mass of fish does not.
A general form of the material balance equation is:
Input − Output + Generation − Consumption = Accumulation Eq. (2.5)
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Section 2.4 Process Flow Calculations
where
Input
Output
Generation
Consumption
Accumulation
=material that enters the system by crossing system
boundaries
=material that leaves the system by crossing system
boundaries
= material that is generated inside the system by
chemical reaction
= material that is used up inside the system by
chemical reaction
= change in material inside the system
Input and output are stream variables, while generation, consumption, and
accumulation are system variables. The dimension of each term in the equation
is either amount of substance [n] or mass [m]. The units of each term in the
equation must be the same for all the variables.
2.4.3
Components
The material balance equation is written for a chosen component. If we apply
the material balance equation to fish, for example, the variables in the equation
would be input of fish, output of fish, generation of fish, consumption of fish,
and accumulation of fish. Alternatively, if we apply the material balance equation to water, the variables in the equation would be input of water, output of
water, generation of water, consumption of water, and accumulation of water.
In process flow calculations, the material balance equation is most commonly used with one of three kinds of components:
Elements: Such as carbon C, oxygen O, hydrogen H, or arsenic As.
Compounds: With defined molecular formulas, such as sucrose C12H22O11,
oxygen O2, water H2O, or gallium arsenide GaAs.
Composite materials: Mixtures of compounds of defined composition, such
as candy bars, air, seawater, or computer chips.
Most of the time, compounds are the most convenient, particularly if there is
a chemical reaction of known stoichiometry. Elements may be more convenient
when there is a chemical reaction of unknown stoichiometry, or when there are
a multitude of chemical reactions to consider. You should write a material balance equation in terms of composite materials only if there is no change in the
composition or chemical makeup of that material anywhere in the process.
2.4.4
Generation, Consumption, Accumulation
For chemical processes, generation and consumption are almost always due to
chemical reaction. Only compounds can be generated or consumed, not elements
(except with nuclear reactions!). Generation and consumption are related by the
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Chapter 2 Process Flows: Variables, Diagrams, Balances
stoichiometric coefficients of the compounds (just as we saw in Chap. 1). For
example, if A and B react to form C, then A and B are consumed by reaction
and C is generated by reaction, then
||
||
​ν​  ​​
moles of A consumed ​​ = ​ _
​​ __________________
​  ​νA​   ​​​ ​
B
moles of B consumed
moles of C generated
​ν​  ​​
​​ __________________
​​ = ​ _
​  C ​ ​
moles of B consumed ​νB​  ​​
These relationships between generation and consumption provide links between
the material balance equations written for compound A, for compound B, and
for compound C. We could write a third equation relating moles of C generated
to moles of A consumed, but it would just be a combination of the two other
equations, so it is not independent.
Illustration: The balanced chemical reaction for ammonia synthesis is
​N2​ ​​ + 3​H2​ ​​ → 2N​H3​ ​​
| |
| |
| |
​ν​H​  ​ ​​​ 3
​moles of H​2​​ consumed ___
​​ ___________________
​​ = ​ ​  ​ν​  2 ​​​ ​ = _
​   ​
​N2​ ​
1
​moles of N​2​​ consumed
​ν​N​  H​3​​​ _
​moles of NH​3​​ generated ____
____________________
​​
​​ = ​ ​  ​ν​   ​
​ = ​  2 ​
​N2​ ​​​
1
​moles of N​2​​ consumed
​νN​  ​H3​ ​​​ _
​moles of NH​3​​ generated ____
____________________
​​
​​ = ​ ​  ​ν​   ​
​ = ​  2 ​
​H2​ ​​​
3
​moles of H​2​​ consumed
The third relationship can be derived from the first two.
Accumulation can be positive, negative, or zero. Accumulation is nonzero
when there is an imbalance between the rate at which materials enter and are
produced and the rate at which materials exit and are consumed. For steady-state
processes, accumulation is zero.
Illustration:
∙ A soup kettle initially contains 6 cups of soup. Charlie Chef adds 2 more
cups. The soup accumulation in the kettle equals 2 cups.
∙ A soup kettle initially contains 6 cups of broth. Charlie Chef adds 2 more
cups, then Hungry Hattie removes 2 cups. The accumulation is zero.
∙ A soup kettle initially contains 6 cups of broth. Hungry Hattie removes
2 cups. The accumulation is negative 2 cups.
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Section 2.4 Process Flow Calculations
2.4.5
A Systematic Procedure for Process Flow Calculations
Now we wish to pull together all these strands—variables, flow sheets, specifications, material balances—to complete process flow calculations. These
calculations form the cornerstone upon which the synthesis and analysis of
chemical processes are built. A systematic approach is the only approach that
will reliably ensure successful and accurate completion of process flow calculations. Here is a highly recommended procedure to follow. Study this
procedure carefully.
Process Flow Calculations in 10 Easy Steps
Step 1.
Draw a flow diagram.
Step 2.
Define a system.
Step 3.Convert all numerical information into consistent units of
mass or moles.
Step 4.Choose components and define stream variables for all
material streams entering or leaving the system.
Step 5.Define a basis. Write an equation describing the basis in
terms of the defined stream variables.
Step 6.Define system variables for generation, consumption, and
accumulation. If there are chemical reactions of known stoichiometry, write equations using system variables, which
relate generation of products to consumption of reactants.
If the system is not at steady state, define a system variable
for accumulation.
Step 7.List all stream composition and system performance specifications. Write these specifications as equations, using the
stream and system variables you defined in steps 4 and 6.
Step 8.Write material balance equations for each component entering
or leaving the system, using the stream and system variables
you defined in steps 4 and 6.
Step 9.Solve the equations you wrote in steps 5 to 8.
Step 10.
Check your solutions.
It goes without saying (but we’ll say it anyway) that this list includes one
additional step.
Step 0. Understand the problem. Solving the wrong problem correctly can
be worse than solving the right problem incorrectly.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
2.4.6
Helpful Hints for Process Flow Calculations
Here are some Helpful Hints that should assist you as you apply the 10 Easy
Steps to solve process flow calculations. Scan the list quickly now, and refer
back to it if you get stuck while working a problem.
Step 1. Draw a diagram.
∙ Don’t skip this step!
∙ Draw one box for each process unit. Label each box as a Mixer,
Reactor, Splitter, or Separator.
∙ Draw one line for each input and output stream. Draw only one
line for a stream that is a mixture of compounds—don’t draw a
line for each compound in the mixture.
Step 2. Define a system.
∙ A system can be a single process unit, a group of units, or an
entire process.
∙ Group together several process units into a single system if you
do not need to know anything about the process streams that
connect the units.
Step 3. Convert all units to moles or mass.
∙ Use moles if there is a reaction of known stoichiometry.
Step 4. Choose components and define stream variables.
∙ Choose compounds as components if there is a chemical reaction
of known stoichiometry.
∙ Choose elements as components if there is a chemical reaction
of unknown stoichiometry.
∙ Choose composite materials as components only if they do not
undergo any changes in composition.
Step 5. Define a basis.
∙ The input of a raw material or the output of a desired product is
often a convenient basis.
∙ You can change a basis if it makes the problem easier to solve,
then scale up or down the solution to get back to the original
basis.
∙ If no basis is specified, choose any convenient basis.
Step 6. Define system variables.
∙ Elements are not consumed or generated, only compounds.
∙ For each chemical reaction, write equations that relate consumption
and generation of compounds to their stoichiometric coefficients.
∙ Accumulation is zero for continuous steady-state processes.
Step 7. List stream composition and system performance specifications.
∙ A “hidden” stream composition specification for a splitter is that
all output streams have the same composition as the input.
∙ A common system performance specification for a mixer is the
ratio of two input streams.
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Section 2.5 A Plethora of Problems
89
∙ A common system performance specification for a reactor is the
fraction or percent conversion of a reactant.
∙ A common system performance specification for a separator is
the fraction of a compound in the input stream that is recovered
in one of the output streams.
∙ A common system performance specification for a splitter is the
fraction of the input stream that goes to a particular output
stream.
Step 8. Write material balance equations.
∙ Write one material balance equation for each component.
∙ A balance equation on total mass can replace one of the component material balance equations.
Step 9. Solve equations.
∙ Solve the equation with the fewest number of unknown variables
first.
Step 10. Check the solution.
∙ Don’t skip this step!
∙ If you’ve used the component material balance equations to solve
the problem, use the total mass balance equation to check the
solution.
2.5
A Plethora of Problems
In this section is a veritable cornucopia of example problems involving process
flow calculations. We strongly recommend that, for each example, you try to
solve the problem by yourself before looking at the worked-out solution. If you
have difficulty, study the solution, then cover it up and try to work it out by
yourself. If you still have trouble, be sure you (1) understand the question,
(2) correctly identify the different terms in the material balance equation,
(3) follow the 10 Easy Steps, and (4) consult the Helpful Hints. The examples
progress from simple to complicated, so be sure you are able to solve on your
own each problem before proceeding to the next.
Warning! The single biggest mistake you can make with process flow
calculations is to think that if you can follow along with the solution then you
have demonstrated an understanding of the material. That is like thinking that
because you have watched Michael Jordan shoot basketballs you too can be an
NBA star (or because you have listened to Beethoven you too can write a
symphony, or. . . . ). The second biggest mistake you can make is to not
follow a systematic procedure like the 10 Easy Steps, even for problems that
are easy and “intuitively obvious.” Intuition is great; joined with logic it’s
unstoppable.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Example 2.1
Mixers: Battery Acid Production
Your job is to design a mixer to produce 200 kg/day of battery acid. The mixer
will operate continuously and at steady state. The battery acid product must contain
18.6 wt% H2SO4 in water. Raw materials available include a concentrated sulfuric
acid solution at 77 wt% H2SO4 and 23 wt% water, and pure (100%) water. What
is the flow rate of each raw material into the mixer?
Solution
Step 1.
Draw a diagram. We draw a block to indicate a mixer. There are two
inputs available, the concentrated sulfuric acid and the pure water, so
we draw two lines with arrows entering the unit. There is one output,
the battery acid, so we draw only one line leaving the unit.
Concentrated acid
Mixer
Battery acid
Pure water
Step 2.
Step 3.
Helpful Hint
For a process
stream that is a
mixture, draw
one line and label
the line to show
all components in
the mixture.
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Define a system. The system is the mixer.
Check units. All numerical information is given in kg/day or wt%.
These units are consistent and therefore no unit conversions are needed.
Step 4.
Choose components, define stream variables. Components might be
elements, compounds, or composite materials. It makes the most sense
to choose two compounds, sulfuric acid (H2SO4) and water (H2O), as
components for the following reasons. The acid solutions are composite materials, but the composition is different in each solution,
therefore we should not choose composite materials as components for
this problem. There is no chemical reaction in the system, so either
elements (H, S, O) or compounds (H2SO4 and H2O) are reasonable
choices. Since there are only two compounds but three elements, and
since information on c­omposition of the acid solutions is given in
terms of the compounds and not the elements, it is simplest to choose
compounds as components. We’ll use S to indicate sulfuric acid and
W to denote water.
We’ll number the process streams 1, 2, and 3. Since there are two
components in stream 1, there should be two stream variables associated
with stream 1, which we will call S1 and W1. By the same reasoning, there
is one stream variable for stream 2, W2, and two for stream 3, S3 and W3,
for a total of five stream variables. We redraw the flow diagram to
reflect our variable-naming scheme.
S1, W1
Mixer
S3, W3
W2
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Section 2.5 A Plethora of Problems
Step 5.
91
Define basis. We are given a desired production rate of 200 kg of
battery acid produced per day, which serves as a convenient basis.
In terms of our stream variables:
S3 + W3 = 200 kg/day
Step 6.
Step 7.
Define system variables. There are no chemical reactions, so no generation or consumption variables are needed (Scons = Sgen = 0, Wcons =
Wgen = 0). The system is at steady state, so the accumulation variable
equals zero (Sacc = 0 and Wacc = 0).
List specifications. There are several stream composition specifications.
The concentrated acid (stream 1) is 77 wt% H2SO4, or the mass fraction
of acid in stream 1 is 0.77. In terms of our stream variables:
​S1​  ​​
_
​
​= 0.77
​S1​  ​​ + ​W​ 1​​
(Alternatively, we can write
​S​  ​​ 77
_
​  1  ​ = ___
​​   ​​
​W1​  ​​ 23
Convince yourself that these equations are equivalent to each other.)
The product quality is also specified—the battery acid must contain 18.6 wt% H2SO4. In terms of our stream variables:
​S3​  ​​
​ _
​= 0.186
​S3​  ​​ + ​W​ 3​​
(Alternatively,
​S​  ​​ 18.6
_
​  3  ​ = ____
​​
​​
​W3​  ​​ 81.4
Helpful Hint
If there are 2
components in a
system, there are
2 material balance
equations.
Either one of these equations is sufficient.)
There are no system performance specifications for the mixer. So
we are done with step 7.
Step 8.
Write material balance equations. We identified two components—
sulfuric acid and water—so there are two independent material balance
equations. Let’s apply the material balance equation, Eq. (2.5), to each
component.
For sulfuric acid:
​S1​  ​​ − ​S​ 3​​ + ​S​ gen ​​ − ​S​ cons ​​ = S​ a​  cc ​​
Since Sgen, Scons, and Sacc are all equal to zero:
​
S1​  ​​ = S​ 3​  ​​
For water, the material balance equation simplifies to
​
W1​  ​​ + ​W​ 2​​ = W
​ 3​  ​​
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Step 9.
Solve the system of equations. We have five equations in five unknowns.
First we combine the basis with the stream composition specification
to find
​
S3​  ​​ = 37.2 kg/day
​
W3​  ​​ = 162.8 kg/day
We use these results along with the material balance on sulfuric acid
to find
S1 = 37.2 kg/day
Next from the stream composition specification for stream 1, we find
W1 = 11.1 kg/day
Finally, the material balance on water yields
W2 = 151.7 kg/day
The updated flow diagram is shown:
S1 = 37.2 kg/day
W1 = 11.1 kg/day
Mixer
S3 = 37.2 kg/day
W3 = 162.8 kg/day
W2 = 151.7 kg/day
Step 10.
Check the answer. To check the answer, use the material balance on
total mass:
S1 + W1 + W2 = S3 + W3
37.2 + 11.1 + 151.7 = 37.2 + 162.8
200 = 200 check!
Example 2.2
Reactors: Ammonia Synthesis
A gas mixture of hydrogen and nitrogen is fed to a reactor, where they react to form
ammonia (NH3). The nitrogen flow rate into the reactor is 150 gmol/h and the hydrogen is fed at a ratio of 4 gmol H2 per gmol N2. The balanced chemical reaction is:
​N​  2​​ + 3​H​  2​​ → 2​NH​  3​​
Of the nitrogen fed to the reactor, 70% is consumed by reaction. The reactor operates
at steady state. What is the flow rate (gmol/h) of N2, H2, and NH3 in the reactor outlet?
Solution
Steps 1 and 2. Draw a diagram and choose a system. The reactor is the system.
Don’t neglect to include the unreacted raw materials in the reactor
outlet stream!
H2
N2
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Reactor
H2
N2
NH3
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Section 2.5 A Plethora of Problems
Helpful Hint
Choose compounds as components if there
is a reaction
of known
stoichiometry.
93
Steps 3 and 4. Check units, choose components, and define stream variables.
All numerical information is given in units of gmol or gmol/h. Since
there is a reaction of known stoichiometry, we’ll choose the compounds N2, H2, and NH3 as the components. We’ll use N to symbolize N2, H to symbolize H2, and A to symbolize ammonia. We’ll call
the streams “in” and “out.” There are two components in the input
stream, so there are two stream variables, Nin and Hin. There are three
components in the output stream, so there are three stream variables:
Nout, Hout, and Aout.
Hin
Nin
Step 5.
Reactor
Hout
Nout
Aout
Define basis. We’ll use as the basis the nitrogen feed rate,
gmol
​
Ni​  n​​ = 150 ​ _
​
h
Step 6.
Define system variables. The chemical reaction involves three
compounds—two reactants and one product—so there are three system
variables related to generation and consumption: Ncons (rate of nitrogen
consumption by reaction), Hcons (rate of hydrogen consumption by reaction), and Agen (rate of ammonia generation by reaction). Using the
known reaction stoichiometric coefficients, we write two equations that
relate these system variables as:
3
​Hc​  ons ​​ __
​​ _____
​​ = ​​   ​​
​Nc​ ons​ 1
​Ag​  en​​ 2
​​ ____ ​​ = __
​​   ​​
​Nc​  ons​​ 1
Quick Quiz 2.10
Suppose the system
performance specification was instead that
30% of the hydrogen
fed is consumed. How
would you write this
specification as an
equation?
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Step 7.
The system is at steady state, so there is no accumulation system variable
(Nacc = Hacc = Aacc = 0).
List specifications. There is one stream composition specification:
Hydrogen is fed at a ratio of 4 gmol H2 per gmol N2, or
​H​  ​​
_
​  in ​= 4
​Ni​  n​​
There is one system performance specification, describing the reaction
taking place inside the system: 70% of the nitrogen fed to the reactor
is consumed, or
​​N​cons​​ = 0.7​​N​in​​
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Step 8.
Write material balances. Since there are 3 components (N2, H2, and
NH3), there are three independent material balance equations, which
simplify to
​Ni​  n​​ − ​N​ out​​ − ​N​ cons​​ = 0
​Hi​  n​​ − ​H​ out​​ − ​H​ cons​​ = 0
− ​Ao​  ut​​ + ​A​ gen​​ = 0
Steps 9 and 10. Solve and check. We substitute in known values for variables
and then proceed to solve, working first with the equations with the
fewest unknown variables. The solution is:
gmol
​Hi​  n​​ = 4​Ni​  n​​ = 4(150) = 600 ​​ _____
​​
h
gmol
​Nc​  ons​​ = 0.7​Ni​  n​​ = 0.7(150) = 105 ​​ _____
​​
h
gmol
​Hc​  ons​​ = 3​Nc​  ons​​ = 3(105) = 315 ​​ _____
​​
h
gmol
​Ag​  en​​ = 2​Nc​  ons​​ = 2(105) = 210 ​​ _____
​​
h
gmol
​No​  ut​​ = ​Ni​  n​​ − ​N​ cons​​= 150 − 105 = 45 ​ _
​
h
gmol
​Ho​  ut​​ = ​Hi​  n​​ − ​H​ cons​​= 600 − 315 = 285 ​​ _____
​​
h
Helpful Hint
Checking your
solution may be as
simple as checking
that the total mass
in equals the total
mass out.
Quick Quiz 2.11
Does the total molar
flow rate into the reactor in Example 2.2
equal the total molar
flow rate out? Why or
why not?
mur83973_ch02_061-154.indd 94
gmol
​Ao​  ut​​ = ​Ag​  en​​ = 210 ​​ _____
​​
h
To check the solution, we see whether the total mass flow in equals
the total mass flow out. The mass flow rate is simply the sum of the
molar flow rate of each compound times its molar mass.
The results can be nicely summarized in table form:
Generation −
Consumption
Input
Output
Molar
mass
gmol/h
g/h
gmol/h
g/h
gmol/h
g/h
Nitrogen
28
150
4200
−105
−2940
45
1260
Hydrogen
2
600
1200
−315
−630
285
570
Ammonia
17
0
0
+210
+3570
210
3570
Compound
Total
5400
5400
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Section 2.5 A Plethora of Problems
Example 2.3
Separators: Fruit Juice Concentration
Fruit juice is a complex mixture of water, fructose (fruit sugar), pulp, citric and
other acids, acetates, and other chemicals. Fresh fruit juice from the Fruity-Fresh
Farm contains 88 wt% water. A fruit juice processor buys a batch of 2680 lb fresh
juice from Fruity-Fresh, and makes concentrated juice by filling an evaporator with
the fresh juice, evaporating 75% of the water, and then removing the concentrated
juice. How much water (lb) must the evaporator remove? If the processor pays
$0.09 per pound for the fresh juice, and sells the concentrated juice for $0.50 per
pound, can he make a profit?
Solution
Steps 1 and 2. Draw a diagram and choose a system. The system is the evaporator, which performs as a separator.
Evaporated
water
Fresh juice
Helpful Hint
Choose a com­
posite material as
a component if
the material acts
as a single entity
throughout the
process.
Evaporator
Concentrated juice
Steps 3 and 4. Check units, choose components, and define stream variables.
Everything is given in mass units, which we will use because there is no
chemical reaction. We do not have much information about the composition of the juice other than that it contains 88 wt% water and unspecified
quantities of a lot of other things. Since the “other things” stay together—
they all come out in the concentrated juice—and they do not undergo any
chemical reaction, then we can lump together the fructose, acids, pulp,
acetates, etc. as one composite material that we’ll call “solids”. (Solids
include dissolved solutes as well as suspended solids.) We’ll indicate
water with W and solids as S. We have three streams, one input and two
outputs, which we will number 1, 2, and 3. Stream variables will be
denoted as, for example, W1 = water flow in stream 1.
W2
S1
W1
Step 5.
Evaporator
S3
W3
Define basis. The basis is the fresh juice into the evaporator, or
​S1​  ​​ + ​W​ 1​​ = 2680 lb
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Steps 6 and 7. Define system variables and list specifications. There are no chemical reactions, so Sgen = Scons = 0 and Wgen = Wcons = 0. All the material put into the evaporator is removed as either water vapor or
concentrated juice, so there is no accumulation of material inside the
evaporator, so Sacc = Wacc = 0. There is one stream composition
specification—the fresh juice is 88 wt% water, or
​​W​1​​= (0.88) × (S1 + W1) = 0.88 × 2680 = 2360 lb
Combining this equation with the basis equation we find:
​
S1​  ​​ = 320 lb
We have one system performance specification: We know that the
evaporator removes 75% of the water in the fresh juice:
​W2​  ​​= 0.75 × ​W1​  ​​= 0.75 × 2360 lb = 1770 lb
Helpful Hint
A system performance specification
may describe a
change between
input and output
streams caused
by the system.
We’ll update our flow diagram, as an easy way to keep track of our
calculations.
W2 = 1770 lb
W1 = 2360 lb
S1 = 320 lb
Step 8.
Evaporator
W3
S3
Write material balances. There are two components, so there are two
independent material balance equations. Because generation, consumption, and accumulation terms are all equal to zero, the material balance
equation simplifies to Input = Output, or
​
W1​  ​​ = ​W2​  ​​ + ​W​ 3​​
​
S1​  ​​ = ​S3​  ​​
Steps 9 and 10. Solve and check. Now we simply plug in known numerical values into the equations and solve:
​
W1​  ​​ = W
​ 2​  ​​ + ​W​ 3​​
2360 lb = 1770 lb + ​W3​  ​​
​
W3​  ​​ = 590 lb
​
S1​  ​​ = ​S3​  ​​ = 320 lb
The total quantity of concentrated juice is W3 + S3 = 590 + 320 =
910 lb.
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Section 2.5 A Plethora of Problems
Quick Quiz 2.12
Now we have a completed block flow diagram.
1770 lb water
Does total mass in
equal total mass out?
Fresh juice
2360 lb water
320 lb solids
590 lb water
320 lb solids
Evaporator
Concentrated juice
Considering only raw material costs, the processor would make a
profit of
$0.50
$0.09
​(910 lb concentrated juice × ​ _
​​ − ​ 2680 lb fresh juice × ​ _
​​
lb ) (
lb )
= $455 − $240 = $215
Example 2.4
Splitter: Fruit Juice Processing
Mr. and Mrs. Fruity squeeze 275 gallons of juice per day at the Fruity-Fresh Farm.
They plan to sell 82% of their juice to a processor, who will make frozen concentrated juice. The processor pays $0.75 per pound of juice solids. Some 17% of the
juice will be bottled for sale as fresh juice at a local farmers’ market, where it
sells for $3 per 2-L bottle. Mr. and Mrs. Fruity will keep the remainder for all the
little Fruitys. What are the total annual sales ($/year) for the Fruity-Fresh Farm?
Solution
Steps 1 and 2. Draw a diagram, define system. The fruit juice producer needs a
splitter, with one input and three outputs.
Fresh juice
Splitter
To processor
To farmer’s market
To the little Fruitys
Steps 3 to 5. Choose components, define stream variables. Check units and define
basis. Since the processor purchases the juice on the basis of its solids
content, we’ll consider two components: water, which we’ll denote as
W, and solids, which we’ll denote as S. There are four streams, identified as 1, 2, 3, and 4. Stream variables will be named accordingly; for
example, S3 is the flow rate of solids in stream 3.
S1, W1
Splitter
S2, W2
S3, W3
S4, W4
The juice flow rate is in gallons per day, which is a volumetric
flow rate. We always work with mass or molar quantities. For this
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98
Chapter 2 Process Flows: Variables, Diagrams, Balances
problem, let’s choose a mass flow rate, because the composition is
given in mass (wt) percent. We’ll work with pounds, because the price
is given as $/lb. To convert volumetric flow rate to mass flow rate, we
need a density. It might be hard to find the density of juice, but we
can find the density of a similar solution—12 wt% fructose in water—
in the CRC Handbook of Chemistry and Physics. The density of a
12 wt% fructose-water solution at 20°C is 1.047 g/mL. Good enough.
The juice flow rate into the splitter, in lb/day, is therefore
gal juice ______
1.047 g _____
L × ​​ ________
1000  ​​ mL × ​​ _______
lb  ​​
275 ​​ _______
​​ × ​​  3.78  ​​  ​​ × ​​  1 lb  ​​ = 2400 ​​ ____
L
mL
day
gal
day
454 g
The juice flow rate is a convenient basis; in terms of stream variables
we write:
​
S1​  ​​ + ​W​ 1​​ = 2400 lb/day
Steps 6 and 7. Define system variables and list specifications. There are no chemical reactions and the system is steady state, so generation, consumption, and accumulation variables are all equal to zero.
For stream composition specifications, we know that the juice is
88 wt% H2O, or in terms of stream variables:
​W1​  ​​
lb ​H​  2​​O
​ _
​ = 0.88 ​ _
​
​S1​  ​​ + ​W​ 1​​
lb juice
or:
lb juice
lb ​H2​ ​O
lb ​H2​ ​O
​​W​1​​ = 0.88 ​​ _______
​​ × 2400 ​​ _______
​​ = 2112 ​​ ______
​​
lb juice
day
day
Helpful Hint
A splitter has
“hidden” stream
composition
specifications: All
streams in and out
of a splitter must
have the same
composition.
Therefore,
lb solids
​​S​1​​ = 288 ​​ _______
​​
day
With a splitter, all input and output streams have the same composition.
In other words,
​W3​ ​
​W2​ ​
​W​4​
lb ​H2​ ​O
​​ _______
​​ = _______
​​
​​ = _______
​​
​​ = 0.88 ​​ _______
​​
​S​2​+ ​W​2​ ​S​3​+ W
​ ​3​ ​S​4​+ ​W​4​
lb juice
The splitter must meet the system performance specifications that 82%
of the juice goes to the processor, and 17% is bottled for sale. In terms
of stream variables,
​
S2​  ​​ + ​W​ 2​​= 0.82 × (​S1​  ​​ + ​W​ 1​​)
​
S3​  ​​ + ​W​ 3​​= 0.17 × (​S1​  ​​ + ​W​ 1​​)
We substitute in our known values of W1 and S1 to get
lb  ​ ​ = 1968 ​ _
lb  ​
​
S2​  ​​ + ​W​ 2​​= 0.82 × ​ 2400 ​ ____
(
day )
day
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99
Section 2.5 A Plethora of Problems
Combining this with the stream composition specification yields
lb  ​
​
W2​  ​​ = 1732 ​ _
day
lb  ​
​
S2​  ​​ = 236 ​ _
day
We proceed in the same manner to find
lb  ​
​
W3​  ​​ = 359 ​ _
day
lb  ​
​
S3​  ​​ = 49 ​ _
day
Steps 8 and 9. Write material balances and solve equations. There are two components, so there are two material balance equations, which simplify
to Input = Output, or
​
S1​  ​​ = ​S2​  ​​ + ​S​ 3​​ + ​S​ 4​​
​
W1​  ​​ = W
​ 2​  ​​ + ​W​ 3​​ + ​W​ 4​​
Now substitute in the known numerical values to get
lb  ​
​
W4​  ​​ = 21 ​ _
day
lb  ​
​
S4​  ​​ = 3 ​ _
day
The problem asked for the total income to the juice producer. To calculate the daily sales receipts of the producer, we sum the sales to the
processor and farmers’ market:
$
$0.75
$177
Sales to processor = ​S2​  ​​ × ​ _ ​ = ___________
​​  236 lb solids
​​ × ​​ _______ ​​ = _____
​​
​​
lb solids
day
lb solids
day
At the farmers’ market, the product is sold per liter of juice, not per
pound of solid, but it is interesting to convert from one cost basis to
the other:
$3.00
L
mL juice
454 g juice ___________
1 lb juice
$5.42
​​ _______
​
​ ​​​​ ________
​
​ ​​​​ ___________
​
​ ​​​​ __________
​
​​​​​
​
​ ​​ = _______
​​
​​
( 2 L juice )( 1000 mL )(1.047 g juice )( lb juice )(0.12 lb solids ) lb solids
(The Fruitys should sell as much juice as possible at the market rather
than to the processor.)
$
$5.42
$266
Sales at farmers’ market = S​
​​ 3​​ × _______
​​
​​ = __________
​​  49 lb solids
​​ × _______
​​
​​ = _____
​​
​​
lb solids
day
lb solids
day
Product sales total $443/day or about $160,000/year, if the family is
able to sell this much product every day of the year.
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100
Chapter 2 Process Flows: Variables, Diagrams, Balances
$
Value of juice consumed by the little Fruitys = S​
​​ 4​​ × _______
​​
​​ = _________
​​  3 lb solids
​​
lb solids
day
$5.42
$16.26
× _______
​​
​​ = ______
​​
​​
lb solids
day
Step 10.
(At nearly $6000/year, the little Fruitys can drink water!)
Check. One way to check the answer is to see if the solids content of
the juice consumed by the little Fruitys is 12 wt%, since we did not
use that information in our calculations yet.
​S​4​
​​ _______
​​ = ______
​​  3  ​​= 0.125 (close enough)
​S4​ ​+ W
​ 4​ ​ 3 + 21
Notice that in this problem we carried along more digits than significant
in our intermediate calculations. This is often a good idea, to avoid roundoff errors. Try recomputing all flow rates carrying just significant digits.
What do you find to be the solids content of the juice in stream 4?
So far, we’ve always set the accumulation term equal to zero. This happens either
when the process is continuous and steady state, or batch over a fixed time
interval with all materials added to the system at the beginning and all materials removed from the system by the end. Now, let’s turn to two problems where
material accumulates in the system during the process.
Example 2.5
Separation with Accumulation: Air Drying
Air is used throughout a process plant to move control valves (special valves that
regulate flow). If the air is humid, it needs to be dried before being used. To produce dry air for instrument use, filtered and compressed humid room air at 83°F
and 1.1 atm pressure, containing 1.5 mol% H2O (as vapor), is pumped through a
tank at a flow rate of 100 ft3/min. The tank is filled with 60 lbs of alumina (Al2O3)
pellets. The water vapor in the air adsorbs (sticks) onto the pellets. Dry instrument
air, containing just 0.06 mol% H2O, exits from the tank. The maximum amount of
water that can adsorb to the alumina pellets is 0.22 lb H2O per lb alumina. How
long can the tank be operated before the alumina pellets need to be replaced?
Solution
Steps 1 and 2. Draw a diagram, choose a system. The system is the separator—
the tank containing the alumina pellets.
Humid air
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Alumina pellets
with adsorbed
water
Dry air
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Section 2.5 A Plethora of Problems
101
Step 3.
Check units. The units are not consistent—the air flow rate is volumetric,
the water content of the air is given as mol%, and the adsorption capacity
of the pellets is given as a mass ratio. We need to convert everything
to consistent mass or mole units—let’s use lbmol.
First let’s convert the volumetric air flow rate to a molar flow rate.
For that, we need a density. We’ll assume air at these conditions
behaves as an ideal gas and calculate the molar density from the ideal
gas law:
(1.1 atm)
lbmol
__
​​  n  ​​ = ___
​​  P  ​​ = _________________________________
​​       ​​ = 0.00278 ​​ _____
​​
V RT (0.7302 ​ft​3​atm∕lbmol °R)(83 + 459 °R)
​ft​3​
The molar flow rate is simply the volumetric flow rate times the molar
density:
​ft​3​  ​​ × 0.00278 ​​ _____
lbmol
lbmol ​​
100 ​​ ____
​​ = 0.278 ​​ _____
min
min
​ft​3​
Step 4.
Step 5.
This is the total molar flow rate of humid air fed to the separator.
Choose components, define stream variables. Air is a composite
material: it contains nitrogen, oxygen, argon, carbon dioxide, water
vapor, and other gases. The alumina pellets remove only water from
the air; all the other gases stay together as the stream passes through
the separator. Therefore, we will choose as our components water (W)
and water-free air (A). In other words, we lump together everything
in the air other than H2O as a single composite material. Stream
variables are: A1 = water-free air into tank, W1 = water vapor into
tank, A2 = water-free air leaving tank, and W2 = water vapor leaving
tank.
Define basis. The total molar flow rate of humid air fed to the separator will serve as our basis. In terms of system variables,
lbmol ​​
​​A​1​​ + W
​​ ​1​​ = 0.278 ​​ _____
min
Step 6.
Define system variables. There are no chemical reactions, so no system
variables of generation or consumption are needed. The air does not
accumulate inside the system, but the water does. Therefore we will
have one system variable, Wacc, which describes the rate of water accumulation inside the tank.
W1
A1
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Wacc
W2
A2
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102
Chapter 2 Process Flows: Variables, Diagrams, Balances
Step 7.
List specifications. Stream composition specifications include that the
humid air is 1.5 mol% water and the dry air is 0.06 mol% water, or
​W1​ ​
lbmol water
​​ _______
​​ = 0.015 ​​ __________
​​
​A​1​+ ​W​1​
lbmol
​W​2​
lbmol water
​​ _______
​​ = 0.0006 ​​ __________
​​
​A​2​+ ​W​2​
lbmol
A system performance specification reflects physical and/or chemical
changes occurring within the system. In this case, we know that the tank
contains 60 lb alumina, which can adsorb, at most, 0.22 lb water/lb alumina. The total allowable accumulation of water in the tank is therefore
0.22 lb water ​​ × ​​ ___________
lbmole water ​​= 0.73 lbmol water
60 lb alumina × ​​ ___________
lb alumina
18 lb water
The stream variables have dimension of [mol/time], so we need to express
the system accumulation variable in the same dimension—as a rate of
accumulation. This will be equal to the total accumulation divided by the
time t over which water is allowed to accumulate in the tank, or
0.73 lbmol water
​​W​acc​​ = ​​ ______________
​​
t (min)
Steps 8 to 10. Write material balances, solve equations, and check. There are
two components, water-free air and water vapor, so two material balance
equations are written:
​​A​1​​ = A
​​ 2​ ​​
​​W​1​​ − W
​​ ​2​​ = ​​W​acc​​
We work through this system of equations (the details are left to you)
to find
​​W​1​​= 0.00417 lbmol∕min
​​A​1​​= 0.2738 lbmol∕min
​​A​2​​= 0.2738 lbmol∕min
​​W​2​​= 0.000164 lbmol∕min
​​Wa​  cc​​​ = 0.00401 lbmol∕min
t = 180 min
Check your answer by checking that the total mass of water adsorbed
plus the instrument air produced equals the mass of humid air fed, over
the 3-h period.
After 3 hours, the separator would no longer have the capacity to
adsorb more water. (Think of a bucket being filled with water—
eventually its capacity is reached.) Two tanks are used in most such
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Section 2.5 A Plethora of Problems
103
operations. The air flows through one of the tanks for 3 hours, then
the flow is switched to the second tank. The wet alumina pellets are
not thrown away, but are regenerated by heating to drive off the
­accumulated water. Then the regenerated pellets are reused.
Example 2.6
Reaction with Accumulation: Light from a Chip
Light-emitting diodes (LEDs) are used in all kinds of lighted displays, from
small handheld electronic games to huge billboards. LEDs are made from semiconductor material; several thin layers of material are built up on a substrate.
By changing the chemical composition of the semiconductor material, different
colors are produced.
A researcher is interested in making blue LEDs. She places a 1 cm × 1 cm
chip of Al2O3 in a reactor. In a process called MOCVD (metalorganic chemical
vapor deposition), trimethyl gallium [(CH3)3Ga] and ammonia (NH3) are pumped
continuously into the reactor at a 1:1 molar ratio, along with a carrier gas. The two
reactants form gallium nitride (GaN), which deposits as an even layer on the Al2O3
chip, and methane (CH4), which is swept out of the reactor continuously by the
carrier gas. The balanced chemical reaction is:
(CH3)3Ga + NH3 → GaN + 3CH4
The researcher would like to develop a method to estimate the rate of growth of
the height of the GaN layer on the chip, in micrometers per hour, by measuring
the methane flow rate, in µmol/h, out of the reactor. Can you help?
Solution
Steps 1 and 2. Draw diagram, define system. The system is the reactor. The sketch
shows the GaN layer growing on the substrate.
Trimethylgallium
Ammonia
Carrier gas
GaN
Methane
Carrier gas
Steps 3 and 4. Check units, choose components, define stream variables. Since a
chemical reaction is involved, we will work in molar units. Since the
researcher will measure the methane flow out of the reactor in µmol/h,
it makes sense to choose these units. Since there is a chemical reaction
of known stoichiometry, we’ll use the four compounds as components:
(CH3)3Ga (T), NH3 (A), GaN (G) and CH4 (M). There is also a carrier
gas, which does not take part in the reaction. We’ll call it I, for inert.
There are two streams that enter and leave the system; we’ll designate
them as streams 1 and 2. The stream variables are therefore: T1 =
trimethylgallium flow into reactor, A1 = ammonia flow into reactor,
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104
Chapter 2 Process Flows: Variables, Diagrams, Balances
Helpful Hint
If no basis is
specified, define
any convenient
basis.
I1 = inert gas flow into reactor, M2 = methane flow out of reactor, and
I2 = inert gas flow out of reactor. There is no stream variable in GaN
because GaN is not present in a stream, only in the system.
Steps 5 and 6. Define basis, define system variables. No basis is specified in the
problem statement. No problem! We just choose any basis that is convenient. We want a relationship involving the methane flow rate out,
so it makes sense to set M2 as the basis. So let’s choose:
​​M​2​​ = 100 μmol∕h
Since there are four reactants and products, there are four system variables for generation and consumption, Tcons, Acons, Ggen, and Mgen. There
are three equations relating these four system variables through their
stoichiometric coefficients.
____
​​  ​Ac​ ons ​​​ = 1
​Tc​ ons​
​G​gen​
​​ ____ ​​ = 1
​Tc​ ons​
​Mg​ en​
​​ ____ ​​ = 3
​Tc​ ons​
The GaN accumulates on the chip, while other compounds do not.
Therefore we have one system variable for accumulation, Gacc.
T1
A1
I1
Icons
Acons
Ggen
Mgen
Gacc
M2
I2
Finally, we chose μmol/h for units, but the researcher requested a relationship involving growth of the GaN layer in micrometers/h. How do
we convert from one unit to the other? First, we recognize that the
layer is three dimensional, with length and width defined by the size
of the Al2O3 chip, so the growth in thickness of the layer is really a
volumetric growth rate. Second, we relate a volumetric growth to a
molar growth rate by finding a density. We look up the density of GaN
in the CRC Handbook of Chemistry and Physics and find that the mass
density is 6.1 g/cm3. That, plus the molar mass of 84 g/gmol for GaN
gives us the conversion factor that we need, where the brackets indicate
the units:
3
84 g
gmol
​10​4​ μm
______
​
​​ _____
​  ​cm​ ​ ​ ​​ _____
​
​ ​​ _________
​  6
​ ​
(
)
(
)
cm
gmol
6.1
g
(
)
(
​10​ ​ μmole )
μm
μmol
___
________________________________
Growth rate​​[​   ​]​​ =
​​
​​ G
​​ a​  cc​​​​​[_____
​   ​
​​
1 cm × 1 cm
h
h ]
μmol
= 0.138​​Ga​  cc​​​​​[_____
​   ​
​​
h ]
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Section 2.6 Process Flow Calculations with Multiple Process Units
105
Steps 7 and 8. List specifications, write material balances. We have one stream
composition specification: The molar ratio in the feed gas is specified
as 1:1, or
​​T​1​​ = ​​A​1​​
There are five components so there are five material balance equations,
simplified from Input − Output + Generation − Consumption = Accumulation:
​​T​  1​​ − T
​​ c​  ons​​​ = 0
​​A​1​​ − ​​Ac​  ons​​​ = 0
​​Gg​  en​​​ = G
​​ a​  cc​​​
−​​M​2​​ + ​​Mg​  en​​​ = 0
​​I1​ ​​ − I​​ 2​ ​​ = 0
Steps 9 and 10. Solve and check. We solve this system of equations by starting
with the methane balance, because this equation has only one unknown.
From the methane balance, we find Mgen = 100 μmol/h. Then we use
the stoichiometric relationships to find Tcons = 33.3 μmol/h = Ggen = Gacc.
Finally, we use the unit conversion factor that we derived to find that,
if Gacc = 33.3 μmol/h, then the growth rate = 4.60 μm/h. (We are unable
to solve for the carrier gas flows I1 and I2 because we don’t have any
information about these streams. That’s OK, as long as the researcher
measures methane flow in the reactor output, and not just total gas flow.)
We calculated the growth rate of 4.6 μm/h given our chosen basis of 100 μmol/h
methane. The researcher wants a relationship that applies for any measured methane
flow, which we get by simply scaling our results:
μm
μmol
growth rate​​[___
​   ​]​​ = 0.046 × ​​M​2​​​​[_____
​   ​
​​
h
h ]
(We could have gotten this answer by choosing any convenient basis, or by symbolic
manipulation of our equations, without choosing a basis at all. Try it!)
2.6
Process Flow Calculations with
Multiple Process Units
In all the previous examples, we worked with a single system. What if we have
the task of calculating process flows for a block flow diagram with many process
units? Although this may seem like a daunting task, when broken down into parts
it is not. The 10 Easy Steps still apply, but there are a few new Helpful Hints.
Step 2. Choose system(s)
∙ You can choose to treat each process unit as a separate system
in turn. As you complete calculations for one system, you gain
information that allows you to proceed to the next system. This
procedure is necessary if you plan to calculate all process stream
variables and all system variables in the block flow diagram.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
∙ You can group two (or more) process units together and choose
the group as your system. If you do this, then draw a box around
the grouped units. Input and output streams are only those
streams that cross the boundaries of your box—not the streams
that are internal to the box. This is advantageous if you have
insufficient information about the streams that are internal to
your box. If you group all of the process units together, you’ve
just converted the block flow diagram into an input-output
diagram! This is often a good place to start.
Step 8. Write material balance equations. For each system in your diagram,
you can write as many material balance equations around that
system as there are components associated with that system.
Step 9. Solve the equations. Set up a table to keep track of your results.
The next few examples illustrate how to apply the 10 Easy Steps to multi-unit
problems.
Example 2.7
Multiple Process Units: Toxin Accumulation
An immunotoxin (IT) is a drug designed to kill cancer cells. An IT is constructed
from two proteins: One protein (usually an antibody) specifically targets cancer
cells and leaves healthy cells alone, while the second protein (the toxin) kills the
cell once it is inside. An experimental IT is internalized into cancer cells at a rate
of 185,000 molecules per minute. Inside the cell, all the IT first enters a compartment called an endosome, which acts like a splitter. 97% of the IT that enters the
endosome is spit back out of the cell. Much of the rest is sent to a compartment
called a lysosome, where it is degraded into harmless byproducts, at a degradation
rate of 5500 molecules/min. Anything remaining moves to another compartment,
called the cytosol, where it accumulates. It is estimated that 500 IT molecules must
accumulate inside the cell before it will be killed. How long will that take?
Solution
Steps 1 and 2. Draw a diagram and define a system.
Cytosol
Immunotoxin
Endosome
Lysosome
Degradation
products
Immunotoxin
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Section 2.6 Process Flow Calculations with Multiple Process Units
107
The cellular compartments are shown as boxes; the endosome is a
splitter, the lysosome is a reactor, and the cytosol is a storage tank. We
can choose each compartment in turn as our system. Or we can group
all compartments together and consider the entire cell as our system,
with the boundary as depicted. Let’s try the latter approach.
Steps 3 to 5. Convert units, choose components and define stream variables,
define basis. IT is our component, and we’ll use units of molecules/
min. Given our choice of system, there are only two streams (we’ll call
them in and out) that have IT in them. (The third stream has degradation products only.) Our stream variables are:
​​IT​in​​= immunotoxin entering the cell
​​IT​out​​= immunotoxin leaving the cell
(We will not worry about the degradation products, as we do not need
to know anything more about them.)
The basis is
molecules
​​IT​in​​ = 185,000 ​​ _________
​​
min
Steps 6 to 10. Define system variables, list specifications, write material balance
equations, solve, check. IT is consumed by degradation reactions inside
the cell. We don’t know anything about the reaction stoichiometry, but
we do know the net rate of degradation, so we write one system
variable as
molecules
​​IT​ cons​​​ = 5500 ​​ _________
​​
min
We wish to know when the total accumulation inside the cell equals
500 molecules. This equals the rate of accumulation ITacc multiplied by
the time of accumulation t.
​​IT​ acc​​​ × t = 500 molecules
We have one system performance specification, because we know the
splitter ratio:
​​IT​ out​​​ = 0.97​​IT​in​​
The material balance equation over the entire cell is
​​IT​in​​ − ​​IT​ out​​​ − ​​IT​ cons​​​ = ​​IT​ acc​​​
We easily combine and solve to find
molecules
​​IT​ acc​​​ = 50 ​​ _________
​​
min
t = 10 minutes
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Example 2.8
Multiple Process Units: Soap Manufacture
Soaps are the sodium salts of various fatty acids, such as stearic acid, which come
from natural products such as animal fat. For example, stearin (a triglyceride, also
called glyceryl tristearate) is one of the main ingredients in tallow (rendered animal
fat). Chemically, stearin is the glyceryl ester of stearic acid and has the molecular
formula (C17H35COO)3C3H5. In the first step of soap-making, stearin is contacted
with hot water at a 10:1 water:stearin molar ratio. Each molecule of fat is cleaved
into three fatty acids (stearic acids) plus glycerol:
(C17H35COO)3C3H5 + 3H2O → 3C17H35COOH + C3H5(OH)3(R1)
100% of the stearin is converted to products. The output from this reactor is sent to a
separator, which has two outputs: all of the stearic acid plus 10% of the water is in one
output, and all the glycerol plus the remaining water is in another. The glycerol/water
mixture is sent to another separator where all of the water is evaporated off and the
glycerol is taken and used for other products. The stearic acid and remaining water are
mixed with caustic (NaOH) at 1:1 molar ratio to produce sodium stearate soap:
C17H35COOH + NaOH → C17H35COONa + H2O(R2)
All of the stearic acid reacts to sodium stearate, and the soap/water mixture leaving the second reactor is heated to remove water by evaporation. The soap is then
taken off to final polishing, where it is aerated, frozen, and cut into bars.
Assume that 890,000 lb/day stearin is processed in one facility to make sodium
stearate soap. Draw the block flow diagram and calculate the flows in all streams,
assuming steady-state operation. Summarize your results in table form and report
flows in both lb/day and lbmol/day.
Solution
At first glance this looks like a more challenging problem than the ones you have
dealt with already. But if you follow the Ten Easy Steps and break the larger
problem down into smaller problems, you will see that it is quite manageable.
Steps 1 to 4. Draw diagram, define system, check units, choose components, and
define stream variables. After reading the step-by-step description of
the process, we generate the block flow diagram which contains two
mixers, two reactors, and three separators.
water
water
glycerol
Separator 1B
water
stearin
Mixer 1
Reactor 1
water
stearic acid
glycerol
glycerol
Separator 1A
water
stearic acid
water
Mixer 2
Reactor 2
sodium stearate
water
Separator 2
sodium stearate
NaOH
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Section 2.6 Process Flow Calculations with Multiple Process Units
As we step through this large problem, we will solve a series of smaller
problems, and the system choice will change. For this particular problem, we will choose Mixer 1 as our first system and then step through
each unit in turn.
The flow rate is expressed in units of lb/day, but we have two
reactions of known stoichiometry, so it will be most convenient to work
in units of lbmol/day, and then convert back to lb/day at the end. The
choice of components is straightforward as is the naming system: water
(W ), stearin (F, for fat), stearic acid (SA), glycerol (G), caustic (C),
and sodium stearate soap (S). To further name stream variables, we
will number the streams, moving from left to right and top to bottom.
Here is the block flow diagram showing the stream numbers.
W
G
5
Separator 1B
W
1
F
2
Mixer 1
W
F
3
Reactor 1
W
SA
G
W
6
G
7
4
Separator 1A
W
SA
C
8
Mixer 2
W
SA
C
10
Reactor 2
W
S
11
Separator 2
W
12
S
13
9
Steps 5 and 6. Define basis, define system variables. The feed rate of stearin is
specified and so can be used as the basis. Converting to molar units,
lb ​​ × ​​ ______
lbmol ​​ = 1000 ​​ _____
lbmol ​​
​
F2​  ​​ = 890,000 ​​ ____
day 890 lb
day
The process is at steady state, so there are no accumulation variables.
When we get to Reactors 1 and 2, we will have generation and consumption variables. There are two reactions, R1 and R2. From the
balanced chemical reactions, we write for R1:
​Wc​  ons1​​ __
​ ______
​ = ​​  3 ​​
​Fc​  ons1​​ 1
S​Ag​  en1​​ 3
​​ _____ ​​ = __
​​   ​​
​Fc​  ons1​​ 1
​Gg​  en1​​ 1
​​ _____ ​​ = __
​​   ​​
​Fc​  ons1​​ 1
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Chapter 2 Process Flows: Variables, Diagrams, Balances
and for R2
​Cc​  ons2​​ __
​​ ______
​​ = ​​  1 ​​
S​Ac​  ons2​​ 1
​Sg​  en2​​
​​ ______ ​​ = __
​​  1 ​​
S​Ac​  ons2​​ 1
​Wg​  en2​​
​​ ______ ​​ = __
​​  1 ​​
S​Ac​  ons2​​ 1
Step 7.
List specifications. The Mixer 2 performance specification is
S​A​  ​​ 1
​ _8 ​ = _
​   ​
​C9​  ​​ 1
Another system performance specification comes from the water:stearin
ratio fed to Mixer 1:
​W​  ​​ 10
​ _1 ​ = _
​   ​
​F2​  ​​
1
There is also a system performance specification which applies to
Separator 1A:
​
W8​  ​​= 0.1 × ​W4​  ​​
Steps 8 and 9. Write material balance equations, solve, check. As mentioned
earlier, we will choose Mixer 1 for our first system. The streams we
need to consider are streams 1, 2, and 3. Mixer 1 is a good choice
because we know the flow rate of one of the streams and the ratio of
the two input streams. There are only two components, so there are
two material balance equations, which are simply In = Out, or:
​
W3​  ​​ = ​W1​  ​​
​
F3​  ​​ = ​​F​2​​
We combine the material balance equations with the basis equation and
the Mixer 1 performance specification. This yields the results (all in
units of lbmol/h):
​
F3​  ​​ = 1,000
​
W3​  ​​ = 10,000
Now that we’ve solved the mixer problem, we move on to choose
Reactor 1 as a system. By solving Mixer 1 first, we now know the
input to Reactor 1. With this choice of system, the relevant streams
are streams 3 and 4. There are four components in Reactor 1, so
there are four material balance equations. Since accumulation is zero
(because of steady-state operation), the material balance equations are
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Section 2.6 Process Flow Calculations with Multiple Process Units
​
W4​  ​​ = ​W3​  ​​ − ​W​ cons1​​
0 = ​F3​  ​​ − ​F​ cons1​​
S​A4​  ​​ = S​Ag​  en1​​
​​G​ 4​​ = ​Gg​  en1​​​
We then use the stoichiometric coefficient ratios along with the four
material balance equations to solve for W4, SA4, and G4.
Next up for choice of system is Separator 1A, with streams 4, 5,
and 8. The analysis of Reactor 1 yielded information on stream 4, so
this result along with the Seperator 1A system performance specification
and the material balance equations allows us to calculate the flows in
streams 5 and 8. In this manner, we step through, one-by-one, all of
the process units. Derivation of the remaining equations is left for the
reader; results are summarized in Table 2.2.
Table 2.2
Component flow in soap manufacture. All flows are in lbmol/day
1
W
2
10,000
F
3
4
10,000
1000
5
6
7
8
7000 6300 6300
9
700
10
11
12
13
700 3700 3700
1000
SA
3000
3000
G
1000 1000
3000
1000
C
3000 3000
S
3000
3000
total 10,000 1000 11,000 11,000 7300 6300 1000 3700 3000 6700 6700 3700 3000
Table 2.3
1
W
Component flow in soap manufacture. All flows are in 1000 lb/day
2
180
F
890
3
4
5
6
180
126
113.4
113.4
7
8
9
10
11
12
12.6
12.6
66.6
66.6
852
852
890
SA
852
G
92
92
92
C
120
120
S
total
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918
180
890
13
1070
1070
205.4
113.4
92
864.6
120
984.6
984.6
918
66.6
918
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Step 10.
Check. For multi-unit processes, it is good practice to check the overall mass balance. Here, we choose the entire process as the system.
The system boundary is shown by the dashed line.
W
G
5
Separator 1B
W
1
F
2
Mixer 1
W
F
3
Reactor 1
W
SA
G
W
6
G
7
4
Separator 1A
W
SA
C
8
Mixer 2
W
SA
C
10
Reactor 2
W
S
11
Separator 2
W
12
S
13
9
With the new choice of system, the only streams are 1, 2, 6, 7, 9, 12,
and 13. Because mass is not generated or consumed, the mass flow in
should equal the mass flow out. (This is NOT necessarily true in molar
units!) Specifically, does the mass flow in (streams 1, 2, and 9) equal
the mass flow out (streams 6, 7, 12, and 13)?
180 + 890 + 120 = 113.4 + 92 + 66.6 + 918
??
1190 = 1190
Check!
2.6.1
Synthesizing Block Flow Diagrams
In Example 2.8, you were handed a block flow diagram and asked to complete
process flow calculations. But what if you need to first synthesize a block flow
diagram, given just a reaction pathway and a basis? Here is one logical approach
that will get you started:
1. Start with the known chemical reactions in the reaction pathway. For each
reaction, draw a reactor process unit, with all reactants entering in one
input stream and with all products and byproducts leaving in one output
stream.
2. Add mixers before each reactor. The output from the mixer is the input to
the reactor. The inputs to the mixer are all the reactants needed, in the
form in which they are available.
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Section 2.6 Process Flow Calculations with Multiple Process Units
113
3. If the raw materials are not pure and contain components that are not
needed for the reaction, consider adding separators before the mixers to
remove unnecessary components.
4. Add separators after the reactors. The input to the separator is the reactor
output. The outputs from the separator include the desired product, any
unreacted reactants, and byproducts.
5. Add splitters if the quantity of a stream is greater than that needed in the
downstream process units.
6. If unreacted reactants leave the reactor and are separated from the product
streams, add a mixer upstream of the reactor, to mix these recycled reactants with fresh reactor feed.
Given this strategy, there are still usually multiple ways to connect together
all the process units. Synthesizing the best preliminary block flow diagram
requires further analysis of costs, feasibility, and safety. A clever engineer uses
heuristics to eliminate clearly undesirable or unworkable options, leaving
fewer options that require more detailed analysis. Heuristics are simply rulesof-thumb—useful guidelines based on experience and logic. They are not
laws, and are not always true. Here is a heuristic that many people find useful
in predicting the weather:
Red sky at night, sailors delight.
Red sky in the morning, sailors take warning.
And here is a heuristic about time management:
A stitch in time saves nine.
Or perhaps you prefer
Better late than never.
Here are a couple of useful heuristics for synthesizing block flow diagrams:
∙ Mix raw materials together just before the reactor, and not earlier.
∙ Remove byproducts and waste products as soon after they are formed as
possible.
∙ If possible, split rather than separate.
∙ If possible, mix together streams of similar composition.
The block flow diagram is simple, yet it is invaluable in facilitating the
generation of alternative schemes. At this stage, we can (and should) consider questions regarding the need for each processing unit, and the appropriate pattern of connecting them together. The preliminary block flow
diagram provides the essential framework for developing a list of key
­questions to be answered. See the Case Study at the end of this chapter for
a specific example.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
2.6.2
The Art of Approximating
Simplifying approximations are frequently made in synthesizing preliminary
block flow diagrams. This might be necessary when we don’t have enough
information about process specifications. These approximations are not made
arbitrarily, but are chosen for good reason. Approximations that are frequently
made early in chemical process synthesis fall into 3 classes:
1. Stream composition approximations. For example,
The raw material is pure.
The product is pure.
Air contains only nitrogen and oxygen.
Reactants are fed at stoichiometric ratio.
2. System performance approximations: For example,
The reactants are fed at stoichiometric ratio.
The reactants are completely consumed by reaction.
No unwanted side reactions take place in the reactor.
The separator separates all components into pure streams.
3. Physical property approximations; For example
Gases behave as ideal gases.
Liquids behave as ideal solutions.
Solid density is independent of temperature.
These approximations make calculations much simpler, and are usually very
appropriate early in the design process. (If you look back through the worked
examples in this chapter, you will find cases where each of these approximations
was made.) Whenever you make an approximation, ask yourself two questions:
∙ Does the approximation have a resemblance to reality? It’s reasonable to
approximate the shape of a strand of uncooked spaghetti as a thin cylinder,
a tortilla as a disk, and even a turkey as a sphere. But it would be unreasonable to approximate a spaghetti strand as a sphere. It’s reasonable to
use the ideal gas law to estimate the density of air at room temperature
and pressure, but it’s unreasonable to use the ideal gas law to estimate the
density of gold at room temperature.
∙ If I made a more realistic approximation, would I make a different decision?
In deciding whether to use benzene or glucose as a reactant, it would be
reasonable to consider each as a pure raw material. But in choosing between
two different manufacturers of glucose, the purity of the material must be
considered carefully.
With experience, you will develop a knack for knowing what approximations
can be safely made, and when. You will also learn when you must reanalyze
a problem, using more stringent conditions.
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Section 2.6 Process Flow Calculations with Multiple Process Units
115
Biological Routes to Specialty Chemicals
In the Case Study of Chapter 1, we compared two 6-carbon compounds,
benzene and glucose, as feedstocks to make products such as adipic acid or
catechol. Here we will look further into developing the process for using glucose to make adipic acid, using a combination of a biological route and a
chemical route. We will first synthesize a block flow diagram, and then complete process flow calculations. Our goal will be to develop a preliminary
process for making adipic acid at a flow rate of 12,000 kg/h.
As described in Chapter 1, adipic acid, along with hexamethylenediamine,
is required in the synthesis of nylon-6,6. Nylon-6,6 is used widely to make
textiles, carpets, injection-molded auto parts and a whole host of other products. Conventionally, benzene, purified from crude oil, is the raw material in
adipic acid production, but there is interest in replacing chemicals from crude
oil with a renewable resource such as glucose.
E. coli was genetically engineered to produce muconic acid (C6H6O4) when
fermented with glucose (C6H12O6) as the feedstock. Because some of the
glucose is used to provide energy, the net reaction is
7C6H12O6 + 25.5O2 → 3C6H6O4 + 24CO2 + 33H2O(R1)
Muconic acid has less hydrogen than adipic acid (C6H10O4), so a second reaction is required, which we plan to carry out using a conventional catalyst.
C6H6O4 + 2H2 → C6H10O4(R2)
Following step 1 of our method for synthesizing block flow diagrams, we start
with the known chemical reactions. These two reactions require two reactors:
Glucose
O2
Muconic acid
H2
Reactor 1
(fermentor)
Reactor 2
(conventional)
Muconic acid
CO2
Water
Adipic acid
Already we have a few things to think about. First consider Reactor 1.
Fermentation requires E. coli, and bacteria grow in water, so besides the raw
materials we will need to have bacteria and water as inputs. Second, bacteria
will reproduce as well as make muconic acid, so an output will be E. coli.
Third, we will need to decide on a source of O2. Air is much cheaper than
pure oxygen and is the preferred source; but because air contains a lot of N2,
there will be N2 in the output stream. Fourth, to keep the bugs happy, we want
to operate with plenty of oxygen, so there will be some O2 in the output. Fifth,
we assumed that all of the glucose is consumed, so there is no glucose in the
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Chapter 2 Process Flows: Variables, Diagrams, Balances
output. We will continue with that approximation in the first step of process
synthesis; at later stages, as the process is further refined and we plunge in
deeper into design of the fermentor, we may revisit that approximation. All of
these considerations lead us to re-draw Reactor 1:
Glucose
Air (O2, N2)
Water
E. coli
Reactor 1
(fermentor)
Muconic acid
O2
N2
CO2
Water
E. coli
Now consider Reactor 2. As drawn, we assume that the muconic acid and
hydrogen are fed at stoichiometric ratio, and that all of the muconic acid and
all the hydrogen is consumed by reaction; we will work with these approximations in this first round of process synthesis. We will also assume that we have
a ready source of H2.
Continuing on with step 2 in block flow diagram synthesis, we add mixers
before each reactor. The output from Mixer 1 will be the input to Reactor 1:
Air (O2, N2)
Water
Glucose
E. coli
Mixer 1
Glucose
Air (O2, N2)
Water
E. coli
Reactor 1
(fermentor)
Muconic acid
O2
N2
CO2
Water
E. coli
Similarly, the output from Mixer 2 will be the input to Reactor 2:
H2
Muconic acid
Mixer 2
Muconic acid
H2
Reactor 2
(conventional)
Adipic acid
The next step is to ascertain if any of the raw materials are impure and, if so,
whether the raw material should be purified in a separator before being fed to
the reactor. Air is impure, so we could consider separating N2 from O2 before
feeding it to Mixer 1. This separation is difficult and expensive, so we decide
not to do it.
In step 4 in the synthesis of block flow diagrams, we add separators after
the reactors. The output from Reactor 1 contains muconic acid, which is an
input to Mixer 2. There are a large number of other materials in Reactor 1
output that are byproducts and not needed. According to our heuristics, it is
generally better practice to remove these byproducts before Mixer 2/Reactor 2.
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Section 2.6 Process Flow Calculations with Multiple Process Units
(The alternative would be to feed all of Reactor 1 output to Mixer 2 and
separate out byproducts downstream of Reactor 2.) We add two separators after
Reactor 1: the first separates gases from liquids, and the second separates
muconic acid from water and bacteria.
Air (O2, N2)
Glucose
Water
E. coli
Mixer 1
Glucose
Air (O2, N2)
Water
E. coli
Muconic acid
O2
N2
CO2
Water
E. coli
Separator 1
Reactor 1
(fermentor)
O2
N2
CO2 Muconic acid
Water
E. coli
Separator 2
Muconic acid
Water
E. coli
Reactor 2 output is pure adipic acid, so no separators are needed. Now we can
put together the pieces in a logical arrangement, and we have our block flow
diagram!
Air (O2, N2)
Glucose
Water
E. coli
Glucose
Air (O2, N2)
Water
E. coli
Reactor 1
Mixer 1
(fermentor)
Muconic acid
O2
N2
CO2
Water
E. coli
Separator 1
O2
N2
CO2
Muconic acid
Water
E. coli
H2
Separator 2
Muconic acid
Mixer 2
Muconic acid
H2
Reactor 2
Adipic acid
(conventional)
Water
E. coli
We will proceed with analysis of this flow diagram, but before we do that, we
recognize that alternative flow diagrams are possible. For example, we might
ask the following questions:
∙ Nitrogen is not needed in Reactor 1. Are there advantages to installing an
O2∕N2 separator on the air input before Mixer 1?
∙ What if all the glucose is not consumed by fermentation? How would the
flow diagram change?
∙ Is all the hydrogen and muconic acid consumed in Reactor 2? If not, how
would the flow diagram change?
∙ We assumed that hydrogen and muconic acid were fed at stoichiometric
ratio. Is there any advantage to operating with excess hydrogen?
∙ Can we re-use the water that leaves Separator 2? Or, will we need to clean
it up before disposal?
∙ What if there is not a readily available source of hydrogen? Normally
hydrogen gas is not readily available unless the process is integrated into
a larger chemical facility, we may need to add a reactor to make hydrogen
from, for example, natural gas.
Answering these questions will take additional information and study. Still, the
synthesis of the preliminary block flow diagram provides a jumping off point
for further rounds of analysis and synthesis.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Now we move to complete process flow calculations for production of
12,000 kg/h adipic acid, based on the flow diagram we just synthesized. Let’s
also specify that, after consulting with fermentation experts, we will feed a
0.1 mol% glucose solution (about 10 mg/mL) to the fermentor, and we will
feed oxygen at 20% more than stoichiometric requirement.
At first glance, this appears to be a challenging problem. If we break the
problem down into smaller chunks and use our Ten Easy Steps, you will see that
it is quite manageable. We have a diagram, so we start by choosing components
and defining stream variables. We will use the following key: glucose G, water W,
oxygen O, nitrogen N, hydrogen H, muconic acid M, carbon dioxide C, adipic
acid A. (Notice that oxygen and nitrogen are both components and we cannot use
air as a composite material; this is because oxygen is consumed by reaction but
nitrogen is not.) We will ignore the bacteria for now, because their mass will be
small. We will number streams, moving from left to right.
O
N
G
W
1
2
Mix 1
3
G
O
N
W
4
React 1
M
O
N
C
W
5
O
N
C
H 10
7
Sep 1
M
W
6
Sep 2
M
8
M
H
Mix 2
React 2
11
A
12
9 W
Since the process includes chemical reactions, we will use molar units. The
desired production rate of adipic acid (stream 12) is 12,000 kg/h; based on the
molar mass of 146 g/gmole, we determine that an appropriate basis is:
​
A1​  2​​ = 82.2 kgmol/h
Since we know stream 12 and we know the stoichiometry of (R2) in Reactor 2,
we choose Reactor 2 as our system. Assuming steady-state operation, and given
that muconic acid and hydrogen are fed to the reactor at stoichiometric ratio,
we write stream composition specifications, system variables, and material balance equations as
1
​M1​  1​​ __
​ _
​ = ​​   ​​
​H1​  1​​ 1
​Mc​  ons,2​​ 1
​Hc​  ons,2​​ 2
​ ______ ​ = __
​   ​, ​ ______ ​ = __
​   ​,
​Ag​  en,2​​
​Ag​  en,2​​ 1
1
​
A1​  2​​ = ​Ag​  en,2​​
​
M1​  1​​ = ​Mc​  ons,2​​
​
H1​  1​​ = ​Hc​  ons,2​​
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Section 2.6 Process Flow Calculations with Multiple Process Units
We solve this set of equations to find
​
M1​  1​​ = 82.2 kgmol/h
​
H1​  1​​ = 164.4 kgmol/h
Working backward, we next choose Mixer 2 as the system. Material balance
equations on the two components simplify to Input = Output, or
​
M8​  ​​ = M
​ 1​  1​​ = 82.2 kgmol/h
​
H1​  0​​ = ​H1​  1​​ = 164.4 kgmol/h
If we continue to work backward unit-by-unit, we would next choose Separator 2
as our system. A difficulty arises, though: we have no direct way to determine
the water flow rate in or out of Separator 2. Choosing a different system gets
us out of this conundrum: we lump together Mixer 1, Reactor 1, Separator 1
and Separator 2 into one system, as indicated by the dashed line:
O
N
G
W
1
2
Mix 1
3
G
O
N
W
4
React 1
M
O
N
C
W
5
O
N
C
H 10
7
Sep 1
M
W
6
Sep 2
M
8
Mix 2
M
H
React 2
11
A
12
9 W
Streams 1, 2, 3, 7, 8, and 9 cross the boundary of this system. From knowing
the molar composition of air, and from the specification that the glucose:water
feed ratio is 1:1000, we write:
​O​  ​​ 21
​ _1 ​ = ___
​​   ​​
​N1​  ​​ 79
1
​G​  ​​
​ _2  ​ = ​​ _____ ​​
​W3​  ​​ 1000
The specification that oxygen is fed at 20% more than what is required for
reaction (R1) can be stated as
​
O1​  ​​= 1.2 × ​Oc​  ons,1​​
From the stoichiometry of (R1) we also have four equations involving the
system variables:
​Cg​  en,1​​ 24
​Wg​  en,1​​ 33
​Gc​  ons,1​​ 7
​Oc​  ons,1​​ 25.5
​ _ ​ = ​​ __ ​​, ​ _ ​ = ​​ ____
​​, ​ _ ​ = ​​ ___ ​​, ​ _ ​ = ​​ ___ ​​
​Mg​  en,1​​ 3
​Mg​  en,1​​
​Mg​  en,1​​
​Mg​  en,1​​
3
3
3
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Chapter 2 Process Flows: Variables, Diagrams, Balances
We write material balance equations as Output = Input + Generation −
Consumption for each of our six components, M, G, O, C, W, and N:
​
M8​  ​​ = ​Mg​  en,1​​
0=G
​ 2​  ​​ − ​G​ cons,1​​
​
O7​  ​​ = ​O1​  ​​ − ​O​ cons,1​​
​
C7​  ​​ = ​Cg​  en,1​​
​
W9​  ​​ = ​W3​  ​​ + ​W​ gen,1​​
​
N7​  ​​ = N
​ 1​  ​​
We can solve this system of equations to obtain the flows in streams 1, 2, 3,
7, and 9. The strategy is (i) use the known value of M8 and the material balance
on muconic acid to find Mgen,1, (ii) use the equations involving the stoichiometric ratios to solve for the four other generation/consumption system variables, (iii) solve the glucose and carbon dioxide balance equations to find
G2 and C7, (iv) find W3 from the ratio of W3 to G2, (v) find O1 from Ocons,1,
(vi) find W9 and O7 using the water and oxygen material balances, respectively,
(vii) use knowledge of the nitrogen:oxygen ratio in air to find N1, and finally
(viii) use the nitrogen material balance to find N7.
To finish this problem, we can move forward, choosing in turn Mixer 1,
Reactor 1, Separator 1, and then Separator 2 as our systems, applying what we
have already solved for along with material balance equations on each system in
turn to calculate the flows in all streams. This is left as an end-of-Chapter problem.
Summary
∙ Chemical processes are represented schematically by three types of
flowsheets:
∙ Input-output diagrams
∙ Block flow diagrams
∙ Process flow diagrams
The diagrams differ in the level of detail, the amount of information needed
to generate them, and the cost to produce them.
∙ Block flow diagrams contain four basic process units: Mixers, Reactors,
Separators, and Splitters. Each process unit represents operations in
which material undergoes physical and/or chemical changes. The block
flow diagram shows how process streams connect the process units by
transferring material between units.
∙ Process operation is categorized according to how input and output streams
are handled. In batch processes, input streams enter the process unit all at
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121
Summary
once, and output streams are removed from the process unit all at once at some
later time. In continuous-flow processes, input streams continuously flow
into the process unit and output streams flow continuously out of the unit.
Semibatch processes are some combination of batch and continuous flow.
∙ Steady-state processes are time independent: Process variables do not
change with time. In transient, or unsteady-state, processes, one or more
process variables change with time. Batch and semibatch processes by their
nature are unsteady state. Continuous-flow processes are usually operated
under steady-state conditions, except during start-up and shutdown.
∙ A basis is a flow rate or quantity that indicates the size of a process. A
system is a specified volume with well-defined boundaries. Streams are
inputs to and outputs from the system. A stream variable describes the
quantity or flow rate of a material in a stream, while a system variable
describes the change in a quantity inside a system. Stream composition
specifications provide information about the composition of a process
stream. System performance specifications describe quantitatively the
extent to which chemical and/or physical changes occur inside the system.
∙ The material balance equation is
Input − Output + Generation − Consumption = Accumulation
The material balance equation is written for a component and around a system. A component can be an element, a compound, or a composite material.
∙ The “10 Easy Steps” is a systematic approach that, if followed, will lead
to successful completion of process flow calculations. The steps are:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
Draw a flow diagram
Choose a system
Convert units
Choose components and define stream variables
Define a basis
Define system variables
List stream composition and system performance specifications
Write material balance equations
Solve
Check
ChemiStory: Guano and the Guns of August
In 1804 the geographer Alexander von Humboldt introduced Europeans to
a marvelous substance from the New World: Peruvian guano. For untold
years, fish-eating sea birds had deposited their nitrogen-rich wastes on rocky
islands off the South American coast. The dry climate preserved the guano
(continued)
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Chapter 2 Process Flows: Variables, Diagrams, Balances
deposits in layers 150 feet deep. Access to these bird droppings was so
important that the United States passed the Guano Island Act in 1856. This
law allowed U.S. citizens who discovered a rock or island covered with
guano to take possession of the land and harvest the material. This was
perhaps the first and only time in which sovereignty over land was determined by chemistry rather than by history or geography.
Enterprising Americans made fortunes selling Peruvian guano to Europe,
as did the European importers. Still, there wasn’t nearly enough of the
nitrogen-rich fertilizer to satisfy the food needs of a rapidly growing population. Nitrogen in air is plentiful but cannot fertilize crops unless it is
converted to liquid or solid substances, like ammonia or ammonium nitrate.
Unfortunately, the triple bond in nitrogen is extremely stable. No one could
figure out how to break the bond and then get N to combine with H to make
liquid ammonia. No one, until Fritz Haber came along.
Fritz Haber was born on Decem­
ber 9, 1858, in Breslau, Germany. His
mother died shortly after his birth; his
father left him to be raised by an
assortment of relatives. Rather aimless
as an adolescent, he attended 6 universities in six years. Although he wanted
to be a chemist, he found chemistry
classes either too boring or too hard.
Chemical synthesis of nitrogenHe finally earned a Ph.D. from the
containing fertilizers from N2
University of Berlin in 1891, and studdramatically increased crop yields.
ied chemical technology: at an alcohol
Pradana/Shutterstock
distillery in Hungary, a Solvay soda
factory in Austria, and a salt mine in Poland. He was particularly interested
in the new field of physical chemistry, and applied to study under the great
Wilhelm Ostwald, but was rejected. (Ostwald didn’t seem to have much of
an eye for young talent—he also rejected an application from Albert
Einstein.) He finally gained a position at the Karlsruhe Institute of
Technology. Chemical engineering had not yet emerged as a separate discipline, but Haber thought and acted as an engineer, solving many practical
problems in chemistry. He was a charming and energetic man who wrote
nonsense poetry in his spare time.
In 1901, Fritz met and married Clara Immerwahr. She was the first woman
to earn a doctorate (in chemistry) from the University of Breslau, and a Jew.
(Fritz was born Jewish, but had converted to Christianity in 1892 because that
was the only way he could get a university position.) Early in their marriage,
Clara kept her hand in science by translating chemical literature and helping
Fritz write his book Thermodynamics of Technical Gas Reactions. But children
came along. Fritz was a thoughtless husband and father, often bringing home
unannounced large groups of friends for dinner parties, and leaving for a
5-month trip to the United States shortly after the birth of their first son.
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Summary
123
Haber had a difficult time gaining the recognition his technical contributions deserved. It could have been his Jewish background, or perhaps
his penchant for moving headlong and recklessly (and successfully) into
research areas already being studied by famous professors. (Ostwald told
him “Achievements generated at greater than the customary rate raise
instinctive opposition amongst one’s colleagues.”) Eventually, in 1906, Fritz
snagged a promotion to an elite German professorship, and became increasingly interested in the nitrogen fixation problem. Germany at the turn of the
century was ripe to solve the problem. Its chemists and chemical technicians
were the best in the world, its chemical industry was large and diversified,
its farms needed fertilizer, and continued access to natural fertilizers was
uncertain. Ostwald and electrochemist Walther Nernst both worked, unsuccessfully, on the problem of nitrogen fixation. Haber had some advantages
over these more established physical chemists: He had experience working
in chemical plants and with mechanical equipment. Haber realized that
higher pressures were needed to drive the reaction toward ammonia production. He and Robert Le Rossignol designed and built a high-pressure experimental chamber. They discovered that hydrogen and nitrogen would convert
to ammonia only under then unheard of conditions: 200°C and 200 atm. At
that time, 7 atm was considered high pressure! Carl Bosch, BASF’s chief
chemist, was intrigued. Three top executives from BASF marched into
Haber’s lab to see for themselves. As luck would have it, one of the seals
on the high-pressure chamber broke, and the experiment was a disaster. But
one of the executives stuck around long enough to see the seal fixed and
was rewarded with the amazing sight of a tiny spoonful of liquid ammonia.
BASF quickly signed a contract with Haber to commercialize the process.
Many problems remained to convert the lab experiment into an industrialscale process. For example, BASF chemists tested 4000 different catalysts,
finally discovering that iron was the best. The process,
patented in 1908, was commercialized within 5 years.
The first plant produced 30 metric tons per day. Haber
became rich beyond belief. The invention of the
Haber-Bosch process ushered in 20th century industrial chemical processing, introducing such concepts as
metallic catalysts and high-pressure, high-temperature
gas reactions.
Germany was freed from its dependence on
imported fertilizer. Haber became a national hero.
He was appointed a director at the Kaiser Wilhelm
Institute in Berlin, and socialized with the wealthy
Fritz Haber with Albert
and powerful of Berlin, including scientists such
Einstein. German
as Einstein and Lise Meitner. Meanwhile, Fritz’s
Photographer,
long-suffering wife Clara moved in very different
(20th century)/German/
social circles—embracing the Reform Movement,
Bridgeman Images.
(continued)
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Chapter 2 Process Flows: Variables, Diagrams, Balances
wearing loose clothing, doing her own marketing, making friends with the
servants, eating simple food. One visiting scientist mistook her for a cleaning woman.
In 1914, Germany invaded Belgium. The war quickly spread to engulf
much of Europe. Germany’s intelligentsia—including Haber and other scientists like Max Planck—saw the war as an “act of purification and a means of
redemption.” Haber directed his scientific talents toward making nitrogen-based
munitions for the war effort. He convinced Carl Bosch and BASF to make
nitric acid from his ammonia. Without the Haber-Bosch process, Germany
would have run out of explosives within 6 months and the war would likely
have ended quickly with German defeat.
Allied forces and Germany fought ferociously for 3 years, at a cost of
millions of lives, without the frontline budging by more than a few miles.
The war’s stalemate led German leaders (as well as leaders in France, Britain,
and the United States) to consider chemical and other unconventional weapons. The Hague Conventions, signed in 1899 and 1907, prohibited the use
of unconventional weapons that would cause unnecessary suffering. But
many (including the United States) did not consider chemical weapons any
worse than shrapnel or explosives. The patriotic Haber agreed to develop
chemical weapons for Germany. He personally supervised the burial of 6000
cylinders of liquid chlorine near the front in Belgium. The cylinders released
150 tons of chlorine and poisoned about 7000 French soldiers. This was the
first systematic use of chemical weapons in warfare. By the end of the war,
both sides had used chemical weapons extensively, although there was no
evidence that their use provided any military advantages. Clara despised her
husband’s work on chemical warfare, and pleaded with him to stop, to no
avail. A week after the first use of chlorine, the morning after hosting a
dinner party, Clara shot herself through the heart. Her 13-year-old son
Hermann found her, alive but near death. Fritz left the teenager alone the
next morning, heading for the Eastern Front.
Germany’s surrender in 1918 pushed Fritz into a deep depression.
Haber’s name reportedly appeared on a draft list of war criminals; he sent
his second wife and their two children to Switzerland and he himself escaped
in disguise. However, his name was not on the final list, and he escaped
prosecution as a war criminal. His fortunes changed dramatically once again,
when he was awarded the Nobel Prize for his development of the ammonia
synthesis process. The award caused a storm of controversy, because of his
work on chemical weapons. Almost all the non-German Nobel Prize winners
boycotted the ceremony.
Despite prohibitions of chemical and biological weapons by the
Versailles Peace Treaty and the 1925 Geneva Protocol, Haber continued to
push for poison gas and chemical warfare research. He helped build poison
gas plants in the Soviet Union and Spain, and remained a dedicated German
nationalist. The rise of the Nazi party took him by surprise. As a leader of
the Kaiser Wilhelm Institute, he was ordered in 1933 to simultaneously fire
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References and Recommended Readings
125
all Jews working there and to keep all important senior scientists. This was
impossible, as many of the leading scientists of the day were Jewish. He
was torn apart by the conflict, and did not want to continue doing poison
gas work for the Nazis, nor did he want to discharge his scientists. His
health deteriorated, his financial situation became shaky, his friends deserted
him, and the chemical industry (except for Carl Bosch) dropped their support. Finally, he acted. In a letter that infuriated the Nazis, he claimed his
right to remain in his post, but his unwillingness to use racial makeup as
a deciding characteristic in employment. The strident German nationalist
left Germany, never to return. He died of a heart attack, in a hotel in
Switzerland, in 1934.
The Nazis, trying to discredit Haber, claimed that others had invented
the ammonia synthesis process. Still, they were willing to use another offshoot of Haber’s work. The pesticide Zyklon B, developed in post–WW I
Germany under Haber’s direction, was used to gas prisoners at concentration
camps, including some of Haber’s own relatives.
Quick Quiz Answers
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
L3, m/L3.
15 wt%.
0°R.
No, the tire is at atmospheric pressure—which means it is flat!
No, because H2O at 77° F and 1 atm is not a gas.
Yes. 1970 tons/day in and 1970 tons/day out.
CH4, H2O, CO, H2.
Because the output streams differ in composition.
Mouth: mixer. Stomach: reactor. Intestine: separator. Semibatch. Unsteady
state.
Hcons = 0.30*Hin.
No, because 150 + 600 ≠ 45 + 285 + 210.
2360 + 320 = 2680. 1770 + 590 + 320 = 2680. Yes!
References and Recommended Readings
1. CRC Handbook of Chemistry and Physics is an invaluable desktop reference containing many pages of physical property data such as molar
masses and formulas of organic and inorganic compounds, and densities
of pure compounds as well as mixtures. It also contains extensive listings
of unit conversion factors. The CRC Handbook is published by CRC
Press, Boca Raton, Florida.
2. Perry’s Chemical Engineers’ Handbook is another invaluable desktop reference. Perry’s has chapters covering unit conversion factors, a review of
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126
Chapter 2 Process Flows: Variables, Diagrams, Balances
mathematics, and some physical and chemical data. The bulk of the handbook is concerned with chemical engineering principles and methods; the
book includes many sketches of chemical process equipment. Perry’s is
published by McGraw Hill, New York.
3. The Kirk-Othmer Encyclopedia of Chemical Technology is a multivolume
compendium of information on chemicals and chemical processes. The
coverage is truly encyclopedic, and includes data on process economics,
market size, physical and chemical properties, and process technology. It
is published by Wiley, New York. Two other books in the same vein are
Shreve’s Chemical Process Industries, McGraw Hill, New York, and the
McGraw Hill Encyclopedia of Science and Technology, McGraw Hill,
New York.
4. The Knovel Engineering and Scientific Online Database is a comprehensive source of searchable information. Perry’s Handbook and CRC
Handbook are a few of the many authoritative references that are searchable from the Knovel database.
Chapter 2 Problems
Warm-Ups
Section 2.2
P2.1 You put a 100-mL volumetric flask on a balance and then tare the balance so it reads 0.00 g. Then you add anhydrous fructose (C6H12O6—the
major sugar in fruit) into the flask until the balance reads 15.90 g. You
fill the flask with water up to the 100 mL line. The balance reads 105.97 g.
Calculate the wt% fructose and the mol% fructose of the solution.
P2.2 Soybean meal is a product made from soybeans after the oil has been
extracted. The meal contains about 48 wt% protein along with carbohydrates and indigestible fiber. In a typical processing plant, about 70%
of that protein can be recovered as “soy protein isolate,” which can then
be spun, mixed, and shaped into soy “bacon,” “burgers,” or other meat
substitutes. For 100 lb of soybean meal, about how many lb of soy
protein isolate can be made? If soybean meal sells for $375/metric ton,
what is an estimate of the cost per lb of soy protein isolate?
P2.3 1000 grams of polystyrene (molar mass = 20,800 g/gmol) is dissolved
in 4000 grams of styrene (C8H8, molar mass = 104 g/gmol). Calculate
the mole percent of polystyrene in the mixture.
P2.4 Air is approximately 79 mol% N2 and 21 mol% O2. Based on this, what
is the wt% N2 and O2 in air?
P2.5 20 g hydrogen (H2) is mixed with 20 g benzene (C6H6) and 20 g cyclohexane (C6H14). What is the mass fraction and the mole fraction of
hydrogen, benzene, and cyclohexane in the mixture?
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Chapter 2 Problems
127
P2.6 100 g of a gas mixture of hydrogen (H2), benzene (C6H6), and cyclohexane (C6H12) contains 20 mol% H2, 40 mol% benzene, and 40 mol%
cyclohexane. How many g of each gas is in the mixture?
P2.7 A liquid solution contains 2.4 mol% sucrose (table sugar, C12H22O11)
dissolved in water. Calculate the wt% sucrose in the solution.
P2.8 A liquid solution contains 22 wt% ethanol (C2H5OH) and 78 wt% water
(H2O). Calculate the mole% ethanol in the solution.
P2.9 I need to make up a 10 mol% glucose solution in water. How many
pounds of glucose do I need to add to 1000 gallons of water? What is
the wt% glucose of this solution? Assume that water density is 1 g/mL.
P2.10 Calculate the molar mass (g/gmol) of aspartic acid, C4H7O4N, 1 of the
20 common amino acids. Round off to the nearest whole number. Then
calculate the mol% aspartic acid in an aqueous solution at its solubility
limit (4.5 g/L). You can assume the density is 1 g/mL.
P2.11 Your biology textbook reports that the human body contains 63% H,
25.5% O, 9.45% C, 1.35% N, 0.31% Ca, and 0.22% P, plus several more
trace elements. Are these mass or mole percents? Explain. If you weigh
65 kg, how many g and gmol of each element do you carry around?
P2.12 A chemical reactor operates at 10 atm pressure (absolute) and 200°C.
Determine the pressure and temperature of this reactor in the following
units: kPa and K, psig and °F, bar and °R.
P2.13 What is the dimension of P × V? Report using only the base dimensions.
P2.14 Calculate the molar volume (cm3/gmol) of an ideal gas at 0°C and
1 atm pressure, using the ideal gas law.
P2.15 2.7 lb of CO2 is held in a vessel at 67°F and 1080 mm Hg. Calculate
the volume (cm3) of the vessel using the ideal gas law.
P2.16 If air is an ideal gas, what is the approximate mass (in tons) of a cubic
mile of air? List any assumptions you needed to make to complete this
calculation.
P2.17 Use the ideal gas law to calculate the molar volume (L/gmol) of water
vapor at 100°C and 1 atm. Then compare to the molar volume of liquid
water, assuming the density of liquid water is 1 g/mL.
P2.18 Calculate the grams of nitrogen in (a) 1 cubic foot of liquid and (b) 1 cubic
foot of gas at STP. The specific gravity of liquid nitrogen is 0.808, and
you can assume nitrogen is an ideal gas at STP.
P2.19 Your road bike tires are properly inflated to 100 psig. The sun is shining,
winds are light, and the temperature is 80°F. At the end of your ride,
you put the bike in your car, leave the car in an asphalt parking lot, and
head out for a picnic lunch. You come back to find that the car interior
is hot and your tires have burst. The tire pressure rating is 120 psi. How
hot did it get inside your car?
P2.20 Turn on the faucet in a kitchen or bathroom full blast, then measure the
water flow rate using a bucket and timer. Report your measurement as:
g/s, lb/h, kgmol/day, and tons/yr.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
P2.21 Oxygen at 100°C and 75 psia flows through a pipe at 115 lb/min.
Calculate the molar flow rate (lbmol/min) and the volumetric flow rate
(m3/min) at both the actual temperature and pressure, and at STP.
P2.22 A gas mixture containing 22 mol% hydrogen cyanide (HCN) and 78 mol%
ethylene oxide (C2H4O) is fed to a reactor at a rate of 215 gmol/min.
The gas is at a pressure of 200 kPa and a temperature of 285°C. What
is the volumetric flow rate, in cm3/min?
P2.23 A gas mixture containing 16 wt% hydrogen cyanide (HCN) and 84 wt%
carbon monoxide (CO) is fed to a reactor at a rate of 57 kg/min. The
gas is at a pressure of 2.5 bar and a temperature of 247°C. What is the
volumetric flow rate, in L/min?
P2.24 A fuel gas mixture (80 mol% CH4, 15 mol% C2H6, 5 mol% N2) is fed
to a burner at a rate of 126 lb/min. The gas is at a pressure of 4.8 atm
and a temperature of 285°C. What is the volumetric flow rate, in cm3/min?
P2.25 Air flows into a reactor at 60,800 cm3/s. The air is at 323 K and 2.7 bar,
and contains 79 mol% N2 and 21 mol% O2. The air behaves as an ideal
gas. Calculate the molar flow rate (gmol/s) and the mass flow rate (g/s)
of air into the reactor.
P2.26 Estimate the metric tons of CO2 produced annually by automotive use
in the United States. Assume there are 300 million people, 8 moles of
CO2 produced per mole of gasoline burned, an average 12,000 miles/
yr/person at 28.6 miles per gallon gasoline, a gasoline specific gravity
of 0.7. An environmental group claims that 5.6 billion metric tons CO2
are pumped into the environment every year, of which about 25% comes
from the United States. Compare these numbers to your estimated CO2
production from automobiles and comment.
Section 2.3
P2.27 List a common household item that functions as a (a) mixer, (b) splitter,
(c) reactor, and (d) separator. Draw a block flow diagram that illustrates each of the items and shows the material flow in and out.
P2.28 Draw an input-output diagram for an automotive gasoline engine.
Identify the raw materials and products/byproducts.
P2.29 Draw a block flow diagram of a milk-producing process plant, also
known as a dairy cow. Indicate grass, water, and oxygen as the input
streams and show the output streams including products and byproducts.
Include mixers, splitters, reactors, and/or separators as needed.
Section 2.4
P2.30 Indicate whether each of the following is a stream composition (“C”)
specification or a system performance (“P”) specification.
(a) 96% of toluene fed is recovered in the separator bottoms.
(b) The CO2:H2O molar ratio in the reactor effluent is 2:1.
(c) The distillate stream leaving the separator contains 98.5 wt% ethanol.
(d) 86% of the glucose fed to the fermentor is consumed.
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Chapter 2 Problems
(e)
(f )
(g)
(h)
P2.31
P2.32
P2.33
P2.34
P2.35
The magic potion should contain 32 wt% toe of frog.
Only 10% of the hydrogen does not react.
99.9% of the hydrogen leaves the flash drum in the vapor stream.
The water and the sulfuric acid are added to the mixer at a 10:1
molar ratio.
(i) One-third of the juice is split off to serve as cutback.
( j) The concentrated juice contains 45 wt% solids.
For each of the following situations, specify what you would choose as
(i) the system and (ii) the component(s), and whether you would choose
mass or molar units.
(a) N2O4 is fed to a reactor, where some of it decomposes to NO2.
(b) LB broth (a mix of tryptone and yeast extract and sodium chloride)
is fed to a fermentor, where bacteria consume the nutrients at a rate
of 1.4 g/h.
(c) Reformate, the number 2 cut from the hydrocracker, along with the
gasoline cut from the fluidized cat cracker, are sent to a storage
tank and then offloaded to a ship for distribution.
(d) 100 μg of a drug is loaded into a controlled-release capsule. Once
injected into the patient, the drug is released at a rate of 8.5 μg/h.
Air (79 mol% N2 and 21 mol% O2) is fed to a separator at a flow rate
of 1000 gmol/h. Two products are made. 80% of the oxygen in the feed
is recovered in one of the products, which is 98 mol% O2. Draw a block
flow diagram and indicate the components in each stream. Identify (a) the
basis, (b) all stream composition specifications, and (c) all system performance specifications. Do not do any calculations.
Write one equation for the following stream composition specifications
using logical choices for stream variables.
(a) Stream 1 contains 75 wt% benzene (B) and 25 wt% toluene (T).
(b) The CO2:H2O molar ratio in the reactor “out” stream is 2:1.
(c) Air in stream 3 contains 79 mol% N2 and 21 mol% O2.
Write one equation for the following system performance specifications,
using logical choices for stream and system variables.
(a) Stream 1 (water) and stream 2 (ethanol) enter the mixer at a 3:1 molar
ratio.
(b) 45% of the ethylene fed to the reactor is consumed.
(c) Stream 3, which contains 50 wt% water and 50 wt% ethanol, is sent
to a splitter, where 30% of the feed is taken off as product in Stream
4 and the rest is sent to a reactor.
(d) 98% of the diethanolamine in the feed to the distillation column is
recovered in the vapor product.
The material balance equation can be written as
Input − Output + Generation − Consumption = Accumulation
For each of the following situations, simplify the material balance equation by crossing out terms that are equal to zero. Also list whether the
system is continuous-flow, semi-batch, or batch.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
(a) Water is pumped into a large tank. System: tank. Component: water
(b) Water is pumped into a large tank that had been preloaded with
sugar crystals. The sugar dissolves, and a sugar solution is pumped
out of the tank. System: tank, Component: sugar.
(c) Same as (b) except component: water
(d) Ethylene and air (a mix of oxygen and nitrogen) are pumped into
a reactor operating at steady state, where 30% of the ethylene reacts
with the oxygen to form ethylene oxide. System: reactor. Component:
ethylene.
(e) Same as (d) except component: ethylene oxide
(f ) Same as (d) except component: nitrogen
P2.36 Your chemical engineering class is taught in Room 1234. At 8:50 am,
six students are already in the classroom. Between 8:50 am and 9 am,
42 students enter the classroom and 3 leave. If the classroom is the
system and students are the component, what is “Input,” “Output,”
“Generation,” “Consumption,” and “Accumulation”?
Drills and Skills
Section 2.2
P2.37 11.2 lb N2 and 2.4 lb H2 are added to a rigid vessel with a volume of
170,000 cm3. The vessel is at 298 K. (a) Calculate the pressure in the
vessel using the ideal gas law. Report your answer in units of psia,
psig, bar, kPa, and atm. (b) The nitrogen reacts completely with the
hydrogen to make ammonia (NH3). If the temperature is still 298 K
and the ammonia is a gas, what is the change in pressure in the vessel
(in bar)?
P2.38 Your company needs on-site storage for 45,000 lb ammonia (NH3).
What is the diameter (in ft) of a spherical vessel needed to store the
ammonia (a) as a gas at 80°F and 5 atm, (b) as a liquid at 80°F and
12 atm, or (c) as a liquid at −30°F and 1 atm? The density of liquid
ammonia is 42.6 lb/ft3 at −30°F and 37.5 lb/ft3 at 80°F. Gas density can
be calculated from the ideal gas law. Which temperature and pressure
would you choose, and why?
P2.39 Seawater contains about 5 grams gold per trillion grams water. About
how much seawater would you need to recover 1.0 ounce gold? If there
are about 3.32 × 108 cubic miles of seawater on the planet, and the
density of seawater is 1.05 g/cm3, how much gold (tons) is dissolved in
the ocean?
P2.40 The following table lists data from the EPA for production and recycling
of plastics in the United States. The annual global production of plastics
is estimated at 78 million metric tons. Assuming that the table includes
the major plastics, calculate (a) the total metric tons of plastics produced
in the United States, (b) the percentage of global production that is in
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Chapter 2 Problems
the United States, and (c) the total metric tons as well as the percent of
all U.S. plastics that are recycled.
Polymer
name
Recycle
number
Annual production,
billion kg
Percent
recycled
Poly(ethylene
terephthalate)
1
PET
4.5
19.5
High density
polyethylene
2
HDPE
5.5
10.3
Poly(vinyl
chloride)
3
PVC
0.9
0.0
Low density
polyethylene
4
LDPE
7.4
5.3
Polypropylene
5
PP
7.2
0.6
Polystyrene
6
PS
2.2
0.9
P2.41 Paper and polystyrene foam containers are used commonly in the food
service industry, and there are debates over the relative environmental
impact of each. For the most part, these materials are not recycled, but
are disposed of by incineration or landfill.
Two studies provide some data regarding the raw materials consumed in making cups of paper or polystyrene, as well as the emissions
created during their manufacture.
Study 1
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Paper
Polystyrene
Raw materials per cup
Wood chips (g)
Petroleum (g)
33.2
5.9
3.25
Finished weight (g)
10.1
1.50
Water emissions
(kg per metric ton)
99
21
Air emissions
(kg per metric ton)
27.2
52.1
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Study 2
Paper
Polystyrene
272.6
38.5
138.9
Finished weight
(lb per 10,000 cups)
229.1
46.9
Water emissions
(lb per 10,000 cups)
2.9
1.2
18.2
5.9
Raw materials
(lb per 10,000 cups)
Bleached pulp
Petroleum
Air emissions
(lb per 10,000 cups)
First convert the data in Study 1 to the same basis (10,000 cups) as the
data in Study 2. Then calculate the atom economy of paper versus polystyrene for Studies 1 and 2, by considering the raw materials and finished product weights. For a fair comparison, you should consider that
it takes about 2 tons of wood chips to make 1 ton of bleached pulp.
Then compare water and air emissions. Are the data from Studies 1 and
2 generally consistent or inconsistent? Do the data lead you to advocate
for paper or polystyrene cups? Explain.
P2.42 Methane (CH4) and oxygen (O2) are mixed and heated before being sent
to a burner, as shown. What is the volumetric flow rate, the mass flow
rate, and the mass fraction of methane in the stream leaving the preheater? You can assume the mixture obeys the ideal gas law.
CH4
75°C
10 atm
100 kgmol/min
O2
25°C
10 atm
400 kgmol/min
Preheater
500 kgmol/min
20 mol% CH4
80 mol% O2
200°C
10 atm
Section 2.3
P2.43 Figure 2.8 is a simplified process flow diagram of an ammonia synthesis facility. Identify units that function as mixers, splitters, reactors, and/
or separators. Also identify the equipment that changes pressure or temperature (e.g., compressors, heat exchangers).
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133
P2.44 Read the following description of a process for making cumene (C9H12).
Propylene (contaminated with propane) is mixed with benzene and
fed to a reactor, where two reactions occur: propylene and benzene
react to make cumene, and some cumene reacts further with propylene to make diisopropylbenzene. The reactor effluent is cooled and
sent to a vapor-liquid separator, where the vapors (unreacted propylene along with propane) are separated from liquids (unreacted
benzene, cumene, and diisopropylbenzene). The vapor stream is sent
to a splitter, where a fraction of the stream is used in another process
and the rest is sent to be burned as fuel gas. The liquid is sent to a
separator, where most of the benzene is removed and blended into
gasoline. The remaining material is sent to another separator, where
the desired product cumene is recovered as a nearly pure product,
and the byproduct diisopropylbenzene is discarded.
Sketch the block flow diagram, showing mixers, reactors, splitters,
and separators. Label each stream as appropriate with P (propylene),
I (propane), B (benzene), C (cumene), and D (diisopropylbenzene).
P2.45 When steel is “pickled,” it is dunked in an acid bath to remove rust on
its surface. This is good for the steel, but a lot of metal-contaminated
acid wastes are produced. Acid waste is an aqueous solution that contains sulfuric, phosphoric, hydrochloric, and nitric acids and dissolved
metal ions. This waste can’t be simply dumped into the nearest river.
Skid-mounted acid-waste processing equipment is used by small pickling operators. All the equipment is mounted on a skid, so it can be
trucked from location to location. In the process, acid waste is heated;
the acids vaporize and the metal ions (mostly iron) remain in the liquid
phase. The acid vapor is collected, condensed, and reused as “clean”
acid. Up to 90% of the spent acid is recovered; the remainder is disposed
of as chemical waste. The metal ion liquid solution is collected and
treated by adding salts, which precipitates the iron and other heavy
metals as insoluble metal salts. The metal salts are sold to other companies, which reclaim the metals. The remaining solution still contains
trace amounts of metals and must be treated as chemical waste.
Sketch a block flow diagram of the process, indicating the mixers,
separators, and splitters as well as the components in each stream.
P2.46 Vinyl chloride (C2H3Cl) is used to make polyvinylchloride (PVC) for
piping and other products. In the direct chlorination process, ethylene
(C2H4) and chlorine (Cl2) are used as raw materials. Liquid chlorine as
received at the plant is contaminated with a viscous liquid, so it is first
sent to an evaporator, where the contaminant is separated from the chlorine. Chlorine vapor and ethylene vapor are then mixed and fed to a reactor, where they are completely reacted to dichlorethylene (C2H2Cl2). The
dichloroethylene is then fed to another reactor, where some of it reacts
to HCl and C2H3Cl. The reaction mix is fed to a separator, where the
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Chapter 2 Process Flows: Variables, Diagrams, Balances
HCl is separated from the two chlorinated ethylene products. The latter
stream is fed to another separator, where unreacted dichloroethylene is
removed as a byproduct. The vinyl chloride product is sent to storage.
From this description, draw a block flow diagram. Identify process units
such as mixers, reactors, or separators. List the compounds present in
each process stream.
P2.47 In the Siemens process for making high-purity electronic-grade silicon,
metallurgical-grade silicon powder, Si, is first converted to tetrachlorosilane and trichlorosilane, which are then purified. The trichlorosilane
is reduced with hydrogen back to Si. The reduction reaction occurs on
a solid silicon rod; thus, the silicon rod grows inside the silicon reaction
chamber. Below is a simplified block flow diagram of a representative
Siemens process. Not all components are shown on all streams. Redraw
the block flow diagram, but with components all explicitly shown on
all streams. All process units operate at steady state except one. Indicate
on the diagram which unit is unsteady state.
H2
Gases
Separator
SiHCl3
Reactor chamber
with Si rod
Mixer
HCl
Metallurgicalgrade silicon
Separator
Liquids
Hydrogen
Separator
SiCl4
Separator
Reactor
Mixer
H2
Separator
Mixer
Impurities
HCl
Makeup
HCl
P2.48 Here is a description of a turkey-parts-to-oil plant from Technology
Review. “Unused turkey parts and water are dumped into a grinder and
pulverized into a slurry that has the consistency of peanut butter. The
slurry is then heated to 260°C and subjected to 275 kilograms of pressure. In the first of two reactor stages, the heated and pressurized mixture
is depolymerized—that is, cooked for between 15 minutes and an hour
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Chapter 2 Problems
135
to break apart the molecular structure of the organic material. The mixture is sent to a flash tank, where the pressure is released. The resulting
steam is recaptured to power the system. Minerals sink to the bottom
and flow into a separate tank. Organic materials move to the second
reactor. Temperatures of almost 500°C further break down the organic
materials. An auger moves carbon particles into a drum. The hot fluid
moves into distillation tanks, where it cools and condenses. Organic
materials and water separate. The water sinks to the bottom. A fuel gas
is taken off the top, leaving a crude oil similar to a mix of diesel fuel
and gasoline. The crude oil moves into storage tanks for later sale.” From
this description, draw a block flow diagram. Identify process units such
as mixers, reactors, or separators. Label the streams to indicate the compounds present in each process stream.
Section 2.4
P2.49 “Hard” water contains calcium, magnesium, and other mineral salts that
tend to deposit on piping, coffeepots, and bathtubs. Hard water is “softened” in water softeners. In a water softener, hard water flows over ion
exchange beads in a vessel. The beads carry Na+ ions. Ca++ and Mg++
from the water preferentially stick to the beads, displacing Na+, which
then dissolves in the water. The “soft” water flows out of the softener.
Is the water softener a Mixer, Reactor, Splitter or Separator? Draw a
flow diagram and indicate the components in the streams. Write material
balance equations for H2O, Ca++, Mg++, and Na+, using the water softener
as the system. Simplify each equation by crossing out terms that are
zero. Explain your reasoning.
P2.50 Read the following description of a process for making cumene (C9H12).
Propylene (contaminated with 5 wt% propane) is mixed with benzene
at a 1.4:1 ratio and fed to a reactor, where two reactions occur:
propylene and benzene react to make cumene, and some cumene
reacts further with propylene to make diisopropylbenzene (C12H18).
In the reactor, 50% of the benzene and 60% of the propylene are
consumed. The reactor effluent is cooled and sent to a vapor-liquid
separator, where the vapors (unreacted propylene along with propane) are separated from liquids (unreacted benzene, cumene, and
diisopropylbenzene). The vapor stream is sent to a splitter, where 30%
of the stream is used as a feed in another process and the rest is sent
to be burned as fuel gas. The liquid is sent to a separator, where 99%
of the benzene fed is removed and blended into gasoline. 100% of the
cumene and the diisopropylbenzene is recovered into the other stream
and sent to another separator, where the desired product cumene is
recovered as a 99.5 wt% pure product, and the byproduct diisopropylbenzene is discarded. The process makes 14,000 lb/day cumene.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
List (a) the basis, (b) all stream composition specifications, and
(c) all system performance specifications.
Section 2.5
P2.51 You’ve got a tank that contains 1910 kg of spent acid (12.43 wt% H2SO4
in water). How much concentrated sulfuric acid (77.7 wt% H2SO4) must
be added to the tank so that it contains a solution meeting battery acid
specifications (18.63 wt% H2SO4)?
P2.52 Lawn and garden fertilizers are labeled to indicate the quantity of nitrogen, phosphorus, and potassium. Gro-Right fertilizer is labeled 5-10-5,
which means that it contains 5 wt% nitrogen as N, 10 wt% phosphorus
as P2O5, and 5 wt% potassium as K2O. The fertilizer is prepared by
mixing ammonium nitrate (NH4NO3), calcium phosphate (Ca(H2PO4)2),
potassium chloride (KCl) and filler. In a 100-lb bag, calculate the mass
(lb) of ammonium nitrate, calcium phosphate, and potassium chloride.
To do this, first draw and label a flow diagram, and identify the basis.
Also calculate the weight percent filler in the bag.
P2.53 Propane-air mixtures will not ignite if the mixture contains more than
11.4 mol% propane, even if exposed to a flame. Per 100 liters air at
STP, how many grams of propane are needed to just exceed the flammability limit and avoid ignition?
P2.54 Self-cleaning window glass is made by coating the glass with nanoparticles of titania (TiO2). The titania works as a catalyst in the presence
of sunlight to react organic dust and grime to CO2 and water. To coat
glass, we need to prepare a bath containing 750 mg titania per kg of
aqueous solution. Fresh coating slurry is available at a concentration of
1.5 g titania per kg solution. 80 kg of spent solution is left in a tank
after a previous coating run; the spent solution is at 400 mg titania per kg
solution. We want to prepare 125 kg solution for another coating run
by combining the spent solution, the fresh coating slurry, and water.
Calculate how much slurry and water we should add to the tank that
contains the spent solution.
P2.55 Semiconductor devices are fabricated from single-crystal silicon rods.
To make these rods, a single-crystal silicon “seed” is dipped into a
container of molten Si. The silicon seed is slowly rotated and pulled up
from the melt as the silicon freezes onto the seed. In one setup, a rod
100 mm in diameter is grown for 50 h to a final length of 60 cm. The
rod diameter is constant. The density of solid silicon is 2.33 g/cm3. Find
the mass rate of accumulation (g/h) of the silicon on the road.
P2.56 You’re a witch in need of a new magic potion. You’ve got three flasks
containing the ingredients listed below. You’d like to mix these together
in your cauldron, heat the cauldron over a fire to evaporate off excess
water, and make 100 g of a liquid potion containing 27 wt% toe of frog,
22 wt% eye of newt, and 11 wt% wool of bat. How many grams from
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Chapter 2 Problems
each flask should you add to your cauldron? How many grams of water
should you evaporate off?
Flask A, wt%
Toe of frog
Flask B, wt%
10 0
Eye of newt 0
Flask C, wt%
50
30 0
Wool of bat
40 0
10
Water
50
40
70
P2.57 Air (assumed to contain 79 mol% nitrogen and 21 mol% oxygen) is separated into two product streams. The separator operates at steady state.
One product stream is 98 mol% oxygen, and it contains 80% of the oxygen in the air fed to the column. The other product stream is mostly
nitrogen. Draw and label a flow diagram. Identify the stream composition
specification, and the system performance specification. Calculate the
quantity of air required (tons/day) to produce 1 ton/day of the oxygen
product. Calculate the mol% nitrogen in the second product stream.
P2.58 A gas stream containing 60 wt% benzene and 40 wt% toluene is fed at
100 g/s to a distillation column operating at steady state. There are two
product streams: the distillate and the bottoms streams. 95% of the benzene
fed is recovered in the distillate stream, which is 98 wt% pure benzene.
Draw and label a block flow diagram. Write equations for the basis,
the stream composition specification(s), and the system performance
specification(s). Write material balance equations for benzene and for
toluene. Calculate the flow rates and compositions of both product streams.
P2.59 A natural gas contains 85 mol% methane and 15 mol% ethane. The gas
must be separated into two product streams, one that is 99 mol% methane,
and the other that is 90 mol% ethane. Draw and label a block flow
diagram. Choose a basis, and list the stream composition specifications.
Calculate the fraction of methane in the feed that goes to the methanerich product, and the fraction of ethane going to the ethane-rich product.
P2.60 To make cherry jam, cherries (18% solids, 82% water) are mixed with
sugar at a 1:2 (lb/lb) cherry:sugar ratio. Then, the mixture is fed to an
evaporator, where 2∕3 of the water is boiled off. Sketch and label a
block flow diagram. Write material balance equations on solids, sugar,
and water. What feed rate of cherries (lb/h) is required to produce
10 lb/h jam?
P2.61 10 lb of cherries (18% solids, 82% water) are mixed with 20 lb sugar in
a pot all at once. Then, water is boiled off at 0.20 lb/min. Sketch and
label a block flow diagram. Write material balance equations on solids,
sugar, and water. How many pounds are in the pot after 30 minutes?
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Chapter 2 Process Flows: Variables, Diagrams, Balances
P2.62 High-fructose corn syrup is a popular and inexpensive sweetener used in
many carbonated beverages. When manufactured, the syrup contains a
small amount of colored impurities; before it is mixed into beverages the
impurities are removed by pumping the syrup over a bed of activated
charcoal. The impurities adsorb to the charcoal while the rest of the syrup
passes through. The charcoal adsorbs 0.4 kg impurities per kg charcoal.
You are in charge of designing a charcoal adsorption facility that can
process 115 lb of high-fructose corn syrup per hour. The syrup typically
contains 0.5 wt% impurities. You’d like the charcoal to last for 7 days
before needing replacement. How much charcoal (lb) should be used?
P2.63 Nitrogen and hydrogen are fed at a 1:3 molar ratio to an ammonia
synthesis reactor operating at 1340 °R and 80 atm. 25% of the N2 fed
is converted to ammonia, and the reactor produces 1000 lbmol/h
NH3. Draw and label a flow diagram. Identify the basis, one stream
composition specification, and one system performance specification.
Write material balance equations on N2, H2, and NH3. Calculate the
volumetric gas feed rate to the reactor (ft3/h), at the reactor temperature
and pressure. Assume an ideal gas, with R = 0.7302 ft3 atm/lb-mol °R.
P2.64 Chlorine dioxide, used to bleach pulp in the paper industry, is produced
by the following reaction:
6NaClO3 + 6H2SO4 + CH3OH → 6ClO2 + 6NaHSO4 + CO2 + 5H2O
3200 kgmol/h of a mixture containing 45 mol% NaClO3, 45 mol%
H2SO4, and 10 mol% CH3OH is fed to a reactor operating at steady
state. 75% of CH3OH fed to the reactor is converted to products. Identify
the basis, the stream composition specifications, and the system performance specification. Calculate the molar flow rate of all compounds in
the reactor output stream.
P2.65 Citral (C10H16O) is extracted from lemongrass oil and is popular in
many consumer products, from dish detergents to ice creams, for its
pleasant lemon-lime fragrance. Alternatively, citral can be made synthetically from butene (C4H8), formaldehyde (CH2O), and oxygen; the
net reaction is:
2​C​4​​​H8​ ​​ + 2​CH​2​​O + 0.5​O2​ ​​ → ​C1​ 0​​​H1​ 6​​O + 2​H2​ ​​O
A gas stream (1200 gmol/h) containing 35 mol% butene and 65 mol%
formaldehyde is mixed with air (79 mol% N2, 21 mol% O2) at an 8:3
gas:air ratio. The mixture is fed to a reactor, where the mixture reacts
over a catalyst to make citral. 90% of the O2 fed to the mixer is consumed by reaction. The reactor operates at steady state. Calculate
(a) the flow rate of all components in the reactor outlet stream, (b) mol%
citral in the reactor outlet stream, and (c) % butene and % formaldehyde
fed that is consumed by reaction.
P2.66 A gas contains 10 mol% propylene (C3H6), 12 mol% ammonia (NH3),
3 mol% water (H2O), and 75 mol% air. You can assume that in air
the N2:O2 molar ratio = 79:21. The gas is fed at 8000 kgmol/day to a
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Chapter 2 Problems
139
reactor. In the reactor, propylene, ammonia, and oxygen react to make
acrylonitrile (C3H3N) by the following chemical reaction:
​C​3​​​H6​ ​​ + ​NH​3​​ + 1.5​O2​ ​​ → ​C3​ ​​​H3​ ​​N + 3​H2​ ​​O
30% of the propylene is consumed by reaction in the reactor. The reactor operates at steady state. Draw and label a block flow diagram.
Identify (i) the basis, (ii) any stream composition specifications, and
(iii) any system performance specifications. Then calculate the molar
flow rate (kgmol/h) of each compound in the reactor outlet stream. Also
report the total molar flow rate and the mol% of each compound in the
reactor outlet. Is the mol% N2 in the outlet stream less than, greater
than, or the same as the mol% N2 in the inlet stream? Explain.
P2.67 Silicon tetrachloride (SiCl4, also called tetrachlorosilane) reacts with
magnesium metal (Mg) to make pure solid silicon Si and magnesium
chloride (MgCl2). A researcher places a mixture of 255 g SiCl4 with
48 g Mg into a small empty laboratory-scale reactor. The next day, the
researcher removes the entire contents of the reactor and finds only Si,
MgCl2, and SiCl4. How many grams of each should he find?
P2.68 A mixture containing 84.2 wt% SiCl4 and 15.8 wt% Zn is fed continuously at a rate of 303 g/h into a small laboratory-scale reactor. The
reactor operates at steady state. Inside the reactor, Si is produced by the
following reaction:
SiCl4 + 2 Zn → Si + 2 ZnCl2
The reactor exit stream contains Si, ZnCl2, and SiCl4. Calculate the flow
rate and composition of the exit stream. (This reaction formed the basis
for the earliest commercial production of electronics-grade silicon, but
has been replaced by the Siemens process.)
P2.69 A typical gas grill burns propane at about 500 g/h. Calculate the
volumetric air flow rate (cm 3/h) needed to completely combust
­propane to CO2 and H2O. Also find the rate of production of CO2, in
g/h and cm3/h. Air can be assumed to be 79 mol% N2 and 21 mol%
O2. The volumetric flow rates can be calculated from the ideal gas
law as STP.
Section 2.6
P2.70 Solve the rest of the Case Study. Derive the necessary equations and
find the flows of all components in all streams. Report in table form,
in units of both kgmol/h and kg/h. Also check: does the mass flow
into the process equal the mass flow out? (should it?) Does the molar
flow into the process equal the molar flow out? (should it?)
P2.71 According to the World Economic Forum, the annual global production of plastic is estimated at 78 million metric tons. Of this total
production, 98% is newly manufactured and 2% is from closed-loop
recycling. Of the 78 million metric tons, 14% is incinerated, 40% ends
up in the landfill, 14% is recycled, and the remainder ends up in the
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Chapter 2 Process Flows: Variables, Diagrams, Balances
environment. Of the 14% recycled, 28% is lost during the process,
58% is processed to lower-quality plastics, and 14% is closed-loop
recycled as high-quality plastics. The block flow diagram below illustrates the different streams. Indicate whether each unit should be
analyzed as a mixer, splitter, separator or reactor. Calculate the quantities of plastics that are in each stream.
Recycled high-quality
Reprocessed to lower-quality
Lost during processing
Newly made
To incinerator
To landfill
To environment
P2.72 As part of the process of producing sugar crystals from sugar cane, raw
sugar cane juice is sent to a series of three evaporators to remove water.
The sugar cane juice, which is 85 wt% water, is fed to the first evaporator at 10,000 lb/h. Equal amounts of water are removed in each
evaporator. The concentrated juice out of the last evaporator is 40 wt%
water. Calculate:
(a) The flow rate of the concentrated juice out of the last evaporator
(b) The flow rate of water removed in each evaporator
(c) The wt% water in the juice fed to the second evaporator
Raw cane
juice
Water
Water
Water
Evaporator 1
Evaporator 2
Evaporator 3
Concentrated
sugar juice
P2.73 Many people have difficulty digesting lactose, the sugar in milk. One
solution is to hydrolyze lactose (C12H22O11) to the simple sugars glucose
and galactose, using an enzyme called lactase:
​C​12​​​H2​ 2​​​O1​ 1​​ + ​H​2​​O → ​C6​ ​​​H1​ 2​​​O6​ ​​ (glucose) + ​C6​ ​​​H1​ 2​​​O6​ ​​(galactose)(R1)
Glucose and galactose are isomers—they have the same molecular formula but different structures (and different tastes—glucose is sweeter
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Chapter 2 Problems
than galactose). Galactose is converted to glucose by another enzymecatalyzed reaction:
​C​6​​​H1​ 2​​​O6​ ​​(galactose) → ​C6​ ​​​H1​ 2​​​O6​ ​​(glucose)(R2)
A process has been proposed for converting lactose to glucose and
galactose. 1000 kgmol/day of a solution containing 31.25 mol% lactose
and the remainder water is fed to Reactor 1. 92% of the lactose fed is
consumed by reaction R1. The output from Reactor 1 is sent to Reactor 2.
60% of the galactose fed to Reactor 2 is consumed by reaction R2. The
output from Reactor 2 is fed to Separator 1, where 3 separate product
streams are taken off: pure glucose, pure galactose, and a lactose-water
solution. The block flow diagram is sketched below, using L for lactose,
W for water, G for glucose, and Ga for galactose. Streams are identified
by numbers.
Demonstrate your understanding of process flow calculations by:
(a) Writing an equation for the basis
(b) Writing one equation that describes the stream composition
specification
(c) Writing two equations for the system performance specifications
(d) Writing a complete set of independent material balance equations
around Reactor 1, Reactor 2, and Separator 1, using appropriate
stream and system variables (use only variable names, no numbers).
4
L
W
Reactor 1
1
L
W
G
Ga
2
Reactor 2
L
W
G
Ga
3
Separator 1
5
6
G
Ga
L
W
P2.74 High-purity silicon can be made from cheap ingredients: sand (SiO2)
and coke (C), which are both solids. There are three reactions: first,
sand and coke are heated in an electric arc furnace to high temperatures,
which reduces SiO2 to Si and produces carbon monoxide gas. Second,
solid Si reacts with chlorine (Cl2) gas to make silicon tetrachloride
(SiCl4) gas. Then SiCl4 reacts with solid magnesium (Mg) to produce
solid magnesium chloride (MgCl2) and solid high-purity Si. The two
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Chapter 2 Process Flows: Variables, Diagrams, Balances
solids can be separated by dissolving MgCl2 in water, because Si is not
soluble in water.
Write the balanced chemical reactions. Sketch out a reasonable
block flow diagram, showing mixers, reactors, and separators. Show
which compounds are in each stream. Then, using the entire process as
a system, calculate the mass of all reactants required as well as the mass
of byproducts per kg of high-quality Si.
P2.75 In a process for making DEA (diethanolamine, C4H11O2N), ethylene is
oxidized to ethylene oxide, then ethylene oxide is reacted with ammonia
to make DEA. The stoichiometrically balanced reactions are:
2​C2​ ​​​H4​ ​​ + ​O​2​​ → 2​C2​ ​​​H4​ ​​O
2​C​2​​​H4​ ​​O + ​NH​3​​ → ​C4​ ​​​H1​ 1​​​O2​ ​​N
Each reaction is carried out in separate reactors. C2H4, air (79 mol% N2,
21 mol% O2), and NH3 are the raw materials available. The first reactor
is operated such that all the oxygen but only 25% of the ethylene is
consumed. In the second reactor, complete conversion of reactants is
achieved. No oxygen is allowed in the second reactor. Ethylene oxide
is readily separated from other gases. Nitrogen and ethylene can be
separated from each other. The separation of oxygen from nitrogen is
prohibitively expensive.
Based on this information, sketch out a block flow diagram showing
what you think is the best process for making DEA. Label all process
units as “mixer,” “reactor,” “separator,” or “splitter,” as appropriate.
Show all components present in each stream. Briefly explain your reasoning. You do not need to do any calculations.
P2.76 Dimethyl carbonate (DMC, C3H6O3) can be synthesized by a process
called oxidative carbonylation of methanol, per the reaction shown:
1 ​ ​O​ ​​ → ​C​ ​​​H​ ​​​O​ ​​ + ​H​ ​​O
2​CH​3​​OH + CO + ​ _
3 6 3
2
2 2
A gas containing 80 mol% CH3OH and 20 mol% CO at 2000 gmol/h is
mixed with air (79 mol% N2 and 21 mol% O2) and then fed to a reactor
operating at steady state, where the reaction takes place. The flow rate
of the stream leaving the reactor (the reactor effluent) is 2264 gmol/h,
and this stream contains no O2.
The reactor effluent is fed to a separator. Two streams leave the
separator. One stream (“gas product”) leaving the separator contains
nitrogen, CO, and 10% of the methanol fed to the separator. The other
stream (“liquid product”) contains the remaining methanol, as well as
all the water and DMC.
Draw and label a block flow diagram. Show components and stream
numbers on your diagram. First choose the mixer plus reactor as the
system. Calculate the molar flow rate (gmol/h) of air fed to the mixer,
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Chapter 2 Problems
143
the mol% DMC in the reactor effluent, and the % of methanol fed to the
reactor that is consumed by reaction. Then choose the separator as
the system. Calculate the flow rate (gmol/h) of liquid product leaving the
separator, and the mol% DMC in the liquid product.
P2.77 Acrylonitrile (C3H3N, used to make carbon fiber, acrylic fibers, nylons,
fumigants, and synthetic rubber) is synthesized by catalytic ammoxidation of propylene (C3H6):
2​C​3​​​H6​ ​​ + 2​NH​3​​ + 3​O2​ ​​ → 2​C3​ ​​​H3​ ​​N + 6​H2​ ​​O
Propylene, ammonia, and air (79 mol% N2, 21 mol% O2) are mixed and
then fed to the reactor, where the mixture reacts over a catalyst to make
acrylonitrile. The reactor operates at steady state. You are the process
engineer in charge of monitoring the performance of the reactor. One day
you determine that the gas flow rate out of the reactor is 7095 gmol/min,
and that the gas contains 28.19 mol% water and 1.88 mol% ammonia,
along with N2, propylene, and acrylonitrile, but no O2.
The acrylonitrile reactor effluent is fed to a separator. Two streams
leave the separator. One stream (“gas product”) leaving the separator contains all of the nitrogen, 85% of the propylene fed to the separator, and
some ammonia. The other stream (“liquid product”) contains 2.1 mol%
propylene, along with ammonia, water, and acrylonitrile.
Draw a flow diagram. Derive material balance equations using the
reactor as system. Calculate: (a) The flow rate (gmol/min) of acrylonitrile leaving the reactor (b) the flow rates (gmol/min) of propylene,
ammonia, and air fed to the reactor, (c) the percent of propylene and of
ammonia fed to the reactor that is consumed by reaction. Then derive
material balance equations using the separator as the system. Calculate:
(d) the percent of ammonia fed to the separator that is recovered in the
liquid product, and (e) the total flow rate of the liquid product.
P2.78 1,3-propanediol (C3H8O2) is a building block in the synthesis of polymers that are used to make fabrics or plastic bottles. 1,3-propanediol
is made commercially by both chemical and biological routes. One
chemical route is called hydroformylation, starting from ethylene oxide
(C2H4O):
​C2​ ​​​H4​ ​​O + CO + 2​H2​ ​​ → ​C3​ ​​​H8​ ​​​O2​ ​​
A gas stream containing 30 mol% C2H4O, 30 mol% CO, and 40 mol%
H2 is fed to a reactor at 900 gmol/min, where the mixture reacts over
a catalyst to make C3H8O2. The reactor operates at steady state. You
are the process engineer in charge of monitoring the performance of the
reactor. One day you sample the gas stream leaving the reactor and
determine that it contains 36 mol% C3H8O2.
The reactor effluent is sent to a separator. Two streams leave the
separator. One stream (“gas product”) leaving the separator contains all
of the H2 and CO, as well as some of the ethylene oxide and some of
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Chapter 2 Process Flows: Variables, Diagrams, Balances
the 1,3-propanediol. 95% of the 1,3-propanediol fed to the separator is
recovered in the other stream (“liquid product”), which is 85 mol%
1,3-propanediol.
Draw and label a flow diagram of the process. Starting with the
reactor as the system, write equations that describe the basis and the
stream composition specifications. Write material balance equations
and then calculate the total flow rate (gmol/min) and molar composition
of the reactor effluent.
Then, choose the separator as the system. Identify basis, stream
composition specifications, and system performance specifications.
Write material balance equations and calculate the flow rates and compositions of both gas and liquid product streams.
P2.79 Sorbitol is an ingredient in “sugar-free” candy. It is sweet, but does not
promote tooth decay because bacteria cannot metabolize it for food, and
it is considered diet food because humans don’t metabolize it well
either. Sorbitol (C6H14O6) is made from glucose C6H12O6 (which does
cause tooth decay) and hydrogen. 100 kg/day of a 30 wt% glucose solution is mixed with a stoichiometric flow rate of hydrogen and sent to a
reactor; 80% of the glucose is converted to sorbitol. The hydrogen is
then separated from the sugar solution as a gas stream. How much
hydrogen (kg/day) is fed to the process? What is the composition (wt%)
and flow rate (kg/day) of the liquid stream leaving the process?
Scrimmage
P2.80 Microorganisms contain a complex mix of proteins, carbohydrates, and
fats that are sometimes lumped together as a single pseudochemical
compound. For example, bacterial biomass has an empirical formula
of CH1.666N0.20O0.27. Under aerobic conditions, bacteria take in glucose (C6H12O2), oxygen (O2), and ammonia (NH3) and make more
bacteria, CO2, lactic acid (C3H6O3), and water. Write three balanced
reactions for glucose reacting to form CO2, lactic acid, or bacteria.
Oxygen and ammonia can also be reactants, and CO2 and water can
be byproducts.
Glucose (18.0 g) is dissolved in buffer and placed in a bottle, to
which 1.0 g bacteria and some ammonia is added. Air is bubbled
through the solution. After some time, the air is stopped and it is
determined that the bottle contains no glucose, 2.10 g bacteria, 3.6 g
lactic acid, and some water. Calculate the grams of CO2 generated.
What fraction of the glucose was consumed to make CO2? lactic acid?
bacteria?
P2.81 Plants need phosphate. Phosphate rock [(CaF)Ca4(PO4)3] is used extensively by organic farmers, but releases phosphate very slowly. If phosphate rock reacts with sulfuric acid (H2SO4), monocalcium phosphate
[CaH4(PO4)2H2O], calcium sulfate (CaSO4), and hydrogen fluoride (HF)
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Chapter 2 Problems
145
are produced. Rapid release superphosphate fertilizer is the mix of
monocalcium phosphate and calcium sulfate produced in this reaction.
(HF is not used.) You are in charge of designing a process to produce
600 tons of superphosphate fertilizer per day. Calculate the tons/day of
phosphate rock and sulfuric acid required. If phosphate rock sells for
$125/ton, sulfuric acid for $60/ton, and superphosphate for $280/ton,
what annual profit can you expect? Assume the process operates 350 days
per year.
P2.82 Cheese whey contains a number of proteins that may have specific uses
when purified. For example, glycomacropeptide (GMP) contains no
­phenylalanine, and is therefore a protein source that could be safely consumed by people with the disease phenylketonuria (PKU). GMP must be
separated from other cheese whey proteins, in particular beta-lactoglobulin
(BLG), before it can be consumed by people with PKU.
In a process under development, whey containing 1.2 g GMP/L and
0.8 g BLG/L is fed to a separator at a flow rate of 150 mL/min. The
separator contains an ion exchange resin, to which some of the protein
adsorbs. 89% of the GMP and 24% of the BLG adsorbs to the resin
while the remainder of the whey passes through. After 30 minutes, the
whey feed is discontinued, and a buffer containing 0.25M NaCl is
pumped through the separator at 150 mL/min for 10 min. During this
time all the protein on the resin is desorbed (“unstuck”) and comes off
with the buffer. The protein-containing buffer is collected and all the
water is evaporated off, producing a dried product. What is the final
purity (wt% GMP) and quantity (grams) of dry product?
P2.83 Vinegar, which is mostly a solution of acetic acid (CH3COOH) in water,
can be made by fermentation of wine or juice. Alternatively, it can be
synthesized chemically by controlled oxidation of ethanol. You are
designing a process for making boutique vinegar, in which you will first
oxidize ethanol to acetic acid in a reactor, and then blend in a special
mix of herbs and spices. The process should produce the vinegar base
(5.0 wt% acetic acid in water) at 4000 kg/day. Oxygen will be supplied
by air and should be fed at stoichiometric quantities. Determine the
balanced chemical reaction. Sketch out a flow diagram, then calculate
the flow rates of air, ethanol, and water into the process, and the flow
rates of any byproducts out of the process.
P2.84 Phenylketonuria (PKU) is a metabolic disorder, where patients cannot
process the amino acid phenylalanine (Phe), which is in almost all proteins.
Patients generally need to avoid eating any foods that contain protein, and
they meet their nutritional needs by consuming a rather bad-tasting amino
acid drink. A protein in cheese whey, called glycomacropeptide (GMP),
is a unique protein that contains almost no Phe, yet it can be used to make
better-tasting foods that PKU patients can eat. However, GMP does not
contain all the essential amino acids, so the protein needs to be supplemented
with some amino acids to meet nutritional needs.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
You are in charge of developing the formula for a powdered
protein-rich product to be sold to food manufacturers. The table
below shows the mg amino acids per gram of GMP. (Not all amino
acids are shown.)
mg/g GMP
Arg
His
Leu
Phe
Trp
Tyr
Lys
3.7
1.2
19.4
0
0
0.5
50.0
Manufacturers sell actual grams of protein, but nutritionists want to know
the “protein-equivalent” (PE) grams, which is calculated based on the
nitrogen (N) content. A PE gram = 6.25 × grams of N in the product.
For GMP, the N content is determined experimentally to be 0.125 g N/g
GMP, and 1 gram of GMP is equal to 0.78 PE grams.
For the powdered protein-rich product, there is a target amino acid
composition, which will be achieved by mixing GMP with pure amino
acids (AA). The target, as well as the N content (moles N per mole amino
acid) and the molecular weight of the amino acid, is listed in the table.
(The 3-letter code for AA is used.)
Arg
His
Leu
Phe
Trp
Tyr
Lys
90
24
200
0
14
93
60
Molar mass (g/gmol)
211
155
131
204
181
183
N content (mol N/mol AA)
4
3
1
2
1
2
Target mg/PE g of product
Calculate the grams of each amino acid, and the grams of GMP, that
should be mixed to make 100 grams of the powdered product. Also
calculate the PE grams of the product. (Hints: draw a diagram, and
think carefully about units and basis, before deriving material balance
equations.)
P2.85 Fresh orange juice contains 12 wt% dissolved solids in water. Most of
the dissolved solids are sugars, but it also includes trace quantities of
volatile and temperature-sensitive compounds that add the unique fragrance and flavor of juice. To concentrate the juice before shipping, some
of the water is removed in special evaporators that operate at low temperature to minimize loss of the juice flavor/fragrance compounds.
However, some of these compounds are lost in the water vapor stream
from the evaporator anyway. To compensate, a “cutback” is used, where
some fresh juice is mixed back in with the concentrated juice.
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Chapter 2 Problems
147
In one such process, fresh juice (10,000 lb/h) is fed to a splitter.
90% of the fresh juice is fed to an evaporator, where water is removed
by evaporation. The liquid from the evaporator contains 80 wt% dissolved
solids. The remaining 10% of the fresh juice from the splitter is combined
with concentrated juice leaving the evaporator in a mixer. Calculate (a) the
rate of water evaporation from the evaporator and (b) the flow rate and
dissolved solids content of the juice leaving the process.
P2.86 Automotive airbags contain a cylinder packed with a mixture of three
solids—sodium azide (NaN3), potassium nitrate (KNO3), and silicon
dioxide (SiO2). In a collision, an electrical signal is sent to the cylinder, causing rapid decomposition of sodium azide to sodium (Na)
and nitrogen (N2). Na metal reacts with KNO3 to produce K2O and
Na2O, as well as more N2. K2O and Na2O fuse with SiO2 to produce
an amorphous glassy solid. Write down the stoichiometrically
balanced reactions for decomposition of sodium azide and for reaction of sodium metal with potassium nitrate. If you want 6 standard
cubic feet of nitrogen in the bag when filled, how much NaN3 (in lb
and in ft3) should be packed into the cylinder? How much potassium
nitrate and silicon dioxide (in lb and in ft3) would you recommend?
What is the increase in volume of the contents of the airbag upon
collision? The densities of solid NaN3, KNO3, and SiO2 are 1.846,
2.1, and 2.6 g/cm3, respectively. To solve this problem, first identify
the basis and write material balance equations, considering the airbag
as a batch reactor.
P2.87 Sulfuric acid (H2SO4) is made by burning sulfur S in air to make SO2.
This step is operated at a 1.2:1 O2:S mole ratio, to ensure that all the
sulfur is oxidized. Then SO2 is further oxidized to SO3 over a catalyst,
using air as the source of oxygen, again at a 1.2:1 O2:S mole ratio to
ensure that all the SO2 is consumed. The effluent from this reactor is
cooled by mixing with water, which also dissolves the SO3 and produces
H2SO4. N2 and O2 do not dissolve appreciably in the water. We want
to build a plant that produces 200 tons/day of concentrated sulfuric acid
(98 wt% H2SO4, 2 wt% water). Water, sulfur, and air are the available
raw materials. Sketch out a block flow diagram showing what mixers,
reactors, splitters, and/or separators that might be used in this process.
Number all the streams. Calculate the raw materials required as: tons sulfur/
day, standard cubic feet of air/h, and tons water/day. Then calculate all the
stream flows, in both molar and mass units. Make any additional approximations or assumptions needed to complete calculations, but list this and
provide a brief justification. Summarize results in table form.
P2.88 Gold is recovered from rock using sodium cyanide (NaCN). Ore containing solid gold (Au) is reacted with sodium cyanide in water to
make NaAu(CN)2 (which is soluble in aqueous solutions) and NaOH.
Then, solid zinc (Zn(s)) reacts with NaAu(CN)2 to produce a soluble
Zn(CN)2 and solid gold, which precipitates as a fairly pure solid nugget.
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Chapter 2 Process Flows: Variables, Diagrams, Balances
NaCN is a byproduct of this reaction. Sketch out a block flow diagram
for processing one ton of ore, containing 0.019 wt% gold, to make pure
gold nuggets. On your diagram, be sure to show all process units needed
(mixers, reactors, splitters, and/or separators). Calculate the quantities
of other reactants required and byproducts produced. Calculate the
­volume of pure gold nuggets produced from 1 ton of rock. The density
of a gold nugget is 19.3 g/cm3. Atomic weights are Na (23), C (12),
N (14), Au (197), O (16), H (1), and Zn (65).
P2.89 Polycarbonates are transparent impact-resistant polymers used to make a
variety of products including contact lens, baby bottles, and compact
discs. Polycarbonates are currently made from the highly toxic gas phosgene (COCl2), in a process with poor atom economy. You are interested
in developing a “greener” process for synthesis of polycarbonates by
using dimethyl carbonate (DMC, C3H6O3) instead of phosgene. You are
searching for an economical and environmentally friendly method to
make DMC. The following set of reactions is of interest:
Syngas (mixture of CO and H2) production from methane and steam:
​CH​4​​ + ​H​2​​O → CO + ​H2​ ​​
Methanol (CH3OH) production from CO and H2:
CO + ​H2​ ​​ → ​CH​3​​OH
Dimethyl carbonate production from methanol, CO, and O2:
​CH​3​​OH + CO + ​O2​ ​​ → ​C3​ ​​​H6​ ​​​O3​ ​​ + ​H​2​​O
Your accounting department tells you the bulk prices for these chemicals
as methane ($0.28/kg), hydrogen ($1.1/kg), methanol ($0.47/kg), oxygen
($0.12/kg), and DMC ($2.65/kg).
Balance the reactions. Use a generation-consumption analysis to
come up with a reaction pathway that makes DMC from methane, water,
and oxygen, with no net generation or consumption of CO or methanol.
Calculate the flows of raw materials, products, and byproducts for
a process that uses your reaction pathway to make 3600 kg/day DMC.
Calculate the profit ($/day) considering only raw material costs and
product values.
Sketch out two or three different block flow diagrams that are reasonable preliminary designs.
P2.90 You are to prepare a broth for a fermenter to grow antibiotic-producing
cells. (A fermentor is just a special reactor vessel for growing microorganisms.) The broth should be an aqueous solution containing 15 wt%
glucose, 6 wt% phosphate, 6 wt% nitrate, and various trace nutrients. A
custom supplier can produce the required blend for $15/kg, but your
boss suspects you can produce the blend in-house for much less. Several
commercial powders are available as raw ingredients, to be mixed with
water as needed. Their mass compositions and costs are:
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Chapter 2 Problems
“Fast-Feed”
Glucose
“Super-Gro”
“Formula N”
45% 0% 0%
Phosphate 2%
35%
12%
Nitrate 1%
15%
58%
Trace nutrients 3% 8% 0%
Filler
49%
42%
30%
Cost/kg
$20
$10
$15
(a) Suggest a combination of these powders that will produce the
desired broth composition. Give the masses of powders and water
required per kilogram of broth.
(b) What is the maximum savings per kilogram broth? (Assume water
is free.) If the mixing equipment and storage tanks for preparing a
20-kg batch cost $2000, how many batches will you need to make
to pay for the equipment before you begin to see real savings?
(c) What is the wt% trace nutrients in the home-made broth? If you
now require 3 wt% trace nutrients, can you mix up a broth from the
available powders that meets all specifications? What combination
of powders would you recommend, and how does meeting this new
requirement affect the economics?
P2.91 Antibiotics are typically produced by fermentation, where the fermenter
is operated under what is called fed-batch culture. In fed-batch culture,
the fermenter is initially filled with fermentation broth and cells. As the
cells grow, they consume nutrients (glucose, phosphate, nitrate, etc.) and
produce products. During cell growth, additional broth is added continuously to feed the cells.
You are planning to run a fermenter to produce antibiotics from fungi,
using the fermentation broth that is an aqueous solution containing 15 wt%
glucose, 6 wt% phosphate, and 6 wt% nitrates. The fermentor is filled
with 6000 mL broth and some antibiotic-producing fungi. Additional
broth is added continuously to the fermenter at a rate of 200 mL/h. The
cells consume glucose at a rate of 35 g/h, phosphates at the rate of 13 g/h,
and nitrates at the rate of 12 g/h. The fermentation is stopped when
the concentration of one of these three nutrients goes to zero (because
the cells can no longer grow). Which nutrient will be depleted first?
How long will the fermentation run? What is the concentration (g/L) of
the other two nutrients in the fermenter at the end of the run?
P2.92 Titanium dioxide (TiO2) is by far the most widely used pigment in white
paint. Specifications for the white pigment powder used in paint making
require that it contain at least 60% TiO2, 5% ZnO, and 25% fillers (such
as SiO2 or CaCO3).
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Chapter 2 Process Flows: Variables, Diagrams, Balances
The following powders are available from suppliers. Sketch out a
block flow diagram for mixing these powders to make white pigment
powder.
Powder A
(wt%)
TiO2
Powder B
(wt%)
Powder C
(wt%)
Powder D
(wt%)
90 0 0 0
ZnO 0
50 0 0
CaCO3 0
50
NaCl
75 0
10
0
25
10
SiO2 0 0 0
90
Come up with a recipe using these powders for making 1000 kg
white pigment powder that meets the specifications. What is the lowest wt% NaCl product you could make? What is the highest wt% NaCl
product?
P2.93 Portland cement is made by mixing a variety of raw materials, grinding
them, then heating the mixture in a kiln. Lots of changes occur in the
kiln, including evaporation of water and decomposition of magnesium
and calcium carbonates. Ignitable materials are burned and exit the kiln
as gases. The cement leaving the kiln must contain 21 wt% SiO2 and
65 wt% CaO. The Fe2O3 content must be between 1 and 5 wt%.
The compositions and costs of the raw materials available are listed
below:
mur83973_ch02_061-154.indd 150
Limestone
(wt%)
Clay
(wt%)
Mill scale
(wt%)
Oyster shells
(wt%)
SiO2
1
68
0
1
CaO
54
0
0
54
Al2O3
0
20
0
0
Fe2O3
0
4
100
0
MgO
4
0
0
0
Ignitable material
41
8
0
45
Cost ($/ton)
88
35
Free
60
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Chapter 2 Problems
151
Your job is to develop a process for making 100 metric tons/day of
Portland cement. Come up with the best recipe for mixing the raw
materials to make a cement product that meets all the specifications.
Calculate the raw materials cost per ton of cement produced ($/ton).
P2.94 In petroleum refining, crude oil is separated into several different mixtures of hydrocarbons. One product stream is called the light alkanes,
which contain mainly methane, ethane, propane, butane, and pentane.
In one facility, the light alkane stream [1000 kgmol/h, 10 mol% methane
(M), 30 mol% ethane (E), 15 mol% propane (P), 30 mol% butane (B),
and 15 mol% isopentane (I)] is processed to produce five different product streams. The separation is done in a series of distillation columns,
as shown. 71.3% of the propane in stream 1 is sent to stream 2. 92.9%
of the propane in stream 3 is sent to stream 4. 99.5% of the methane in
stream 5 is sent to stream 6. Identify (a) basis, (b) all stream composition specifications, and (c) all system performance specifications. Write
material balance equations, using each unit in turn as the system. (Use
6 systems in total: 5 separators and 1 mixer.) Report the total number
of material balance equations. Calculate flow rates (kgmol/h) and compositions (mol%) for all streams. Report your results in both mole and
mass units, using a table format, with stream numbers as column headings and components as row headings.
6
M
E
P
2
M
E
P 1
B
I
M
E
P
Mixer
M
E 0.5%
8
Separator
5
7
Separator
M
E
P 2%
Separator
P 4
3
P
B
I
9
Separator
11
E 1%
P
P
B
10
P
B
I
Separator
12
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B 2%
I
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Chapter 2 Process Flows: Variables, Diagrams, Balances
P2.95 The sugar cane industry is a big industry on Hawaii. Raw sugar cane is
first cut from the fields, then chopped and shredded. The raw cane contains
15 wt% sucrose, 25 wt% solids, and water, along with some additional
impurities that can be assumed to have negligible wt%. To produce raw
sugar for shipment to California, the chopped sugar cane is mixed with
some water and macerated in a mill. About 93% of the sugar juice in the
cane is recovered in the mill. The spent cane (called bagasse) contains
about 20 wt% water and is burned for fuel along with the unrecovered
juice. The recovered juice is sent to a clarifier, where lime is added to
precipitate impurities. You can assume that all the lime precipitates with
the settled-out “mud,” and that essentially no juice is lost to the mud. The
juice leaving the clarifier, which contains 85 wt% water, is sent to an
evaporator, where the water content of the juice is reduced to 40 wt%. The
thickened juice is sent to an evaporator/crystallizing pan, where more water
is removed and sugar crystals start to form. The crystals and syrup leaving
the pan contain 10 wt% water. The crystals are separated from the syrup
in a centrifuge. The raw sugar crystals are 97.8 wt% sucrose, and the syrup
(called blackstrap molasses) is 50 wt% sugar.
Assume that 10,000 lb/h of sugar cane is processed. Sketch a simplified process flow diagram. Calculate the flow rate of raw sugar crystals and molasses, the flow rate of water added to the cane fed to the
mill, and the flow rate of water removed in the evaporator.
Game Day
P2.96 Methanol (CH3OH), originally made by distillation of wood, is now
typically synthesized from methane and water. It is an important commercial solvent and feedstock for production of other chemicals such as
formaldehyde.
A typical methanol manufacturing facility has two reactors. In the
first reactor, called a steam reformer, two reactions occur:
​CH​4​​ + ​H​2​​O → CO + 3​H2​ ​​
(R1)
​CH​4​​ + 2​H2​ ​​O → ​CO​2​​ + 4​H2​ ​​
(R2)
The steam:methane molar ratio fed to the reformer is typically 2:1 to
suppress unwanted side reactions. The relative importance of (R1) and (R2)
depends on the reactor temperature: high temperatures (1500–1800°F)
favor (R1) while lower temperatures (600–700°F) favor (R2). Typical
reactor pressure is 300 psig. At these conditions, all the methane is
converted to products.
In the methanol synthesis reactor, two reactions take place over a catalyst at high pressures (4500 psig) and moderate temperatures (500–600°F):
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Chapter 2 Problems
CO + 2​H2​ ​​ → ​CH​3​​OH
(R3)
​​CO​2​​ + ​​3H​2​​ → ​​CH​3​​OH + ​​H2​ ​​O
(R4)
About 15% of the CO/CO2 fed to the reactor is converted to methanol.
The methane available is contaminated with 2% N2. Methanol and
water can easily be separated from CO, CO2, N2, and H2 by cooling and
condensing. CO and CO2 can be separated from H2 and N2 by absorption. It is very expensive to compress gases to the high pressures
required for the second reactor.
Your job is to design a process for making 60 million lbs/year
methanol based on this information. First sketch out an input-output
diagram and calculate the overall feed rate of steam and methane. Then
sketch out a block flow diagram showing your preferred design. Indicate
the chemical species in each stream. Write one or two paragraphs
explaining why you think your block flow diagram is superior to other
alternative arrangements. Indicate whether you would choose to run the
steam reformer at high or low temperatures. Finally, sketch out a preliminary process flow diagram, including pumps, compressors, and heat
exchangers. You do not have to calculate process flows.
P2.97 Mr. Big, a senior executive at ABC Industrial Alcohols Inc., has decided
that new products are needed for continuing corporate growth. The market for ethyl acetate (a commercial solvent) is expanding, and your
group, as the process engineering team for ABC, has been assigned the
job of coming up with a process for production of 700 million lb/yr
ethyl acetate (assuming 330 days/yr operation). After an initial discussion with company chemists and environmental engineers, you’ve come
up with the following information:
Raw materials available:
Solution containing 70 mol% ethanol and 30 mol% water.
Air (79 mol% nitrogen, 21 mol% oxygen).
Reaction pathways available:
Oxidation of ethanol to produce acetic acid and water
C2H5OH + O2 → CH3COOH + H2O
Reaction takes place at high pressure in the vapor phase over a catalyst.
At least 50 mol% nitrogen is needed in the feed as a diluent. Ethyl
acetate must not be present in the feed. Water can be present in the
feed. Oxygen must be in a 20% excess of the stoichiometric amount
to allow for complete consumption of the ethanol.
Esterification of ethanol and acetic acid to produce ethyl acetate
C2H5OH + CH3COOH → CH3COOC2H5 + H2O
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Chapter 2 Process Flows: Variables, Diagrams, Balances
Reaction takes place in liquid solution at ambient conditions. Only
60% conversion can be achieved because of equilibrium constraints.
Oxygen is prohibited in the feed. Water and nitrogen are allowed
contaminants.
Waste products:
No acetic acid, ethanol, or ethyl acetate is allowed to leave the plant
in a waste stream.
Come up with a process flow sheet for ethyl acetate production, using
this information. You should submit your design in the form of a
proposal to Mr. Big. Your report should include: (1) a 1 to 2 page
executive summary describing the key features of your process, the
assumptions that went into your design, any uncertainties or additional
information you would need to finalize the design, and what action you
recommend Mr. Big should take, (2) completed flow sheet showing all
flows and compositions that you were able to specify, and (3) an appendix showing detailed supporting calculations. (Mr. Big is too busy to
look at this, but documentation of your results would be important for
any follow-on engineering work.)
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CHAPTER THREE
3
Mathematical Analysis of
Material Balance Equations and
Process Flow Sheets
In This Chapter
We revisit some key ideas from Chap. 2, but take a more mathematical and
rigorous approach. We develop generalized expressions for the material balance equation, and use these expressions to solve problems of increasing complexity. We construct linear models of steady-state process flow sheets, and
illustrate how to determine if a process flow sheet is correctly specified.
Some questions we address include:
∙
∙
∙
∙
What is a mathematical way to write the material balance equation?
How do I handle transients in the material balance equation?
What are useful specifications for performance of process units?
How do I develop a system of linear independent equations to describe a
process flow sheet?
∙ How can I determine whether a system of equations has a solution?
Words to Learn
Watch for these words as you read Chapter 3.
Material balance equations
Extent of reaction
Linear equations
Linear model
Fractional split
Fractional conversion
Fractional recovery
Degree of freedom analysis
155
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
3.1
Introduction
Chemical process synthesis is evolutionary. We start with the basics, as we did
in Chap. 1, asking: What product do I want to make? What raw materials are
available? What reaction chemistries are feasible? Then we move on to sketches
of simple block flow diagrams, as we did in Chap. 2. We identify the key
process units required, and we consider how to connect these units together.
We make simplifying approximations, quickly complete process flow calculations,
and make preliminary assessments of alternative arrangements.
Once one (or a few) preliminary block flow diagrams have been sketched,
we move further into the details. At this stage in the tour de process synthesis,
we continue to work with block flow diagrams. We include more realistic
specifications of stream composition and system performance. We consider
time-varying processes. We determine if we have sufficient information to
completely solve a process flow problem. We develop methods that allow
examination of how the overall performance of a process might change with
changes in specific process variables or specifications. In short, we take a more
rigorous, mathematical, and systematic look at process flow calculations.
3.2
The Material Balance Equation—Again
In Chap. 2, we introduced the material balance equation as
Input − Output + Generation − Consumption = Accumulation
Eq. (2.5)
You learned the importance of clearly choosing a system, identifying components, and defining stream and system variables. In this section we will revisit
Figure 3.1 Until the advent of pocket calculators in the 1970s, engineers relied on tools
like slide rules and their own numerical prowess. Modern scientific calculators and personal
computers place incredible calculational power at the engineer’s fingertips, making it much
easier to find mathematically rigorous solutions to engineering problems.
Source: Left: richcano/Getty Images; Right: George Marks/Retrofile/Getty Images
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Section 3.2 The Material Balance Equation—Again
157
these ideas with four goals in mind: (1) to reiterate the importance of mastering these concepts, (2) to develop more complete and rigorous expressions for
the material balance equation, (3) to describe a systematic method for determining if a process calculation problem is solvable, and (4) to illustrate use
of the material balance equation in solving more challenging process flow
problems.
3.2.1
Stream Variables
Recall from Chapter 2 that a system is a specified volume with well-defined
boundaries: a system is three-dimensional, and it is enclosed by a surface area.
If material enters or leaves the system, it does so in streams that cross the
system boundary. Stream variables describe the quantity or flow rate of a
material in a stream.
In the mathematical approach we employ in Chapter 3, a stream variable
is denoted using a very compact nomenclature, which indicates the dimensions,
and identifies the component and the stream. Stream variables might have
dimension of mass or of moles, which we indicate using m or n as
​
m [mass]​
​
n [moles]​
Alternatively, stream variables might have dimension of mass per unit time,
or moles per unit time. We indicate flow rates by placing a “dot” above the
letter:
​​ṁ ​ [mass/time]​
​​n ​ ̇ [moles/time]​
We use subscripts i and j to identify the component and stream, respectively:
​
i ≡ component​
​
j ≡ stream​
For example, the compact notation ​​​ṁ ​​ij​​​indicates the mass flow rate of component i in stream j.
Mole and mass stream variables for a particular compound are related to
each other by the compound’s molar mass M​
​​ i​​​:
​​m​ ij​​ = ​Mi​  ​​ ​ni​  j​​​
​​​ṁ ​​ij​​ = M
​ i​  ​​ ​​n ​​i̇ j​​​
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Eq. (3.1)
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Illustration: A gas mixture of nitrogen N2 (N ) and hydrogen H2 (H ) is fed
to a reactor in Stream 1. The nitrogen flow rate in Stream 1 is 150 gmol/h and
the hydrogen flow rate is 600 gmol/h.
​​​n ​​Ṅ 1​​= 150 gmol/h​
28 g
​​​ṁ ​​N1​​ = M
​ N​  ​​ ​​n ​​Ṅ 1​​ = ​ ​  _​ (​​ 150 gmol/h)​= 4200 g/h​
( gmol )
​​​n ​​Ḣ 1​​= 600 gmol/h​
2g
​​​ṁ ​​H1​​ = ​MH​  ​​ ​​n ​​Ḣ 1​​ = ​ ​  _​ ​​(600 gmol/h)​= 1200 g/h​
( gmol )
The total quantity or flow rate in a stream is indicated by eliminating the subscript for a component. For example, ​​n​ j​​​denotes the total moles in stream j.
In any given stream j, the sum of all of the individual components will equal
the total. Mathematically, this is expressed as
​​ ∑ ​​​ ​mi​  j​​ = ​mj​  ​​, ​ ∑ ​​​ ​ni​  j​​ = ​nj​  ​​​
all i
all i
​​ ∑ ​​​ ​​ṁ ​​ij​​ = ​​ṁ ​​j​​, ​ ∑ ​​​ ​​n ​​i̇ j​​ = n​​  ​​j̇ ​​​
all i
Eq. (3.2)
all i
where the summation is taken over all components.
Illustration: A gas mixture of nitrogen N2 (N ) and hydrogen H2 (H ) is fed
to a reactor in Stream 1. The nitrogen flow rate in Stream 1 is 150 gmol/h and
the hydrogen flow rate is 600 gmol/h.
​​​n ​​Ṅ 1​​= 150 gmol/h​
​​​n ​​Ḣ 1​​= 600 gmol/h​
​​​n ​​1̇ ​​ = n​​  ​​Ṅ 1​​ + ​​n ​​Ḣ 1​​= 750 gmol/h​
In process flow calculations, it is frequently useful to use mass or mole
fractions. Unless otherwise specified, we will use w​
​​ ij​​​for mass fraction and
​​zi​  j​​​ for mole fraction:
​​m​ ij​​ = ​wi​  j​​ ​mj​  ​​, ​n​ ij​​ = z​ i​  j​​ ​nj​  ​​​
Eq. (3.3)
​​​ṁ ​​ij​​ = ​wi​  j​​ ​​ṁ ​​j​​, ​​n ​​i̇ j​​ = z​ i​  j​​ ​​n ​​j̇ ​​​
Notice that the mass or mole fractions in a given stream j must sum up to one:
​​ ∑ ​​​ ​wi​  j​​ = 1, ​ ∑ ​​​ ​zi​  j​​ = 1​
all i
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all i
Eq. (3.4)
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Section 3.2 The Material Balance Equation—Again
159
Illustration: Stream 1 is a solution of glucose in water. The mass fraction of
glucose (G) is 0.12 and the total mass of the solution is 100 g.
​​wG​  1​​ = 0.12​
​​m1​  ​​= 100 g​
​​mG​  1​​= 0.12(100 g) = 12 g​
Illustration: Stream 2 contains 12 mol% glucose (G) and 3 mol% sodium
chloride (S) along with some water (W).
​​zG​  2​​ = 0.12​
​​zS​  2​​ = 0.03​
​​zW
​  2​​= 1 − 0.12 − 0.03 = 0.85​
3.2.2
Helpful Hint
Never use m
​​​  ​​ ̇ i,sys​​​,
​​​m ​​ ̇ sys​​​,​ n
​​  ​​ i̇ ,sys​​​, or n
​​​  ​​ ṡ ys​​​.
Mass and molar
flow rates are used
only for streams,
never for systems.
System Variables
Recall from Chapter 2 that system variables describe a change in quantity of
a material inside a system. There are two ways in which a change in quantity
can occur inside a system: accumulation, and generation or consumption by
chemical reaction.
In the mathematical notation that we employ in Chapter 3, the quantity
inside a system is indicated by the subscript sys:
​
sys ≡ system​
The quantity inside a system can have units of mass or moles, and can identify
the quantity of a specific component, or the total quantity. For example, ​​m​ i,sys​​​
is the mass of component i in the system, while m​
​​ sys​​​is the total mass of all
components in the system.
System boundary
Stream 1
Stream 2
ṁA1
ṁB1
mA,sys
mB,sys
ṁA2
ṁB2
Figure 3.2 (Top) A system containing A (larger shaded spheres) and B (smaller dark spheres),
with one stream carrying material into the system and another stream carrying material out of
system. (Bottom) Block flow representation.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
3.2.2.1
eneration and Consumption by Chemical Reaction, and
G
Extent of Reaction Concept
As you learned in Chapter 2, compounds can be generated or consumed inside
a system by chemical reaction. Just like flows across system boundaries, generation or consumption by chemical reaction can be described in mass or moles,
or in mass per time or moles per time. To indicate which, we employ the following notation:
​
R [mass]​
​
r [moles]​
​​R ​ ̇ [mass/time]​
​​r ​ [̇ moles/time]​
R and r are the quantities generated or consumed, whereas ​​R ​​̇ and ​​r ​​̇ are the
rates of reaction.
We wish to indicate the specific compound that is generated or consumed,
and, for systems where multiple reactions can occur, we wish to identify the
specific reaction of interest. We use subscripts i and k to identify the component
and reaction, respectively:
​
i ≡ component​
​
k ≡ reaction​
For example, ​​R​ A2​​​is the mass of compound A that is generated or consumed
in reaction 2, and r​​​ ​​Ḃ 1​​​is the molar rate of generation or consumption of compound B by reaction 1. How do we know whether the compound is generated
or consumed? This is indicated simply by sign: for example, R​
​​ A2​​​ is positive if
A is generated and negative if A is consumed by reaction 2. (As you learned
in Chapter 2, components can be elements, compounds, or composite materials.
In general, we use subscript i to identify a component. However, because only
compounds are generated or consumed by reaction, the discussion in this section is restricted to the case where the component is a compound.)
Quick Quiz 3.1
In this ammonia
synthesis illustration,
what is the sum of
​​​RN​​̇ 1​​ + R
​​ H​​̇ 1​​ + R
​​ A​​̇ 1​​​?
What is the sum of
​​​r ​​ Ṅ 1​​ + r​​  ​​ Ḣ 1​​ + ​​r ​​ Ȧ 1​​​?
Explain why the sum
is zero in mass units
but not in mole units.
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Illustration: A gas mixture of nitrogen N2 (N ) and hydrogen H2 (H ) is fed
to a reactor in Stream 1. The nitrogen flow rate in Stream 1 is 150 gmol/h and
the hydrogen flow rate is 600 gmol/h. Within the reactor, 50 gmol/h N2 and
150 gmol/h H2 are consumed by reaction to make 100 gmol/h ammonia (A)
by the reaction (R1)
​​N2​ ​​​ + 3H​2​​ → ​2NH​3​​​
​​​r ​​Ṅ 1​​= −50 gmol/h​
​​​r ​​Ḣ 1​​= −150 gmol/h​
​​​r ​​Ȧ 1​​= 100 gmol/h​
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Section 3.2 The Material Balance Equation—Again
161
Illustration: In the first reaction step in the synthesis of the painkiller ibuprofen, 134 g acetic anhydride (A) is mixed with 134 g isobutylbenzene (B) in a
pot. 93.8 g of A and 123.3 g of B are consumed by reaction, to make 161.9 g
isobutylacetophenone (C ) and 55.2 g acetic acid (D).
​​RA​  1​​= −93.8 g​
​​RB​  1​​= −123.3 g​
​​RC​  1​​= +161.9 g​
​​RD​  1​​= +55.2 g​
There are two useful points to make. First, the mass quantity or rate of reaction
of a compound i in reaction k is related to the molar quantity or rate of reaction
of that compound through its molar mass Mi:
​​R​ ik​​ = ​Mi​  ​​ ​ri​  k​​​
​​​R ​​i̇ k​​ = ​Mi​  ​​ ​​r ​​i̇ k​​​
Eq. (3.5)
Second, the quantity or rate of reaction of one compound in a reaction is related
to that of another compound in the same reaction, by the stoichiometric coefficients. For example, suppose the stoichiometrically balanced chemical reaction is
​2A → 3B​
For every 2 moles of A consumed, 3 moles of B are generated, or
​rA​  ​​ _
−2
​​ _
​rB​   ​​​ = ​  3 ​​
More generally, for any two compounds A and B that take part in a chemical
reaction k,
​rA​  k​​ _
​νA​  k​​
​​ _
​rB​  k ​​​ = ​  ​νB​  k ​​​​
where νik is the stoichiometric coefficient for compound i in reaction k, νik is
negative for reactants and positive for products. We can rewrite and generalize as
​rA​  k​​ _
​rB​  k​​ _
​ri​  k​​
​​ _
​νB​  k ​​​ = ​  ​νB​  k ​​​ = ​  ​νi​  k ​​​​
This ratio, of reaction rate divided by stoichiometric coefficient, is so useful
that it has a name: the extent of reaction ξ:
​ri​  k​​
​​ _
​νi​  k ​​​ = ​ξk​  ​​​
Eq. (3.6a)
The units of ξ​ ​(moles) are the same as those of the reaction r​ ​​ik​​​.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
We can apply the same logic to rates of reaction, where the “dot” indicates
per unit time.
​​r ​​i̇ k​​
̇
​​ _
​ν​   ​​​ = ​​ξ ​​k​​​
ik
Eq. (3.6b)
The units of ξ​​ ̇ ​​(moles/time) are the same as those of the reaction rate r​​​ ​​i̇ k​​​.
Helpful Hint
The extent of reaction has only one
subscript, to indicate the reaction k,
and never a subscript to indicate a
compound or a
stream!
Illustration: In the first step of synthesis of ibuprofen, 134 g acetic anhydride
(A, MA = 102 g/gmol) is mixed with 134 g isobutylbenzene (B, MB = 134 g/gmol)
in a pot. 93.8 g of A and 123.3 g of B are consumed by reaction, to make
161.9 g isobutylacetophenone (C, MC = 176 g/gmol) and 55.2 g acetic acid
(D, MD = 60 g/gmol).
​​R​  A1​​= −93.8 g, ​r​  A1​​ = ​R​  A1​​∕​M​  A​​ = ​(−93.8 g)​∕(​ 102 g/gmol)​= −0.92 gmol​
​​R​  C1​​= 161.9 g, ​r​  C1​​ = ​R​  C1​​∕M
​ ​  C​​ = ​(161.9 g)​∕​(176 g/gmol)​= 0.92 gmol​
(Calculate r​​ B​  1​​​ and r​​ D​  1​​​ on your own.)
Illustration: A gas mixture of nitrogen N2 (N ) and hydrogen H2 (H) is fed to
a reactor in Stream 1. The nitrogen flow rate in Stream 1 is 150 gmol/h and
the hydrogen flow rate is 600 gmol/h. Within the reactor, 50 gmol/h N2 and
150 gmol/h H2 are consumed by reaction to make 100 gmol/h ammonia (A) by
the reaction (R1)
​​N2​ ​​​ + 3H​2​​ → ​2NH​3​​​
​​​ξ ​​1̇ ​​ = ​​r ​​Ṅ 1​​∕​νN​  ​​ = ​(−50 gmol/h)​∕​(−1)​= 50 gmol/h​
​​​ξ ​​1̇ ​​ = r​​  ​​Ḣ 1​​∕ν​ H​  ​​ = ​(−150 gmol/h)​∕(​ −3)​= 50 gmol/h​
Quick Quiz 3.2
For the reaction
2C2H4 + O2 →
2C2H4O, if
​​​re​​̇ thylene​​= −4 gmol/s,
what is ξ​​ ​​?̇
For the reaction
C2H4 + 0.5O2 →
C2H4O, if
​​​r​​ė thylene​​= −4 gmol/s,
what is ξ​​ ​​?̇
​​​ξ ​​1̇ ​​ = ​​r ​​Ȧ 1​​∕ν​ A​  ​​ = ​(100 gmol/h)​∕​(+2)​= 50 gmol/h​
The extent of reaction is a very useful number because it links together the
consumption and generation of all compounds that participate in the reaction.
Note that the value of ​​ξ ​​̇ depends on the way in which the stoichiometrically
balanced equation is written. For 2A → 3B, if ξ​​  ​​̇ = 2 gmol/s then r​​​ ​​Ḃ ​​ = 6 gmol/s,
but for A → _​​ 32 ​​ B and ξ​​  ​​̇ = 2 gmol/s, then r​​​ ​B​̇ ​​ = 3 gmol/s.
3.2.2.2
Accumulation
Accumulation is the net change in the material inside the system, due to material crossing into and out of the system plus any generation or consumption by
chemical reactions occurring within the system. Accumulation can have units
of either mass or moles, and can apply to either a specific component, or the
total material. Just as with generation or consumption, accumulation can be
expressed either in quantities, or in rates.
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Section 3.2 The Material Balance Equation—Again
163
Imagine a bucket that initially contains 3 pounds of potatoes. Somebody
adds some more potatoes. At the end of this, the bucket contains 11 pounds
of potatoes. The bucket has accumulated (11 − 3) or 8 pounds of potatoes.
Accumulation as a quantity is the difference between the final amount and the
initial amount. To denote this difference, we need to add another subscript that
indicates the specific time at which the quantity is ascertained; for example,
we can use the subscript f to indicate the final time (or finish of operation) and
0 to indicate the initial time (or beginning of operation). In mathematical notation, accumulation is expressed as this difference, or
​​m​ i,sys, f​​ − ​mi​  ,sys,0​​​
(change in mass of component i)
​​m​ sys, f​​ − ​ms​  ys,0​​​
(change in total mass)
​​n​ i,sys, f​​ − n​ i​  ,sys,0​​​
(change in moles of component i)
​​n​ sys, f​​ − ​ns​  ys,0​​​
(change in total moles)
Be sure you understand the difference between, e.g., m​
​​ sys​​​(the mass in the
system) versus m​
​​ sys, f​​ − ​ms​  ys,0​​​(the change in mass in the system over time).
At steady state, there is no change in the system over time and accumulation
(​​ms​  ys, f​​ − m
​ s​  ys,0​​​) is zero, but m
​​ s​  ys​​​ can be nonzero.
Now imagine the bucket again, initially containing 3 pounds of potatoes.
Someone adds 2 potatoes to the bucket every minute. The rate of accumulation
is 2 potatoes/minute. Mathematically this rate is expressed as a derivative:
Helpful Hint
Express accumulation as a difference
when you are interested in what happens over a period
of time. Express
accumulation as a
derivative when
you are interested
in a single point in
time.
Helpful Hint
Never use m
​​​  ​​̇ i,sys​​​,
​​​m ​​̇ sys​​​, n
​​​  ​​i̇ ,sys​​,​ or n
​​​  ​​ṡ ys​​​ for
accumulation. The
rate of accumulation is expressed as
the derivative of
mass or moles with
respect to time.
mur83973_ch03_155-230.indd 163
d​mi​  ,sys​​
​​ _​​
dt
(rate of change in mass of component i)
d​ms​  ys​​
​​ _​​
dt
(rate of change in total mass)
d​ni​  ,sys​​
​​ _​​
dt
(rate of change in moles of component i)
d​ns​  ys​​
​​ _​​
dt
(rate of change in total moles)
At steady state, there is no change in the system over time and accumulation
d​m​  ​​
(e.g., ___
​​  dtsys ​​) is zero, but the total quantity in the system (e.g., m​
​​ sys​​​) can be nonzero.
Illustration: Water flows into a tank at a rate of 14 gmol/s and out of the
tank at a rate of 12 gmol/s.
​​​n ​​Ẇ ,in​​= 14 gmol/s​
​​​n ​​Ẇ ,out​​= 12 gmol/s​
d​nW
​  ,sys​​
_
​​
​= 2 gmol/s​
dt
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Illustration: A bucket initially contains 10 g sucrose. Susy Sweet-tooth
pours 100 g water into the bucket, and stirs. At the end of the process, the sugar
is completely dissolved in the water, and the bucket contains 110 g of a sweet
solution.
​​mS​  ,sys,0​​= 10 g, ​m​ S,sys, f​​= 10 g​
​​mS​  ,sys, f​​ − ​mS​  ,sys,0​​= 10 − 10 = 0 g​
​​mW
​  ,sys,0​​= 0 g, ​m​ W,sys, f​​= 100 g​
​​mW
​  ,sys, f​​ − ​mW
​  ,sys,0​​= 100 − 0 = 100 g​
​​ms​  ys,0​​= 10 g, ​m​ sys, f​​= 110 g​
​​ms​  ys, f​​ − ​ms​  ys,0​​= 110 − 10 = 100 g​
3.2.3
The Differential Material Balance Equation: Molar Units
Recall from Chapter 2 that the material balance equation is
Input − Output + Generation − Consumption = Accumulation
Eq. (2.5)
Now we have the notation we need to write the familiar material balance equation in mathematical format. In this section we will derive the differential (or
rate-based) material balance equation. In other words, we will replace the
words in Eq. (2.5) with notation indicating the rate of flow in, the rate of flow
out, the rate of generation or consumption by chemical reaction and the rate
of accumulation. In molar units, with one input stream ( j = in), one output
stream ( j = out), and one reaction (k = 1), we write the differential material
balance equation for compound A as
d​nA​  ,sys​​
​​​n ​​Ȧ in​​ − ​​n ​​Ȧ out​​ + r​​  ​​Ȧ 1​​ = ​ _​​
dt
Quick Quiz 3.3
Write the differential
material balance
equation in mass units
for compound C,
considering two input
streams ( j = 1 and
j = 2), one output
stream ( j = 3), and
no reactions.
mur83973_ch03_155-230.indd 164
Eq. (3.7a)
If we wish, we can use the extent of reaction notation rather than the reaction
rate notation, or
d​nA​  ,sys​​
​​​n ​​Ȧ in​​ − n​​  ​​Ȧ out​​ + ​νA​  1​​ ​​ξ ​​1̇ ​​ = ​ _​​
dt
Eq. (3.7b)
Be sure you can see the direct correspondence between Eq. (2.5) and Eq. (3.7).
Notice that all of the variables on the left-hand side of Eq. (3.7) indicate rate
by putting a “dot” above the variable, but accumulation on the right-hand side
is expressed on a rate basis by a derivative.
Eq. (3.7) applies to one specific compound, but if we have two compounds
we write two Eq. (3.7) twice, one for each compound. For example, suppose
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Section 3.2 The Material Balance Equation—Again
we have compounds A and B, one input stream ( j = in), one output stream
( j = out), and one reaction (k = 1). Then:
d​nA​  ,sys​​
​​ _​ = n​​  ​​Ȧ in​​ − ​​n ​​Ȧ out​​ + ​νA​  1​​ ​​ξ ​​1̇ ​​​
dt
Eq. (3.8a)
d​nB​  ,sys​​
​​ _​ = ​​n ​​Ḃ in​​ − ​​n ​​Ḃ out​​ + ν​ B​  1​​ ​​ξ ​​1̇ ​​​
dt
Eq. (3.8b)
Furthermore we can add these two equations together:
d​nA​  ,sys​​ d​nB​  ,sys​​
​​ _​ + _
​
​ = ​​n ​​Ȧ in​​ + ​​n ​​Ḃ in​​ − ​(​​n ​​Ȧ out​​ + ​​n ​​Ḃ out​​)​+ ν​ A​  1​​ ​​ξ ​​1̇ ​​ + ν​ B​  1​​ ​​ξ ​​1̇ ​​​
dt
dt
But ​​nA​  ,sys​​ + n​ B​  ,sys​​ = ​ns​  ys​​, n​​  ​​Ȧ in​​ + n​​  ​​Ḃ in​​ = ​​n ​​i̇ n​​​, and so forth, so we can derive
d​ns​  ys​​
​​ _​ = n​​  ​​i̇ n​​ − ​​n ​​ȯ ut​​ + ​(​νA​  1​​ + ​νB​  1​​)​ ​​ξ ​​1̇ ​​​
dt
Eq. (3.8c)
Eqs. (3.8a) and (3.8b) are balances on compounds A and B, and Eq. (3.8c) is
a total balance. Notice that we derived Eq. (3.8c) from (3.8a) and (3.8b) from
simply knowing that compounds A and B are the only compounds in this
system. Any equation can be derived from the other two; only two of the three
equations are independent. We use whichever equations are most convenient
for a specific problem.
Example 3.1
Decomposition Reactions
N2O4 is fed to a reactor at a flow rate of 84 gmol/min, where some of it decomposes to NO2. The reactor operates at steady state. The stream leaving the reactor
flows at 126 gmol/min. What is the extent of reaction? What is the molar flow
rate of each component in the reactor outlet stream?
Solution
We start as always by sketching and labeling a flow diagram.
N2O4
n1 = 84 gmol/min
Reactor
N2O4
NO2
n2 = 126 gmol/min
The stoichiometrically balanced reaction is
​​N2​ ​​​O4​ ​​ → 2​NO​2​​​
If N2O4 is identified as compound A and NO2 as B, then νA = −1 and νB = 2.
We start with Eq. (3.8c). Since the reactor operates at steady state, there is no
change in the number of moles in the system, or dnsys∕dt = 0. Using the information
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
provided, that ​​n​​1̇ = 84 gmol/min, and ​​n​​2̇ =126 gmol/min, we substitute into
Eq. (3.14) to find:
​0 = 84 − 126 + (−1 + 2) ξ​  ​​̇
​​ξ ̇ ​= 42 gmol/min​
To find the molar flow rate of each component in the outlet stream we return to
Eq. (3.8a–b), applied first to N2O4 and then to NO2.
d​nA​  ,sys​​
​​ _​ = ​​n ​​Ȧ 1​​ − ​​n ​​Ȧ 2​​ + ​νA​  ​​​ξ ​​̇
dt
​0 = 84 − ​​n ​​Ȧ 2​​ − 42​
​​​n ​​Ȧ 2​​= 42 gmol/min​
d​nB​  ,sys​​
​​ _​ = ​​n ​​Ḃ 1​​ − n​​  ​​Ḃ 2​​ + ν​ B​  ​​​ξ ​​̇
dt
​0 = 0 − ​​n ​​Ḃ 2​​ + 2(42)​
​​​n ​​Ḃ 2​​= 84 gmol/min​
More generally, there can be many components, many streams in and out of a
system, and many reactions occurring in the system (Fig. 3.3). Suppose for
example that compound A flows into a system through two streams, stream 1
and stream 2 ( j = 1 and j = 2), and out of the system in stream 3 and stream
4 ( j = 3 and j = 4). Suppose also that compound A is consumed by two reactions, reaction R1 and reaction R2 (k = 1 and k = 2). We sum up all the flows
A+B
→C+
D
2A → E + F
D+
G
F→
+H
Figure 3.3 In the general case, many streams enter and leave the system, carrying many
components. Multiple reactions occur within the system.
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Section 3.2 The Material Balance Equation—Again
167
in and out, and the rates of consumption and generation to write the differential
material balance equation for compound A:
Input − Output + Generation − Consumption = Accumulation
Eq. (2.5)
d​nA​  ,sys​​
​​(​​n ​​Ȧ 1​​ + n​​  ​​Ȧ 2​​)​− ​(​​n ​​Ȧ 3​​ + n​​  ​​Ȧ 4​​)​+ ​(​​r ​​Ȧ 1​​ + r​​  ​​Ȧ 2​​)​ = _
​
​​
dt
Or, in terms of extents of reaction:
d​nA​  ,sys​​
​​(​​n ​​Ȧ 1​​ + ​​n ​​Ȧ 2​​)​− ​(​​n ​​Ȧ 3​​ + ​​n ​​Ȧ 4​​)​+ (​ ​νA​  1​​ ​​ξ ​​1̇ ​​ + ν​ A​  2​​ ​​ξ ​​2̇ )​​ ​ = _
​
​​
dt
To generalize further, for any component i, assuming many streams in (many
values of jin), many streams out (many values of jin), and many reactions (many
values of k), the differential material balance equation in molar units becomes
Quick Quiz 3.4
In Eq. (3.9), why is
there no summation
sign on d​ ​ni​  ,sys​​∕dt​?
And no “dot”?
Quick Quiz 3.5
How is ∑
​ all k ​νi​  k​​ ​​ξ ​​k̇ ​​​
related to the “net”
column in generationconsumption analysis
of Chapter 1?
Helpful Hint
For a system at
steady state,
d​ni​  ,sys​​ d​mi​  ,sys​​
​​  _​ = ​  _​​
dt
dt
=0
Example 3.2
d​ni​  ,sys​​
​​ _​ = ​  ∑ ​​​ ​​n ​​i̇ j​​ − ​  ∑ ​​​ ​​n ​​i̇ j​​ + ​  ∑ ​​​ ​​r ​​i̇ k​​​
dt
all j​i​  n​​
all j​o​  ut​​
all k
Eq. (3.9a)
or equivalently
d​ni​  ,sys​​
_
​​
​ = ​  ∑ ​​​ ​​n ​​i̇ j​​ − ​  ∑ ​​​ ​​n ​​i̇ j​​ + ​  ∑ ​​​ ​νi​  k​​ ​​ξ ​​k̇ ​​​
dt
all j​i​  n​​
all j​o​  ut​​
all k
Eq. (3.9b)
where the summation is taken over all streams or all reactions as indicated.
(We flipped the equation and put accumulation on the left-hand side, just to
put Eq. (3.8) in the conventional format for differential equations.) Eq. (3.9)
is the differential mole balance equation for component i. If there are a total
of I components, we can write I independent equations of the form of Eq. (3.9).
If the system is operating at steady state, there is no change with time and
the derivative is equal to zero. In this case, Eq. (3.9) is often rewritten as
​​  ∑ ​​​ ​​n ​​i̇ j​​ = ​  ∑ ​​​ ​​n ​​i̇ j​​ + ​  ∑ ​​​ ​​r ​​i̇ k​​​
Eq. (3.10a)
or equivalently
​​  ∑ ​​​ ​​n ​​i̇ j​​ = ​  ∑ ​​​ ​​n ​​i̇ j​​ + ​  ∑ ​​​ ​νi​  k​​ ​​ξ ​​k̇ ​​​
Eq. (3.10b)
all j​o​  ut​​
all j​o​  ut​​
all j​i​  n​​
all j​i​  n​​
all k
all k
Eq. (3.10) is the steady-state differential mole balance equation for component i.
Differential Mole Balances: Manufacture of Urea
Urea, (NH2)2CO, is a widely used fertilizing agent made from ammonia. (It is also
a component of urine, made during metabolism of proteins and amino acids.)
Commercially, urea is manufactured from ammonia and carbon dioxide:
​2​NH​3​​ + ​CO​2​​ → (​NH​2​​​)2​ ​​CO + ​H2​ ​​O​
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Two gas streams, one at 230 gmol/min and containing 85 mol% NH3 and 15 mol%
CO2, and the other at 100 gmol/min and containing 80 mol% CO2 and 20 mol%
H2O, are fed to a reactor operating at steady state. Inside the reactor, ammonia is
consumed at a rate of 180 gmol/min. (See sketch.) Use Eqs. (3.9) or Eq. (3.10) to
find the flow rate (gmol/min) of each component out of the reactor.
85% NH3
15% CO2
80% CO2
20% H2O
n1 = 230 gmol/min
Reactor
n3
n2 = 100 gmol/min
NH3
CO2
Urea
H2O
Solution
We’ll use subscripts A, C, U, and W to indicate components ammonia, carbon
dioxide, urea, and water, respectively. Streams are indicated by number, as shown
on the flow diagram. First we calculate the extent of reaction (R1), ​​​ξ ​​1̇ ​​,​ from r​​​  ​​Ȧ 1​​,​
the rate of consumption of ammonia by reaction (R1):
​​​r ​​Ȧ 1​​= −180 gmol/min = ν​ A​  1​​ ​​ξ ​​1̇ ​​ = (−2) ​​ξ ​​1̇ ​​​
​​​ξ ​​1̇ ​​= 90 gmol/min​
Noting that the reactor is operating at steady state, we use Eq. (3.10) to derive four
balance equations, one for each compound:
​​​n ​​Ȧ 3​​ = n​​  ​​Ȧ 1​​ + ​νA​  1​​​​ξ ​​1̇ ​​= (0.85)230 + (−2)90 = 15.5 gmol ​NH​3​​/min​
​​​n ​​Ċ 3​​ = ​​n ​​Ċ 1​​ + ​​n ​​Ċ 2​​ + ​νC​  1​​​​ξ ​​1̇ ​​= (0.15)230 + (0.8)100 + (−1)90 = 24.5 ​gmol CO​2​​/min​
​​​n ​​U̇ 3​​ = ​νU​  1​​​​ξ ​​1̇ ​​= (+1)90 = 90 gmol urea/min​
​​​n ​​Ẇ 3​​ = ​​n ​​​Ẇ 2​  ​​​​ + ​νW
​  1​​​​ξ ​​1̇ ​​= (0.2)100 + (+1)90 = 110 gmol ​H2​ ​​O/min​
Example 3.3
Differential Mole Balances: Urea Manufacture from Cheaper
Reactants
Urea can be manufactured from methane (CH4), water, and nitrogen via a pathway
requiring four chemical reactions:
​​CH​4​​ + ​H2​ ​​O → CO + 3​H2​ ​​​
(R1)
​CO + ​H​2​​O → ​CO​2​​ + ​H2​ ​​​
(R2)
​​N​2​​ + 3​H2​ ​​ → 2​NH​3​​​
(R3)
​​NH​3​​ + 0.5​CO​2​​ → 0.5​(​NH​2​​)​2​​CO + 0.5​H2​ ​​O​
(R4)
We’d like to design a process to make 90 gmol/min urea at steady state, from
the raw materials methane, water, and nitrogen. Furthermore, there should be no
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Section 3.2 The Material Balance Equation—Again
169
reactants (CH4, H2O, or N2) nor any CO, CO2, or NH3 leaving the process. (See
sketch.) What should be the reactant feed rates? What are the byproducts?
CH4
H2O
N2
n1
Process
n2
(NH2)2CO
H2
Solution
In Chap. 1, we would have used a generation-consumption analysis to solve this
problem. Here we’ll use differential mole balance equation instead. We’ll use subscripts M, W, CO, CD, H, N, A, and U to indicate components methane, water,
carbon monoxide, carbon dioxide, hydrogen, nitrogen, ammonia, and urea, respectively. Because the process is at steady state, ​d​ni​  ,sys​​∕dt = 0​for all components.
Eq. (3.10) simplifies to:
Helpful Hint
With flow rates n
​​ ​​i̇ j,
the second subscript is the stream
number. With
stoichiometric
coefficients νik, the
second subscript
is the reaction
number.
​0 = ​​n ​​Ṁ 1​​ + ​νM
​  1​​​​ξ ​​1̇ ​​ = ​​n ​​Ṁ 1​​ − ​​ξ ​​1̇ ​​​
​0 = ​​n ​​Ẇ 1​​ + ν​ W
​  1​​​​ξ ​​1̇ ​​ + ​νW
​  2​​​​ξ ​​2̇ ​​ + ​νW
​  4​​​​ξ ​​4̇ ​​ = ​​n ​​Ẇ 1​​ − ​​ξ ​​1̇ ​​ − ​​ξ ​​2̇ ​​ + 0.5​​ξ ​​4̇ ​​​
​0 = ν​ C​  O1​​​​ξ ​​1̇ ​​ + ν​ C​  O2​​​​ξ ​​2̇ ​​ = ​​ξ ​​1̇ ​​ − ​​ξ ​​2̇ ​​​
​0 = ν​ C​  D2​​​​ξ ​​2̇ ​​ + ν​ C​  D4​​​​ξ ​​4̇ ​​ = ​​ξ ​​2̇ ​​ − 0.5​​ξ ​​4̇ ​​​
​​​n ​​Ḣ 2​​ = ​νH​  1​​​​ξ ​​1̇ ​​ + ​νH​  2​​​​ξ ​​2̇ ​​ + ν​ H​  3​​​​ξ ​​3̇ ​​ = 3​​ξ ​​1̇ ​​ + ​​ξ ​​2̇ ​​ − 3​​ξ ​​3̇ ​​​
​0 = ​​n ​​Ṅ 1​​ + ν​ N​  3​​​​ξ ​​3̇ ​​ = n​​  ​​Ṅ 1​​ − ​​ξ ​​3̇ ​​​
​0 = ​νA​  3​​​​ξ ​​3̇ ​​ + ​νA​  4​​​​ξ ​​4̇ ​​ = 2​​ξ ​​3̇ ​​ − ​​ξ ​​4̇ ​​​
​​​n ​​U̇ 2​​= 90 gmol/min = ​νU​  4​​​​ξ ​​4̇ ​​ = 0.5​​ξ ​​4̇ ​​​
Now, we work backwards, first to find the extents of reaction and then to find the
flow rates:
​90 gmol/min = 0.5​​ξ ​​4̇ ​​
⇒​​ξ ​​4̇ ​​= 180 gmol/min​
​2​​ξ ​​3̇ ​​ = ​​ξ ​​4̇ ​​
⇒​​ξ ​​3̇ ​​= 90 gmol/min​
​​​ξ ​​2̇ ​​ = 0.5​​ξ ​​4̇ ​​
⇒​​ξ ​​2̇ ​​= 90 gmol/min​
​​​ξ ​​1̇ ​​ = ​​ξ ​​2̇ ​​
⇒​​ξ ​​1̇ ​​= 90 gmol/min​
​​​n ​​Ṅ 1​​ = ​ξ3​  ​​
⇒​​n ​​Ṅ 1​​= 90 gmol/min​
​​​n ​​Ẇ 1​​ = ​​ξ ​​1̇ ​​ + ​​ξ ​​2̇ ​​ − 0.5​​ξ ​​4̇ ​​
⇒​​n ​​Ẇ 1​​= 90 gmol/min​
​​​n ​​Ṁ 1​​ = ξ​​  ​​1̇ ​​
⇒​​n ​​Ṁ 1​​= 90 gmol/min​
​​​n ​​Ḣ 2​​ = 3​​ξ ​​1̇ ​​ + ​​ξ ​​2̇ ​​ − 3​​ξ ​​3̇ ​​
⇒​​n ​​Ḣ 2​​= 90 gmol/min​
Now we want to generalize further to consider many components, many
streams and many reactions. The total flow of a single stream is simply the sum
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of the flows of all components of that stream. If there are I components in
stream 1, then
​​​n ​​Ȧ 1​​ + ​​n ​​Ḃ 1​​ + n​​  ​​Ċ 1​​ + ⋯ ​​n ​​İ 1​​ = ​  ∑ ​​​ ​​n ​​i̇ 1​​ = ​​n ​​1̇ ​​​
all i
and if we sum over all Jin streams we obtain the total flow into the system:
​​ ∑ ​​​ ​​n ​​i̇ 1​​ + ​  ∑ ​​​ ​​n ​​i̇ 2​​ + ​  ∑ ​​​ ​​n ​​i̇ 3​​+ ⋯ = ​ ∑ ​​​ ​  ∑ ​​​ ​​n ​​i̇ j​​ = ​​n ​​1̇ ​​ + ​​n ​​2̇ ​​ + ​​n ​​3̇ ​​+ ⋯ = ​ ∑ ​​​ ​​n ​​j̇ ​​​
all i
all i
all i
all j​​ in​​ all i
all j​​ in​​
and similarly the total flow out is simply ∑
​ al​l ​​j​ out​ ​​​​ n ​​j̇ ​​​.
In Eq. (3.9) we considered the case of a single compound that participates in
multiple reactions. If a compound A is involved in K total reactions, then the
generation and consumption of compound A is the sum of all of its reactions, or
​​​r ​​Ȧ 1​​ + ​​r ​​Ȧ 2​​ + r​​  ​​Ȧ 3​​ + ⋯ r​​  ​​Ȧ K​​ = ​  ∑ ​​​ ​​r ​​Ȧ k​​​
all k
or in terms of extents of reaction
​​ν​ A1​​​​ξ ​​1̇ ​​ + ​νA​  2​​​​ξ ​​2̇ ​​ + ​νA​  3​​​​ξ ​​3̇ ​​+ ⋯ ν​ A​  K​​​ ​ξ ​​K̇ ​​ = ​  ∑ ​​​ ​νA​  k​​ ​​ξ ​​k̇ ​​​
all k
Similarly, if a compound B is involved in the same reactions, then
​​​r ​​Ḃ 1​​ + ​​r ​​Ḃ 2​​ + ​​r ​​Ḃ 3​​ + ⋯ r​​  ​​Ḃ K​​ = ​  ∑ ​​​ ​​r ​​Ḃ k​​​
all k
​​ν​ B1​​​​ξ ​​1̇ ​​ + ​νB​  2​​​​ξ ​​2̇ ​​ + ​νB​  3​​​​ξ ​​3̇ ​​+ ⋯ ν​ B​  K​​​​ξ ​​K̇ ​​ = ​  ∑ ​​​ ​νB​  k​​ ​​ξ ​​k̇ ​​​
all k
If we want to sum up all generation and consumptions of compounds A and B:
​​​r ​​Ȧ 1​​ + ​​r ​​Ȧ 2​​ + ​​r ​​Ȧ 3​​ + ⋯ r​​  ​​Ȧ K​​ + ​​r ​​Ḃ 1​​ + ​​r ​​Ḃ 2​​ + ​​r ​​Ḃ 3​​ + ⋯ r​​  ​​Ḃ K​​ = ​  ∑ ​​​ ​​r ​​Ȧ k​​ + ​  ∑ ​​​ ​​r ​​Ḃ k​​​
all k
all k
What if, instead of just 2 compounds A and B, we had I compounds? Then we
would need I summation signs on the r.h.s. We write this compactly as
​​  ∑ ​​​ ​​r ​​Ȧ k​​ + ​  ∑ ​​​ ​​r ​​Ḃ k​​+ ⋯ + ​ ∑ ​​​ ​​r ​​İ k​​ = ​  ∑ ​​​ ​  ∑ ​​​ ​​r ​​i̇ k​​​
all k
all k
all k
all k all i
The double summation indicates that we index from i = 1 to i = I for each
reaction k, from k = 1 to k = K, and then sum up everything. It doesn’t matter
the order in which we do the sums.
In terms of extents of reaction, the summation looks a little simpler because
we don’t need to include ξ​​​ ​​k̇ ​​​in the summation over i:
​​  ∑ ​​​ ​νA​  k​​ ​​ξ ​​k̇ ​​ + ​  ∑ ​​​ ​νB​  k​​ ​​ξ ​​k̇ ​​ + ⋯​  ∑ ​​​ ​νI​  k​​ ​​ξ ​​k̇ ​​ = ​  ∑ ​​​ ​​ξ ​​k̇ ​​ ​  ∑ ​​​ ​νi​  k​​​
all k
all k
all k
all k
all i
Putting all of this together, we get Eq. (3.11), the total mole differential material balance equation:
mur83973_ch03_155-230.indd 170
d​ns​  ys​​
​​ _​ = ​  ∑ ​​​ ​​n ​​j̇ ​​ − ​  ∑ ​​​ ​​n ​​j̇ ​​ + ​  ∑ ​​​ ​  ∑ ​​​ ​​r ​​i̇ k​​​
dt
all j​​ in​​
all j​​ out​​
all i all k
Eq. (3.11a)
d​ns​  ys​​
​​ _​ = ​  ∑ ​​​ ​​n ​​j̇ ​​ − ​  ∑ ​​​ ​​n ​​j̇ ​​ + ​  ∑ ​​​ ​​ξ ​​k̇ ​​ ​  ∑ ​​​ ​νi​  k​​​
dt
all j​​ in​​
all j​​ out​​
all k
all i
Eq. (3.11b)
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Section 3.2 The Material Balance Equation—Again
Example 3.4
Total Mole Differential Balance: Urea Manufacture from
Cheaper Reactants
In Example 3.3 we described a pathway to manufacture urea that requires four
chemical reactions. Referring back to that example and its solution, now calculate
​​​n ​​1̇ ​​​, ​​​n ​​2̇ ​​​, and ∑
​​ ​all k​ ​​ξ ​​k̇ ​​ ​∑​all i​ ​νi​  k​​​. Then check if consistent with Eq. (3.11b).
Solution
​​​n ​​1̇ ​​ = ​​n ​​Ṁ 1​​ + ​​n ​​Ẇ 1​​ + ​​n ​​Ṅ 1​​= 90 + 90 + 90 = 270 gmol/min​
​​​n ​​2̇ ​​ = ​​n ​​U̇ 2​​ + ​​n ​​Ḣ 2​​= 90 + 90 = 180 gmol/min​
​​  ∑ ​​​ ​​ξ ​​k̇ ​​ ​  ∑ ​​​ ​νi​  k​​ = ​​ξ ​​1̇ ​​​(​νM
​  1​​ + ν​ W
​  1​​ + ​νC​  O1​​ + ​νH​  1​​)​ + ​​ξ ​​2̇ ​​​(​νC​  O2​​ + ​νW
​  2​​ + ​νC​  D2​​ + ​νH​  2​​)​
all k
all i
̇
̇
+ ​​ξ ​​3(​​​ ​νN​  3​​ + ​νH​  3​​ + ​νA​  3​​)​ + ​​ξ ​​4​​​(​νA​  4​​ + ​νC​  D4​​ + ν​ U​  4​​ + ν​ W
​  4​​)​​
​​  ∑ ​​​ ​​ξ ​​k̇ ​​ ​  ∑ ​​​ ​νi​  k​​= 90((−1) + (−1) + 1 + 3) + 90((−1) + (−1) + 1 + 1)
all k
all i
+ 90((−1) + (−3) + 2) + 180((−1) + (−0.5) + 0.5 + 0.5)
= 90(2) + 90(0) + 90(−2) + 180(−0.5) = −90 gmol/min​
Now plug into Eq. (3.10b), recalling that the system is at steady state:
d​ns​  ys​​
_
​​
​ = ​  ∑ ​​​ ​​n ​​j̇ ​​ − ​  ∑ ​​​ ​​n ​​j̇ ​​ + ​  ∑ ​​​ ​​ξ ​​k̇ ​​ ​  ∑ ​​​ ​νi​  k​​​
dt
all j​​  in​​
all j​​ out​​
all k
all i
​0 = 270 − 180 − 90​??
​0 = 0​ Check!
3.2.4
The Differential Material Balance Equation: Mass Units
In some process calculations, it is more convenient, or even necessary, to work
in mass rather than molar units. It is straightforward to derive the differential
material balance equation in mass units from the equations in mole units, by
using the fact that mass and moles of component i are related by its molar
mass Mi. Using Eqs. (3.1) and (3.5) to replace molar variables with mass variables in Eq. (3.9a), we derive the differential mass balance equation for
component i.
d​mi​  ,sys​​
​​ _​ = ​  ∑ ​​​ ​​ṁ ​​ij​​ − ​  ∑ ​​​ ​​ṁ ​​ij​​ + ​  ∑ ​​​ ​​R ​​i̇ k​​​
dt
all j​​ in​​
all j​​ out​​
all k
Eq. (3.12a)
or, using extents of reaction:
d​mi​  ,sys​​
​​ _​ = ​  ∑ ​​​ ​​ṁ ​​ij​​ − ​  ∑ ​​​ ​​ṁ ​​ij​​ + ​  ∑ ​​​ ​νi​  k​​ ​Mi​  ​​ ​​ξ ​​k̇ ​​​
dt
all j​​ in​​
all j​​ out​​
all k
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Eq. (3.12b)
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
At steady state, Eq. (3.12) can be rewritten in a useful form:
​​  ∑ ​​​ ​​ṁ ​​ij​​ = ​  ∑ ​​​ ​​ṁ ​​ij​​ + ​  ∑ ​​​ ​​R ​​i̇ k​​​
all j​​ out​​
all j​​ in​​
Eq. (3.13a)
all k
or equivalently
​​  ∑ ​​​ ​​ṁ ​​ij​​ = ​  ∑ ​​​ ​​ṁ ​​ij​​ + ​  ∑ ​​​ ​νi​  k​​ ​Mi​  ​​ ​​ξ ​​k̇ ​​​
all j​​ out​​
all j​​ in​​
Eq. (3.13b)
all k
What about total mass? This turns out to be easier than total moles, because
total mass is conserved but total moles are not! The consequence is that the
sum of all reaction terms in mass units ​∑all k ​​R ​​i̇ k​​​must equal zero! You showed
this already in Quick Quiz 3.1. Therefore, the total mass differential material
balance equation is simply
d​ms​  ys​​
​​ _​ = ​  ∑ ​​​ ​​ṁ ​​j​​ − ​  ∑ ​​​ ​​ṁ ​​j​​​
dt
all j​​ in​​
all j​​ out​​
Example 3.5
Eq. (3.14)
Differential Mass Balance: Sugar Dissolution
A bucket is initially filled with a large quantity of sucrose (table sugar). Then,
water at 3 kg/min is pumped continuously into the bucket, and the contents of the
bucket are mixed so that sugar dissolves into the water. Sugar water (84 wt%
sucrose) is pumped out of the bucket at 3 kg/min. Apply Eqs. (3.12) and (3.14) to
this situation.
Water in
3 kg/min
Sugar water out
3 kg/min
Solution
Starting with the total mass differential balance equation,
d​ms​  ys​​
_
​​
​ = ​  ∑ ​​​ ​​ṁ ​​j​​ − ​  ∑ ​​​ ​​ṁ ​​j​​ = ​​ṁ ​​in​​ − m
​​ ̇ ​​out​​= 3 − 3 = 0 kg/min​
dt
all j​​ in​​
all j​​ out​​
The total mass in the bucket does not change over time.
There are two components in the system: water W and sucrose S. In the bucket,
the sugar dissolves in the water, but there are no chemical reactions. Applying
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Section 3.2 The Material Balance Equation—Again
Eq. (3.12) to each component in turn, and using wSout = 0.84 and wWout = 1 − 0.84 =
0.16, yields
d​mW
​  ,sys​​
_
​​
​ = ​​ṁ ​​Win​​ − m
​​ ̇ ​​Wout​​ = ​​ṁ ​​Win​​ − ​wW
​  ​​ ​​ṁ ​​out​​= 3 − (0.16)3 = +2.52 kg/min​
dt
d​mS​  ,sys​​
_
​​
​ = ​​ṁ ​​Sin​​ − m
​​ ̇ ​​Sout​​ = ​​ṁ ​​Sin​​ − w
​ S​  ​​ ​​ṁ ​​out​​= 0 − (0.84)3 = −2.52 kg/min​
dt
d​mW
​  ,sys​​
d​mS​  ,sys​​
d​ms​  ys​​
Notice that ____
​​  dt ​+ ____
​  dt ​= ___
​  dt ​​, as it should.
Example 3.6
Differential Mass Balance: Glucose Consumption in a Fermentor
A broth containing 20 wt% glucose is fed to a fermentor continuously at 100 g/h.
Yeast in the fermentor consume glucose at a rate of 12.9 g/h. The fermentor
becomes contaminated with bacteria, which consume glucose at 1.4 g/h. What is
the rate of change in glucose in the fermentor?
Broth, 20%
glucose
Solution
Glucose (G) is the only component of interest. There is one input stream, no output streams, and two chemical reactions in which glucose is consumed, by yeast
and by bacteria. Applying Eq. (3.12) to glucose,
d​mG​  ,sys​​
_
​​
​ = ​  ∑ ​​​ ​​ṁ ​​ij​​ − ​  ∑ ​​​ ​​ṁ ​​ij​​ + ​  ∑ ​​​ ​​R ​​i̇ k​​ = ​wG​  ​​ ​​ṁ ​​in​​ + ​​R ​​Ġ 1​​ + ​​R ​​Ġ 2​​​
dt
all j​​ in​​
all j​​ out​​
all k
d​mG​  ,sys​​
_
​​
​= (0.2)100 + (−12.9) + (−1.4) = 5.7 g/h​
dt
Equations (3.9), (3.11), (3.12), and (3.14) may look complicated, but they are
simply mathematical ways to express concepts you are already very familiar
with. Table 3.1 summarizes these four differential material balance equations
and compares them to the material balance equation that you learned in Chap. 2.
3.2.5
The Integral Material Balance Equation
Whether for a single component or total, or in units of moles or mass, the differential material balance equation applies to a snapshot of the system—a single
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Table 3.1
Material Balance Equations: Differential Form
Accumulation =
Input
− Output
+ Generation − Consumption
d​ms​  ys​​
Total mass​​ _​ =​​​  ∑ ​​​ ​​ṁ ​​j​​​​− ​  ∑ ​​​ ​​ṁ ​​j​​​
dt
all j​​in​
all j​​out​
d​
m
​
​​
i
,sys
Mass of i​​ _​ =​​​  ∑ ​​​ ​​ṁ ​​ij​​​​− ​  ∑ ​​​ ​​ṁ ​​ij​​​​+ ​  ∑ ​​​ ​Mi​  ​​ ​νi​  k​​ ​​ξ ​​k̇ ​​​
dt
all j​​in​
all j​​out​
all k
d​
n
​
​​
sys
Total moles​​ _​ =​​​  ∑ ​​​ ​​n ​​j̇ ​​​​− ​  ∑ ​​​ ​​n ​​j̇ ​​​​+ ​  ∑ ​​​​​ ​​ξ ​​k̇ ​​​​​ ​  ∑
dt
all j​​in​
all j​​out​
all k
all i
​ ​ )
(​  compounds
​​​​νi​  k​​ ​​
d​ni​  ,sys​​
Moles of i​​ _​ =​​​  ∑ ​​​ ​​n ​​i̇ j​​​​− ​  ∑ ​​​ ​​n ​​i̇ j​​​​+ ​  ∑ ​​​ ​νi​  k​​ ​​ξ ​​k̇ ​​​
dt
all j​​in​
all j​​out​
all k
instant in time. But what if we want to consider the system over a finite time
interval, say between t = t0 and t = tf? (See Figure 3.4.) We still begin with:
Accumulation = Input − Output + Generation − Consumption
Eq. (2.5)
But now we are interested in the input, output, reaction and accumulation that
occur over the defined finite time interval—quantities rather than rates. We
use units of moles or mass, rather than moles/time or mass/time. For example,
imagine that compound A is pumped into a system in stream 1 over the time
interval between t0 and tf. Then the moles of A input into the system via
stream 1, nA1, is just the total amount of A added during the specified time
interval. We are not specifying ( just yet) whether A is pumped into the system
System at t = t0
System at t = tf
Figure 3.4 A system with multiple inputs and outputs, shown at two different times. The
system is a three-dimensional enclosed volume, and the system boundary is a two-dimensional
surface area. The system is shaded, and the system boundary is shown as a dark line. The
darkness of the shading corresponds to the quantity of the material in the system, which might
change with time. Material flows in and out are shown as lines with arrows. The thickness
of the lines corresponds to the quantity of material flow, which might change with time.
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Section 3.2 The Material Balance Equation—Again
175
at a steady rate between t0 and tf, or if A is added in one large bolus. Similarly,
if A reacts inside the system by reaction 2, RA2 is the moles of A consumed
by reaction 2 over the time interval between t0 and tf. Accumulation is no
longer a derivative but a difference: ​​nA​  sys, f​​ − ​nA​  sys,0​​​ is the change in the moles
of A in the system over the time interval between t0 and tf. Given what you
have already learned about the differential material balance equations, we can
quickly write general forms of the integral material balance equations:
For component i in molar units:
​​n​ i,sys, f​​ − n​ i​  ,sys,0​​ = ​  ∑ ​​​ ​ni​  j​​ − ​  ∑ ​​​ ​ni​  j​​ + ​  ∑ ​​​ ​ri​  k​​​
all j​​ in​​
all j​​ out​​
Eq. (3.15)
all k
For total moles:
​​n​ sys, f​​ − ​ns​  ys,0​​ = ​  ∑ ​​​ ​nj​  ​​ − ​  ∑ ​​​ ​nj​  ​​ + ​  ∑ ​​​ ​  ∑ ​​​ ​ri​  k​​​
all j​​ in​​
all j​​ out​​
Eq. (3.16)
all i all k
For component i in mass units:
​​m​ i,sys, f​​ − ​mi​  ,sys,0​​ = ​  ∑ ​​​ ​mi​  j​​ − ​  ∑ ​​​ ​mi​  j​​ + ​  ∑ ​​​ ​Ri​  k​​​
all j​​ in​​
Helpful Hint
Use the integral
total mass balance
equation to evaluate changes in total
mass over a specified time interval.
Table 3.2
all j​​ out​​
Eq. (3.17)
all k
For total mass:
​​m​ sys, f​​ − ​ms​  ys,0​​ = ​  ∑ ​​​ ​mj​  ​​ − ​  ∑ ​​​ ​mj​  ​​​
all j​​ in​​
Eq. (3.18)
all j​​ out​​
The integral material balance equations are summarized, using extents of reaction, in Table 3.2.
Material Balance Equations: Integral Form
Accumulation =
Input
− Output
+ Generation − Consumption
Total mass​​m​ sys, f​​ − ​ms​  ys,0​​ =​  ∑ ​​​ ​mj​  ​​
− ​  ∑ ​​​ ​mj​  ​​​
Mass of i​​m​ i,sys, f​​ − ​mi​  ,sys,0​​ =​  ∑ ​​​ ​mi​  j​​
− ​  ∑ ​​​ ​mi​  j​​
+ ​  ∑ ​​​ ​Mi​  ​​ ​νi​  k​​ ​ξk​  ​​​
Total moles​​n​ sys, f​​ − n​ s​  ys,0​​ =​  ∑ ​​​ ​nj​  ​​
− ​  ∑ ​​​ ​nj​  ​​
+ ​  ∑ ​​​ ​  ∑ ​​​ ​νi​  k​​ ​ξk​  ​​​
Moles of i​​n​ i,sys, f​​ − ​ni​  ,sys,0​​ =​  ∑ ​​​ ​ni​  j​​
− ​  ∑ ​​​ ​ni​  j​​
+ ​  ∑ ​​​ ​νi​  k​​ ​ξk​  ​​​
all j​​ in​​
all j​​ in​​
all j​i​  n​​
all j​​ in​​
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all j​​ out​​
all j​​ out​​
all j​​ out​​
all j​​ out​​
all k
all i all k
all k
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Example 3.7
Integral Equation: Blending and Shipping
At a blending and shipping facility at a large integrated refinery, gasoline from
three different processes—called reformer, isomax, and FCC—is pumped into a
large storage tank. The processes run continuously, producing 15,400, 18,200, and
10,500 barrels of gasoline per day, respectively. (A barrel is 42 gallons, and gasoline has a density of about 6.6 lb/gal.) A large tanker comes to port to load up.
The tanker holds 61.0 million lb of gasoline, and it takes 54 hours to pump it
full. When pumping to the tanker is first started, the storage tank contains 154,000
barrels. The ship’s captain is worried that there isn’t enough gasoline to fill the
tanker, but the supervisor at the blending and shipping facility tells him not to
worry. Who’s right?
Solution
The flow diagram is shown, with streams numbered.
Gasoline from reformer
Gasoline from isomax
Gasoline from FCC
1
2
Storage tank
4
Tanker
3
The storage tank is the system. We will do all calculations on a mass basis, which
requires us to convert from volumetric flow to mass flow. The conversion factor is
42 gal 6.6 lb _
1 day
lb/h  ​​
​​ _ ​ × _
​
​ × ​
​ = 11.55 ​ _
barrel
gal
24 h
barrel/day
Since we are interested not in a specific component but in the total quantity of
material, we work in total mass units. We calculate the mass flow rates in each
stream from the provided information.
lb/h  ​ = 178,000 ​ _
barrels
lb ​​
​​​ṁ ​​1​​ = 15,400 ​ _
​ × 11.55 ​ _
day
barrel/day
h
lb/h  ​ = 210,000 ​ _
barrels
lb ​​
​​​ṁ ​​2​​ = 18,200 ​ _
​ × 11.55 ​ _
day
barrel/day
h
lb/h  ​ = 121,000 ​ _
barrels
lb ​​
​​​ṁ ​​3​​ = 10,500 ​ _
​ × 11.55 ​ _
day
barrel/day
h
61,000,000 lb
lb ​​
​​​ṁ ​​4​​ = ____________
​
​ = 1,130,000 ​ _
h
54 h
Since we are interested in what happens over a specified time interval (54 hours),
we use the integral total mass balance equation (Eq. 3.18):
​​ms​  ys, f​​ − ​ms​  ys,0​​ = ​  ∑ ​​​ ​mj​  ​​ − ​  ∑ ​​​ ​mj​  ​​​
all j​​ in​​
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Section 3.2 The Material Balance Equation—Again
177
which simplifies to
​​ms​  ys, f​​ − ​ms​  ys,0​​ = ​m1​  ​​ + ​m2​  ​​ + ​m3​  ​​ − ​m4​  ​​​
For this problem we need to know the total quantity that enters or leaves the storage tank via each stream, over the specified time interval of 54 hours, but we are
given only the flow rates in each stream. We can easily calculate the total quantity
in each stream by integrating the flow rate from t0 = 0 to tf = 54 h, or
​tf​  ​​
54
​​m​ 1​​ = ​  ∫​  ​​​​ṁ ​​1​​dt = ​ ∫​  ​​178,000 dt​
​t0​  ​​
0
lb ​× (​ 54 − 0)​= 9,612,000 lb​
​= 178,000 ​ _
h
Similarly, we find:
​​m​ 2​​= 11,340,000 lb​
​​m​ 3​​= 6,534,000 lb​
​​m​ 4​​= 61,000,000 lb​
Or
​​ms​  ys, f​​ − ​ms​  ys,0​​= 9,612,000 + 11,340,000 + 6,534,000 − 61,000,000 = −33,500,000 lb​
Accumulation is negative, so the quantity of gasoline in the storage tank is being
depleted.
It is important to recognize that the integral material balance equations allow
us to evaluate the change in material in the system, but not the absolute quantity
in the system. For that, we need additional information, such as the initial quantity
of material.
Luckily, we know the quantity of gasoline in the storage tank at the beginning
of the pumping operation:
Helpful Hint
Use an integral
material balance
equation to analyze
the performance of
batch and semibatch processes
over a specified
time interval.
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42 gal 6.6 lb
​​ms​  ys,0​​= 154,000 barrels × ​ _ ​ × _
​
​= 42,700,000 lb​
barrel
gal
Therefore:
​​ms​  ys, f​​= −33,500,000 + 42,700,000 = 9,200,000 lb​
After filling up the ship, the storage tank still contains over 9 million lb gasoline,
or over 33,000 barrels, so the ship’s captain doesn’t need to worry.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Example 3.8
Integral Equation with Unsteady Flow: Jammin’ with Cherries
At the award-winning Jumpin’ Jam Factory, cherry jam is produced by chopping
up 100 lb cherries and mixing the chopped cherries all at once with 200 lb sugar
in a pot. Then, water is boiled off. As the mixture thickens, the water evaporation
rate decreases. Charlie Cherrypit, the engineer at Jumpin’ Jam, estimated that the
evaporation rate can be modeled as ​​ṁ ​​w,evap = 2.0 − 0.03t with t in units of minutes
and m
​​ ̇ w​​ ,evap in units of lb/min. How long will it take to make 240 lb of jam?
Solution
The flow diagram, with the pot as the system, is shown.
Cherries, sugar
Water
Jam
t < t0
t0 < t < tf
t = tf
This is a semibatch process: The cherries and sugar are charged all at once to the
pot initially, the water vapor is removed continuously, and the jam is collected
from the pot at the end. Notice some features of this problem: There is an initial
charge of material to the system, there is a specified time interval (tf − t0, which
we are asked to find), and we want to know the total mass left in the system at the
end of the time interval. Furthermore, there is no chemical reaction. These clues
together indicate that the easiest way to solve this problem is to use the integral
total mass balance equation:
​​ms​  ys, f​​ − ​ms​  ys,0​​ = ​  ∑ ​​​ ​mj​  ​​ − ​  ∑ ​​​ ​mj​  ​​​
all j​​ in​​
all j​​ out​​
From the information given, at t = t0,
​​ms​  ys,0​​= 100 lb cherries + 200 lb sugar = 300 lb​
At the end of jam making, there should be 240 lb jam in the pot, or
​​ms​  ys, f​​= 240 lb​
There is no flow of material into the pot from t = t0 to t = tf, so
​​  ∑ ​​​ ​mj​  ​​ = 0​
all j​​ in​​
Material leaves the pot in a single output stream of water vapor, with the rate
decreasing with time.
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Section 3.3 Linear Models of Process Flow Sheets
We are given the rate of evaporation as a function of t, so to calculate the total
amount of mass evaporating over a time interval of interest, we need to integrate
​tf​  ​​
​tf​  ​​
mout = ​​m​ evap​​ = ​  ∫​  ​​ ​​ṁ ​​evap​​ dt = ​ ∫​  ​​(2.0 − 0.03t)dt​
​t0​  ​​
​t0​  ​​
​= 2(​tf​  ​​ − ​t0​  ​​) − 0.015​(t​ f​  ​​ − ​t0​  ​​)​​  2​​
Inserting these expressions into the material balance equation gives
​240 − 300 = −[2(​tf​  ​​ − ​t0​  ​​) − 0.015​(t​ f​  ​​ − ​t0​  ​​)​​  2​]​
This is a quadratic equation, with two solutions: tf − t0 = Δt = 45.6 minutes
or 87.7 minutes. Which is right? Only one answer makes physical sense. If Δt =
87.7 minutes, the water evaporation rate would be negative (−0.063 lb/min), which
would mean that water would be entering the pot. The answer of Δt = 45.6 minutes
is reasonable; the water evaporation rate at the end of the jam-making session is
+ 0.063 lb/min, about 80% less than the evaporation rate initially.
3.3
Linear Models of Process Flow Sheets
In Chap. 2 we completed process flow calculations on several preliminary
block flow diagrams. In those examples, we often made simplifying approximations: that the separation was perfect, for example, or that the reactants were
completely consumed. Now we would like to analyze more realistic process
flow sheets, with fewer simplifying approximations.
What we’re after is, in mathematical terms, a linear model of a process
flow sheet. A process flow sheet such as a block flow diagram is a visual
representation of process units, process streams, and how they are all connected. A linear model is a mathematical representation of that flow sheet.
To develop a linear model of a process flow sheet, we derive a system of
linear equations by combining steady-state material balance equations with
appropriate system performance specifications for each process unit. These
linear equations express outlet flows and system variables as linear functions
of inlet flows and/or system performance specifications.
Not all process flow problems are amenable to linear analysis. Timedependent processes, for example, cannot be modeled by linear equations.
Only certain kinds of system performance descriptors give rise to linear equations; we will see in later chapters that analysis of reactors and separators
sometimes requires nonlinear functions. Still, when a process flow problem
can be cast as a set of linear equations, there are powerful mathematical tools
at our disposal. If we are able to write linear algebraic equations to describe
all the process units in a complex process flow sheet, then mathematical analysis of that flow sheet is simplified. Furthermore, once we set up a model, it
is easier to examine the impact of various design decisions on the overall
process flow.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
3.3.1
ystem Performance Specifications and Linear
S
Models of Process Units
As we learned in Chap. 2, there are only four fundamental kinds of process
units in block flow diagrams: mixers, splitters, reactors, and separators. It is
amazing that we can make so many different products by combining these four
simple building blocks in myriad ways. This diversity in outcome is achieved
through the diversity in chemical reactions, reactor designs, and separation
technologies, and through the diversity in how the units are connected.
In this section we will apply the steady-state differential component mole
balance equation (Eq. 3.10) for each of these four kinds of process units. Then
we’ll write linear equations describing system performance specifications appropriate for that type of process unit. Together, the material balance equations and
the system performance equations constitute a linear model of a process unit
operating at steady state. The linear models are structured assuming full knowledge of the input streams, and provide a means for calculating all output steams.
These models yield a set of linear equations that can be written in matrix form,
Ax = b, where x contains the unknown variables. For simplicity, in this section
we will work only in mole units, but mass units can be used equally well.
Mixers Mixers are designed to combine multiple (Jin) input streams into a
single output stream. The sketch shows a block flow diagram for a mixer where
Jin = 3.
ni1
ni2
Mixer
ni,out
ni3
There is no chemical reaction, so the steady-state component mole balance
equation Eq. (3.10) for a mixer simplifies to
​​​n ​​i̇ ,out​​ = ​  ∑ ​​​ ​​n ​​i̇ j​​​
all j​​ in​​
Eq. (3.19)
If there are I components passing through the mixer, then we can write I
independent balance equations of the form of Eq. (3.19).
Mixer performance is often specified as a ratio of flows of different input
streams, either total flows or flows of specific components. We can provide at
most Jin − 1 independent specifications of this type.
A linear model of a mixer can be built by combining I material balance equations (of the form of Eq. (3.19)) with Jin − 1 mixer performance
specifications.
Example 3.9
Linear Models of Mixers: Sweet Mix
Two aqueous solutions are mixed together in a continuous steady-state mixer.
One of the solutions (solution A) contains 15 mol% glucose, 12 mol% fructose,
and 73 mol% water. The flow rate of solution A is 180 kgmol/h, and the
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Section 3.3 Linear Models of Process Flow Sheets
181
design calls for 40% of the mixer output to be supplied by solution A. The other
solution (solution B) contains 6 mol% glucose, 3 mol% fructose, and 91 mol%
water. What is the flow rate of each component in the stream leaving the mixer?
What is the total flow rate of solution B as well as the total flow rate out of
the mixer?
Solution
A
B
180 kgmol/h
15% g
12% f
73% w
Mixer
?
6% g
3% f
91% w
We’ll use g to indicate glucose, f for fructose, and w for water. We’ll use subscripts
A, B and out to indicate the streams. We write three material balance equations
because there are three components:
​​​n ​​ġ ,out​​ = ​​n ​​ġ A​​ + n​​  ​​ġ B​​​
​​​n ​​ḟ ,out​​ = ​​n ​​ḟ A​​ + n​​  ​​ḟ B​​​
​​​n ​​ẇ ,out​​ = ​​n ​​ẇ A​​ + ​​n ​​ẇ B​​​
There are 2 input streams (Jin = 2), so we provide (Jin − 1) or one system performance equation:
​​​n ​​Ȧ ​​∕n​​  ​​Ḃ ​​​ = 40∕60
Since
​​​n ​​Ȧ ​​= 180 kgmol/h​
We can easily find that
​​​n ​​Ḃ ​​ = ​(_
​  60 ​)​(180) = 270 kgmol/h​
40
The component material balance equations, combined with specified stream compositions, become:
​​​n ​​ġ ,out​​ = ​zg​  A​​ ​​n ​​Ȧ ​​ + z​ g​  B​​ ​​n ​​Ḃ ​​= (0.15)180 + (0.06)(270) = 42 kgmol/h​
​​​n ​​ḟ ,out​​ = ​zf​  A​​ ​​n ​​Ȧ ​​ + ​zf​  B​​ ​​n ​​Ḃ ​​= (0.12)180 + (0.03)(270) = 29 kgmol/h​
​​​n ​​ẇ ,out​​ = ​zw​  A​​ ​​n ​​Ȧ ​​ + ​zw​  B​​ ​​n ​​Ḃ ​​= (0.73)180 + (0.91)(270) = 359 kgmol/h​
Written in matrix form, these equations are simply:
1
​​ 0​ ​
[0
mur83973_ch03_155-230.indd 181
0
1​
0
42
0 ​​n ​​ġ ,out​​
0​ ​​ ​  ​​n ​​ḟ ,out​​​​ ​ = ​ ​  29​​ ​​
]
[359]
​​
1 [​​n ​​ẇ ,out]
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182
Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
The solution is, of course,
​​n ​​ġ ,out​​
42
n
​​
​​
​​
̇
​​ ​  f,out​​ ​ = ​ ​  29​​ ​​
[359]
[​​n ​​ẇ ,out]
​​
Notice that the matrix A for a mixer is simply the identity matrix! This will always
be true for mixers, so the matrix equation can be written down immediately.
Splitters Splitters take a single input stream and divide it into two or more
output streams of identical composition. The sketch is an example of a splitter
with three outputs (Jout = 3). The key feature of a splitter is that the mole (or
mass) fraction of every component is the same in the input stream and in all
of the output streams.
ni,1
ni,in
Splitter
ni,2
ni,3
There is no chemical reaction, so the steady-state component mole balance
equation (Eq. 3.10) simplifies to
​​  ∑ ​​​ ​​n ​​i̇ j​​ = ​​n ​​i̇ ,in​​​
all j​​ out​​
Eq. (3.20)
If there are I components passing through the splitter, then we can write I
independent balance equations of the form of Eq. (3.20).
The performance of a splitter can be defined by the fractional split fSj
​​n ​​​j̇ ​ ​​​​
moles leaving in stream ​jo​  ut​​ _
_______________________
​​f​ Sj​​ = ​
​ = ​  out ​​
​​n ​​i̇ n​​
moles in
Eq. (3.21a)
or, written as a linear equation relating output to input:
​​​n ​​​j̇ o​  ut​​​​ = ​fS​  j​​ ​​n ​​i̇ n​​​
Eq. (3.21b)
since the composition of all streams in or out of the splitter are the same, it
must also be true that
​​​n ​​​i̇ j​ out​​​​ = ​fS​  j​​ ​​n ​​i̇ ,in​​​
Eq. (3.21c)
for all I components.
There are Jout values of fSj, one for each output stream. But, there is one
restriction that must be satisfied: the fractional splits must sum up to equal 1, or
​​  ∑ ​​​ ​fS​  j​​ = 1​
all j​o​  ut​​
Therefore, we can independently specify at most only (Jout − 1) values of fSj.
A model of a splitter, then, can be built by combining I material balance
equations (of the form of Eq. 3.20), plus I × (Jout − 1) equations for splitter
performance (Eq. 3.21c).
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Section 3.3 Linear Models of Process Flow Sheets
Example 3.10
183
Linear Model of a Splitter: Sweet Split
A solution containing 9.8 mol% glucose, 6.6 mol% fructose, and 83.6 mol% water
is fed at a rate of 430 kgmol/h to a splitter operated continuously and at steady
state. The splitter produces three outlet streams, A, B, and C. Stream A is 25% of
the inlet stream ( fSA = 0.25) and Stream B is 35% of the inlet stream ( fSB = 0.35).
Calculate the flow rates of the output streams.
430 kgmol/h
9.8% g
6.6% f
83.6% w
A, fSA = 0.25
B, fSB = 0.35
C, fSC = 0.40
Splitter
Solution
There are three components and three outlet streams, so I = 3 and Jout = 3.
Therefore there are three material balance equations, plus 3 × (3 − 1) or 6 splitter
performance equations.
Material balance equations:
​​​n ​​ġ A​​ + ​​n ​​ġ B​​ + n​​  ​​ġ C​​ = ​​n ​​ġ ,in​​​
​​​n ​​ḟ A​​ + n​​  ​​ḟ B​​ + n​​  ​​ḟ C​​ = ​​n ​​ḟ ,in​​​
​​​n ​​ẇ A​​ + ​​n ​​ẇ B​​ + n​​  ​​ẇ C​​ = ​​n ​​ẇ ,in​​​
Splitter performance equations:
​​​n ​​ġ A​​ = ​fS​  A​​ ​​n ​​ġ ,in​​​
​​​n ​​ḟ A​​ = ​fS​  A​​ ​​n ​​ḟ ,in​​​
​​​n ​​ẇ A​​ = ​fS​  A​​ ​​n ​​ẇ ,in​​​
⎢
​​​n ​​ġ B​​ = ​fS​  B​​ ​​n ​​ġ ,in​​​
​​​n ​​ḟ B​​ = ​fS​  B​​ ​​n ​​ḟ ,in​​​
​​​n ​​ẇ A​​ = ​fS​  B​​ ​​n ​​ẇ ,in​​​
⎥
⎢⎥⎢ ⎥
⎢⎥⎢ ⎥
Written in matrix form as Ax = b, these equations become:
⎡1
0
0
1
0
0
1
0
0⎤ ⎡ ​​n ​​ġ A​​ ⎤ ⎡ ​​n ​​ġ ,in​​ ⎤
​​n ​​ḟ ,in​​
0
1
0
0
1
0
0
1
0 ​​n ​​ḟ A​​
​​n ​​ẇ ,in​​
n
​​
​​
​​
̇
0
0
1
0
0
1
0
0
1
wA
​fS​  A​​ ​​n ​​ġ ,in​​
1
0
0
0
0
0
0
0
0 ​​n ​​ġ B​​
n
​​
​​
​​
̇
​​ 0​
​
1​
0​
0​
0​
0​
0​
0​
0​ ​​ ​​ ​  fB​​ ​​ = ​​ ​​fS​  A​​ ​​n​​ ​​f​̇ ,in​​​ ​ ​​ ​​
​fS​  A​​ ​​n ​​ẇ ,in​​
0
0
1
0
0
0
0
0
0 ​​n ​​ẇ B​​
n
​​
​​
​​
​fS​  B​​ ​​n ​​ġ ,in​​
̇
gC
0
0
0
1
0
0
0
0
0
​fS​  B​​ ​​n ​​ḟ ,in​​
0
0
0
0
1
0
0
0
0 ​​n ​​ḟ C​​
n
​​
​​
​​
⎣0
⎦
̇
⎣
⎦
​fS​  B​​ ​​n ​​ẇ ,in​​⎦
⎣
wC
0
0
0
0
1
0
0
0
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184
⎢
⎥ ⎢⎥
Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
⎢⎥
⎢⎥
If we now specify the three input streams plus the two fractional splits, we will have
nine equations in nine unknowns. The matrix equation for this problem becomes
⎡1
0
0
1
0
0
1
0
0⎤ ⎡ ​​n ​​ġ A​​ ⎤ ⎡ 42.1 ⎤
28.4
0
1
0
0
1
0
0
1
0 ​​n ​​ḟ A​​
359.5
0
0
1
0
0
1
0
0
1 ​​n ​​ẇ A​​
10.5
1
0
0
0
0
0
0
0
0 ​​n ​​ġ B​​
​​ 0​
​
1​
0​
0​
0​
0​
0​
0​
0​ ​​ ​​ ​  ​​n ​​ḟ B​​​​ ​​ = ​​ ​  7.1​​​​ ​​
89.9
0
0
1
0
0
0
0
0
0 ​​n ​​ẇ B​​
14.8
0
0
0
1
0
0
0
0
0 ​​n ​​ġ C​​
n
​​
​​
​​
̇
9.9
fC
0
0
0
0
1
0
0
0
0
⎣0
⎣125.8⎦
0
0
0
0
1
0
0
0⎦ ⎣​​n ​​ẇ C⎦​​
⎢⎥
⎢⎥
⎢⎥
The solution is easy to obtain on a calculator with matrix function keys, by solving
x = A−1b or
⎡ ​​n ​​ġ A​​ ⎤ ⎡ 10.5 ⎤
​​n ​​ḟ A​​
7.1
​​n ​​ẇ A​​
89.9
​​n ​​ġ B​​
14.8
​​ ​  ​​n ​​ḟ B​​​​ ​​ = ​​ ​  9.9​​​​ ​​
​​n ​​ẇ B​​
125.8
​​n ​​ġ C​​
16.8
​​n ​​ḟ C​​
11.4
⎣143.8⎦
⎣​​n ​​ẇ C⎦​​
Notice the general structure of the splitter matrix equation. In the b vector are the
known input flows and fractional splits. The unknown output flows are in the x
vector. The A matrix is a combination of 3 × 3 identity matrices and 3 × 3 null
matrices. The pattern in the A matrix can be quickly adjusted to incorporate a
different number of components or different number of output streams.
convert reactants to products by chemical reaction. An idealized
reactor has only one inlet and only one outlet flow.
Reactors
ni,in
Reactor
ni,out
The steady-state component mole balance equation is
​​​n ​​i̇ ,out​​ = ​​n ​​i̇ ,in​​ + ​  ∑ ​​​​νi​  k​​ ​​ξ ​​k̇ ​​​
all k
Eq. (3.22)
Reactors are so important in chemical processes that we’ve devoted two
chapters to the topic. All we want right now is a straightforward way to write
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Section 3.3 Linear Models of Process Flow Sheets
185
reactor performance specifications that is useful for linear models. In this chapter
we will use fractional conversion to specify reactor performance. In Chap. 4
we will discuss fractional conversion in greater detail and also describe other
useful measures for specifying reactor performance.
The fractional conversion of reactant i, fCi, is the fraction of the reactant
fed to the reactor that gets consumed by chemical reaction, and is defined as
− ​  ∑ ​​​ ​νi​  k​​ ​​ξ ​​k̇ ​​
net moles of i consumed by reaction _________________
____________________________
​​f​ Ci​​ = ​
​ =​​ ​  all k  ​​
​​n ​​i̇ ,in​​
moles of i fed to reactor
Rearranging, we get a linear equation describing reactor performance:
​​f​ Ci​​ ​​n ​​i̇ ,in​​ = − ​  ∑ ​​​ ​νi​  k​​ ​​ξ ​​k̇ ​​​
all k
Eq. (3.23)
Note that 0 ≤ fCi ≤ 1.
A fractional conversion can be defined for any reactant fed to the reactor.
It is not used for compounds that are solely products of reaction. Notice that
the fractional conversions of different compounds are related through their
stoichiometry. Fractional conversions of different compounds can be different
even if they are reactants in the same reaction, if their feed rates are different.
Illustration: For the reaction 2​ A + B → C + D​, f​​ C​  A​​ = 2​​ξ ​​1̇ ​​∕n​​  ​​Ȧ ,in​​, and ​fC​  B​​ =
​​ξ ​​1̇ ​​∕​​n ​​Ḃ ,in​​ ​.
Illustration: Suppose compounds A and B are fed to a reactor, with n​​​  ​​Ȧ ,in​​ =
100 gmol/min​ and n​​​  ​​Ḃ ,in​​= 100 gmol/min​, where they react by R1: ​2A + B →
C + D​. ​ ​​ξ ​​1̇ ​​= 20 gmol/min​. Then ​​f​ CA​​ = 2 ​​ξ ​​1̇ ​​∕​​n ​​Ȧ ,in​​= 2(20)∕100 = 0.4, and
​fC​  B​​= 20∕100 = 0.2.​
Illustration: Compounds A and B are fed to a reactor, where two reactions
take place:
​2A + B → C + D​
(R1)
1 ​ B + C → E​
​​ _
(R2)
2
Therefore,
​​f​ CA​​ = 2​​ξ ​​1̇ ​​∕n​​  ​​Ȧ ,in​​​
​​f​ CB​​ = ​(​​ξ ​​1̇ ​​ + 0.5​​ξ ​​2̇ ​​)​∕​​n ​​Ḃ ,in​​​
If we have K independent chemical reactors, we can specify up to K independent
fractional conversions. A linear model of a reactor may contain I material
balance equations (Eq. 3.22) and K equations of reactor performance (Eq. 3.23).
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Example 3.11
Linear Model of a Reactor: Glucose-Fructose Isomerization
A solution of 9.8 mol% glucose, 6.6 mol% fructose, and 83.6 mol% water is fed
to a reactor at a rate of 172.3 kgmol/day. Glucose and fructose are isomers: They
both have the same molecular formula, C6H12O6, but they have different chemical
structures, and fructose is much sweeter-tasting than glucose. In the reactor, 53.25%
of the glucose is converted to fructose:
C6H12O6 (glucose) → C6H12O6 (fructose)
Calculate the reactor output flow rate and composition. The reactor operates at
steady state.
172.3 kgmol/day
9.8% g
6.6% f
83.6% w
Reactor
n out
fC,g = 0.5325
Solution
There are three components and one reaction. We write three material balance
equations and one reactor performance equation, putting the unknowns on the lefthand side.
​​​n ​​ġ ,out​​ + ξ = ​​n ​​ġ ,in​​​
​​​n ​​ḟ ,out​​ − ξ = ​​n ​​ḟ ,in​​​
​​​n ​​ẇ ,out​​ = ​​n ​​ẇ ,in​​​
​ξ = ​fC​  ,g​​ ​​n ​​ġ ,in​​​
In matrix form this set of equations, which constitutes a linear model of this reactor, is written
⎥⎢ ⎥ ⎢ ⎥
⎢
⎡1 0
0
1 ⎤ ⎡ ​​n ​​ġ ,out​​ ⎤ ⎡ ​​n ​​ġ ,in​​ ⎤
​​n ​​ḟ ,in​​
0 1
0
− 1 ​​n ​​ḟ ,out​​
​​
​
​
​
​
​ ​ ​ ​ ​   ​
​
​​ ​= ​ ​
​​ ​ ​​
​​n ​​ẇ ,out​​
​​n ​​ẇ ,in​​
0 0
1
0
⎣0 0
0
1 ⎦ ⎣ ​ξ ​̇ ⎦ ⎣​ f​ C,g​​ ​​g ​​ġ ,in⎦​​
The output streams and extent of reaction are the four unknown variables, all in
the x vector. The known fractional conversion of glucose, fC,g = 0.5325, and the
known input flow rates are all in the b vector. Inserting the numerical values into
the matrix equation yields
⎢
⎥⎢ ⎥ ⎢ ⎥
⎡1
0
0
1 ⎤ ⎡ ​​n ​​ġ ,out​​ ⎤ ⎡16.9⎤
0
1
0
− 1 ​​n ​​ḟ ,out​​
11.4
​​
​ ​
​
​
​ ​​ ​ ​ ​   ​
​​ ​= ​ ​  ​​​ ​​
n
​​
​​
​​
̇
w,out
0
0
1
0
144
⎣0
0
0
1 ⎦ ⎣ ​ξ ​̇ ⎦ ⎣ 9.0 ⎦
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Section 3.3 Linear Models of Process Flow Sheets
187
The matrix is already in its reduced form, the set of linear equations is independent,
and solution is straightforward.
⎢ ⎥⎢ ⎥
⎡ ​​n ​​ġ ,out​​ ⎤ ⎡ 7.9 ⎤
​​n ​​ḟ ,out​​
20.4
​​ ​   ​
​​ ​= ​ ​
​​​​ ​​
​​n ​​ẇ ,out​​
144
⎣ ​ξ ​̇ ⎦ ⎣ 9.0 ⎦
Notice how we set up the matrix equation. The x vector contains the reactant output
flows, followed by the product output flows, then the output flows of any inerts,
and finally the extent of reaction. This vector can be built using this pattern for any
number of flows and extents of reaction. In the b vector, we placed the known input
flows of the components in the same order, followed by the known fractional conversions times the input reactant flows. The matrix A contains four parts: at the top
left there is a 3 × 3 identity matrix, at the top right is a column containing the
negative of the stoichiometric coefficients of each reactant and product in the reaction, in the bottom right is the negative of the stoichiometric coefficient of the
reactant only, and the remaining bottom left is filled with zeros to complete the
square matrix. With more reactants and products, the identity matrix expands, and
with more reactions, the number of columns of stoichiometric coefficients increases.
Thus, you can write down the matrix A simply by following this pattern.
always have at least one input stream and at least two output
streams. The output streams differ in composition from each other, and from
the input stream. Separators achieve changes in composition through physical
means, not by chemical reactions. Chaps. 6 and 7 are devoted to the topic of
separation technologies.
Separators
ni,in
Separator
ni,1
ni,2
The steady-state differential material balance equation for a separator with
one input stream and two or more output streams is:
​​  ∑ ​​​​​n ​​i̇ j​​ = ​​n ​​i̇ ,in​​​
all jout
Eq. (3.24)
We build separators to recover desired components from mixtures, and to
produce pure products. For the purpose of building a linear model of a separator, we will specify separator performance by using fractional recovery, fRij,
which is the fraction of component i in the inlet stream that is recovered in
outlet stream j
​​n ​​i̇ j ​​
moles of i leaving in stream jout _
________________________
​​f​ Rij​​ = ​
​ = ​  out ​​
​​n ​​i̇ ,in​​
moles of i fed to separator
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Rearranging, we get a linear equation that describes the performance of the
separator by relating an output stream to its input:
​​​n ​​i̇ jout​​ = f​ R​  ij​​ ​​n ​​i̇ ,in​​​
Helpful Hint
Don’t get
confused—
fractional recoveries
of a stream
summed up over all
components does
not equal one!
Example 3.12
Eq. (3.25)
We can write similar equations for fractional recovery of all the components.
For a separator, the fractional recovery of one component is not the same as
that for any other component: f​ ​​RAj​​ ≠ ​fR​  Bj​​ ≠ ​fR​  Cj​​ ≠ ⋯​.
This is an important difference between fractional recovery fRij in a separator, and fractional split fSj in a splitter: fRij is a unique value for each component
and each output stream, while fSj is unique only for each output stream. The
fractional recoveries of a component i summed up over all output streams must
equal one, or:
​​  ∑ ​​​ ​fR​  ij​​ = 1​
all j​o​  ut​​
For a separator with I components and Jout outlet streams, we can specify up
to I × (Jout − 1) independent fractional recoveries.
Linear Model of Separators: Sweet Solutions
A solution of 9.8 mol% glucose, 6.6 mol% fructose, and 83.6 mol% water is fed
to a separator at a rate of 172.3 kgmol/day. Three product streams leave the separator, which operates at steady state. Stream A contains most of the glucose, stream
B contains most of the fructose, and stream C contains most of the water. 94% of
the glucose is recovered in stream A, and 4% is recovered in stream B. 85% of the
fructose is recovered in stream B, and 10% in stream A. 70% of the water is
recovered in stream C, and 15% in each of stream A and stream B. Calculate the
flows of all output streams.
172.3 kgmol/day
9.8% g
6.6% f
83.6% w
A
Separator
B
C
Solution
There are three compounds g, f, and w, and three output streams A, B, and C, so
we write three steady-state differential mole balance equations:
​​​n ​​ġ A​​ + n​​  ​​ġ B​​ + ​​n ​​ġ C​​ = ​​n ​​ġ ,in​​​
​​​n ​​ḟ A​​ + ​​n ​​ḟ B​​ + ​​n ​​ḟ C​​ = ​​n ​​ḟ ,in​​​
​​​n ​​ẇ A​​ + ​​n ​​ẇ B​​ + ​​n ​​ẇ C​​ = n​​  ​​ẇ ,in​​​
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Section 3.3 Linear Models of Process Flow Sheets
189
We also write six separator performance equations:
⎢
​​​n ​​ġ A​​ = f​ R​  gA​​ ​​n ​​ġ ,in​​​
​​​n ​​ḟ A​​ = f​ R​  fA​​ ​​n ​​ḟ ,in​​​
​​​n ​​ẇ A​​ = ​fR​  wA​​ ​​n ​​ẇ ,in​​​
​​​n ​​ġ B​​ = ​fR​  gB​​ ​​n ​​ġ ,in​​​
​​​n ​​ḟ B​​ = ​fR​  fB​​ ​​n ​​ḟ ,in​​​
​​​n ​​ẇ B​​ = ​fR​  wB​​ ​​n ​​ẇ ,in​​​
In matrix form, this set of 9 equations becomes:
⎥
⎢⎥⎢ ⎥
⎢⎥⎢ ⎥
⎡1
0
0
1
0
0
1
0
0⎤ ⎡ ​​n ​​ġ A​​ ⎤ ⎡ ​​n ​​ġ ,in​​ ⎤
​​n ​​ḟ ,in​​
0
1
0
0
1
0
0
1
0 ​​n ​​ḟ A​​
​​n ​​ẇ ,in​​
0
0
1
0
0
1
0
0
1 ​​n ​​ẇ A​​
f
​
​
n
​​
​​
​​
̇
Rg, A​​ ​​n ​​ġ ,in​​
gB
1
0
0
0
0
0
0
0
0
​​
​ ​​​ ​ ​​ ​​
0​ ​
1​
0​
0​
0​
0​
0​
0​
0​ ​​ ​​ ​  ​​n ​​ḟ B​​​​ ​​ = ​​ ​  ​fR​  f,A​​ ​​n ​​ḟ ​,in
n
​​
​​
​​
f
​
​
​​ ​​
n
​​
̇
̇
wB
Rw, A w,in​​
0
0
1
0
0
0
0
0
0
​fR​  g,B​​ ​​n ​​ġ ,in​​
0
0
0
1
0
0
0
0
0 ​​n ​​ġ C​​
n
​​
​​
​​
̇
​fR​  f,B​​ ​​n ​​ḟ ,in​​
fC
0
0
0
0
1
0
0
0
0
⎣0
⎣ ​fR​  w,B​​ ​​n ​​ẇ ,in​​⎦
0
0
0
0
1
0
0
0⎦ ⎣​​n ​​ẇ C⎦​​
⎢
⎥ ⎢⎥
⎢⎥
⎢⎥
We input the known values of stream flow rates and fractional recoveries:
⎡1
0
0
1
0
0
1
0
0⎤ ⎡ ​​n ​​ġ A​​ ⎤ ⎡16.9⎤
11.4
0
1
0
0
1
0
0
1
0 ​​n ​​ḟ A​​
144
0
0
1
0
0
1
0
0
1 ​​n ​​ẇ A​​
n
​​
​​
​​
̇
15.9
gB
1
0
0
0
0
0
0
0
0
​​
​ ​​​ ​​
0​ ​
1​
0​
0​
0​
0​
0​
0​
0​ ​​ ​​ ​  ​​n ​​ḟ B​​​​ ​​ = ​​ 1.14​
n
​​
​​
​​
̇
21.6
w
B
0
0
1
0
0
0
0
0
0
n
​​
​​
​​
0.68
ġ C
0
0
0
1
0
0
0
0
0
n
​​
​​
​​
̇
9.7
f
C
0
0
0
0
1
0
0
0
0
n
​​
​​
​​
⎣0
⎦
0
0
0
0
1
0
0
0 ⎣ ẇ C⎦ ⎣21.6⎦
and readily find the solution:
Quick Quiz 3.6
What’s the difference
between saying “94%
of the glucose is recovered in stream A”
and “stream A is 94%
glucose”?
mur83973_ch03_155-230.indd 189
⎢⎥
⎢⎥
⎢⎥
⎡ ​​n ​​ġ A​​ ⎤ ⎡ 15.9 ⎤
​​n ​​ḟ A​​
1.14
​​n ​​ẇ A​​
21.6
​​n ​​ġ B​​
0.68
n
​​
​​
​​
̇
​​ ​  fB​​ ​​ = ​​ ​  9.7​​​ ​​
​​n ​​ẇ B​​
21.6
​​n ​​ġ C​​
0.32
​​n ​​ḟ C​​
0.56
​​⎣n ​​ẇ C​​⎦ ⎣100.8⎦
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
3.3.2
Diverging tree
Converging tree
Loop
Figure 3.5 General
topological patterns
observed in many process flow diagrams.
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Process Topology
We’ve just described the four individual process units that are common to all
process flow sheets, and shown how to define performance specifications for
each. An important feature of a process flow sheet is its topology—the way
in which the process units are connected together, and the direction of flow
from one unit to another. In a mathematical model of a process flow sheet,
connections between process units are shown simply by appropriately equating
output flows from one unit to input flows of another.
There are three basic topologies we’ll discuss—diverging trees, converging trees, and loops. Figure 3.5 illustrates the general shape of these patterns.
On any particular flow sheet you will usually see some variation or combination of these. The diverging tree is a common pattern for refineries or other
similar processes, where a mixture (such as crude oil) is separated into its
component parts (such as gasoline, diesel, and asphalt). The converging tree
is common in synthesis of complex organic chemicals such as polymers or
pharmaceuticals, where a single product is produced as a result of multiple
chemical reactions involving many reactants. Loops are observed often in
chemical processes, particularly where low fractional conversions in reactors
are advantageous.
Observing the general topological patterns helps in devising strategies for
solving process flow calculations for complicated process flow sheets. For
diverging trees, it is usually easiest to start at the beginning (feed rate of raw
material) and work forward. For converging trees, it is often easiest to start at
the end (flow rate out of desired product) and work backward.
Loops on process flow sheets present challenges. One useful way to attack
loops is to “tear” the loop by choosing a “tear stream” that is internal to the
loop. We then follow a component as it moves around the loop, using performance specifications to relate one stream to another. We do this until we’ve
looped all the way back to the beginning. By doing this, we derive a new
relationship between outlet and inlet streams.
Let’s illustrate with a typical loop that contains a mixer, reactor, and separator. There are two reactants, A and B, one product C, and one reaction:
​A + 0.5B → 2C​
The flow diagram is shown in Fig. 3.6. The fractional conversion of A in the
reactor, fCA, the fractional recovery of A in stream 5, fRA5, and the molar flow
rates of all components in stream 1, n​​ ​​Ȧ 1 and n​​ ​​Ḃ 1 are known.
The conundrum that arises in the loop is that we can’t solve for stream 2
without knowing stream 5, but we can’t solve for stream 5 without knowing
stream 2. What is needed is an equation that relates the flow in stream 5 to
stream 1. To demonstrate how to come up with this equation, we’ll trace component A around the loop.
We can start at any stream that is internal to the loop; the reactor inlet is
usually a good choice. Now we start marching around the loop. From the reactor
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Section 3.3 Linear Models of Process Flow Sheets
191
5
Mixer
2
3
Reactor
Separator
A, B, C
A, B
1
4
A, B, C
Figure 3.6 Typical block flow diagram with a recycle loop. Streams 2, 3, and 5 are internal to the loop.
performance specification, we relate the flow of component A in stream 3 to
the flow of A in stream 2:
​​n ​​Ȧ 3​​= (1 − f​C​  A​​) ​​n ​​Ȧ 2​​​
​
Continuing on the loop, we relate the flow of A in stream 5 to that in stream 3:
​​​n ​​Ȧ 5​​ = ​fR​  A5​​ ​​n ​​Ȧ 3​​​
Combining these two equations gives the flow of A in stream 5 related to that
in stream 2:
​​​n ​​Ȧ 5​​ = ​fR​  A5​​(1 − f​ C​  A​​) ​​n ​​Ȧ 2​​​
Continuing again around the loop, we get back to where we started!
​​​n ​​Ȧ 2​​ = n​​  ​​Ȧ 1​​ + n​​  ​​Ȧ 5​​​
Combining these two last equations gives
​​​n ​​Ȧ 5​​ = ​fR​  A5​​(1 − f​ C​  A​​)(​​n ​​Ȧ 1​​ + ​​n ​​Ȧ 5​​)​
or, rearranging, we get
​fR​  A5​​(1 − f​ C​  A​​)
​​​n ​​Ȧ 5​​ = ​ ______________
​
​ n​​​  ​​̇ ​​​
[ 1 − ​fR​  A5​​(1 − f​ C​  A​​) ] A1
Now we have what we wanted: ​​n​​Ȧ 5 expressed in terms of known system performance specifications and known flows.
Example 3.13
Multiple Process Units and Recycle: Taking an old Plant
out of Mothballs
A customer is considering your company as a new supplier of an ethylene oxide
product. The customer requires that the product contain at least 98 mol% ethylene
oxide, and he would like to purchase 1.7 million kgmol product per year. The
contract is potentially quite lucrative, and your company could use the business.
Luckily, your company owns an ethylene oxide plant that has been in mothballs for several years, because of low demand for the product. You propose
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
recommissioning the plant now that there is a new customer. Many of the records
from the plant were lost, but you are able to dig up an old block flow diagram.
The plant produced ethylene oxide by reaction of oxygen and ethylene:
2​​ C2​ ​​​H4​ ​​ + ​O2​ ​​ → 2​C2​ ​​​H4​ ​​O​
From handwritten notes on yellowed pages you find that the reactor was designed
to operate at 6% conversion of ethylene feed, and that the separator was designed
to recover 97% of the ethylene oxide as the product stream, along with 98% of the
unreacted ethylene and 99.95% of the unreacted oxygen for recycle. From a few
old plant operating records, you find that the ethylene flow rate into the plant is
set at a maximum of 196 kgmol ethylene/h and the maximum oxygen feed rate is
84.5 kgmol O2/h.
Can the facility meet the customer’s requirements? If not, can you determine
which system performance specifications most affect the overall performance of
the process, in order to come up with proposals for modifying the plant? Assume
that the facility will be operating at maximum feed flow rates.
Solution
Here’s the block flow diagram, with streams labeled.
E, O, EO
5
Mixer
E, O
E, O, EO
2
Reactor
E, O, EO
3
1
Separator
4
E, O, EO
The equations for each process unit are (in kgmol/h):
Mixer: Three material balance equations
​​​n ​​Ė 2​​= 196 + ​​n ​​Ė 5​​​
​​n​  ​​Ȯ 2​​= 84.5 + n​​  ​​Ȯ 5​​​
​​​n ​​Ė O2​​ = n​​  ​​Ė O5​​​
Reactor: Three material balance equations plus one reactor performance specification equation
​​​n ​​Ė 3​​ = ​​n ​​Ė 2​​ − 2​ξ ​​̇
​​​n ​​Ȯ 3​​ = ​​n ​​Ȯ 2​​ − ​ξ ​​̇
​​​n ​​Ė O3​​ = ​​n ​​Ė O2​​ + 2​ξ ​​̇
​f​  ​​ ​​n ​​Ė 2​​
​​ξ ​̇ = _
​  CE  ​
= 0.03​​n ​​Ė 2​​​
2
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Section 3.3 Linear Models of Process Flow Sheets
Separator: Three material balance equations plus three separator performance
specification equations
​​​n ​​Ė 4​​ + ​​n ​​Ė 5​​ = ​​n ​​Ė 3​​​
​​​n ​​Ȯ 4​​ + n​​  ​​Ȯ 5​​ = ​​n ​​Ȯ 3​​​
​​​n ​​Ė O4​​ + ​​n ​​Ė O5​​ = ​​n ​​Ė O3​​​
​​​n ​​Ė 5​​ = ​fR​  E5​​ ​​n ​​Ė 3​​ = 0.98​​n ​​Ė 3​​​
​​​n ​​Ȯ 5​​ = ​fR​  O5​​ ​​n ​​Ȯ 3​​ = 0.995​​n ​​Ȯ 3​​​
​​​n ​​Ė O4​​ = ​fR​  EO4​​ ​​n ​​Ė O3​​ = 0.97​​n ​​Ė O3​​​
We work around the loop to derive:
​fR​  E5​​(1 − f​ C​  E​​)
0.98 × 0.94  ​ 196 = 2291​
______________
​​n​  ​​Ė 5​​ =
​
​ ​​n ​​̇ ​​ = ________________
​
1 − ​fR​  E5​​(1 − f​ C​  E​​) E1 1 − (0.98 × 0.94)
This set of equations constitutes a mathematical model of the ethylene oxide plant.
Now we can proceed to solve this set of equations, either by working through the
equations one by one, or by going to an equation-solver program. We find ​​ξ​​ ̇ =
74.6 kgmol/h and
​​n ​​i̇ 1​​n ​​i̇ 2​​n ​​i̇ 3​​n ​​i̇ 4​​n ​​i̇ 5
Ethylene
196
Oxygen 84.5
2487
2338 47
2291
2051
1976 10
1966
Ethylene oxide 0 4.5 154
149
Total
206
280
4542
4468
4.5
4262
Stream 4 is the product stream. Figuring that the plant operates 24 h/day and
350 days/year, 206 kgmol product per hour is equal to 1.73 million kgmol product
per year, just barely enough to satisfy our customer. But our product purity is only
72 mol%, well below what our customer demands.
What would be good ways to change the process to get closer to customer
requirements? Here are some ideas:
1.
2.
3.
4.
Increase fractional conversion of ethylene
Improve fractional recovery of ethylene oxide to product
Improve fractional recovery of ethylene to recycle
Reduce oxygen fresh feed to the mixer
The next step is to figure out which of these changes would have the greatest effect
on process operation. Once we have the equations set up, it is easy to explore. If
we are able to find changes in operation that might get us closer to our goal, then we
can focus efforts on redesigning the equipment to allow such changes.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
3.4
Degree of Freedom Analysis
Now that you have learned how to solve process flow calculations using a
more mathematical approach, we turn to a simple but important question: is
there a quick way to predict whether a solution to a process flow problem
exists, without actually solving the problem? Or perhaps there is more than
one solution? A problem is going to be “solvable,” with one unique solution,
if there is just the right amount of information, and just the right kind of
information. There are rigorous mathematical methods for ascertaining
whether a problem has a unique solution. What we’ll present here is a simpler
method, called degree of freedom (DOF) analysis. DOF analysis provides
a rapid means for determining if a specific process flow calculation problem
is solvable, without actually solving the problem or even setting up the equation. If you can count, you can complete a DOF analysis. What’s more, DOF
analysis helps you select the right equations to choose, after you determine
that a solution exists! It takes a little practice, but once the steps are mastered,
you will find DOF analysis to be a very rapid and useful problem-solving
technique.
3.4.1
Degree of Freedom Analysis for Single Process Units
Here is the basic procedure (you will see overlap with the Ten Easy Steps!)
1.
2.
3.
4.
Draw a flow diagram and choose a system.
Choose components.
Label the flow diagram to show components in each stream.
Note whether the problem is unsteady-state or there is any accumulation
in the system.
5. Write down chemical reactions, if any.
6. Then…
Count the number of independent variables:
Determine the
number of…
mur83973_ch03_155-230.indd 194
By counting…
in mathematical notation
stream variables
components in each​​n​ ij​​​, ​​mi​  j​​​, ​​​n ​​i̇ j​​​, and/or m
​​​ ̇ ​​ij​​​
stream
reaction variables
independent chemical​​ξ ​​k̇ or ξk
reactions
accumulation
variables
components that ​​n​ isys, f​​ − ​ni​  sys,0​​​,
accumulate or are​​m​ isys, f​​ − m
​ i​  sys,0​​​,
depleted inside system​d​ni​  ,sys​​∕dt​, and/or d​ ​mi​  ,sys​​∕dt​
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Section 3.4 Degree of Freedom Analysis
195
Count the number of independent equations:
Determine the
number of⋯
comments
equations of the type
specified flows
component or total flow​​n​ ij​​ = ⋯​, n​​​  ​​i̇ j​​ = ⋯​, n​​ j​  ​​ = ⋯​,
​​​n ​​j̇ ​​= ⋯​, or equivalent in
mass units
specified stream
compositions
at most ​(I − 1)​​​z​ ij​​ = ⋯​, or w
​​ i​  j​​ = ⋯​,
independent specified
​​ni​  j​​∕​nj​  ​​= ⋯​, or similar
stream compositions
specified system
performances
mixer feed ratio; fractional ​​n ​1̇ ​​∕n​​  ​2̇ ​​ = ⋯, ​​fS​  j​​​ = ⋯, f​​ C​  i​​ = ⋯​,
split, conversion or recovery​​fR​  ij​​ = ⋯​
splitter restrictions
total of ​(I − 1) × (​Jo​  ut​​ − 1)​​​z​ i​j​ out​​​​ = ​z​i​  j​in​​​​
equations
or w
​​ i​  ​j​ out​​​​ = ​w​i​  j​in​​​​
material balance
equations
Total of I equations
Table 3.1 or Table 3.2
7. Calculate
DOF = number of independent variables − number of independent equations.
If DOF = 0: The problem is correctly specified. There are an equal number of variables and equations, and there is probably a solution.
If DOF < 0: The problem is overspecified. There are more equations than
variables. Some of the equations are either redundant or inconsistent.
If DOF > 0: The problem is underspecified. There are more variables than
equations. This may be an opportunity for optimization. For example,
you could add additional constraints that describe desirable outcomes,
such as minimizing costs.
One of the biggest difficulties in DOF analysis is establishing whether
equations are independent. An independent equation provides unique information
that cannot be obtained by combining two or more other equations or specifications. If you find DOF < 0, look carefully at your equations and specifications.
Here are two common pitfalls:
∙ If a reactant is completely consumed ( fCi = 1.0), then there is no stream
variable for that reactant in the reactor outlet, and therefore fCi = 1.0 is not
an independent specification and should not be counted.
∙ Even if all I stream compositions are given in the problem statement, only
(I − 1) are independent.
In the next examples, we illustrate DOF analysis for several examples; you
have already seen and solved some of these.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Example 3.14
DOF Analysis: Fruit Juice Processing
Mr. and Mrs. Fruity squeeze 275 gallons of orange juice every day at the FruityFresh Farm. Their delicious juice contains 89 wt% water, 8 wt% carbohydrate, plus
several other compounds in trace quantities. They sell 82% of the juice to a processor, to make frozen concentrated juice. Some 17% of the juice is bottled for sale
at the local farmer’s market, and 1% is kept to feed all the little Fruitys. Determine
if this problem is correctly specified.
Solution
This is a splitter problem, with three output streams. The juice contains three
components: water W, carbohydrates C, and other compounds, lumped together as O.
W
C
O
W, C, O
to processor
W, C, O
to market
W, C, O
to little Fruitys
Count the number of independent variables:
No.
explanation
stream variables
12
3 in inlet, 3 in each of 3 outlets
reaction variables
0
no reaction
accumulation variables
0
steady-state
Count the number of independent equations:
No.
explanation
specified flows
1
275 gallons/day fed
specified stream compositions
2
89 wt% water, 8 wt% carbohydrates
specified system performances
2
82% to processor, 17% to market
splitter restrictions
4
3 component, 3 outlets so
(3 − 1)(3 − 1) = 4
material balance equations
1 for each component, W, C, O
3
No. of independent variables = 12
No. of independent equations = 1 + 2 + 2 + 4 + 3 = 12
DOF = 12 − 12 = 0. Problem is correctly specified!
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Section 3.4 Degree of Freedom Analysis
Example 3.15
197
DOF Analysis: Air Drying
Humid air (1.5 mol% H2O vapor, 83°F, 1.1 atm) is pumped through a tank at
100 ft3/min. The tank is filled with alumina pellets. The water vapor adsorbs to
the pellets and drier air, containing 0.06 mol% H2O vapor, leaves the tank. What
is the rate of accumulation of water in the tank? Determine the DOF.
Solution
The system is the tank with the pellets, and there are two components: water (W)
and air (A). There is no chemical reaction, but there is accumulation of water inside
the tank.
W
A
Pellets with
adsorbed W
W
A
Count the number of independent variables:
No.
explanation
stream variables
4
2 in inlet, 2 in outlet
reaction variables
0
no reaction
accumulation variables
1
W in tank
Count the number of independent equations:
No.
specified flows
1
specified stream compositions
2
specified system performances
0
splitter restrictions
0
material balance equations
2
explanation
100 ft3/min air fed
1.5 mol% water in inlet, 0.06 mol%
water in outlet
1 for each component
No. of independent variables = 4 + 1 = 5
No. of independent equations = 1 + 2 + 2 = 5
DOF = 5 − 5 = 0. Problem is correctly specified!
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Example 3.16
DOF Analysis: Ammonia Synthesis
A gas mixture of hydrogen and nitrogen is fed to a reactor, where they react to
form ammonia. The nitrogen flow rate into the reactor is 150 gmol/h, and the
hydrogen is fed at a molar ratio of 4:1 H2:N2. 70% of the N2 fed to the reactor is
converted to product. The reactor operates at steady state. Determine if this problem is correctly specified, and identify the set of variables and equations needed
to solve the problem.
Solution
Here is a diagram, labeled to indicate the three components, H, N, and A, and the
two streams, 1 and 2.
H 1
N
2
H
N
A
The process operates at steady state, and there is one reaction:
​​N2​ ​​​ + 3H​2​​ → ​2NH​3​​​
Count the number of independent variables:
No.
explanation
variables
stream variables
5
2 in stream 1​​​n ​​Ḣ 1​​​,
3 in stream 2​​​n ​​Ḣ 2,​​​
reaction variables
1
1 reaction​​​ξ ​​1̇ ​​​
accumulation variables
0
steady state
​​​n ​​Ṅ 1​​​
​​​n ​​Ṅ 2​​​, n​​​  ​​Ȧ 2​​​
Count the number of independent equations:
No. explanation
equations
specified flows
1
150 gmol/h N2 in stream 1​​​n ​​Ṅ 1​​= 150 gmol/h​
specified stream
compositions
1
4:1 H2:N2 ratio in stream 1​​​n ​​Ḣ 1​​∕n​​  ​​Ṅ 1​​ = 4​
specified system
1 70% N2 is converted in​​f​ CN​​= 0.7 = (1 × ξ​​ ​​1̇ )​​ ∕​​n ​​Ṅ 1​​​
performancesreactor
splitter restrictions
0
material balance
3 1 for each component,
​​​n ​​Ḣ 2​​ = ​​n ​​Ḣ 1​​ − 3​​ξ ​​1̇ ​​​
equations
H, N, A​​​n ​​Ṅ 2​​ = ​​n ​​Ṅ 1​​ − ξ​​  ​​1̇ ​​​
​​​n ​​Ȧ 2​​ = 2​​ξ ​​1̇ ​​​
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Section 3.4 Degree of Freedom Analysis
199
No. of independent variables = 5 + 1 = 6
No. of independent equations = 1 + 1 + 1 + 3 = 6
DOF = 6 − 6 = 0. Problem is correctly specified!
In the table, besides counting, we wrote down the variables and equations needed
to solve this problem. We were careful to write the equations in the second table
using the variables identified in the first. You can see how the DOF analysis helps
you to identify the correct set of variables and equations that will allow you to
solve the problem!
Example 3.17
DOF Analysis: Battery Acid Production
Your job is to design a mixer to produce 200 kg/day battery acid. The mixer will
operate at steady state. The battery acid product must contain 18.6 wt% sulfuric
acid in water. Raw materials available include a concentrated solution (77 wt%
sulfuric acid), a dilute solution (4.5 wt% sulfuric acid), and pure water. Is this
problem correctly specified? What variables and equations should be used?
Solution
Here is a diagram, labeled to indicate the two components, S (sulfuric acid) and
W (water), the three feed streams and one product stream.
conc
dilute
pure
S
W
S
W
S
W
battery acid
W
The process operates at steady state, and there are no chemical reactions.
Count the number of independent variables:
No.
mur83973_ch03_155-230.indd 199
explanation
in
in
in
in
variables
“conc”​​​n ​​Ṡ ,conc​​​, ​​​n ​​Ẇ ,conc​​​
“dilute”​​​n ​​Ṡ ,dil​​​, ​​​n ​​Ẇ ,dil​​​
“pure”​​​n ​​Ẇ ,pure​​​
“battery”​​​n ​​Ṡ ,batt​​​, ​​​n ​​Ẇ ,batt​​​
stream variables
7
2
2
1
2
reaction variables
0
no reaction
accumulation variables
0
steady state
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Count the number of independent equations:
No. explanation
specified flows
1
specified stream
3
compositions
specified system
performances
0
splitter restrictions
0
equations
200 kg/day battery acid​​​n ​​Ṡ ,batt​​ + n​​  ​​Ẇ ,batt​​= 200 kg/day​
77 wt% S in conc’d​​​n ​​Ṡ ,conc​​∕(​​n ​​Ṡ ,conc​​ + n​​  ​​Ẇ ,conc​​) = 0.77​
4.5 wt% S in dilute​​​n ​​Ṡ ,dil​​∕(​​n ​​Ṡ ,dil​​ + ​​n ​​Ẇ ,dil​​) = 0.045​
18.6 wt% S in battery​​​n ​​Ṡ ,batt​​∕(​​n ​​Ṡ ,batt​​ + ​​n ​​Ẇ ,batt​​) = 0.186​
material balance
2
1 for each component​​​n ​​Ṡ ,batt​​ = ​​n ​​Ṡ ,conc​​ + ​​n ​​Ṡ ,dil​​​
equations​​​n ​​Ẇ ,batt​​ = ​​n ​​Ẇ ,conc​​ + ​​n ​​Ẇ ,dil​​ + n​​  ​​ṗ ure​​​
No. of independent variables = 7
No. of independent equations = 1 + 3 + 2 = 6
DOF = 7 − 6 = 1. Problem is under specified!
An underspecified problem provides an opportunity for optimization. As the
designer of the battery acid production tank, you could adjust the relative flows
from the different sources to, for example, minimize cost. This would add one
extra constraint and allows the problem to be solved.
3.4.2
egree of Freedom Analysis for Block Flow Diagrams with
D
Multiple Process Units
DOF analysis is particularly useful when there are block flow diagrams with
multiple process units, because of the increased complexity and information
content of these problems. We can determine not only if we have enough
information, but also where information might be lacking. DOF analysis can
also guide us in setting up a strategy for completing process flow calculations.
We proceed in a similar manner, but we count stream variables in all streams,
and we count material balance equations for each process unit and then sum
them up. We need to take care to count stream composition and system performance specifications only once.
The process is illustrated in the next examples. We first complete DOF
analysis for each individual process unit in a block flow diagram, and you will
see that almost all of the units, when considered individually, are underspecified. Then we calculate DOF for the entire process and find that it is correctly
specified. Finally we show how the process DOF analysis yields a strategy for
generating a self-consistent system of equations and variables that will yield a
solution.
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Section 3.4 Degree of Freedom Analysis
Example 3.18
DOF Analysis: Adipic Acid Production
In the Case Study of Chap. 2, we developed a preliminary block flow diagram for
the production of 12,000 kg/h adipic acid from glucose. Briefly, air (21 mol% O2,
79 mol% N2) and a glucose–water solution (10 mg/mL glucose) are mixed and then
fed to Reactor 1, where the glucose and oxygen react to muconic acid, CO2, and
water. Reactor 1 outlet is sent to Separator 1, where the gases (N2 and CO2) are
removed in one stream, the water is removed in another stream, and the muconic
acid is sent to a mixer to which H2 is added. Muconic acid and hydrogen then are
sent to Reactor 2, where they are converted to adipic acid. For this example, we will
further stipulate that (a) both oxygen and glucose are 100% converted in Reactor 1;
(b) 95% of muconic acid is recovered in Separator 1 to be fed to Reactor 1, while
the remainder is lost in the water stream; (c) a 3:1 molar ratio of H2 to muconic acid
is fed to the second mixer; (d) 70% of muconic acid is converted in Reactor 2; and
(e) the outlet of Reactor 2 is sent to Separator 2, where all of the adipic acid is
recovered as pure product, and unreacted hydrogen and muconic acid are discarded.
Solution
We start by drawing and labeling the block flow diagram. Components include
glucose (G), water (W ), oxygen (O), nitrogen (N ), muconic acid (M ), carbon
dioxide (C ), hydrogen (H ), and adipic acid (A). There are two mixers, two reactors,
and two separators. The process description guides us in determining which components are in which streams. For example, since 100% of glucose and oxygen are
consumed in Reactor 1, there is no G or O in Reactor 1 outlet. It’s very important
to get the diagram correct!
O
N
G
W
1
2
mix 1
O
N
G
W
3
reactor 1
C
N
M
W
4
5
C
N
7
M
6
M
W
sep’r 1
H
8
mix 2
H
M
9
reactor 2
H
M
A
10
H
M
11
sep’r 2
12
A
Now we will go through each process unit to count variables and equations. For
each unit, we count only the streams that are input or output for that unit. We
count only the information that you know about that particular unit. Don’t count
anything that you need to solve for (even if you can solve it easily in your head).
Note the following specific points:
∙ The compositions of streams 1 and 2 are known. Since these streams each have
2 components, only one of those compositions in each stream is independent
and is counted.
∙ 100% of the glucose and oxygen are consumed by reaction in Reactor 1. We
have accounted for that by not including glucose and oxygen in the Reactor 1
output. Therefore we do not also count 100% conversion as a system performance specification.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
∙ The ratio of the hydrogen and muconic acid flows in stream 6 and 8 counts
as a system performance specification. We do not count that ratio also as a
stream composition for stream 9.
∙ The adipic acid flow rate in stream 12 counts as a specified flow for Separator 2.
The flow rate does not also count as specified for reactor 2 (stream 10),
because we know that only by solving the material balance.
∙ Complete recovery of adipic acid into a pure product stream, in separator 2,
does not count as an independent specification of recovery or composition,
because that information is accounted for by the facts that stream 12 is pure
adipic acid and stream 11 contains no adipic acid.
The DOF analysis for each unit is summarized, without further explanation,
in the table.
Count the number of independent variables:
mixer 1 reactor 1 sep’r 1 mixer 2 reactor 2 sep’r 2
stream
8
8
9
4
5
6
reaction
0
1
0
0
1
0
accumulation
0
0
0
0
0
0
Count the number of independent equations:
mixer 1 reactor 1 sep’r 1 mixer 2 reactor 2 sep’r 2
flows
0
0
0
0
0
1
compositions
2
0
0
0
0
0
performances
0
0
1
1
1
0
splitter restriction
0
0
0
0
0
0
material balances
4
6
4
2
3
3
DOF:
mixer 1 reactor 1 sep’r 1 mixer 2 reactor 2 sep’r 2
total variables
8
9
9
4
6
6
total equations
6
6
5
3
4
4
DOF
2
3
4
1
2
2
All of the process units are underspecified! Is it hopeless? Let’s complete
the DOF analysis for the entire process, where we count all the variables and all
the specifications.
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Section 3.4 Degree of Freedom Analysis
203
Count the number of independent variables:
No.
explanation
stream variables
26Count each component in every stream on the
block flow diagram, and sum
reaction variables
2
Count both reactions
accumulation variables
0
Steady state
Count the number of independent equations:
No.
specified flows
1
explanation
12,000 kg/day adipic acid in stream 12
specified stream
2
compositions
21% O2 in stream 1
10 mg/ml glucose in stream 2
specified system
3
performances
95% recovery muconic acid in separator 1
3:1 ratio of feeds to mixer 2
70% conversion of muconic acid in reactor 2
splitter restrictions
No splitters
0
material balance
22
equations
4 (mixer 1) + 6 (reactor 1) + 4 (separator 1)
+ 2 (mixer 2) + 3 (reactor 2) + 3 (separator 2)
No. of independent variables = 26 + 2 = 28
No. of independent equations = 1 + 2 + 3 + 22 = 28
DOF = 28 − 28 = 0
The process is correctly specified even though all of the individual process
units in isolation are underspecified! This is a solvable problem.
Now consider how the DOF analysis helps us to select the correct set of
variables and equations so that we can find the solution. The variables are
straightforward: we need 26 stream variables: n​​​  ​​Ȯ 1​​, ​​n ​​Ṅ 1​​, ​​n ​​Ġ 2​​, … . ​​n ​​Ȧ 12​​​, and two
reaction variables, ​​​ξ ​​1̇ ​​ and ​​ξ ​​2̇ ​​​. We need 26 equations, all expressed in terms of
these 26 variables, and from the DOF analysis we know what those equations
should be!
Specified flow: n​​​  ​​Ȧ 12​​= 12,000 kg/day​
Specified stream compositions: n​​​  ​​Ȯ 1​​∕​​n ​​Ṅ 1​​ = 21∕79, n​​  ​​Ġ 2​​∕n​​  ​​Ẇ 2​​ = 0.001​
Specified system performance: f​ ​​RM6​​= 0.95 = n​​  ​​Ṁ 6​​∕n​​  ​​Ṁ 4​​​, ​​​n ​​Ḣ 8​​∕n​​  ​​Ṁ 6​​ = 3​, and
​​fC​  M​​= 0.7 = ​​ξ ​​2̇ ​​∕n​​  ​​Ṁ 9​​​
Material balance equations: 22! All written as ∑​
​​ all ​j​ out​ ​​​​ n ​​i̇ j​​ = ​∑​all ​j​ in​ ​​​​ n ​​i̇ j​​ + ​∑​all k​ ​νi​  k​​ ​​ξ ​​k̇ ​​​
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Manufacture of Nylon-6,6
Nylon-6,6 was the first completely synthetic polymer to be manufactured and
sold on a large scale. Worldwide production of nylons is currently around 8 billion
pounds per year. Nylon fibers are used in carpeting, clothing, and tires; nylon
resins are used in making injection-molded automotive and electrical components,
photographic film, wires and cables.
Nylons are polymers, large molecules (macromolecules) with molar masses
in the thousands or even millions, made by repeatedly linking the same basic
chemical structure together. Nylons belong to a specific class of polymers
called polyamides. (Proteins are also polyamides.) Polyamides contain carbonyl
(CO) and amine (NH) groups next to each other. The repeating structure of
nylon-6,6 is
H
H
H2N
N
(CH2)6
C
(CH2)4
O
C
H
N
N
(CH2)6
C
(CH2)4
O
n
O
C
OH
O
Nylon-6,6
The “6,6” simply means that the nylon is built from two different building
blocks, or monomers, each containing 6 carbons. The bracket indicates the
repeat unit, and the n indicates the number of repeat units. Commercial grades
of nylon-6,6 have average molecular weights of about 12,000 to more than
20,000. Since the molecular weight of the repeat unit is 226, nylon-6,6 polymers
typically contain about 50 to 100 repeat units.
Nylon-6,6 is built by linking together a dicarboxylic acid, adipic acid
(abbreviated AA), and a diamine, hexamethylenediamine (HD):
HO
C
(CH2)4
C
OH
H2N
O
O
Adipic acid
(CH2)6
NH2
Hexamethylenediamine
(Can you pick out these two 6-carbon groups from the nylon-6,6 repeat unit?)
Each of the polymer building blocks is “difunctional”—that is, it has
reactive groups at both ends (carboxylic acids for adipic acid and amines for
HD). This allows, at least theoretically, unlimited growth by linking the two
building blocks end to end. Water (W) is a byproduct. The reaction is called
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Section 3.4 Degree of Freedom Analysis
205
polycondensation, and it is an important class of reaction for production of
many natural and synthetic polymers:
​n HOOC(C​H2​ ​​)4​ ​COOH + n ​H2​ ​N(C​H2​ ​​)6​ ​N​H2​ ​ →​
HO[OC(C​​H2​ ​​​​)4​ ​​CONH(C​​​H2​ ​)​6​​NH​​]n​ ​​​H + (2n − 1) ​​H2​ ​​O
(R5)
In this case study we’ll come up with a preliminary process flow sheet for
manufacture of nylon-6,6. This will take several iterations, as we go from the
simplest design to a more realistic block flow diagram. Observe that these
iterations roughly follow the route we’ve taken from Chap. 1 through Chap. 3.
First we address the question of raw material source. Adipic acid and
hexamethylenediamine are the building blocks for nylon synthesis. Do we want
to buy these from outside suppliers, or make them ourselves? Most nylon
producers make AA and HD in-house for two reasons: (1) there is essentially
no other market for HD, and (2) it is absolutely essential to control the quality
of the building blocks in order to get good quality nylon products.
How should we make these two chemicals? First, let’s consider HD synthesis. Here is one reaction pathway used commercially. First, butadiene and
hydrogen cyanide are combined to make adiponitrile:
​​C​4​​​H6​ ​​+ 2HCN → ​NC(CH​2​​​)4​ ​​CN​
(R1)
Then adiponitrile is hydrogenated to make HD:
​​NC(CH​2​​​)​4​​CN + 4​H2​ ​​ → ​H2​ ​​​N(CH​2​​​)6​ ​​​NH​2​​​
(R2)
Adipic acid can be synthesized starting with cyclohexane. (In Chap. 1, we
described an alternative process for making adipic acid from glucose. That
process has not been commercialized yet.) First, cyclohexane is partially oxidized with oxygen to cyclohexanone, with water as a byproduct. (In reality
other reactions occur simultaneously; we’ll ignore them for simplicity.)
​​C​6​​ ​H1​ 2​​ + ​O2​ ​​ → ​C6​ ​​​H1​ 0​​O + ​H2​ ​​O​
(R3)
Then, nitric acid is used as a strong oxidant to produce adipic acid (AA), with
nitric oxide (NO) and water as byproducts:
​​C​6​​​H1​ 0​​O + 2​HNO​3​​ → ​HOOC(CH​2​​​)4​ ​​COOH + 2NO + ​H2​ ​​O​
(R4)
Let’s put together a generation-consumption table for production of 1 mole of
nylon-6,6 with an average molecular weight of about 15,000 (n = 66), using
these reaction pathways, and with no net production of AA or HD.
Let’s assume that we want to build a process capable of producing
100,000 lb/day nylon-6,6. Raw material requirements and byproduct generation
are calculated from the generation-consumption analysis and summarized in
Table 3.4.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Table 3.3
Generation-Consumption Table for Nylon-6,6
Manufacture
Compound
Abbreviation
νi1
νi2
νi3
νi4
νi5
νi,net
C4H6
B −66 −66
HCN
CN
NC(CH2)4CN
AN +66 −66
H2
H −264−264
H2N(CH2)6NH2
HD +66 −66
C6H12
CH−66 −66
O2
O−66 −66
C6H10O
CK+66
−66
H2O
W+66
+66 +131 +263
HNO3
NA−132−132
HO2C(CH2)4CO2H
AA+66
NO
NO+132+132
Nylon 6,6
N66
−132−132
−66
1
1
We’ve already learned a few things. First, we need large quantities of a
strong acid (nitric acid) and of a highly toxic material (hydrogen cyanide). As
we develop our process, we need to keep in mind how safety concerns might
influence design choices. Second, we’ll be generating a lot of wastewater,
which will likely be contaminated with organic chemicals. We will need to
design systems for handling this wastewater. Third, we’ll be generating a lot
of NO, something we’ll want to minimize because of its negative environmental impact. At this early stage we would complete a preliminary economic
analysis by comparing reactant costs and product values, and we might investigate alternative raw materials and reaction schemes.
But let’s move forward and make our first attempt at a block flow diagram.
We incorporate mixers whenever two or more reactants are needed. We use
the heuristics of Chap. 2 (e.g., introduce reactants as late as possible, remove
byproducts as soon as possible) to come up with a very preliminary block flow
diagram (Fig. 3.7). The process topology is basically that of a converging tree,
common for polymer production plants.
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Section 3.4 Degree of Freedom Analysis
Raw Material and By-Products Flows for Production of
100,000 lb/day Nylon-6,6
Table 3.4
Mi
Compound
νi,net
lb/lbmol
Flow rate, lb/day
(SF = 100,000/14,934)
C4H6
−66
54
−23,900
HCN
−132
27
−23,900
H2
−264
2
−3,500
C6H12
−66
84
−37,100
O2
−66
32
−14,100
H2O
+263
18
+31,700
HNO3
−132
63
−55,700
NO
+132
30
+26,500
+1
14,934
+100,000
Nylon 6,6
Before we attempt any process flow calculations, let’s complete a DOF
analysis of the block flow diagram in Fig. 3.7. Let’s begin by making a number of simplifying approximations, as discussed in Chap. 2. We’ll assume that
the reactions go to completion, that the separators work perfectly, and that
reactants are fed in stoichiometric ratio.
H2
1
2
HCN
7
M1
C4H6
8
R1
9
M2
10
R2
3
19
4
M5
20
R5
22
S5
HNO3
5
17
C6H12
O2
6
M3
11
R3
12
S3
14
13
M4
15
R4
16
S4
21
18
Figure 3.7 Preliminary block flow diagram for nylon-6,6 production. Mixers (M), reactors (R), and separators (S)
are indicated, as are stream numbers.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Table 3.5
Enumeration of Stream Variables of Figure 3.7
× indicates that the component is present in that stream, with the simplifying
approximations that the reactions go to completion, that the separators work perfectly,
and that reactants are fed in stoichiometric ratio.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
B××
CN××
AN× ×
H
××
HD××
CH××
O××
CK×
××
W×××××
×
NA××
AA×
××
NO××
N66××
Table 3.5 shows how, under these approximations, components are distributed among the process streams. Notice how components appear and disappear on the table—this charts the progression from raw material to product.
Table 3.5 shows the count of stream variables.
In Table 3.6, we list all process units, and indicate whether the component
passes through each unit. This is really a way to enumerate the number of
material balance equations.
Now let’s complete a DOF analysis:
Count the Number of Variables
Step
stream variables
Answer
Comment
32
The number of ×’s in Table 3.5
reaction variables 5
Total
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5 chemical reactions
37
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Section 3.4 Degree of Freedom Analysis
209
Enumeration of Material Balance Equations of Figure 3.7
Table 3.6
× indicates that a material balance equation for that component
in that process unit is needed.
M1 R1 M2 R2 M3 R3 S3 M4 R4 S4 M5 R5 S5
B
×
×
CN
×
×
AN ×
×
×
H ×
×
HD××
CH ×
×
O ×
×
CK ×
×
×
W×
××
NA×
×
×
××
×
×
AA×
×
NO×
×
×
×
N66×
×
Count the Number of Equations
Step
Answer
Specified flows 1
Comment
Nylon-6,6 production rate of
100,000 lb/day
Specified stream compositions 0
Specified system performances 0
Material balance equations
36
Total
37
See Table 3.6
The problem is completely specified.
This analysis tells us exactly what we need to set up a linear model of this
system: 36 material balance equations (identified in Table 3.6), written in terms
of 5 reaction variables and 32 system variables (identified in Table 3.5), plus
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
one equation for the known flow rate. We notice that the flow diagram has the
general pattern of a converging tree. Thus, starting at the trunk (the product)
and working backward is the most straightforward strategy. The nylon-6,6 production rate of 100,000 lb/day is chosen as the basis. We move through the
material balance equations, starting with the material balance equation on
nylon-6,6 for separator S5. We leave the detailed calculations as an end-ofchapter exercise.
Whew! We’ve done a lot of work. But all we really have is the skeleton
of a block flow diagram. Still, even at this early stage, the flow diagram
aids in raising important design questions such as: What if we use air instead
of pure oxygen? Will the nitric acid be pure or in water? What if we eliminate S3 and let S4 do the job? What if we eliminate S3 and S4 and get
by with only one separator? Should we add a separator to remove water
before R4?
To make a more realistic model of this process, we remove all the simplifying approximations about complete conversion of reactants ( fC = 1.0),
perfect separation ( fR = 1.0), and stoichiometric reactant feed ratio. This
Table 3.7
Enumeration of Stream Variables of Figure 3.7
× indicates that the component is present in that stream. No simplifying approximations
were made.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
B
×
× × ×
××
×
×
×
CN
×
× × ×
××
×
×
×
AN
× ×
××
×
×
×
H
××
×
×
×
HD××
×
×
×
CH× ×
×
×
×
×
×
×
×
×
×
×
×
O× ×
×
×
×
×
×
×
×
×
×
×
×
CK ×
×
×
×
×
×
×
×
×
×
×
W ×
×
×
×
×
×
×
×
×
×
×
NA× ×
×
×
×
×
×
×
×
AA×
×
×
×
×
×
×
NO×
×
×
×
×
×
×
N66×
×
×
×
×
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Section 3.4 Degree of Freedom Analysis
Table 3.8
211
Enumeration of Material Balance Equations of Figure 3.7
× indicates that the a material balance equation for that component in that process
unit is needed. No simplifying approximations were made.
M1
R1
M2
B
×
×
×
CN
×
×
AN×
R2
M3
R3
S3
M4
R4
S4
M5
R5
S5
××
×
×
×
××
×
×
×
××
×
×
H×
××
×
×
HD××
×
×
CH ×
×
×
×
×
×
×
×
×
O ×
×
×
×
×
×
×
×
×
CK ×
×
×
×
×
×
×
×
W ×
×
×
×
×
×
×
×
NA×
×
×
×
×
×
AA×
×
×
×
×
NO×
×
×
×
×
N66×
×
generalizes the problem. Essentially, this means that once a component enters
the process its presence must be accounted for in all downstream streams. This
greatly increases the complexity of the problem, but it is actually quite straightforward to trace the components through the process—we simply enter an ×
in all cells in Tables 3.7 and 3.8 that are connected downstream of the entry
point for that component. We know this information by knowing the process
topology.
(Compare Table 3.5 to 3.7, and Table 3.6 to 3.8.) Now let’s complete a
DOF analysis:
Count the Number of Variables
Step
Stream variables
Answer
Comment
111
The number of ×’s in Table 3.7
System variables 5
Total
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Still 5 chemical reactions
116
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Count the Number of Equations
Step
Answer
Specified flows 1
Comment
Nylon-6,6 production rate of
100,000 lb/day
Specified stream compositions 0
Specified system performances 0
See explanation following.
Material balance equations
81
The number of ×’s in
Table 3.6.
Total
82
DOF = 116 − 82 = 34. Greatly underspecified!
What kind of specifications should we add to make this completely specified? We can use our tables to guide us:
∙ Mixers. Each mixer has two inputs, so we can provide one specification
for the ratio of input flows per mixer. There are 5 mixers, so 5 independent
mixer performances can be specified.
∙ Reactors. One fractional conversion per reactor per reaction can be specified, for a total of 5.
∙ Separators. There are I × (Jout − 1) independent fractional recoveries per
separator. Each separator has 2 outlet streams. S3 has 4 components,
S4 has 7, and S5 has 13 components (easily seen from Table 3.8). So,
we specify 4 fractional recoveries for S3, 7 for S4, and 13 for S5, or a
total of 24.
This adds up to 34 specifications, exactly the number we needed!
This preliminary analysis provides a systematic outline for deriving all the
equations that we need to build a linear model of this process flow sheet. Thus,
the model consists of
∙
∙
∙
∙
∙
81 material balance equations
1 product flow rate
5 mixer performance specifications (relative flow of input streams)
5 reactor performance specifications (fractional conversions)
24 separator performance specifications (fractional recoveries)
Engineers use process simulators to generate the model equations automatically, given a flow sheet along with a basis and stream composition and
system performance specifications.
Once the hard work is done, we can solve the linear model of this flow
sheet for any number of variations in the specifications. Such an effort is an
essential feature of process synthesis. Let’s illustrate by discussing a few cases.
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Section 3.4 Degree of Freedom Analysis
213
Case 1: How does cyclohexane consumption change with reactor fractional
conversion? If all reactors and separators work perfectly, 442.2 lbmol
cyclohexane/day is required. We examined several cases where the fractional
conversion dropped; we fixed the nylon production rate at 100,000 lb/day
and assumed that the fractional recoveries in the separators were still 1.0 and
that reactants were fed at stoichiometric ratio.
(a) If fractional conversion in R3 drops to 0.9, cyclohexane consumption
rises to 491.3 lbmol/day.
(b) If fractional conversion in R1 drops to 0.9, cyclohexane consumption
also rises, to 491.3 lbmol/day! This is surprising at first glance, because
cyclohexane is not in the branch with R1. This result is seen because
we specified stoichiometric feed ratios of all raw materials. If conversion in one of the branches is higher than that in another, we likely
will redesign our process to adjust feed ratios. We can use our model
to find the optimum feed ratios as a function of reactor conversion.
(c) If fractional conversion in all the reactors drops to 0.9, cyclohexane
consumption rises nearly 40% over the base case, to 606.6 lbmol/day.
(d) A drop in fractional conversion in all reactors to 0.8 is about equivalent to a drop in conversion in just R5 to 0.5—cyclohexane consumption nearly doubles in either case.
Case 2: Which system performance specifications most affect NO production?
Because of the chemical’s adverse contribution to smog, we wish to hold
NO production near the minimum possible. In the base case, 884.4 lbmol
NO/day is generated.
(a) Reducing fractional conversion in R4 has no effect on NO production
rates!
(b) Reducing fractional recovery of adipic acid in S4 to 0.9 increases NO
production, to 983 lbmol/day.
(c) Reducing fractional recovery of adipic acid in S4 to 0.9 and reducing
fractional conversion in R5 to 0.9 increases NO production even more,
to 1092 lbmol/day.
Case 3: How do separator efficiencies affect the quantity of water in the final
product? If all three separators work perfectly, no water is in the nylon
product. We considered four other conditions:
(a) If 99% of the water is removed in each of S3, S4, and S5, the nylon
product contains 0.16 wt% water.
(b) If 99% of the water is removed in S4 and S5, but only 90% in S3, the
nylon product contains, again, 0.16 wt% water.
(c) If 99% of the water is removed in S3 and S5, but only 90% in S4, the
water content of the nylon product increases slightly, to 0.17 wt%
water.
(d) If 99% water removal is achieved in S3 and S4, but only 90% in S5,
the nylon product is contaminated with 1.6 wt% water.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Clearly, water removal is most important in the last separator. We will
want to pay closest attention to careful design of this separator.
Such exploration of the flow sheet model may lead us to changes in design.
This analysis is a prelude to further detailed design of each process unit. For
example, we might ask: Are there unwanted side reactions in any of the reactors? Is it possible to design a reactor that achieves high fractional conversion,
or will separation and recycle be required? What kind of separation technology
can achieve the necessary purity? Can these separations be achieved in a single
separator, or will multiple pieces of equipment be required? These are the sorts
of questions that the next chapters aim to address.
Summary
∙ The material balance equation derives from the Law of Conservation
of Mass.
∙ Material balance equations in differential form are summarized in the
following table (i is a component, j is a stream, and k is a chemical reaction).
Accumulation =
Input
− Output
+ Generation − Consumption
d​ms​  ys​​
Total mass​​ _​ =​​​  ∑ ​​​ ​​m ​​̇ j​​​​− ​  ∑ ​​​ ​​m ​​̇ j​​​
dt
all j​​in​
all j​​out​
d​mi​  ,sys​​
Mass of i​​ _​ =​​​  ∑ ​​​ ​​ṁ ​​ij​​​​− ​  ∑ ​​​ ​​ṁ ​​ij​​​​+ ​  ∑ ​​​​ ν​ ik​​ ​Mi​  ​​ ​​ξ ​​k̇ ​​​
dt
all j​​in​
all j​​out​
all k
d​ns​  ys​​
Total moles​​ _​ =​​​  ∑ ​​​ ​​n ​​j̇ ​​​​− ​  ∑ ​​​ ​​n ​​j̇ ​​​​+ ​  ∑ ​​​​​ ​​ξ ​​k̇ ​​ ​ ∑ ​​​​ ν​ ik​​​
dt
all j​​in​
all j​​out​
all k
all i
d​ni​  ,sys​​
Moles of i​​ _​ =​​​  ∑ ​​​ ​​n ​​i̇ j​​​​− ​  ∑ ​​​ ​​n ​​i̇ j​​​​+ ​  ∑ ​​​ ​νi​  k​​ ​​ξ ​​k̇ ​​​
dt
all j​​in​
all j​​out​
all k
The differential material balance equation describes a single point in time.
At steady state, the accumulation term is set equal to zero. Steady-state
continuous-flow processes are analyzed by using the steady-state differential equation.
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Summary
215
∙ Material balance equations in integral form are summarized in the following table (i is a component, j is a stream, and k is a chemical reaction).
Accumulation =
Input
− Output
+ Generation − Consumption
Total mass​​m​ sys, f​​ − m
​ s​  ys,0​​ =​  ∑ ​​​ ​mj​  ​​
− ​  ∑ ​​​ ​mj​  ​​​
Mass of i​​m​ i,sys, f​​ − ​mi​  ,sys,0​​ =​  ∑ ​​​ ​mi​  j​​
− ​  ∑ ​​​ ​mi​  j​​
+ ​  ∑ ​​​ ​Mi​  ​​ ​νi​  k​​ ​ξk​  ​​​
Total moles​​n​ sys, f​​ − ​ns​  ys,0​​ =​  ∑ ​​​ ​nj​  ​​
− ​  ∑ ​​​ ​nj​  ​​
+ ​  ∑ ​​​ ​  ∑ ​​​ ​νi​  k​​ ​ξk​  ​​​
Moles of i​​n​ i,sys, f​​ − ​ni​  ,sys,0​​ =​  ∑ ​​​ ​ni​  j​​
− ​  ∑ ​​​ ​ni​  j​​
+ ​  ∑ ​​​ ​νi​  k​​ ​ξk​  ​​​
all j​i​  n​​
all j​i​  n​​
all j​i​  n​​
all j​i​  n​​
all j​o​  ut​​
all j​o​  ut​​
all j​o​  ut​​
all j​o​  ut​​
all k
all i all k
all k
The integral material balance equation describes the system over a defined
time interval. Batch systems are usually analyzed using the integral balance.
∙ Extent of reaction ​​ξ​​ ̇ is a measure of the number of reaction events per
unit time. ​​ξ​​ ̇ relates the rates of consumption and generation of compounds in a chemical reaction. For any reaction k and reactant or product
i, ξ​​​  ​​k̇ ​​ = ​ri​̇ k​∕​νik​ ​ = ​R​i̇ k​∕​νik​ ​​Mi​​,​ where ​​r​​i̇ k is the molar rate (​​R ​​i̇ k is the mass rate)
of reaction of compound i by reaction k and νik is the corresponding stoichiometric coefficient (negative for reactants, positive for products).
∙ There are four fundamental types of process units: mixers, splitters, reactors,
and separators.
∙ Mixer performance is characterized by a ratio of input streams
∙ Splitter performance is characterized by fractional split fSj
​​n ​​j̇ ​​
moles leaving in stream j ___
​​fS​  j​​ = ​  _____________________
​ = ​   ​​
​​n ​​i̇ n​​
moles fed to splitter
∙ Reactor performance is characterized by fractional conversion fCi
− ​  ∑ ​​​ ​vi​  k​​ ​​ξ ​k​̇ ​​
moles
of
i
consumed
by
reaction
​​fC​  i​​ = ​ ____________________________
​ = _____________
​  all k  ​​
​​n ​​i̇ ,in​​
moles of i fed to reactor
∙ Separator performance is characterized by fractional recovery fRij
​​n ​​i̇ j​​
moles of i leaving in stream j _
________________________
​​fR​  ij​​ = ​
​ = ​   ​​
n
​​
​​i̇ ,in​​
moles of i fed to separator
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
ChemiStory: Of Toothbrushes and Hosiery
The Roaring 20s was a wild, exciting time in U.S. history—a time of bootleg booze and speakeasies, rising skirts and rising fortunes. The DuPont
family was one of several fabulously wealthy families of the time. The
DuPont Company started as a gunpowder manufacturer, and had grown to
become the major supplier of explosives to the Allied forces in World War I.
With the end of WWI and the beginning of the peacetime economic expansion, the company wisely moved from explosives to consumer goods.
DuPont illustrated its new consumer focus through their famous motto:
“Better Things for Better Living through Chemistry.” Using their expertise
in cellulose and nitrocellulose chemistry, the company developed and sold
a host of new consumer products: cellophane packaging, rayon stockings,
lacquers for painting cars bright colors.
Cellulose and nitrocellulose are plant-derived polymers, although at the
time little was known about their true nature. Debates raged among European
chemists: Were polymers true molecules, albeit very large, or were they
aggregates of small molecules held together by some as-yet-unknown noncovalent force? Virtually every well-respected chemist believed the latter—
they could not fathom the idea of a molecule with a molecular weight of
100,000, any more than they could imagine “an elephant. . . 1500 ft long
and 300 ft high.” At a conference held in Europe in 1926, Hermann
Staudinger (later awarded the Nobel Prize in Chemistry) stood virtually
alone as he argued that polymers were true molecules. A young theoretical
organic chemist named Wallace Carothers was one of the small minority
who agreed with Staudinger.
Wallace Carothers, born in 1896, had an inauspicious start; he attended
his father’s secretarial school and studied typing and penmanship. Only later
would he study chemistry at the University of
Illinois and Harvard. In 1928 Carothers was
wooed to DuPont by Charles Stine, a man who
believed that corporations should have fundamental research groups for the prestige they
would bring to the company. This was a revolutionary idea at the time. DuPont was interested in Carothers because Carothers was
interested in polymers. Carothers wanted to
prove that Staudinger was right about the
molecular nature of polymers, and DuPont
seemed to be the place to do it.
In Carothers’ first attempts to make polyWallace Carothers.
mers,
he exploited the well-known chemical
Photo Researchers/
reaction between an alcohol and an organic acid
Science History Images/
Alamy Stock Photo
to produce an ester. He reasoned that if both the
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Summary
217
alcohol and the acid were difunctional (that is, possessed two reactive
groups, one on each end) he could link them together in an infinite chain,
as “poly-esters.” This worked, to a point. Polyesters were indeed produced,
but their molecular weights were only 5000 to 6000, too short to have any
commercial value. Carothers eventually realized that the huge amounts of
water produced during the ester reaction might be limiting the extent of
polymerization. His group adapted a “molecular still” apparatus to continuously remove water during the condensation reaction. Julian Hill, a chemist
in the Carothers lab, set up the molecular still and distilled water from an
acid/alcohol reaction mixture. After 12 days, Julian poked the resultant mass
with a glass rod. When he pulled the rod back, much to his surprise and
delight, along came a long thin filament. By chance, the group had discovered a polymeric material that could be spun into fibers—of great interest
for clothing, carpeting, and the like. Although these fibers were strong and
pliable, they had one serious drawback as a fabric: they melted at low temperatures, a real problem for ironing clothing. The group attempted to synthesize polyamides, reasoning that polyamides should have higher temperature
stability than polyesters, were unable to make anything of commercial interest,
and abandoned the project.
The Great Depression of the 1930s changed DuPont. The company laid
off workers and cut wages. Charles Stine was promoted and replaced by
Elmer Bolton, who was much more interested in applied than fundamental
research. Wallace Carothers became deeply depressed and suffered from
constant mood swings. Still, his scientific output was prodigious.
Bolton pushed Carothers to work once again on polyamides. In 1934,
Donald Coffman in Carothers’ group dipped a glass stirring rod into a
molten mass made from pentamethylene diamine and sebacic acid and
pulled out a fine filament. The product was lustrous, stronger than silk,
and impervious to hot water or dry-cleaning solvents. Despite the excitement surrounding Coffman’s find, it could not be commercialized—the
starting materials were too expensive and the fibers were difficult to spin.
The following year, Gerard Berchet (also in Carothers’ group) came up
with a method of making polyamide fibers from cheaper benzene-derived
chemicals—the first nylon-6,6. Significant engineering challenges lay
ahead: producing hexamethylenediamine and adipic acid in large quantities and of sufficient purity, controlling the polymer length, melt-spinning
a polymer that was insoluble in all common solvents. In the course of the
next few years these problems were solved and nylon-6,6 became the first
totally synthetic fiber to be sold commercially.
Wallace Carothers had accomplished what he set out to do: He collected
irrefutable evidence that supported Staudinger’s molecular theory of polymers. In the process, he launched a huge new industry. However, his health,
especially his mental health, drastically worsened. He began to doubt his
scientific abilities. In February 1936, he surprised everyone by getting
(continued)
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
married. He was elected to the National Academy
of Science, collapsed, and spent months recuperating in the Alps. On April 29, 1937, Wallace checked
into a hotel, emptied the contents of a cyanide capsule into a glass of lemon juice, and drank. Seven
months after his death, his daughter was born.
Nylon-6,6 was put on the market in 1940; its
first commercial use was as toothbrush bristles, to
replace the Chinese pig bristles that became unavailable after the Japanese invaded Manchuria. The key
to nylon’s commercial success, however, was not
toothbrushes, but women’s stockings: 5 million
pairs of nylon stockings went on the market in 1940
Universal Images Group/
and sold out in a single day. During World War II,
SuperStock
nylon was diverted to the manufacture of parachutes, tire cords, and tents. It is now used for clothing, carpeting, upholstery, and myriad other items, to the tune of about 1.5 lb for every person
on earth.
In the 1960s, Julian Hill, the chemist who first discovered a polyester
fiber, and Paul Flory, a prominent polymer physical chemist working at
DuPont, began to raise concerns about the huge waste problem brought
about by nylon and other plastics, and about the rise of an entire industry
based on making single-use throwaway products from cheap petroleum.
Increasingly, companies now are exploring the use of agricultural materials
to make biodegradable polymers. In a sense, these companies are moving
full circle, back to the early days of cellulose-based polymers.
Quick Quiz Answers
3.1
3.2
3.3
3.4
3.5
3.6
mur83973_ch03_155-230.indd 218
0, −100. Mass is conserved but moles are not.
2 gmol/s, 4 gmol/s
​nC​̇ 1 + ​​nC​̇ 2 − ​​n​Ċ 3 = dnC,sys∕dt
There is no summation because there is only one system and therefore
only one quantity that can be ni,sys. In the differential balance, accumulation is expressed as a derivative and not a flow.
This term is equal to the net column, scaled to the desired basis.
“94% of glucose is recovered” means that, of all the glucose fed to the
separator, 94% goes to stream A and the rest goes to other streams.
“94% glucose in stream A” means that stream A is a mixture, containing
94% glucose and 6% other materials.
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Chapter 3 Problems
219
References & Recommended Reading
1. For more on Wallace Carothers and the story of nylon, read “The Nylon
Drama” by D. A. Houshell and J. K. Smith Jr., American Heritage of
Invention and Technology, Fall 1988, pp. 40−55, or Prometheans in the
Lab, S. B. McGrayne, McGraw Hill.
2. A good readable introduction to the basics of linear algebra is contained
in Chap. 1 of Linear Algebra and Its Applications, 3rd ed., by Gilbert
Strang, Harcourt, Brace, Jovanovich, San Diego (1988).
Chapter 3 Problems
Hein’s Law: Problems worthy of attack prove their worth by hitting back.
Warm-Ups
Section 3.2
P3.1 You mix 10 gmol polystyrene (PS, average molecular weight 66,000)
with 1000 gmol benzene (C6H6). Calculate wPS and zPS.
P3.2 A 5 wt% salt/95 wt% water solution flows into a tank at 15 g/min, where
it mixes with some salt already in the tank. A 10 wt% salt/90 wt% water
solution flows out of the tank at 15 g/min. What is m
​​​ ̇ ​​w,in​​, m
​​​ ̇ ​​w,out​​, m
​​​ ̇ ​​s,in​​,
​​​ṁ ​​s,out​​, ​​​ṁ ​​in​​, ​​​ṁ ​​out​​?
P3.3 Given the reaction
​C​H4​ ​ + 2​O2​ ​→ C​O2​ ​ + 2​H2​ ​O​
If ξ​​  ​​̇ = 4 gmol/min, determine: r​​​ ​​​Ȯ 2​ ​​,​​ ​​r ​​Ċ ​O2​ ​​,​​ ​​r ​​​Ḣ 2​ ​​O​​, ​​r ​​Ċ ​H4​ ​​,​​ ​​R ​​Ċ ​H4​ ​​,​​ ​​R ​​​Ȯ 2​ ​​,​​ ​​R ​​Ċ ​O2​ ​​​,​​ ​​​R ​​​Ḣ 2​ ​​O​​​.
P3.4 100 gmol/min hydrogen and 100 gmol/min nitrogen are fed to a reactor
at steady state, where they react to ammonia:
​N
​ 2​ ​​ + 3​H2​ ​​ → 2​NH​3​​​
If the ammonia flow rate out of the reactor is 45 gmol/min, what is ξ​​ ​​?̇
P3.5 A salt water solution (5 wt% salt) flows into a tank at 15 g/min where
it mixes with some salt already in the tank. A salt solution (10 wt% salt)
flows out of the tank at 15 g/min. Write the differential mass balance
equation for salt (S). Solve for d​ ​mS​  ,sys​​∕dt​.
P3.6 For the following situations, state whether you would use the differential
or integral balance equation. Then simplify the balance equation for the
component indicated.
(a) Water is pumped into a large tank. The tank fills up over several
hours. System: tank, component: water.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
(b) A bucket is initially filled with some salt. Water is pumped into the
bucket, where it dissolves the salt. After the bucket is completely
filled, a salt solution is pumped out until all of the salt is out of the
bucket. System: bucket, component: salt.
(c) Ethylene and air are pumped into a reactor operating at steady state,
where part of the ethylene reacts with oxygen to form ethylene
oxide. System: reactor, component: ethylene.
(d) Ethylene and air are pumped into a reactor operating at steady state,
where part of the ethylene reacts with oxygen to form ethylene
oxide. System: reactor, component: ethylene oxide.
P3.7 Refer to Example 2.2 in Chap. 2. Write the material balance equations
using the mathematical notation of Chap. 3.
Section 3.3
P3.8 Define fractional split, fractional conversion, and fractional recovery.
Draw flow diagrams for a splitter, reactor, and separator. Use the flow
diagram to illustrate your definitions.
P3.9 100 lb/h of a 10 wt% glucose/90 wt% water solution is fed to a splitter.
The splitter produces two output streams, streams 2 and 3. The flow
rate in stream 3 is 75 lb/h glucose. What is fS3? fS2? wG3?
P3.10 1000 lbmol/h of a 10 mol% glucose solution is fed to an isomerization
reactor, where part of the glucose (C6H12O6) is converted to its isomer,
fructose. The extent of reaction is 40 lbmol/h. What is fCg?
P3.11 100 lb/h of a 6 wt% glucose/4 wt% fructose solution is fed to a separator.
Two product streams are produced: one stream is 50 lb/h of a 9 wt%
glucose/1 wt% fructose solution. What is the fractional recovery of glucose
in this product stream?
Section 3.4
P3.12 Refer to Example 3.1. What is the number of stream variables, number
of reaction variables, and number of accumulation variables? What is
the number of specified flows and number of independent material balance equations? Show that DOF = 0.
P3.13 Refer to Example 3.2. How many independent stream variables are in
this problem? How many independent specified stream compositions,
and what are they? How many specified flows, and what are they?
P3.14 Refer to Example 3.3. How many independent stream variables are in
this problem? How many reaction variables?
P3.15 “Air (assumed to contain 79 mol% nitrogen and 21 mol% oxygen) is
processed in a cryogenic distillation column to produce a 98 mol% oxygen product, and a nitrogen-rich byproduct. 80% of the oxygen in the
air feed is recovered in the oxygen-rich product.” From this description,
identify the (a) stream composition specifications and (b) system performance specifications. Write equations that express these specifications using the notation of Chap. 3.
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Chapter 3 Problems
221
Drills and Skills
Section 3.2
P3.16 100 gmol ethane (C2H6) and 400 gmol O2 are mixed, and the ethane is
completely burned to CO2 and H2O. What is ​​R​​C2​ ​​H6​ ​​​, ​​R​​O2​ ​​​, ​​R​C​O2​ ​​​ and R
​​ ​​H2​ ​O​​?
What is ​​r​​C2​ ​​H6​ ​​​, ​​r​O​ 2​ ​​​, ​​r​C​O2​ ​​​, and r​​ ​H​ 2​ ​O​​? Use your results to show that mass is
conserved but moles are not.
P3.17 Two streams are sent to a mixer. One stream, flowing at 64 kg/h, contains 8 wt% methanol, 24 wt% ethanol, and the remainder water. The
other stream, flowing at 128 kg/h, contains 15 wt% acetic acid, 10 wt%
methanol, and the remainder water. The mixer operates at steady state.
Selecting an appropriate version of the equations on Table 3.1 or 3.2,
simplify the equation(s) to write balances on (a) methanol, (b) ethanol,
(c) acetic acid, (d) water, and (e) total. Write equations first using only
variables of the form m
​​​ ̇ ​​ij​​​. Then substitute in numerical values and solve
for any unknowns.
P3.18 A tank is initially empty. Water flows into a tank. (a) Suppose the water
flow rate into the tank is constant at 1 kg/h. How much water enters
the tank in 2 hours? (b) Now suppose the water flow rate (kg/h) increases
as m
​​ ̇ w​​ = 1 + 2t where t is in hours. How much water enters the tank
in 2 hours?
P3.19 A solution containing 5.4 mol% ethanol, 8.3 mol% acetic acid and water
is fed to a reactor at 97 kgmol/h. In the reactor, operating at steady-state,
ethanol and acetic acid react to form ethyl acetate and water:
​​C2​ ​​H5​ ​OH + C​H3​ ​COOH → C​H​3​COO​C2​ ​​H5​ ​ + ​H2​ ​O​
The molar rate of reaction of ethanol is 4.8 kgmol/h. Draw and label a
flow diagram. Write the steady-state differential mole balance equations
for ethanol, acetic acid, water, and ethyl acetate. Calculate the flow rates
(kgmol/h) and composition (mol%) of the output stream.
P3.20 Sugar beet juice, which contains 18 wt% sugar and 82 wt% water, is
contained in a large vessel. Water is removed by evaporation at a
constant rate. If the vessel is initially loaded with 100 kg sugar beet
juice, and the process is stopped when concentrated sugar beet juice
(in the vessel) is at 65 wt% sugar, what is the total mass of water that
must be removed? What is the change in mass of sugar in the vessel?
To solve, first draw and label a flow diagram, and derive integral mass
balance equations for sugar and for water.
P3.21 Fed-batch fermentation is used to produce lysine, an essential amino
acid. The fermentor is initially filled with 600 g of a broth that contains
60 g glucose in water (plus a number of trace nutrients). Lysineproducing cells are charged to the fermentor at the beginning of the
manufacturing period. Additional broth of the same composition is fed
continuously to the fermentor at a rate of 200 g/h, and the cells consume
glucose at a rate of 25 g/h. What is the concentration of glucose
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
(g glucose/g broth) in the fermentor at the end of 6 h? Use an integral
mass balance equation to solve this problem.
P3.22 You are planning to run a fermentor to produce antibiotics from fungi,
using a broth that contains 25 wt% glucose, 6 wt% phosphate, 6 wt%
nitrate, and water. At 8 am, the fermentor is filled with 5000 g broth
and some antibiotic-producing fungi. Over the next several hours,
additional broth of the same composition is added to the fermentor at a
steady rate of 200 g/h. The cells consume glucose at a rate of 35 g/h,
phosphates at a rate of 13 g/h, and nitrates at a rate of 12 g/h. The
fermentation is stopped when the concentration of one of these three
nutrients goes to zero, because the cells can no longer survive. Which
nutrient is depleted first? At what time? What is the concentration (g/g
broth) of the other two nutrients when the fermentation is stopped?
Draw a flow diagram and use the integral mass balance equation to
solve this problem.
Section 3.3
P3.23 A solution containing 3.7 mol% ethyl acetate, 2.6 mol% acetic acid,
5.4 mol% ethanol, and the remainder water is fed to a splitter at 97 kgmol/h.
The splitter operates at steady state and has three output streams. 27%
of the flow exits in output stream 1, 54% in output stream 2, and the
remainder output stream 3. Draw and label a flow diagram. What are
the values for fS1, fS2, and fS3? Derive equations that relate the flows in
the 3 output streams to the input stream and fS1, fS2, and fS3. Then solve
for all values of n​​​  ​​​j̇ ​ out​​​​​ and n​​​  ​​i̇ ​j​ out​​​​​.
P3.24 A solution containing 6.2 mol% ethanol, 5.4 mol% acetic acid, and
water is fed to a separator at 97 kgmol/h. The separator operates at
steady state. Three product streams leave the separator. 94% of the
ethanol fed to the separator leaves in stream A and 4% leaves in
stream B. 85% of the acetic acid fed leaves in stream B and 10% in
stream A. 70% of the water fed to the separator leaves in stream C
and 15% in stream A. Draw and label a flow diagram. What are all
the values of fRij? Derive equations that relate the flows in the 3 output streams to the input stream and fRij. Then solve for all values of ​​​n ​​​j̇ ​ out​​​​​
and ​​​n ​​i̇ ​j​ out​​​​​.
P3.25 A solution containing 9.8 mol% glucose, 6.6 mol% fructose, and 83.6
mol% water is fed to a reactor at 172.3 mol/min. Glucose and fructose
are isomers—they have the same molecular formula, but different
structures, and fructose is much sweeter than glucose. In the reactor,
which operates at steady state, 53.25% of the glucose is converted to
fructose. What is the mole rate of reaction of glucose? What is ​​ξ ​​?̇
Derive three differential mole balance equations for glucose, fructose,
and water. Solve for the flow rates of all components in the reactor
output.
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P3.26 Chlorine dioxide gas is used to bleach pulp in the paper industry. The
gas is produced by the following reaction:
​6NaCl​O3​ ​ + 6​H2​ ​S​O4​ ​ + C​H3​ ​OH → 6Cl​O2​ ​ + 6NaHS​O4​ ​ + C​O2​ ​ + 5​H2​ ​O​
3000 kgmol/h of an equimolar mixture of NaClO3 and H2SO4 are mixed
with 200 kgmol/h CH3OH in a lead-lined reactor. 90% conversion of
the methanol is achieved. What is the composition and flow rate of the
reactor outlet stream?
P3.27 Formaldehyde (HCHO) is produced by partial oxidation of methanol
(CH3OH). Two unwanted side reactions can also occur when methanol
reacts with O2: the production of formic acid (HCOOH) as well as
complete combustion to CO2. Water is a product of all three reactions.
First write down the three balanced reactions. Suppose 100 kgmol/h
methanol and 20 kgmol/h O2 are fed to a reactor operating at steady
state, where 40% of the methanol and 95% of the O2 are converted to
products. The molar ratio of formaldehyde:formic acid is 10:1 in the
reactor output stream. Draw and label a flow diagram. Write down
equations relating the fractional conversion of methanol, fCM, and the
fractional conversion of oxygen, fCO, to the three extents of reaction.
Then derive differential mole balance equations for all components and
solve for the extents of reaction as well as the flow rates of all components in the reactor output.
P3.28 A stream containing 65 mol% ethylene (C2H4) and 35 mol% oxygen is
fed to a reactor at 31,000 kgmol/h. The reactor operates at steady state.
25% of the ethylene and 90.9% of the oxygen are converted to products.
The reactor output contains ethylene oxide, CO2, and water as well as
ethylene and oxygen. Draw and label a flow diagram. Find the two
chemical reactions that must be occurring in the reactor, and relate the
fractional conversions of ethylene and of oxygen to the extents of reaction.
Derive differential mole balance equations for all components and solve
for the reactor output.
Section 3.4
P3.29 “Air (79 mol% N2 and 21 mol% O2) is fed at 1000 kg/day to a cryogenic
distillation column to produce an oxygen-rich product and a nitrogen-rich
byproduct. 80% of the oxygen in the feed is recovered in the oxygenrich product.” Draw a flow diagram corresponding to this description
and correctly label all streams. Complete a DOF analysis and determine
if the problem is correctly specified. You do not have to calculate any
flows.
P3.30 Select an example problem from Chap. 2. Complete a DOF analysis and
show that the problem is correctly specified. You do not have to do any
calculations.
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
P3.31 Salad greens are washed to remove dirt, bugs, and other debris (dbd)
before being packaged for sale. A facility processes 1500 16-oz bags of
greens every day. The fresh-picked greens contain 1 pound of dbd per
12 lb greens. These greens are mixed with 150 gallons of water per day
and washed, then spun to separate the dirty water from the washed
greens. The process removes 99.9% of the dbd and all of the wastewater from the greens, and the washed greens are sent for packaging. The
wastewater is dumped to a river. The plant is limited by environmental
constraints to dumping a maximum of 4 barrels of dirty water per day,
at a maximum 1.5 vol% dbd. Complete a DOF analysis and determine
whether this problem is correctly specified. You do not need to set up
any equations or complete any flow calculations.
P3.32 Your job is to design a mixer to produce 200 kg/day battery acid. The
batter acid must contain 18.6 wt% sulfuric acid in water. Raw materials
available include a concentrated sulfuric acid solution at 62 wt% acid,
a dilute solution at 10.8 wt% sulfuric acid, and waste acid solution that
contains 0.5 wt% acid. The concentration solution costs 10 cents/kg, the
dilute solution costs 2 cents/kg, and the waste acid is free. Draw a flow
diagram and complete a DOF analysis. Derive an equation that relates
the product cost to the cost and flow rates of the raw materials. Then
determine the optimum flows of the concentrated acid, dilute acid, and
water into the mixer.
Scrimmage
P3.33 You’re a witch in need of a new magic potion. You’ve got three flasks,
containing the ingredients listed below. You’d like to mix these
together in your cauldron, heat the cauldron over a fire to evaporate
off excess water, and make 100 g of a liquid potion containing
27 wt% toe of frog, 22 wt% eye of newt, and 11 wt% wool of bat.
How many grams from each flask should you add to your cauldron?
How many grams of water should you evaporate off? The rate of water
evaporation from the cauldron decreases with time, as the potion
becomes thicker. If the evaporation rate (in grams per minute) is
30 − 2t, where t is in minutes, how long will it take to evaporate off
the right amount of water?
Flask A, wt%
Toe of frog
10 0
Eye of newt 0
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Flask B, wt%
Flask C, wt%
50
30 0
Wool of bat
40 0
10
Water
50
40
70
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P3.34 A pesticide product containing the active ingredient d-phenothrin is
sprayed within passenger aircraft cabins on planes flying international
routes to comply with the World Health Organization’s International
Health Regulations. An airline cabin of 30,000 ft3 containing 10,000 ppm
d-phenothrin is to be flushed with fresh air until the d-phenothrin concentration is reduced to less than 100 ppm. At that concentration of
d-phenothrin the cabin is considered safe. If the flow rate of air into the
cabin is 600 ft3/min, for how many minutes must the cabin be flushed
out? Assume that the flushing operation is conducted so that the air in
the cabin is well mixed.
P3.35 A 12-oz. mug of coffee contains 200 mg of caffeine. Caffeine is eliminated from the body at a rate of dmc,sys∕dt = −0.116mc,sys, where mc,sys
is the mass of caffeine in the body, mc,sys is in mg and t is in h. After
you chug down a mug of coffee, how long will it take for the caffeine
in your body to drop to 100 mg? You drink one 12-oz. mug at 6 a.m.
and another at 2 p.m. Plot the caffeine content of your body as a function of time for one 24-h interval. If you are having trouble falling
asleep at 11 p.m., would it help much to cut out that second mug?
P3.36 Titanium dioxide (TiO2) is by far the most widely used pigment in
white paint. Specifications for the white pigment powder used in paint
making require that it contain 70 wt% TiO2, 5 wt% ZnO, and 25 wt%
SiO2. Each of these powders is received at the paint factory in 50-kg
sacks. At 7 a.m., an operator fills a large empty tank, equipped with a
mixer, with 28 sacks of TiO2, 2 sacks of ZnO, and 10 sacks of SiO2.
Then, he starts making paint by continuously drawing off 500 kg/h
white powder to mix with the latex paint. At 10 a.m. and again at noon,
he adds another 14 sacks of TiO2, 1 sack of ZnO, and 5 sacks of SiO2
to the tank. At 3 p.m. he shuts off the paint-mixing operation and goes
home.
Divide the work day into separate time intervals, and apply the
material balance equation to each interval. What is the lowest amount
of powder in the tank, and when does that occur? Does the tank ever
run out? How much pigment powder is left in the tank when the operator leaves work for the day?
P3.37 A chemical plant has an accidental spill of acrylaldehyde, a volatile
liquid. The concentration of acrylaldehyde in the outside air rapidly reaches
10 ppm (parts per million). Acrylaldehyde is extremely toxic: Exposure
to concentrations above 4 ppm pose immediate dangers to human health.
Operators at the chemical plant work inside a control house, which is
located near the spill site. The control house has a volume of 10,000 ft3
and there are three air exchanges per hour with the outside air (in other
words, air flow rate through the control house is 30,000 ft3/h). The
control house normally contains no acrylaldehyde vapors. Assume that
the air in the control house is well mixed and that the outside air concentration remains steady at 10 ppm. Calculate how long the operators
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
P3.38
P3.39
P3.40
P3.41
P3.42
have to put on protective breathing equipment. (Hint: Use the control
house as the system.)
Fresh fruit juice contains 88 wt% water and 12 wt% solids. A fruit
juice processor buys fresh juice every week and makes concentrated juice by evaporating most of the water off. When the evaporator is clean, it removes water at a rate of 1770 lb/day. Over the course
of a week, however, the evaporator performance worsens because
of fouling. At the end of the week, the evaporator is shut down and
cleaned. The plant engineer estimates the evaporation rate decreases
by 10% per day. The concentrated juice must be 44 wt% solids.
Derive an equation that expresses the fresh feed rate as a function
of the day. How much fresh juice should the processor buy per
week?
A 5000 L tank is filled to capacity with a solution that contains 40 wt%
nitric acid in water. The density of this solution is 1.256 g/mL. A small
hole develops in the bottom of the tank at a corroded spot, and, as the
nitric acid leaks out, the diameter of the hole increases. This causes
the flow rate through the hole to increase linearly with time. Assume
the flow rate through the leak is initially 5 L/min and is 55 L/min after
10 min. Calculate how many grams of HNO3 have spilled on the floor
in 20 minutes, when you discover the problem.
A 15 wt% Na2SO4 solution is fed at the rate of 12 lb/min into a mixer
that initially holds 100 lb of a 50-50 mixture (by weight) of Na2SO4
and water. The exit solution leaves at the rate of 12 lb/min. Assume
uniform mixing, which means that the concentration of the exit solution
is the same as the concentration in the mixer. What is the total mass in
the mixer at the end of 10 minutes? What is the concentration of Na2SO4
in the mixer at the end of 10 minutes?
Complete the process flow calculations described in Tables 3.5 and 3.6
of the case study.
As part of a start-up process, a liquid solution containing a very hazardous material (compound X) is pumped into an empty feed tank.
The cylindrical feed tank is 1 m in diameter and 3 m high. The solution (density = 1000 kg/m3, 0.1 kg X/kg solution) is pumped in at a
rate of 40 kg/min. At a point 1 m up the tank wall, a corroded spot
gives way, and a leak develops. The leak rate gets worse as the tank
fills, with the rate increasing as the square root of the height of the
liquid above the leak:
​(kg/min) = 4 × (liquid height in tank − tank height at leak point​)0.5
​ ​​
How much compound X has leaked out, when you walk by the tank
40 minutes after the start of the tank filling process and notice the leak?
P3.43 Some diseases are treated by injecting proteins into the bloodstream of
the patient. The problem with this is that there is a sudden increase in
the protein concentration in the blood, and then a rapid fall-off. This
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227
means patients require several injections per day. A steadier blood concentration, with fewer injections, could be achieved by using controlledrelease technology. Let’s look at one example. Protein C is a blood
protein important in clotting. Researchers encapsulated protein C in a
polymeric particle. The particle is designed to slowly release the
encapsulated protein C. 100 “units” of protein C were encapsulated per
100 mg polymeric particles. The researchers placed 100 mg of encapsulated protein C in a beaker containing a blood-like solution, and
measured the amount of protein C released as a function of time. Here
are some data:
Time, (h)
Total amount of protein C
released into beaker, units
0 0
0.34 5
0.56 8
1.0
14
2.0
25
3.0
35
The researchers, who are great polymer synthetic chemists but not too
good with engineering, need your help to analyze the data. They want
you to determine:
(a) how much protein C is left in the particles as a function of time
and
(b) the rate of protein C release (units/h) as a function of time. Then,
they want you to use these data to come up with a model equation
for how the rate of protein C release depends on the quantity of
protein C left in the particle, and use this equation to determine how
long it will take for 90% of the protein C to be released. Can you
help? Start by calculating Δmsys∕Δt for each time interval, then
plotting Δmsys∕Δt versus msys, and then applying an appropriate
material balance equation.
P3.44 As part of the process of producing sugar crystals from sugar cane, raw
sugar cane juice is sent to a series of evaporators to remove water. The
sugar cane juice, which is 85 wt% water, is fed to the first evaporator
at 10,000 lb/h. The concentrated juice out of the last evaporator is
40 wt% water. First examine a system with two evaporators. Calculate
the water evaporated in each evaporator, assuming that the fraction of
water in the feed removed in each evaporator is the same. Then,
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
develop a linear model of a flow sheet with N evaporators, assuming
that the fractional recovery of water fR,w in each evaporator is the same.
Use your model to develop a plot of N versus fR,w, letting N vary from
1 to 6.
P3.45 Most pharmaceutical products are complex organic chemicals that are
made by multistep synthesis; that is, there are many reactions in series
required to convert the raw materials to the desired product. Let’s consider how the number of reactors and the fractional conversion per reactor affect the drug production rate. Suppose we feed 1000 kg/day of a
reactant to a process requiring multiple reactions. Develop an equation
of a flow sheet with N reactors, where the fractional conversion in each
reactor is fC. Use the model to plot the rate of production of the drug
product (kg/day) as a function of N and fC, letting N vary from 1 to 10
and fC vary from 0.1 to 0.9.
P3.46 A waste gas contains 55 mol% DMF (dimethylformamide—a common
solvent) in air. A purification unit is available that can remove a fraction
of the DMF in the feed to the unit. Some of the material leaving the
purification unit is recycled back to the inlet. First complete a DOF
analysis of this process. Then develop an equation that relates the DMF
in the product stream to DMF in the feed through fS, the fractional split,
and fR, the fractional recovery in the separator. Calculate the purity of
the final product, and the composition of the stream fed to the purification unit, as a function of fS and fR. Plot your results. What is the
required fractional split if fR = 0.67 and the DMF content of the exit
gas must be reduced to 10 mol%?
55% DMF
45% air
Mixer
Separator
Splitter
10% DMF
90% air
DMF
P3.47 In petroleum refining, crude oil is separated into several different
mixtures of hydrocarbons. One product stream is the light fraction
C1–C5 alkanes (methane, ethane, propane, butane, and pentane).
Before these hydrocarbons can be sold they are further separated into
five different products using a series of separators. Each separator
produces two product streams: one called “overhead” and the other
called “bottoms.”
The light alkane stream that we are in charge of processing has
a flow rate of 1000 kgmol/h and contains 10% methane, 30% ethane,
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229
15% propane, 30% butane, and 15% pentane. This stream is sent to
Separator 1. 100% of the methane and ethane, and 44.6% of the propane
fed to Separator 1 is recovered in an overhead product. All other material is recovered in the bottoms product. The overhead from Separator 1
is sent to Mixer 1, where it is mixed with the overhead from Separator 4.
Output from Mixer 1 is sent to Separator 2. 99.5% of the methane fed
to Separator 2 is recovered in the overhead product; 99.83% of the
ethane and all the propane fed to Separator 2 is recovered in its bottoms
product. The bottoms from Separator 2 is fed to Separator 3. 100% of
the methane and 99.5% of the ethane fed to Separator 3 is recovered as
overhead; 95.8% of the propane fed to Separator 3 is recovered in the
bottoms product.
The bottoms from Separator 1 is fed to Separator 4. 96.4% of the
propane fed to Separator 4 is recovered as overhead, which is sent to
Mixer 1 as mentioned earlier. 100% of the butane and pentane fed
to Separator 4 is recovered in the bottoms product, which is sent to
Separator 5. 100% of the propane and 99% of the butane sent to Separator
5 is recovered as overhead; 100% of the pentane sent to Separator 5 is
recovered as bottoms.
Draw and label a block flow diagram. Is this an example of a diverging or converging tree structure? Then, complete a DOF analysis of the
process. Give the system of variables and equations that you could use
to solve for all flow rates. You do not need to do the calculations.
P3.48 We want to design a system that produces 1000 tons/day of freshwater
from saltwater and recovers 30% of the water in the saltwater fed to the
system as freshwater. The salt removed from the feed leaves the process
as a brine byproduct. (Brine is a concentrated aqueous salt solution.)
Explore the following design variations by first setting up a general
system of equations, then simplifying by applying the specifications for
each of the variations.
(a) The seawater contains 3.5 wt% salt and the remainder water.
Compare the seawater feed rate, and the production rate and salt
concentration of the briny byproduct, assuming first that concentration of salt in the freshwater product is 0 ppm salt and then assuming it is 10,000 ppm salt.
(b) Now assume that the salt concentration in the freshwater product is
fixed at 1000 ppm salt. Examine the effect of using different saltwater feeds. Calculate the saltwater feed rate, the production of
brine, and the salt concentration of the brine, as a function of the
concentration of salt in the feed, from 1 wt% salt to 10 wt% salt.
Plot your results.
What affects the process flow calculations more, the amount of salt in
the feed or the amount of salt in the product? Why? Would it be a
reasonable approximation to assume that the freshwater is pure water,
even if there was some salt in it?
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Chapter 3 Mathematical Analysis of Material Balance Equations and Process Flow Sheets
Game Day
P3.49 Refer to the block flow diagram. A 95 mol% propylene (C3H6)∕5 mol%
propane (C3H8) feed is mixed with benzene B (C6H6) feed at a molar
ratio of 1.2:1 propylene:benzene (P:B). These fresh feeds are mixed
with recycled streams, then fed to a reactor. Two reactions occur simultaneously in the reactor, producing the desired product cumene C
(C9H12) and an undesired byproduct diisopropylbenzene D (C12H18):
​​C​3​​H6​ ​ + ​C6​ ​​H6​ ​ → ​C9​ ​​H1​ 2​​
(R1)
​​C​3​​H6​ ​ + ​C9​ ​​H1​ 2​ → ​C1​ 2​​H1​ 8​​
(R2)
Under the reaction conditions, propane I is an inert. The reactor effluent is
cooled and sent to a separator, where a vapor stream containing propylene
and propane is taken off the top and a liquid stream containing benzene,
cumene, and diisopropylbenzene is taken off the bottom. A fraction of the
vapor stream leaving the separator is purged and the remainder is recycled
to be mixed with the incoming feed stream. The liquid stream leaving the
reactor is sent to a series of two distillation columns, where benzene is
recovered and recycled, and cumene and diisopropylbenzene are separated
and sent to storage tanks. The cumene production rate is 25 gmol/s.
Develop a model of this flow sheet, where the fractional split and
the fractional conversions of propylene and benzene in the reactor are
initially unspecified. All separators work perfectly. Then, use your
model to explore how the flow rate through the reactor (stream 2) is
affected by these three performance specifications.
10
11
Splitter
P
I
9
1
P
I
B
Mixer
2
P
I
B
Reactor
3
P
I
B
C
D
P
I
P
I
8
V/L
Separator
4
B
C
D
Distillation
column 1
6
C
5
C
D
Distillation
column 2
7
D
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CHAPTER FOUR
4
Synthesis and Analysis of
Reactor Flow Sheets
In This Chapter
We look more closely inside chemical reactors. We quickly review methods
to select chemical reaction pathways that were discussed in Chap. 1. We revisit
material balances with chemical reactors, but in a more comprehensive manner.
We introduce three ways to think about chemical reactor performance: conversion, yield, and selectivity, and we show the important role of recycle and
purge in synthesizing reactor flow sheets. You won’t become an expert, but
you will gain deeper insight into the inner workings of reactors.
The questions we’ll address in this chapter include:
∙
∙
∙
∙
∙
What are the main classes of industrially important chemical reactions?
What are the different kinds of chemical reactors?
How do I describe reactor performance?
Why don’t all reactors achieve complete conversion of reactants?
When is it useful to insert recycle or purge streams into reactor flow sheets?
Words to Learn
Watch for these words as you read Chapter Four.
Extent of reaction
Limiting reactant
Excess reactant
Catalyst
Conversion
Yield
Selectivity
Recycle
Purge
231
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
4.1
Introduction
Chemical reactors are at the heart of any chemical process (Fig. 4.1). Reactors provide the conditions that allow chemical reactions to occur, so raw
materials are converted into products. In fact, you might say that the ability
to deal with reacting systems is one of the distinct skills that differentiates
chemical engineers from other kinds of engineers. The challenge in engineering chemical reactions are many: raw materials and reaction pathways must
be selected; reactor shape, size, and operating conditions must be chosen; the
reactor flow sheet must be designed; and the chemistry, equipment, and flow
sheet must be combined in a way that is safe, economical, and environmentally sound.
4.1.1
Industrially Important Chemical Reactions
It’s impossible to list all the myriad chemical reactions that humans employ to
convert the raw materials we have into the products we want. Here, we list
some categories of chemical reactions that are important industrially. Most of
these chemical reactions are also important in the natural world—for the functioning of everything from single-cell organisms to ecosystems.
Oxidation. Oxidation was probably the first chemical reaction to be exploited
by humans. Complete oxidation of carbon- and hydrogen-containing materials provides heat for cooking and warmth. Oxidizing agents range from
oxygen to hydrogen peroxide, widely used for bleaching and disinfecting, to potassium nitrate, used in explosives. Partial oxidation allows
introduction of oxygen groups into hydrocarbons derived from fossil
fuels and is an important step in production of a huge array of industrial
chemicals, including alcohols and organic acids.
Figure 4.1 Chemical reactors come in all shapes and sizes. In commodity chemical plants, multistory reactors
in outdoor facilities are common. In a brewery, the fermentors are human size. Mammalian cells are tiny
chemical reactors just a few micrometers in diameter.
Left: Belish/Shutterstock; Middle: Tanya Sakharova/Shutterstock; Right: Dr. Dennis Emery/Iowa State University/McGraw Hill
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Section 4.1 Introduction
233
Hydrogenation and dehydrogenation. These reactions are of utmost
importance in organic and inorganic chemistry. Crude oil is hydrogenated to remove sulfur and nitrogen, thus avoiding release of harmful
acid gases upon burning of gasoline or other fuels. Dehydrogenation
of fats to oils changes the material from solid to liquid as carboncarbon single bonds are converted to double bonds. Hydrogenation
of nitrogen produces ammonia; discovery of this reaction pathway led
to huge increases in agricultural output. Either hydrogen gas or stronger reducing agents, like sodium borohydride, are commonly used
reactants in hydrogenation processes.
Polymerization. In polymerization reactions, one or two types of small molecules with reactive ends are linked together to form chains that can reach
molecular weights in the millions. Rubber, cellulose, starch, proteins, and
DNA are all naturally occurring polymers. Nylon, polyester, Teflon, polycarbonate, and other synthetic polymers are ubiquitous in modern life.
Hydrolysis and dehydration. Water is added to compounds in hydrolysis
reactions and removed in dehydration reactions. Hydrolysis often leads
to the breakdown of larger molecules to smaller—for example, hydrolysis
of starch produces sugars—and is a key chemical reaction in biodegradation. Conversely, dehydration is frequently a key step in polymerization
reactions.
Halogenation and other substitution reactions. Halogens have strong electronwithdrawing power; chlorine and fluorine in particular are added to
hydrocarbons to tune their physicochemical properties. Halogenated
hydrocarbons include refrigerants like the Freons and polymers such as
polyvinylchloride (PVC). They are generally quite resistant to chemical
and biological degradation. This resistance to degradation makes them
very useful—PVC is popular for underground pipe for example—but also
means that these compounds persist for long times in the environment.
Isomerization. Isomers are chemicals with identical molecular formulas but
different spatial arrangements of the constituent elements. This spatial
arrangement can dramatically alter the properties of isomers. Glucose
and fructose are both simple carbohydrates (C6H12O6), but fructose is
much sweeter than glucose. Conversion of glucose to fructose is big
business—check out the ingredients listed on a bottle of nondiet soda.
N-octane and iso-octane (2,2,4-trimethylpentane) are both alkanes of the
same molecular formula (C8H18), but perform drastically different in
automobile engines: A sedan will run like a race car on isooctane but
knock and ping on n-octane.
Ring opening/ring closing. Cyclic aromatics are mainstays of dyes and
pharmaceuticals. Benzene (a cyclic aromatic) and cyclohexane (a cycloalkane) are examples of cyclic compounds that serve as raw materials for
the synthesis of a large diversity of compounds. Ring-opening reactions
are an important class of reactions leading to polymer synthesis.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
4.1.2
Heuristics for Selecting Chemical Reactions
Given the plethora of possible chemical reaction pathways, it is useful to have
a few heuristics (rules of thumb) to guide initial selection of raw materials and
chemical reaction pathways. Heuristics are guidelines, not laws. Experienced
engineers use heuristics to eliminate clearly unsafe or unworkable schemes and
to quickly generate a few reasonable choices that can be evaluated in more
detail. Some heuristics for synthesis of block flow diagrams were introduced
in Section 2.6.1. Some useful heuristics specifically for reactor design are:
1. Aim to maximize incorporation of reactant atoms into the final product.
Choose raw materials that are as close as possible in chemical structure to
the final product. Avoid chemical syntheses that use temporary chemical
modification of the reactants (e.g., protection/deprotection schemes). Avoid
introducing elements that are not incorporated into the final product.
2. Choose reactants to minimize risk of explosions, fires, or release of toxic
materials. If use of hazardous materials is unavoidable, design for minimum
reactor volume. In syntheses requiring multiple reactors, avoid storage of
hazardous intermediates.
3. Use high-purity raw materials to minimize unwanted side reactions. Consider
purifying raw materials before introduction into a reactor, if possible.
4. Favor reaction schemes requiring fewer steps.
5. Use a catalyst (a material that speeds up the reaction rate) if at all possible.
6. Choose reactions that proceed spontaneously at temperatures and pressures
as close to ambient conditions as possible. Temperatures and pressures above
ambient are preferable to those below ambient.
The rationale behind many of these heuristics will become clearer as we delve
into chemical reactor design and analysis.
4.1.3 A
Brief Review: Generation-Consumption Analysis
and Atom Economy
Selection of appropriate raw materials and reaction pathways is the first step in
design of chemical reactor flow sheets. These ideas were introduced in Chap. 1
and are reviewed here.
A generation-consumption analysis is a systematic way to analyze chemical reaction pathways. To complete a generation-consumption analysis, start
with a set of balanced chemical reactions, make a table, and
1. List all compounds (reactants or products) in the first column.
2. Using a new column for each chemical reaction, write νik for each compound
involved in that reaction.
3. Add up the numbers in each row and put the sum in the last column.
4. If there is an unwanted nonzero entry in the last column, find multiplying
factors for the reactions involving that species such that the row will sum
to zero.
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Section 4.1 Introduction
235
Atom economy is a simple indicator of the efficiency of utilization of raw
materials in a given reaction pathway and is easily calculated, once the generationconsumption analysis is complete, as
​νP​ ​​ M​P​
Fractional atom economy = ____________
​
​
−​  ∑ ​​​ ​ν​i​ ​M​i​
Eq. (1.4)
all reactants
where νP and MP are the stoichiometric coefficient and molar mass, respectively, of the product. Generation-consumption analysis and calculation of atom
economy tell you the best you can do, given the chosen reaction pathway.
A real process will never achieve quite as good utilization of raw materials.
All else being equal, reaction pathways with high atom economy are preferable;
these should have fewer waste products and, by making good use of the raw
materials, should be more cost-efficient.
Example 4.1
Generation-Consumption and Atom Economy: Improved Synthesis
of Ibuprofen
Ibuprofen [2-(p-isobutylphenyl)propionic acid, C13H18O2] is an over-the-counter
drug used to treat minor aches and pains. About 30 million lb of the medicine are
produced per year. The traditional synthesis of ibuprofen involves six steps, starting with isobutylbenzene (C10H14) and acetic anhydride (C4H6O3):
C10H14 + C4H6O3 + AlCl3 + 6H2O → C12H16O + CH3COOH
+ AlCl3°6H2O
(R1a)
C12H16O + C4H7O2Cl + NaOC2H5 → C16H22O3 + C2H5OH + NaCl
(R1b)
C16H22O3 + HCl → C13H18O + C2H5OOCCl
(R1c)
C13H18O + NH2OH → C13H19ON + H2O
(R1d)
C13H19ON → C13H17N + H2O
(R1e)
C13H17N + 2H2O → C13H18O2 + NH3
(R1f)
In the early 1990s, when the patent for ibuprofen expired, a new process was
developed and commercialized. The new process involves three reaction steps over
catalysts, again starting with isobutylbenzene and acetic anhydride:
C10H14 + C4H6O3 → C12H16O + CH3COOH
(R2a)
C12H16O + H2 → C12H18O
(R2b)
C12H18O + CO → C13H18O2
(R2c)
What is the difference in atom economy between the traditional and the newer
process?
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Solution
First let’s complete a generation-consumption analysis and calculate the atom
economy of the traditional scheme. For the atom economy calculation, we consider
all net reactants but only the desired product.
Compound​
νi​1a​​ν​i1b​​ν​i1c​​ν​i1d​​ν​i1e​​ν​i1f​​ν​i,net​
Mi
νiMi
C10H14
−1 −1
134
−134
C4H6O3
−1 −1
102
−102
AlCl3
−1 −1
133.5
H2O
−6 +1
C12H16O
+1
CH3COOH
+1 +1
AlCl3°6H2O
+1 +1
+1
−2
−6
18
−133.5
−108
−1
C4H7O2Cl
−1 −1
122.5
−122.5
NaOC2H5
−1 −1
68
−68
C16H22O3 +1
−1
C2H5OH
+1 +1
NaCl
+1 +1
HCl
−1 −1
C13H18O +1
36.5
−36.5
−1
C2H5OOCCl
+1 +1
NH2OH−1 −1
C13H19ON +1
33
−33
206
+206
−1
C13H17N +1
−1
C13H18O2
+1
+1
NH3
+1
+1
The fractional atom economy is
206
​νP​ ​ ​MP​ ​
___________________________________________
​ _____________
​ =
​        ​
−​  ∑ ​​​ ​ν​i​ ​M​i​ 134 + 102 + 133.5 + 108 + 122.5 + 68 + 36.5 + 33
all reactants
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= 0.28
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Section 4.1 Introduction
237
Now let’s analyze the new scheme:
Compound​
νi​2a​​ν​i2b​​ν​i2c​​ν​i,net​
Mi
νiMi
C10H14
−1
−1
134
−134
C4H6O3
−1
−1
102
−102
C12H16O
+1
CH3COOH
+1
−1
2
−2
CO−1
−1
28
−28
C13H18O2+1
+1
206
+206
−1
H2
−1
C12H18O
+1
+1
−1
206
​νP​ ​ ​MP​ ​
​ ____________
​ = _________________
​     ​= 0.77
−​  ∑ ​​​ ​ν​i​ ​M​i​ 134 + 102 + 2 + 28
all reactants
This is an incredible improvement.
4.1.4
Reactor Design Variables
In a chemical reactor, chemical reactions take place under controlled conditions.
The engineer exercises substantial control in selecting reactor design variables
to optimize the performance of the process and the quality of the product. Some
of the key choices that must be made include:
Reactor temperature and pressure. Reactor temperature and pressure are
manipulated to maximize the conversion of raw material to desired
product while reducing or eliminating undesired reactions. A fermentor
will usually operate at about 37°C and 1 atm pressure. Reactors processing hydrocarbon gases might operate at temperatures as high as 500 to
600°C and pressures as high as 400 bar. Higher temperatures and pressures are feasible, but require special materials of construction. Some
reactions, such as those involving semiconductor materials, are carried
out under vacuum. Rigorous control of temperature and pressure are
required for optimal performance.
Reactor volume. Chemical reactors differ in size by orders of magnitude.
A single yeast cell, for example, is about 1 μm in diameter, yet carries
out a huge number of chemical reactions, including the remarkable reactions involved in self-replication. On the other hand, a commodity
chemical reactor might process a million tons per year using a single
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
chemical reaction, and could easily be 30 ft high and 15 ft wide. The
reactor size is chosen on the basis of the volume of material to be processed as well as the time required for the desired reaction.
Residence time. The residence time is the time that the material stays in
the reactor. Residence time varies from less than a second to several
days, depending on the rate of the reaction. For continuous reactors,
residence time is the reactor volume divided by the volumetric flow rate
through the reactor. A fast reaction rate means a shorter required residence time, which translates into lower reactor volume and therefore
lower equipment cost.
Reactant addition. Most chemical reactions require two or more reactants.
The reactants can be mixed and fed to the reactor at exact stoichiometric ratio. This isn’t always the best choice; nonstoichiometric feed ratios
might be chosen to minimize unwanted byproducts, for example, or to
ensure complete conversion of the most expensive reactants, or to control heat release. In semibatch reactors, one of the reactants may be
slowly dripped into a pool of the other reactants, for similar reasons.
Catalysts. A catalyst is a material that speeds up the rate of a reaction.
Speeding up reactions reduces the size of reactors and saves money.
Catalysts may allow the reactions to occur at reactor temperatures and/or
pressures closer to ambient, thus increasing the safety of a process. A
good catalyst will speed up the rate of a desired reaction without speeding
up the rate of undesired reactions. In this way, more of the desired product and less of undesired byproducts are made. Catalysts are not consumed by reaction, so theoretically they can be used indefinitely. Acids
and bases are used quite commonly as catalysts, both in the laboratory
and in the plant. In large reactors producing commodity organic chemicals, solid metal catalysts are often preferred. These are often expensive,
but solid catalysts are easily separated from fluid process streams for
recovery and reuse. Enzymes are protein catalysts. The yeast cell requires
hundreds of enzymes to control its metabolism and growth. Enzymes are
used commercially, especially in the food, pharmaceutical, and biotechnology industries, because they are highly selective and specific catalysts.
Mode of operation. Reactors can operate in batch, semibatch, or continuous
mode. Batch reactors have several advantages: The initial capital investment is lower, and operation is more flexible because the same equipment can be used to make many different products. They are preferred
for small-volume and specialty chemicals, and are used widely for biochemical, specialty polymer, and prescription pharmaceutical applications. Continuous-flow reactors are cheaper to operate in the long run,
and provide greater quality control. They are ubiquitous in the manufacture of large-volume products such as commodity chemicals. Semibatch
reactors are specialty reactors, used sometimes for example in polymer
synthesis or fermentation reactions.
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Section 4.2 Reactor Material Balance Equations
(a)
(b)
239
(c)
Figure 4.2 Typical flow patterns in chemical reactors include (a) stirred tank batch, (b) stirred
tank continuous flow, and (c) plug flow.
Mixing patterns. The extent of mixing inside the reactor is carefully controlled. Batch, semibatch and continuous reactors can be operated as
stirred-tank reactors, with complete mixing of their internal contents. The
concentration inside a stirred tank reactor may change with time, but it is
the same at every location inside the reactor. Continuous-flow reactors are
sometimes designed as plug-flow reactors. In plug-flow reactors, the fluid
moves as a “plug” through the reactor, and the concentration changes with
distance as the reaction proceeds. There are many other variations in flow
pattern that lie in between the stirred-tank and plug-flow reactor (Fig. 4.2).
Throughout Chapters 4 and 5, we will discuss how the choice of various
reactor design variables affects reactor performance. By knowing how reactor
design variables influence reactor performance, we make better engineering
choices. But first we will review and discuss reactor process flow calculations.
4.2
Reactor Material Balance Equations
Reactors exist for one purpose: to provide the conditions that allow a desired
chemical reaction to occur. At its simplest, a reactor has one input and one output.
In this section, we review the use of material balance equations with reacting
systems. As you learned in Chap. 3, there are several different forms of the material balance equation. Our primary goal in this section is to discuss which of the
material balance equations is most suitable for particular types of problems.
4.2.1
Reactors with Known Reaction Stoichiometry
If the stoichiometric coefficients are known, then the extent-of-reaction concept is very useful in process flow calculations for reacting systems. Recall
that reaction rates are characterized by the extent of reaction ​ξ ​̇ (moles/time).​
ξ ​̇ relates the reaction rates of reactants and products through their stoichiometric coefficients. If ​​r ​​Ȧ ​is the rate of consumption of reactant A and ​​r ​​Ḃ ​ is
the rate of generation of product B, then r​​ ​​Ȧ ​ and ​​r ​​Ḃ ​are related by:
​​r ​​Ȧ ​ __
​​r ​​Ḃ ​
̇
​ __
​ν​  ​​ = ​  ​ν​  ​​ = ​ξ ​
A
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B
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
or, in general, the extent of reaction is defined such that:
​​r ​​i̇ k​= ν​ ​ik​ ​​ξ ​​k̇ ​
Eq. (3.6b)
where ​​r ​​i̇ k​is the molar rate of reaction of compound i in reaction k, ​ν​ik​is the
stoichiometric coefficient for compound i in reaction k, and ξ​​ ​​k̇ ​ is the extent of
reaction k.
Similarly, if the extent of reaction is expressed in terms of quantity rather
than rate, as ξ (moles), then
​
r​ik​= ν​ ​ik​ ​ξ​k​
Eq. (3.6a)
Assuming that the stoichiometric coefficients are known, then the best
choice for the material balance equation depends on the nature of the reactor.
Is the reactor continuous-flow, batch, or semibatch? Steady state or unsteadystate? We will consider several variations.
4.2.1.1 Continuous-Flow Steady-State Reactors Continuous-flow steady-state
reactors are the workhorses of commodity chemical processes. Assuming one
input and one output stream, the material balance equation simplifies to
​​ṅ ​​i,out​ = n​​ ̇ ​​i,in​ +​  ∑ ​​​ ​ν​ik​ ​​ξ ​​k̇ ​
all k
n˙ i,in
Helpful Hint
Don’t forget step 1 in
the 10 Easy Steps—
always draw a
diagram!
Example 4.2
Σνikξ˙k
Eq. (4.1)
n˙i,out
Given information about the output stream, then we solve for the extents of
reaction; alternatively, given information about the reaction rates, we solve for
the output stream. We first apply the material balance equation to the compounds about which most is known, then move on to the material balances on
other compounds.
Continuous-Flow Steady-State Reactor with Known Reaction
Stoichiometry: Sustainable Synthesis of Acetic Acid
Acetic acid is an important bulk chemical that is typically synthesized from
methanol and CO, where the reactants are sourced from fossil fuels. There is
great interest in developing alternative processes using readily available CO 2,
along with methanol made from biomass, a renewable raw material. A team of
process development engineers is researching acetic acid synthesis using the
reaction:
CH3OH + CO2 + H2 → CH3COOH + H2O(R1)
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Section 4.2 Reactor Material Balance Equations
241
The team feeds a mix of 25 mol% CH3OH, 25 mol% H2, and 50 mol% CO2 at
20.0 gmol/h into a pilot-scale reactor operating at steady-state. The stream leaving
the reactor is analyzed and shown to contain 17.6 mol% acetic acid. What is the
flow rate and composition (mol% of all compounds) of the reactor output, and what
is the extent of reaction ξ​​ ​​1̇ ​?
Solution
For review, we will proceed using the Ten Easy Steps explicitly.
Steps 1–4. Draw diagram, choose system, check units, choose components, and
define stream variables. The reactor is our system, components are
methanol (M), CO2 (C), H2 (H), acetic acid (A), and water (W). Units
are all in gmol/h and mol% so no unit conversion is needed. Streams
are labeled 1 and 2.
M
C
H
1
Reactor
2
M
C
H
A
W
The stream variables are n​​ ̇ ​​M1​, ​​n ​​̇C1​, ​​n ​​̇H1​, ​​ṅ ​​M2​, ​​ṅ ​​C2​, ​​ṅ ​​H2​, ​​ṅ ​​A2​, ​​ṅ ​​W2​.
Step 5. Define basis. A flow rate is given; in terms of stream variables we
write:
​​ṅ ​​M1​ + ​​ṅ ​​C1​ + ​​ṅ ​​H1​= 20.0 gmol/h
Step 6.
Step 7.
Define system variables. There is one reaction, so we define one extent
of reaction ξ​​ ​​1̇ ​, which we will need to solve for. The reactor operates
at steady state so we do not need to worry about accumulation variables.
List all specifications. We are given stream composition information
about both the input and output streams. Writing these in terms of our
stream variables
​​ṅ ​​M1​
_______________
​
​= 0.25
​​ṅ ​​M1​ + ​​ṅ ​​C1​ + ​​ṅ ​​H1​
​​ṅ ​​C1​
_______________
​
​= 0.50
​​ṅ ​​M1​ + ​​ṅ ​​C1​ + n​​ ̇ ​​H1​
​​ṅ ​​A2​
_________________________
​
​= 0.176
​​ṅ ​​M2​ + ​​ṅ ​​C2​ + ​​ṅ ​​H2​ + n​​ ̇ ​​A2​ + ​​ṅ ​​W2​
(We can also write an equation for the H2 mole fraction in the input
stream, but this is not an independent specification.)
There are no reactor performance specifications.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Step 8.
Write material balance equations. We have 5 components so there are
5 equations, all which follow the form of Eq. (4.1) and written to
incorporate the known stoichiometric coefficients of reaction (R1):
​​ṅ ​​M2​ = ​​ṅ ​​M1​ − ​​ξ ​​1̇ ​
​​ṅ ​​C2​ = ​​ṅ ​​C1​ − ξ​​  ​​1̇ ​
​​ṅ ​​H2​ = ​​ṅ ​​H1​ − ​​ξ ​​1̇ ​
​​ṅ ​​A2​ = +​​ξ ​​1̇ ​
​​ṅ ​​W2​ = +​​ξ ​​1̇ ​
Before we proceed to solve this system of equations, let’s step back
and look at the DOF analysis:
Count the number of independent variables:
No.
Explanation
stream variables
8
3 in stream 1, 5 in stream 2
reaction variables
1
One reaction
accumulation variables
0
steady state
Count the number of independent equations:
No.
Explanation
specified flows
1
20 gmol/h fed
specified stream compositions
3
25 mol% methanol, 25 mol% hydrogen,
17.6 mol% acetic acid
specified system performances
0
splitter restrictions
0
material balance equations
5
1 for each component
No. of independent variables = 8 + 1 = 9
No. of independent equations = 1 + 3 + 5 = 9
DOF = 9 − 9 = 0. Problem is correctly specified!
Notice how the DOF analysis maps onto the variables and equations
that we identified. It is now straightforward to march through the equations and solve for the flow rates of all components in the output stream
as well as for the extent of reaction. A convenient way to organize the
calculations is through the use of a “mole table.” The “input” column
is calculated from the known flow rate and composition of stream 1,
while the “output” is from the material balance equation.
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243
Section 4.2 Reactor Material Balance Equations
Compound
Input (gmol/h)
Output (gmol/h)
methanol M
5
5 − ξ​​​  ​​1̇ ​​
carbon dioxide C
10
10 − ξ​​​  ​​1̇ ​​
hydrogen H
5
5 − ξ​​​  ​​1̇ ​​
acetic acid A
0
0 + ξ​​​  ​​1̇ ​​
water W
0
0 + ξ​​​  ​​1̇ ​​
total
20
20 − ξ​​​  ​​1̇ ​​
We use the known acetic acid composition of the output stream to solve
for ​​​ξ ​​1̇ ​​:
​​ξ ​​1̇ ​
0.176 = ________
​
​
(20 − ξ​​  ​​1̇ ​)
​​ξ ​​1̇ ​= 3.0 gmol/h
The remaining calculations are left for the reader. The reactor output flow
rate is 17 gmol/min, and it contains 11.8 mol% methanol, 41.2 mol% CO2,
11.8 mol% H2, 17.6 mol% acetic acid, and 17.6 mol% water.
Example 4.3
Continuous-Flow Steady-State Reactor with Multiple Chemical
Reactions: Combustion of Natural Gas
Natural gas, produced over eons by decay of ancient plants, is recovered from wells
and used widely as a source of heat and energy. The composition of natural gas
varies from well to well. Here are analyses of natural gas from three different
sources Well NM (New Mexico), Well B (Brazil), and Well TX (Texas):
Composition of Natural Gas (mol%)
Component
mur83973_ch04_231-320.indd 243
Well NM
Well B
Well TX
CH4 (methane)
96.91
81.57
67.0
C2H6 (ethane)
1.33
9.17
3.8
C3H8 (propane)
0.19
5.13
1.7
C4H10 (butane)
0.05
2.66
0.8
C5H12 (pentane)
0.02
0.56
0.5
CO2 (carbon dioxide)
0.82
0.39
0.0
N2 (nitrogen)
0.68
0.52
26.2
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
A furnace operating at steady state burns 1.00 MMSCFD (1 million standard cubic
feet per day) of natural gas from Well NM. Mixed with the fuels is 23.0 MMSCFD
air (assumed to be 79 mol% N2, 21 mol% O2). The only detectable compounds in
the flue gas are CO2, H2O, O2, and N2.
Complete combustion of hydrocarbons is described by the general chemical
equation
y
y
​C​x​ ​Hy​ ​ + ​(x + __
​   ​)​​O2​ ​ → x​CO​2​ + __
​   ​ ​H2​ ​O
4
2
What is the flow rate (kgmol/h) and composition (mol%) of the flue gas?
Solution
In a furnace, natural gas is fed to burners that line the floor or the lower walls of
the firebox. Air is mixed in and the gas is combusted. The hot gases rise by convection, and as the gases rise, they cool by heat exchange with other fluids. For
example, there may be tubes placed in the furnace through which water flows; as
the gases rise, the water is heated to steam. The cooled gases exit through a flue
to the atmosphere.
Flue gas to atmosphere
CO2, H2O, O2, N2
Natural gas:
CH4, C2H6, C3H8,
C4H10, C5H12,
CO2, N2
Air:
N2, O2
The furnace is modeled as a mixer plus reactor, as shown in the block flow diagram.
Natural gas
Air
mur83973_ch04_231-320.indd 244
M
E
Pr
B
Pe
C
N
O
N
Mixer
Reactor
O
N
C
W
Flue gas
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Section 4.2 Reactor Material Balance Equations
245
Let’s quickly complete a DOF analysis. There are 13 stream variables and 5 reaction
variables (one combustion reaction for each of the hydrocarbons). Flows of all
compounds into the furnace are readily calculated from the 9 specifications
(2 specified flows and 7 independent specified compositions). There are 9 different
compounds in the system, so we can write 9 independent material balance equations. Thus, DOF = (13 + 5) − (9 + 9) = 0.
Next we need to convert volumetric flow rates to molar flow rates. We’ll
assume that natural gas and air behave as ideal gases. The volumetric flow rate is
reported at standard temperature and pressure: 0°C and 1 atm (Sect. 2.2.4).
Therefore, the molar flow rate of natural gas to the furnace is
P​V ​̇
(1 atm)(​10​6​ ​ft​3​/day)(1 day/24 h)
​ṅ ​ = ___
​   ​ = ​ ___________________________________________________________
​
RT (0.082057 L atm/gmol K)(1000 gmol/kgmol)(0.03531467 ​ft​3​/L)(273.15 K)
​ṅ ​= 52.6 kgmol/h
An equivalent calculation gives the air flow rate to the furnace as 1210 kgmol/h
(956 kgmol/h N2 and 254 kgmol/h O2).
Since the system is continuous-flow and steady-state, we use Eq. (4.1). Given
the large number of compounds, a table format is convenient for organizing the
information. All rates are in kgmol/h.
Component​​​ṅ ​​i,in​​​​∑​ ​​​ ​​ν​ik​​ ​​​ξ ​​k̇ ​​​​​ṅ ​​i,out​​
Natural
gas
CH4
Air
50.98
Total
50.98
− ​​​ξ ​​1̇ ​​
0
− ​​​ξ ​​2̇ ​​
0
− ​​​ξ ​​3̇ ​​
0
− ​​​ξ ​​4̇ ​​
0
C2H6
0.70
0.70
C3H8
0.10
0.10
C4H10
0.03
0.03
C5H12
0.01
0.01
− ​​​ξ ​​5̇ ​​
CO2
0.43
0.43​​​ξ ​​1̇ ​​
5 ​​​ξ ​​5̇ ​​​​ṅ​​​CO​2​,out​
N2
0.36
O2
52.6
3 ​​​ξ ​​3̇ ​​
4 ​​​ξ ​​4̇ ​​
956
956.4​​
ṅ ​​​N​2​,out​
254
254
H2O
Total
2 ​​​ξ ​​2̇ ​​
0
1210
0
−2 ​​​ξ ​​1̇ ​​
−3.5 ​​​ξ ​​2̇ ​​
−5 ​​​ξ ​​3̇ ​​
−6.5 ​​​ξ ​​4̇ ​​
−8 ​​​ξ ​​5̇ ​​​​ṅ ​​​O​2​,out​
2 ​​​ξ ​​1̇ ​​
3 ​​​ξ ​​2̇ ​​
4 ​​​ξ ​​3̇ ​​
5 ​​​ξ ​​4̇ ​​
6 ​​​ξ ​​5̇ ​​​​ṅ ​​​H​2​O,out​
1262.6
We use the balances on each of the hydrocarbons to determine ​​ξ​​k̇ , then use those
numbers to calculate the flow rates of the remaining compounds. We find that the
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
flue gas flow rate is 1263.1 kgmol/h and that it contains 75.7 mol% N2, 11.8 mol%
O2, 8.3 mol% H2O and 4.2 mol% CO2.
4.2.1.2 Batch Reactors Batch reactors are chosen for small-volume specialty
chemicals (e.g., prescription drugs, specialty plastics) and sometimes for processes dating back to antiquity (e.g., fermentation of grapes to wine). For batch
reactors, the integral mole balance equation is particularly useful when we are
interested in the change in the system due to reaction over a defined time
interval, from t = t0 to t = tf .
Since by definition no materials are added to or removed from the reactor
during that time interval, the material balance simplifies to:
​
n​i,sys, f​− ​n​i,sys,0​ = ​  ∑ ​​​ ​ν​ik​ ​ξ​k​Eq. (4.2)
all k
Generally speaking, with batch reactors we specify the initial contents of the
system (​n​i,sys,0​). We might know the extent of reaction, and solve for the final
contents; alternatively, we might know the final contents and solve for the
extent of reaction.
Example 4.4
Batch Reactor with Known Reaction Stoichiometry:
Ibuprofen Synthesis
Ibuprofen is a well-known painkiller and fever reducing pharmaceutical. The drug,
whose chemical name is 2-( p-isobutylphenyl)propionic acid (C13H18O2), is synthesized from isobutylbenzene (C10H14), acetic anhydride (C4H6O3), H2, and CO in
a three-step reaction scheme:
C10H14 + C4H6O3 → C12H16O + CH3COOH
(R1)
C12H16O + H2 → C12H18O
(R2)
C12H18O + CO → C13H18O2
(R3)
The reaction was carried out in a batch reactor. Initially the reactor was charged
with 1.4 gmol C10H14, 1.4 gmol C4H6O3, 3 gmol H2, and 2 gmol CO. The reactor
was brought to reaction temperature and held there for 2.3 h. Then the gases were
vented off and the liquid was recovered. Chemical analysis shows that the liquid
contained 0.5 gmol isobutylbenzene and 0.5 gmol acetic anhydride, some ibuprofen, and some other compounds that were not identified. 3.51 moles of gas was
vented, and it contained 37.5 mol% CO and H2. Determine the identities and quantities of these other compounds.
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Section 4.2 Reactor Material Balance Equations
Solution
This process is a batch reactor followed by a gas-liquid separator.
H2
CO
H2
4
C 10H 1
O3
C 4H 6
CO
C1
2H
16 O
Liquid
t0 < t < tf
Quick Quiz 4.1
Given the same initial
charge as Example 4.4,
what would be the
maximum quantity of
ibuprofen that could be
produced?
The liquid remaining in the separator after the gases are vented off is the same
as the liquid in the batch reactor at t = tf . Therefore we simply choose the batch
reactor, indicated by the dashed line, as our system, and analyze using Eq. (4.2).
First we consider the reactants in order to find the extents of reaction. The
calculations are presented in table form (all in units of gmol).
​​n​i,sys,0​​
​​n​i,sys, f​​
​​n​i,sys, f​​ − ​n​i,sys,0​​
​​  ∑ ​​​​ ​​ν​ik​​ ​​ξ​k​​
C10H14
1.4
0.5
−0.9
(−1)​​ξ​1​​
C4H6O3
1.4
0.5
−0.9
(−1)​​ξ​1​​
H2
3
2.2
−0.8
(−1)​​ξ​2​​
CO
2.0
1.31
−0.69
(−1)​​ξ​3​​
Compound
all k
We easily solve to find ​ξ​1​= 0.9 gmol, ​ξ​2​= 0.8 gmol, and ​ξ​3​= 0.69 gmol. Now
we use the extents of reaction to determine the quantities of ibuprofen and other
materials in the liquid, recognizing that for all these compounds, ni,sys,0 = 0.
​​  ∑ ​​​​ ​​ν​ik​​ ​​ξ​k​​
​​n​i,sys, f​​
(+1)0.9 + (−1)0.8
0.1
(+1)0.9
0.9
C12H18O
(+1)0.8 + (−1)0.69
0.11
C13H18O2
(+1)0.69
0.69
Compound
C12H16O
CH3COOH
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
4.2.1.3 Semibatch Reactors
Semibatch reactors display features of both the
continuous-flow and the batch reactor. They usually require more attention
from plant operators than their simpler counterparts. They are used only when
the semibatch nature provides some important advantage. One common situation is where the reactor contents are liquid or solid, but one of the reactants
or products is a gas. In this case, one might choose to charge the reactor initially
with the liquid or solid, and then continuously add the gaseous reactant (or
remove the gaseous product). Semibatch reactors are sometimes used when a
reaction is potentially explosive; by trickling in a small amount of one reactant
over a long period of time into a pot containing the other reactant, the reaction
can be controlled more safely. Whether the differential or integral form of the
material balance equation is more useful depends on the question to be
answered. For analysis of the state of a system at a single point in time, the
differential balance is best; the integral balance is appropriate for analysis of
processes over a defined time interval.
Recall that the integral balance on a component, written in molar terms, is
​
n​i,sys, f​− ​n​i,sys,0​ = ​  ∑ ​​​ ​n​ij​ −​  ∑ ​​​ ​n​ij​ + ​  ∑ ​​​ ​r​ik​
all j​​in​
all j​​out​
all k
Eq. (3.15)
or equivalently (Table 3.2):
​
n​i,sys, f​− n​ ​i,sys,0​ = ​  ∑ ​​​ ​n​ij​ −​  ∑ ​​​ ​n​ij​ + ​  ∑ ​​​ ​νi​k​​ξ​k​
all j​​in​
all j​​out​
all k
It is important to recognize that nij indicates the total moles that cross the
system boundary, either entering or leaving the system, over the time interval
from t = t0 to t = tf . In a semibatch process, there is typically a flow across
the boundary, which could be a constant flow rate or more generally could
change with time. We express this idea as:
​t​f​
​
n​ij​ = ​​  ∫​  ​​​​n ​̇​ij​​dt
​t​0​
Eq. (4.3)
Additionally, ​r​ik​ (or equivalently, ​νi​k​​ξ​k​) indicates the moles generated or consumed by reaction k over the time interval from t = t0 to t = tf . The rate of
the reaction rather than the total quantity may be given in a problem; the reaction rate could be constant or more generally could change with time. We
express this idea as:
​t​f​
​
r​ik​ = ​​  ∫​  ​​​​r ​​i̇ k​​dt
​t​0​
Eq. (4.4a)
or
​t​f​
​
νi​k​​ξk​ ​ = ​​  ∫​  ​​​νi​k​ ​​ξ ​​k̇ ​ ​dt
​t​0​
Eq. (4.4b)
If ​​ṅ ​​ij​, ​​r ​​i̇ k​, and ​​ξ ​​k̇ ​ are constant or are expressed as functions of time t, then
Eqs. (4.3) and (4.4) can be evaluated for use in the integral balance.
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Section 4.2 Reactor Material Balance Equations
Example 4.5
249
Semibatch Reactor with Known Reaction Stoichiometry:
Ibuprofen Synthesis
The last reaction in the three-step ibuprofen synthesis is
C12H18O + CO → C13H18O2(R3)
Because CO is a highly toxic gas, the reactor is designed to be operated in semibatch
mode. Initially, 1.5 gmol C12H18O is charged to the reactor. CO is added at a rate
of 0.30 gmol/h. If the reaction rate under these conditions is 0.27 gmol/h, how long
should the reactor be operated to produce 1.2 gmol ibuprofen? What is in the reactor at the end of operation?
CO
O
H 18
C 12 C
13 H
18 O
2
Solution
We are interested in what happens over an interval of time (although we don’t yet
know what that time interval is!), so we use the integral balance, where we have
explicitly incorporated Eqs. (4.3) and (4.4) into Eq. (3.15):
​t​f​
​t​f​
​t​f​
​n​i,sys, f​− ​n​i,sys,0​ = ​ ∫​  ​​ ​​ṅ ​​ij​ dt − ​ ∫​  ​​ ​​ṅ ​​ij​ dt + ​ ∫​  ​​ ​ν​ik​ ​​ξ ​​k̇ ​dt
​t​0​
​t​0​
​t​0​
We start with a balance on ibuprofen (I, C13H18O2). We know that there is no
ibuprofen initially in the reactor (​n​I,sys,0​= 0) and that we wish to have 1.2 gmol
at the end of the run (​n​I,sys, f​= 1.2). No ibuprofen enters or leaves the reactor over
​t​​
the time interval of interest, so ​ ∫0​  f​​ n​​ ̇ ​​Ij​ dt = 0 for both input and output. The reaction
rate for reaction R3 is given as r​​ ​​İ 3​= ν​ I​3​ ​​ξ ​3​̇ ​= 0.27 gmol/h. Inserting these values
into the material balance equation gives
​t​f​
1.2 − 0 = 0 − 0 + ​ ∫​  ​​ 0.27 dt
0
Solving, we find that the reactor run time t​​f​= 4.44 h.
For C12H18O (A): The integral balance simplifies to
4.44
​​ṅ ​​A,sys, f​= 1.5 − ​ ∫​  ​​0.27 dt = 0.30 gmol
0
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Helpful Hint
Take particular
care to distinguish between
quantity (mass,
moles) and rate
(mass/time,
moles/time) in analyzing semibatch
reactors.
For CO we need to consider the flow rate in as well as the reaction rate. There
is no initial charge of CO to the reactor, and no CO output stream, so the balance
equation simplifies to
​t​f​
​t​f​
​n​CO,sys, f​ = ​  ∫​  ​​ ​​n​​Ċ O,in​ dt + ​  ∫​  ​​ ​ν​​CO​3​ ​​​ξ ​​3̇ ​ dt
​t​0​
​t​0​
4.44
4.44
0
0
​n​CO,sys, f​ = ​ ∫​  ​​ 0.30 dt − ​ ∫​  ​​ 0.27 dt = 0.13 gmol
4.2.2
Independent Chemical Reactions
As part of completing a DOF analysis, the number of independent reaction
variables (or equivalently, the number of independent chemical reactions) is
counted. But how do we know the number of independent reactions, and how
do we know that a set of proposed reactions is independent?
A simple rule-of-thumb that works almost all the time is that the number
of independent chemical reactions in any given system equals the number of
compounds I minus the number of elements H.
Illustration: A reacting gas mixture contains NO, NO2, NH3, H2, H2O, O2,
CO, and CO2. There are 8 compounds and 4 elements (N, O, H, C), so there
are I − H or 8 − 4 = 4 independent chemical reactions.
Using some linear algebra tricks, there is a method that determines the
number of independent reactions and moreover, yields an independent set of
stoichiometrically balanced reactions, starting simply from a list of compounds
with no prior knowledge of the reactions! Since we don’t invoke any knowledge
of chemistry, the reactions do not provide any mechanistic insight and may not
describe how the compounds really combine and reform. But chemical mechanisms are not our quest, we simply want to know enough to complete process
flow calculations.
Here are the mechanics of the method, without explaining the underlying
linear algebra that makes this method work. Briefly, the method incorporates
the constraints that elements are neither generated or consumed across the
entire system of compounds, and that each reaction must be balanced.
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Section 4.2 Reactor Material Balance Equations
251
To find a set of independent chemical reactions for a system of I compounds made up of H different elements:
Step 1. Write a matrix A that contains I columns and H rows, where each
entry is the number of atoms of element h in compound i.
Step 2. Find the “reduced row-echelon” form of A. This is a standard
matrix operation. The reduced matrix will still have I columns and
H rows, made up of an H × H identity matrix plus an H × (I − H)
matrix of other numbers.
Step 3. Eliminate any rows that are all zeros. (Usually there are no such rows.)
Step 4. Create a new matrix A′ by (a) erasing the identity matrix,
(b) multiplying the remaining entries by −1, and (c) adding a new
(I − H) × (I − H) identity matrix below such that A′ has I rows
and (I − H) columns.
Step 5. The number of independent chemical reactions is equal to the number of columns of A′ that have at least two entries. Read off the
stoichiometric coefficients of each reaction by reading down.
This method is demonstrated in the following Example.
Example 4.6
Independent Chemical Reactions
Katie Chemist is investigating a new catalyst for methanol synthesis from methane
and oxygen. She samples her system and identifies seven compounds in the gas
mixtures: CH4, CO, CO2, O2, H2O, H2, and CH3OH. Find a system of independent
chemical reactions that describes this mixture.
Solution
There are 7 compounds comprised of 3 elements, C, H, and O. There are most
likely 7 − 3 or 4 independent chemical reactions. We will show that and find a
set of independent reactions.
Step 1:
Write a matrix A that contains 7 columns and 3 rows, where each entry
is the number of atoms of element h in compound i. For bookkeeping,
we write the 7 compounds above each column and the 3 elements
beside each row.
CH4 CO CO2 O2 H2O H2 CH3OH
1
1
0
0
0
1⎤
C ⎡1
​  ​ ​ ​
H
4​ ​
0​
0​
0​
2​
2​
4​ ​
⎣
O 0
1
2
2
1
0
1⎦
⎢
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⎥
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Step 2:
Find the row reduced-echelon form. This is easily accomplished using
a matrix-based equation-solving program such as MATLAB, or a scientific calculator with matrix functions. The result is
CH4 CO CO2 O2 H2O H2 CH3OH
C
1
0
0
H
0
1
0
O
0
0
1
0 0.5 0.5
1
−1
−1
2 1.5 0.5
1
−2
Identity matrix
−2
Remaining entries
Steps 3 and 4: There are no rows with all zeros. We erase the identity matrix,
multiply the remaining entries by −1 and then add a 4 × 4 identity
matrix to yield A′:
Four independent reactions
CH4
CO
CO2
0 −0.5 −0.5
−1
1
1
−2 −1.5 −0.5
−1
2
2
O2
1
0
0
0
H2O
0
1
0
0
H2
0
0
1
0
CH3OH
0
0
0
1
Negative of remaining
entries from matrix A
Identity matrix
There are indeed 4 independent reactions. Reading down each column,
we find this set:
2CO2 → 2CO + O2
0.5CH4 + 1.5CO2 → 2CO + H2O
0.5CH4 + 0.5CO2 → CO + H2
CH4 + CO2 → CO + CH3OH
This set is not the only possible set of independent reactions that you
could write. But we only need one. Notice that compounds O2, H2O,
H2, and CH3OH appear in only one reaction. This is another clue that
these reactions are independent—for example, there is no combination
of the first three reactions that could give you the fourth, because
methanol does not appear in any of the first three reactions.
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Section 4.2 Reactor Material Balance Equations
4.2.3
253
Reactors with Unknown Reaction Stoichiometry
The extent of reaction concept is very useful when the stoichiometric coefficients are known. But what if we don’t know ​ν​ik​? And, why wouldn’t we know​
ν​ik​anyway? There are a couple of common situations: (1) The raw material
is highly complex and the molecular formulas are unknown or uncertain. This
is true for many natural materials, like wood or coal. (2) The reaction network is
highly complex and the reaction products are unknown or uncertain. Combustion
and degradation reactions may fall into this category. In these situations, we
need a different strategy.
4.2.3.1 Balances with Elements as Components
Helpful Hint
Write balance
equations on
elements rather
than compounds
if the molecular
formulas are
unknown or if
the number of
reactions exceeds
the number of
elements.
Example 4.7
If the elemental composition of the raw material and/or reaction products is known, then we use elements, rather than compounds, as our components. Since elements do not
undergo chemical reactions (and we are ignoring nuclear reactions), ξ​​ ​​k̇ ​ = 0,
and the material balance equations are simplified whether the reactor is steadystate continuous-flow, batch, or semibatch.
Even when we know the molecular formulas of the compounds and the
chemical reactions, if there are many chemical reactions but only a few elements it may be easier to write balances on elements rather than on compounds.
To do element balances when the molecular formula is known, we convert the
moles (or molar flows) of compounds to moles (or molar flows) of the elements
in that compound using
​n​h = εhini
Eq. (2.2)
or
​​n ​​ḣ = εhi​​ni​​ ̇
where εhi is number of atoms at element h in compound i, and nh (​​nh​​ ̇ ) is the
moles (molar flow) of element h.
Material Balance Equation with Elements: Combustion of Natural Gas
Redo Example 4.3, but use elements rather than compounds as components in the
material balance equation.
Solution
The feed to the furnace of Example 4.3 contains eight compounds, and there are
five combustion reactions to consider. However, there are only four elements: C,
H, O, and N. We need only four element balance equations. The steady-state mole
balance equation for each element h, where h is C, H, O, or N, simplifies to:
​​n ​​h,in
̇ ​ = ​​n ​​ḣ ,out​
Since we know the molar flow rates of the compounds but not of the elements, the
big job is to calculate the molar flow rate of each element, using Eq. (2.2) adapted
for flows:
​​n ​h,in
​̇ ​= ​εh​ i​ ​​n ​​i̇ ,in​
This is readily implemented by using a spreadsheet: Results are shown in tabular
form. All flows are given in kgmol/h.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Molar flow,
C
H
O
​​​ṅ ​​i,in​​​​ε​Ci​​
​​n
​ ̇ ​​C,in​​
ε​​ ​Hi​​
CH4
50.98
1
50.98
4
203.92
C2H6
0.70
2
1.40
6
4.20
C3H8
0.10
3
0.30
8
0.80
C4H10
0.03
4
0.12
10
0.30
C5H12
0.01
5
0.05
12
0.12
CO2
0.43
1
0.43
N2
0.36
Total
​​n
​ ̇ ​​H,in​​
N
ε​​ ​Oi​​
2
52.6 53.28209.34
​​n
​ ̇ ​​O,in​​
ε​​ ​Ni​​
n
​​​  ​​̇ N,in​​
0.86
2
0.86
0.72
0.72
Similar calculations show that 1912 kgmol N/h and 508 kgmol O/h enter in the
air stream.
Now we proceed to evaluate the material balance equation for each element.
N atom balance:
kgmol N in air
kgmol N in gas
kgmol
_____________
​​n ​​N,in
​+ 0.72 ​
​= 1912.7 ​ ______
​ = ​​n ​​Ṅ ,out​
̇ ​= 1912 ​ ____________
h
h
h
Helpful Hint
Solve the balance
involving the
fewest number of
unknowns first.
Since all N leaves as N2, the N2 flow rate out with the flue gas is
​​n ​​Ṅ ,out​ ______
kgmol
1912.7
______
​​n ​​​Ṅ 2​ ​,out​ = ​ _____
​ε​N,​N2​  ​​​ = ​  2 ​= 956.4 ​  h ​
H atom balance:
209.34 kgmol
____________
​​n ​​H,in
​ = ​​n ​​Ḣ ,out​
̇ ​ = ​
h
Since all the H leaves as water:
​​n ​​Ḣ ,out​ ______
kgmol
​​n ​​​Ḣ 2​ ​O,out​ = _____
​  ​ε​
​​ = ​  209.34
​= 104.67 ​ ______
​
H,​H2​ ​O
2
h
C atom balance:
kgmol
​​n ​​C,in
​ = ​​n ​​Ċ ,out​
̇ ​= 53.28 ​ ______
h
All the C leaves as CO2:
​​n ​C​̇ ,out​ 53.28
kgmol
​​n ​​​Ċ O​2​,out​ = _____
​  ​ε​  ​​ = _____
​   ​= 53.28 ​ ______
​
C,​CO​2​
1
h
O atom balance:
508 kgmol O in air _________________
0.86 kgmol O in gas
kgmol
​​n ​​O,in
​ + ​
​= 508.86 ​ ______
​ = ​​n ​​Ȯ ,out​
̇ ​ = ​ ________________
h
h
h
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Section 4.2 Reactor Material Balance Equations
255
O leaves in several different forms: as H2O, CO2, and O2. From the H and C balances,
we know the flow rates of H2O and CO2. (Good thing we left the O balance for last!)
​​n ​​Ȯ ,out​= ε​ ​O,​H2​ ​O​​​n ​​​Ḣ 2​ ​O,out​+ ε​ ​O,​CO​2​​​​n ​​​Ċ O​2​,out​+ ​ε​O,​O2​ ​​​​n ​​​Ȯ 2​ ​,out​
kgmol
kgmol
kgmol
508.86 ​ ______
​= 1​(104.67 ​ ______
​
​+ 2​(53.28 ​ ______
​​ + 2​​n ​​​Ȯ 2​ ​,out​
)
h
h
h )
kgmol
​​n ​​​Ȯ 2​ ​,out​= 148.8 ​ ______
​
h
To summarize:
Total flue gas flow rate = 956.4 + 104.67 + 53.28 + 148.8 = 1263.1 kgmol/h
Flue gas contains: 11.8 mol% O2
4.2 mol% CO2
8.3 mol% H2O
75.7 mol% N2
4.2.3.2 Balances with Mass Rates of Reaction In some situations where
stoichiometric coefficients are unknown, mass rates of reaction are useful. This
strategy is applicable, for example, when the component i is a mixture of related
species without a unique molecular formula, such as a polymer (where there
is a distribution of molecular weights) or a protein, where the material undergoes
a common reaction, such as hydrolysis. Example 4.8 illustrates this approach.
Example 4.8
Mass Rates of Reaction: Microbial Degradation of Soil Contaminants
At an abandoned gasoline station, material from the old underground storage tank
has leaked into the ground. After many years, the soil has become contaminated,
primarily with aromatics: benzene, toluene, ethylbenzene and xylene (called BTEX).
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Benzene
Toluene
Ethylbenzene
o-xylene
m-xylene
p-xylene
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Your job is to see if the contaminated soil can be spiked with bacteria that
degrade aromatics to less noxious, more volatile compounds. The idea is to turn
the soil itself into a batch reactor, and decontaminate the soil in situ. The alternative is to dig out the soil and dispose of it as hazardous waste.
You need to determine how long it will take to decontaminate the soil, so
you dig up some information from the scientific literature. One study reports that
the microbial degradation rate for BTEX is 0.22 μg BTEX per day per gram of
soil. In another study, degradation rates were reported of 6 × 10−5 μmol benzene/g
soil/day, 2 × 10−3 μmol toluene/g soil/day, 6 × 10−4 μmol ethylbenzene/g soil/day,
and 1.8 × 10−3 μmole xylene/g soil/day.
The contaminated soil contains 8 μg BTEX/g soil. The soil must be decontaminated to 0.1 μg BTEX/g soil. Are the data from the two studies consistent?
How long will it take to decontaminate the soil?
Solution
To compare the two studies, we convert the molar rates of degradation from the
second study to mass rates of degradation, by multiplying by the molar mass of
each component.
78 μg benzene
μmol benzene ____________
​​R ​​Ḃ ​= 6 × ​10​−5​ ____________
​
​ ×
​     ​
g soil/day
μmol benzene
= 0.0047 μg benzene/g soil/day
92 μg toluene
μmol toluene ____________
___________
​​R ​​Ṫ ​= 2 × ​10​−3​ ​
​ × ​
​
g soil/day
μmol toluene
= 0.18 μg toluene/g soil/day
μmol ethylbenzene _________________
106 μg ethylbenzene
​​R ​​Ė ​= 6 × ​10​−4​ ​  ________________
​ × ​
​
g soil/day
μmol ethylbenzene
= 0.064 μg ethylbenzene/g soil/day
μmol xylene ____________
106 μg xylene
​​R ​​Ẋ = 1.8 × ​10​−3​ ​ ___________
​ × ​
​
g soil/day
μmol xylene
= 0.19 μg xylene/g soil/day
These rates are reasonably consistent with the rate of 0.22 μg BTEX per day per
gram reported in the first study. The first study is more useful for our purposes:
We have a mixture of related compounds and we know the total mass of those
compounds, but we don’t know the exact composition. The second study does point
out, however, that the degradation rates observed will depend strongly on the
identity of the contaminants.
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Section 4.2 Reactor Material Balance Equations
257
The soil is a batch reactor, and we are interested in the degradation that occurs
over a specified time interval, so we use an integral mass balance equation. Let’s
choose as a basis 1 g of soil.
​t​f​
​m​BTEX,sys, f​− ​m​BTEX,sys,0​ = ​  ∫​  ​​ ​​R ​​Ḃ TEX​ dt
​t​0​
​t​f​
μg BTEX
0.1 μg BTEX − 8 μg BTEX = ​ ∫​  ​​ −0.22 ​ ________
​ dt = −0.22​t​f​
day
0
​
t​f​= 36 days
Using the degradation rate for pure benzene we find something sobering: If the
soil were contaminated with pure benzene, the decontamination time would be
much longer: about 1580 days!
In the next two examples, you will see how the integral equations are used
with mass rates of reaction, where the rate of reaction is a function of time.
Example 4.9
Helpful Hint
If you get stymied,
remember
accumulation =
input − output +
(generation −
consumption),
check for dimensional consistency,
and recall the
10 Easy Steps!
Integral Equation with Unsteady Flow and Chemical Reaction:
Controlled Drug Release
Patients with certain diseases are treated by injection of proteins or drugs into their
bloodstream. Upon injection there is a sudden increase in the protein or drug concentration in the blood to very high levels, but then the concentration rapidly falls.
A steadier blood concentration is often desirable, to reduce toxic side effects and
increase therapeutic efficacy. Controlled-release technology reduces the variability
in drug concentration in the blood. With controlled release, the protein or drug is
encapsulated in a polymer and is released slowly into the bloodstream. This maintains the concentration of drug or protein in the bloodstream at a lower, more
constant level.
1. 100 μg of a drug is loaded into a controlled-release capsule and then injected
into a patient. The drug is released at a rate of 8e−0.1t μg/h, where t is hours
after injection. What fraction of the drug has been released after 24 h?
2. Once in the bloodstream, the drug is lost at a rate of 3.1 μg/h as a result of
degradation reactions and elimination processes. How does the mass (μg) of
drug in the bloodstream vary as a function of time after injection? At what time
is the drug concentration the highest?
Solution
1. We choose the capsule as the system. The capsule releases the drug in all directions into the bloodstream.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Drug into bloodstream
Controlledrelease
capsule
Drug into bloodstream
Drug into bloodstream
We are interested in what happens to the mass of drug D in the capsule over a
24-h time interval, so we use the integral mass balance applied to the drug as
the component:
​mD​  ,sys, f​​ − ​mD​  ,sys,0​​ = mD,in − mD,out + RD
The initial quantity of drug in the capsule is ​m​D,sys,0​= 100 μg. There is no new
drug entering the capsule, so
mD,in = 0
There is no chemical reaction inside the capsule, so
RD = 0
The mass flow rate of drug from the capsule as a function of time is known:
​​ṁ ​​D,out​​ = 8​e​​  −0.1t​
We now evaluate the total mass flow rate of drug from the capsule over the
24-h interval:
​tf​  ​​
24
8  ​ (​e​​  −2.4​− ​e​​  0​) = 72.7 μg​
​  ∫​  ​​​ṁ ​​D,out​​ dt​= ​ ∫​  ​8​e​​  −0.1t​ dt = − ​ _
0.1
​t0​  ​​
0
These terms are inserted into the integral mass balance equation to yield:
​mD​  ,sys, f​​− 100 = −72.7
The drug remaining in the capsule after 24 h is
​mD​  ,sys, f​​= −72.7 + 100 = 27.3 μg
In other words, 27.3% of the drug remains in the capsule and 72.7% has been
released 24 hours after injection.
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Section 4.2 Reactor Material Balance Equations
259
8
80
7
70
6
60
5
50
4
40
3
30
2
20
1
10
0
0
4
8
12
16
20
Time since injection, h
Total drug leaked, 𝜇g
Rate of leakage of drug, μg/h
We can use a similar procedure to evaluate drug release profiles at any time t,
simply by integrating from 0 to t rather than from 0 to 24 h. Results of these
calculations are shown.
0
24
2. Now we are interested in the drug quantity in the bloodstream at any time t
after injection. Our system is the blood.
Loss by
degradation
+
elimination
Encapsulated
drug
An integral mass balance is still useful, because we are interested in the net
accumulation of drug in the system over a finite time interval from t = 0 to t.
Assuming that there is no drug in the blood initially, m
​ ​D,sys,0​= 0, and
​mD​  ,sys, f​​ = mD,in − mD,out + RD
The mass flow rate of drug into the blood equals the mass flow rate of drug
out of the capsule, or
​tf​  ​​
t
mD,in = ​ ∫​  ​​​ṁ ​​D,in​​ dt​= ​ ∫​  ​8​e​​  −0.1t​ dt = 80 (1 − e​ ​​  −0.1t​)​
​t0​  ​​
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Drug is lost from the blood at a rate of 3.1 μg/h by elimination (a mass flow
rate out) and degradation reactions. We don’t have enough information to separate these two loss mechanisms, but we can lump them together:
t
t
mD,out + RD = ​ ∫​  ​​(m
​ ̇ ​​D,out​​ + ​R ​Ḋ )dt​= ​ ∫​  ​3.1dt = 3.1t​
0
0
Inserting these expressions into the integral mass balance equation yields
​mD​  ,sys, f​​= 80(1 − e​ ​​  −0.1t​) − 3.1t
Mass of drug in bloodstream, 𝜇g
A plot of m
​ D​  ,sys, f​​ versus t is illuminating. The rapid initial release of the drug
from the capsule causes a rise in the blood concentration. As the release
slows, degradation reactions and elimination eventually become faster than
the release, and the drug concentration decays. No drug is left 24 h after
injection. The maximum concentration is reached at about 9.5 h after
injection.
Example 4.10
80
Without elimination/
degradation
70
60
50
40
Including elimination/
degradation
30
20
10
0
0
4
8
12
16
20
Time after injection, h
24
Differential Equation with Unsteady Flow and Chemical Reaction:
Glucose Utilization in a Fermentor
Yeast metabolize glucose (C6H12O6) and make ethanol (C2H5OH). Humans have
exploited this process for thousands of years to make wine, beer, and other alcoholic
beverages. Although the chemical reactions are highly complex, the overall reaction
can be written simply as
​C6​ ​​​H1​ 2​​​O6​ ​​ → ​2C​2​​​H5​ ​​OH + ​2CO​2​​
Besides making ethanol, yeast grow and reproduce, consuming some of the glucose
for maintenance and growth. The rate of glucose consumption depends on the
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Section 4.2 Reactor Material Balance Equations
261
number of yeast in the fermentor as well as the rate of growth of the yeast. So, the
glucose consumption rate increases as the yeast multiply.
We start with a 10-L fermentor containing 1000 g glucose and some yeast.
The fermentor is operated in semibatch mode. During fermentation, additional
glucose is added continuously at a rate of 20 g/h. CO2 is continuously vented to
prevent pressure buildup. No other products or byproducts are removed. The mass
rate of glucose consumption ​​​R ​​ġ ​​ (g glucose consumed per hour) increases with time
as the number of yeast increases:
​​R ​​ġ ​​ = −2.8​e​​  0.1t​
where t is in hours. About 90% of the glucose consumption goes toward ethanol
production, with the rest going to support yeast growth.
1.
2.
3.
4.
Plot the rate at which glucose in the fermentor changes with time.
Plot the grams of glucose and ethanol in the fermentor at any time.
Calculate the CO2 flow rate out of the fermentor as a function of time.
How long will it take for the glucose concentration in the fermentor to drop
to zero (at which point the fermentation is stopped)? How much ethanol is in
the fermentor at that point?
Solution
We start, as always, with a diagram. The fermentor is the chosen system. The fermentor operates in semibatch mode; the glucose and ethanol concentrations inside
the vessel change with time, and the CO2 is continuously removed. All information
is given as mass and mass flows, so we’ll stick with grams and hours for units.
CO2
Fermentor
Glucose
Fermentor
Glucose + yeast
t = t0
t > t0
1. We want to know the rate of change of glucose in the fermentor at any specified
time, so we’ll use a differential equation. We’ll use a mass balance, because all
information is given in mass units, with glucose as the component. There is no
flow of glucose out, so the differential component mass balance simplifies to:
d​mg​  ,sys​​
​ _​ = ​​ṁ ​​g,in​​ + ​​R ​​ġ ​​= 20 − 2.8​e​​  0.1t​
dt
where m
​​  ​̇ ​g,in​​= mass flow of glucose into fermentor.
We’ll plot this equation versus time:
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
dmg,sys
, g glucose/h
dt
100
Accumulation
0
Depletion
–100
–200
–300
–400
0
10
20
30
Time, h
40
50
As we see from the graph, initially there is a net accumulation of glucose in
the fermentor (dmg,sys∕dt > 0). At 20 h, however, as the yeast population increases,
the rate of consumption of glucose equals the flow rate in (dmg,sys∕dt = 0);
above 20 h, the rate of consumption exceeds the flow rate in and glucose is
depleted (dmg,sys∕dt < 0).
2. To determine the accumulated quantity of glucose in the fermentor at a given
time, we need to find an expression for mg,sys as a function of time. We get that
by integrating the differential material balance equation:
​mg​  ,sys​​= ∫ (20 − 2.8​e​​  0.1t​) dt = 20t − 28​e​​  0.1t​ + C
where C is a constant of integration. We know that at t = 0, ​mg​  ,sys​​= 1000 g.
From this we find that C = 1028 g, and
​mg​  ,sys​​= 1028 + 20t − 28​e​​  0.1t​
No ethanol enters or leaves the fermentor, and there is no ethanol in the system at t = 0. The differential mass balance equation for ethanol simplifies to:
d ​me​  ,sys​​
​ _​ = ​​R ​​ė ​​
dt
where me,sys = mass of ethanol in the system and ​​R ​​ė ​​= mass rate of reaction
of ethanol.
What is ​​R ​​ė ​​? From the stoichiometry of the glucose-to-ethanol reaction, we
know that 2 moles of ethanol are produced per mole of glucose consumed by
this reaction, or νe∕νg = −2. We calculate the molar masses: Me = 46 g/gmol
and Mg = 180 g/gmol. Finally, from the problem statement we know that, of
all the glucose consumed, 90% of it goes toward making ethanol. Therefore:
​ν​  ​​ ​M​  ​​
​​R ​​ė ​​ = _
​  e e  ​ (0.9​​R ​​ġ ​​) = −2​(_
​  46  ​)​[0.9(−2.8​e​​  0.1t​)] = 1.29​e​​  0.1t​
​νg​  ​​ ​Mg​  ​​
180
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Section 4.2 Reactor Material Balance Equations
263
We substitute this expression into the ethanol differential mass balance equation,
and then integrate (with me,sys = 0 at t = 0) to find:
​me​  ,sys​​ = 12.9​e​​  0.1t​− 12.9
Glucose and ethanol in fermentor, g
We plot these two expressions versus time:
1200
Glucose
1000
800
600
400
Ethanol
200
0
0
10
20
Time, h
30
40
Glucose rises from its initial value of 1000 g, peaking around 20 h into the
fermentation before it rapidly drops. Ethanol increases slowly at first, but then
at an accelerating rate as the yeast proliferate (until the glucose runs out).
3. CO2 is generated by reaction along with the ethanol. From the stoichiometry
we calculate that
​ν​C​  O​2​​​​ _
​M​C​  O​2​​​​ ̇ _
​​R ​​​Ċ O​2​​ ​​ (ethanol) = ​ _
​​ ​​ = ​  2 ​​ _
​  44 ​ ​(1.29​e​​  0.1t​) = 1.23​e​​  0.1t​
​νe​   ​ ​
​​ ​Me​   ​ ​​R
​​ e 2 ( 46 )
Additional CO2 is generated when glucose is consumed for maintenance and
growth of the yeast. We don’t have sufficient information to calculate this quantity exactly. However, we can calculate a limiting case where we assume that
all the remaining glucose (not consumed in the ethanol production reaction)
reacts to CO2 and H2O
​C6​ ​​​H1​ 2​​​O6​ ​​ + ​6O​2​​ → ​6CO​2​​ + ​6H​2​​O
This gives us the maximum amount of CO2 production possible. Since 10% of
the glucose is consumed for maintenance and growth, we calculate that the
maximum CO2 production from this reaction is
​ν​C​  O​2​​​​ _
​M​C​  O​2​​​​
6 _
0.1t
0.1t
44
_
̇
​​R ​​​Ċ O​2​​​​ (maintenance) = ​ _
​νg​   ​​​ ​  ​Mg​   ​ ​​ (0.1​​R ​​g​​) = − ​  1 ​​(​  180 ​)​[0.1(−2.8​e​​  ​)] = 0.41​e​​  ​
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
The maximum mass rate of reaction of CO2 is obtained by summing the rate of
CO2 generation from the ethanol reaction and the rate of CO2 generation from
reactions associated with maintenance and growth
​  ∑ ​​​​​R ​​​Ċ O​2​​,k​​ = 1.23​e​​  0.1t​+ 0.41​e​​  0.1t​= 1.64​e​​  0.1t​
all k
CO2 is continuously vented so that it does not build up in the fermentor. The
differential mass balance equation for CO2 is set up and then integrated:
d​m​C​  O​2​​,sys​​
​ _
​= 0 = − ​​ṁ ​​​CO​2​​,out​​ + ∑ ​​R ​​​Ċ O​2​​​​
dt
​​ṁ ​​​CO​2​​,out​​ = 1.64​e​​  0.1t​
4. To get the time at which glucose drops to zero in the fermentor, we can solve
the integral material balance equation with mg,sys, f = 0:
​tf​  ​​
​tf​  ​​
​mg​  ,sys, f​​ − ​mg​  ,sys,0​​ = ​  ∫​  ​​​ṁ ​​g,in​​ dt​+ ​ ∫​  ​​​R ​​ġ ​​ dt​
​t0​  ​​
​tf​  ​​
​t0​  ​​
​tf​  ​​
0 − 1000 = ​ ∫​  ​20 dt​ + ​ ∫​  ​− 2.8 exp (0.1 t) dt​
0
0
Evaluating the definite integrals produces:
0 − 1000 = 20​tf​  ​​ − 28[​e​​  0.1​tf​  ​​​− e​ ​​  0​]
​​t​f​​= 42 h
The ethanol in the fermentor at 42 h is found, by a similar strategy, to be 847 g.
4.3
Stream Composition and System
Performance Specifications for Reactors
In a perfect world, we choose reactants and a reaction pathway with a high
atom economy. We design the perfect chemical reactor, providing exactly the
right combination of size, residence time, temperature, pressure, catalyst,
mixing pattern, and reactant addition to: (1) completely and rapidly convert
all the raw materials to useful products by the desired chemical reaction
and (2) completely prevent unwanted chemical reactions. As you can imagine, the perfect reactor rarely exists.
Figure 4.3 compares an example of a perfect reactor to an example of a real
reactor. In the perfect reactor, reactants A and B are fed at the correct stoichiometric ratio and are 100% converted to desired product D by a single reaction:
A+B→D
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Section 4.3 Stream Composition and System Performance Specifications for Reactors
A
B
Perfect
reactor
D
A
B
C
Real
reactor
265
A
B
C
D
E
F
G
Figure 4.3 In the perfect reactor, reactants A + B are fully converted to desired product
D and nothing else. In the real reactor, reactants A and B and contaminant C undergo
multiple reactions producing D, E, F, and G.
In a more realistic reactor, reactants A and B, along with contaminant C, are
fed at a nonstoichiometric ratio, and are partially converted to desired product D.
Along with the desired reaction, a couple of undesired reactions take place:
A+C→E+F
A+D→G
A comparison of the degrees of freedom of the perfect and real reactors
is enlightening (Table 4.1). We assume for this analysis that the production
rate of D is specified and that the reactor operates at steady-state and with
continuous flow. For the perfect reactor, no additional specifications are
required! Indeed, when we completed generation-consumption analysis in
Chap. 1 and then scaled up to a desired production rate, we were able to
Table 4.1
DOF Analysis of the Perfect Reactor
and the Real Reactor of Fig. 4.3
Perfect reactor
Real reactor
Stream variables
3
10
Reaction variables
1
3
Total variables
4
13
Specified flows 1 1
Specified stream composition 0 0
Specified system performance 0 0
Material balances 3 7
Total equations 4 8
DOF
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
complete all calculations because we implicitly assumed that the reactor was
perfect. The real reactor is grossly underspecified, however. To proceed, we
need additional information about stream composition and/or system performance. In this section we describe commonly used reactor specifications.
For convenience we describe everything in terms of differential balances,
flow rates, and reaction rates, but equivalent specifications can be applied
to integral balances and quantities.
4.3.1 Stream
Composition Specification:
Excess and Limiting Reactants
In the perfect reactor of Figure 4.3, there is only one reaction and the reactants
are fed at exactly the right stoichiometric ratio. But sometimes reactants are
fed at nonstoichiometric ratio. (We’ll discuss reasons why this is sometimes
a good choice later.) A reactant fed at less than its stoichiometric ratio
(relative to the other reactants) is called the limiting reactant. Reactants
fed at greater than stoichiometric ratio relative to the limiting reactant are
called excess reactants. If A and B are reactants, then the reactants are fed
at stoichiometric ratio if
​​ṅ ​​A,in​ ​ν​A​
​​ ____ ​​ = ​​ __
​​
​​ṅ ​​B,in​ ​ν​B​
Reactant A is the excess reactant if
​​ṅ ​​A,in​ ​νA​ ​
​​ ____ ​​ > ​​ __
​​
​​ṅ ​​B,in​ ​ν​B​
Eq. (4.5a)
Similarly, reactant A is the limiting reactant if
​​ṅ ​​A,in​ ​ν​A​
​​ ____ ​​ < ​​ __
​​
​​ṅ ​​B,in​ ​νB​ ​
Eq. (4.5b)
If there are more than two reactants, than the limiting reactant would be
the reactant for which Eq. (4.5b) was true relative to all other reactants. If
there are two or more reactions involving the same reactants, only the
desired reaction is used in assessing whether or not the reactants are fed at
stoichiometric ratio.
The percent excess reactant indicates the percent by which the feed of the
excess reactant exceeds what would be required for stoichiometry. We define
percent excess as
​ν​  ​​
​​ṅ ​​A,in​​ − ​(_
​  ​νA​   ​​​ )​ ​​ṅ ​​B,in​​
B
______________
​
​× 100% = % excess
​νA​  ​​
_
​(​  ​ν​   ​​​ )​ ​​ṅ ​B​ ,in​​
Eq. (4.6)
B
where A is an excess reactant and B is the limiting reactant.
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Section 4.3 Stream Composition and System Performance Specifications for Reactors
Illustration: The reaction is
2A + B → D
150 gmol/min A and 100 gmol/min B are fed to a continuous-flow reactor
operating at steady state. Since
​​ṅ ​​A,in​
____
​
​ =
​​ṅ ​​B,in​
​νA​ ​
150 ​ < ​ __
​ ____
​ =
100 ν​ ​B​
2 ​
​ __
1
A is the limiting reactant and B is the excess reactant.
​ν​  ​​
100 − ​(__
​  1 ​)​ 150
​​ṅ ​​B,in​​  − ​(_
​  ​νB​   ​​​ )​ ​​ṅ ​​A,in​​
A
2  ​​× 100% = 33.3% excess B
______________
____________
​
​× 100% = ​​
​νB​  ​​
1
_
__
​(​  ​ν​   ​​​ )​ ​​ṅ ​​A,in​​
​(​   ​)​ 150
A
2
Quick Quiz 4.2
12 gmol SiCl4 and
20 gmol H2 are reacted
to make pure solid silicon Si, with hydrogen
chloride HCl as the
byproduct. Is SiCl4
or H2 the limiting
reactant?
What is the % excess
for the excess reactant?
Illustration: 150 gmol/min A and 100 gmol/min B are fed to a continuousflow reactor operating at steady state. The desired reaction (R1) is
2A + B → D
(R1)
An undesired reaction (R2) also occurs in the reactor.
A + 3B → E + F
(R2)
150 gmol/min A and 100 gmol/min B are fed to a continuous-flow reactor
operating at steady state.
​​ṅ ​​A,in​ ____
​ν​A1​ __
2
____
​
​ = ​  150 ​ < ​ __
​
ν
​B1 ​​ = ​  1 ​
​​ṅ ​​B,in​ 100
and A is still the limiting reactant. B is fed at 33.3% excess.
Percent excess has a special meaning with combustion reactions. Combustion
reactions are considered to go to completion if:
∙
∙
∙
∙
all
all
all
all
C is converted to CO2
H is converted to H2O
S is converted to SO2
N is converted to N2
Even if part of the fuel is incompletely combusted (e.g., some CO is produced),
the percent excess is calculated on the assumption of complete combustion.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Illustration: Mercaptan (methanethiol, CH4S) is a chemical whose odor has
been compared to unwashed socks or rotting cabbage. 100 gmol/min are fed to
a furnace where it is combusted with 25% excess oxygen. The stoichiometric
requirement for oxygen is calculated by assuming complete combustion to
CO2, H2O, and SO2:
CH4S + 3O2 → CO2 + 2H2O + SO2
Stoichiometric O2 flow rate would be
​ν​O​ ​ ​​
​​​ṅ ​​​O2​ ​,in​​ = ​​ _____
​  ν​ ​ 2  ​ ​ ​​​​​ṅ ​​​CH​4​S,in​​ = ​​(__
​  3 ​)​​ 100 = 300 gmol/min
( ​CH​4​S )
1
25% excess O2 is calculated from
​νO​ ​
n​​ ̇ ​​O2,in​− ​(_____
​  ​ν​ 2  ​​)​ ​​ṅ ​​CH4S,in​
​​ṅ ​​O2,in​ − ​(__
​  3 ​)​ 100
CH4S
1  ​​× 100% = 25% excess O
____________________
_____________
​​
​​ × 100% = ​​
2
​ν​O2​
3
__
_____
​(​   ​)​ 100
​(​  ​ν​  ​​)​ ​​ṅ ​​CH4S,in​
1
CH4S
Or n​​​  ​​̇​O2​ ​,in​​= 375 gmol/min
Example 4.11
Excess Reactants: A Badly Maintained Furnace
Natural gas from Well NM is fed to an industrial furnace at 1.00 MMSCFD along
with 25% excess air. The flue gas is tested and found to contain both CO and CO2,
at a 1:10 mole ratio. Also in the flue gas is N2, O2, and H2O. Calculate the flow
rate and composition (mol%) of the flue gas.
Solution
In Example 4.7, we calculated that Well NM natural gas at 1.00 MMSCFD was
equivalent to a molar flow rate to the furnace of: 53.28 kgmol/h C, 209.34 kgmol/h H,
0.86 kgmol/h O, and 0.72 kgmol/h N. First we will calculate the stoichiometric
amount of oxygen required for complete combustion of these elements. We require
1 kgmol O2 per kgmol C and ​​1⁄4​​ kgmol O2 per kgmol H. A general combustion
reaction for hydrocarbons is CxHy + (x + _​​ 14​​ y)O2 → xCO2 + y​​⁄ 2​​H2O.
Complete combustion of 53.28 kgmol/h C requires 53.28 kgmol/h O2. Complete
combustion of 209.34 kgmol/h H requires (209.34/4) or 52.34 kgmol/h O2. The
fuel gas itself supplies (0.86/2) or 0.43 kgmol/h O2 (equivalent). Therefore, the
stoichiometric oxygen flow for complete combustion is 53.28 + 52.34 − 0.43 or
105.2 kgmol/h O2. The total oxygen flow to the reactor is therefore the stoichiometric quantity plus 25% more, or (105.2)(1.25) = 131.5 kgmol/h. Since air is 21 mol%
O2 and 79 mol% N2, the nitrogen flow rate from the air is 131.5(0.79∕0.21) =
494.7 kgmol/h.
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269
Flue gas
CO2, CO, H2O, O2, N2
Natural gas:
53.28 C
209.34 H
0.86 O
0.72 N
Air:
131.5 O2
494.7 N2
We then quickly derive element balance equations. Coupled with the information
that the CO2:CO molar ratio is 10:1, we solve for the flue gas flow rate and
composition.
kgmol
​​ṅ ​​N,in​​= 2(494.7) + 0.72 = 990.12 ​ _
​ = ​​ṅ ​​N,out​​ = 2​​ṅ ​​​N2​ ​​,out​​
h
kgmol
​​ṅ ​​​N2​ ​​,out​​ = 495 ​ _
​
h
kgmol
​​ṅ ​​H,in​​ = 209.34 ​ _
​ = ​​ṅ ​​H,out​​ = 2​​ṅ ​​​H2​ ​​O,out​​
h
kgmol
​​ṅ ​​​H2​ ​​O,out​​ = 104.67 ​ _
​
h
kgmol
​​ṅ ​​C,in​​ = 53.28 ​ _
​ = ​​ṅ ​​C,out​​ = ​​ṅ ​​CO,out​​ + ​​ṅ ​​​CO​2​​,out​​
h
​​ṅ ​​CO,out​​ = 0.1​​n ​​​Ċ O​2​​,out​​
kgmol
kgmol
​​ṅ ​​CO,out​​ = 4.84 ​ _
​, ​​ṅ ​​​CO​2​​out​​ = 48.44 ​ _
​
h
h
kgmol
​​ṅ ​​O,in​​= 2(131.5) + 0.86 = 263.8 ​ _
​
h
= n​​ ̇ ​​O,out ​​= 104.67 + 4.84 + 2(48.44) + 2​​n ​​​Ȯ 2​ ​​,out​​
kgmol
​​ṅ ​​​O2​ ​​,out​​ = 28.7 ​ _
​
h
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
As a quick check, we know that if the furnace was working well, the oxygen in the
flue gas should be exactly the excess oxygen in the air feed, or (0.25)(105.19) =
26.3 kgmol/h. Since there is some partial combustion, the O2 in the flue gas is
slightly higher.
The flue gas flow rate is 681.6 kgmol/h; the flue gas composition is 72.6% N2,
4.2 mol% O2, 0.7 mol% CO, 7.1 mol% CO2, and 15.4 mol% H2O.
4.3.2
System Performance Specifications
Commonly used measures of the performance of real reactors are: conversion,
selectivity, and yield. These system performance specifications provide information about the change in component flows or quantities between the reactor
output and the reactor input, and can be related to the extents of reaction in
the reactor. For convenience, all performance specifications will be described
in terms of flow rate n​​ ̇ ​i​j​and reaction rate r​​ ​i​̇ k​= ν​ i​k​ ​​ξ ​​k̇ ​, but equivalent expressions
can be derived in terms of quantities n​ i​j​or r​ i​k​= ν​ i​k​ξ. We will describe each of
these performance measures in turn.
4.3.3 System
Performance Specification:
Fractional Conversion
The fraction of reactant converted to products is one of the most commonly
used measures of reactor performance. Fractional conversion is defined succinctly in words as
of reactant consumed
Fractional conversion = ________________________
​ moles
​
moles of reactant fed
Let’s now define fractional conversion mathematically. Suppose reactant i
enters a reactor operating at steady-state at flow rate ​​​ṅ ​​i,in​​and leaves at a flow
rate n​​​ ̇ ​i​,out​​. The steady-state mole balance equation for reactant i is
​​ṅ ​​i,out​​ = ​​ṅ ​​i,in​​ + ​  ∑ ​​​ ​νi​  k​​ ​​ξ ​​k̇ ​​
all k
Eq. (4.1)
Now, subtract n​​​  ​̇​i,in​​from both sides:
​​ṅ ​​i,out​​ − ​​ṅ ​​i,in​​ = ​  ∑ ​​​ ​νi​  k​​ ​​ξ ​​k̇ ​​
all k
The left-hand side is simply the difference between what comes out of
the reactor and what goes in. The right-hand side is the net reaction rate
(generation − consumption). If we divide both sides by the flow rate into the
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Section 4.3 Stream Composition and System Performance Specifications for Reactors
271
reactor, ​​​ṅ ​​i,in​​, and multiply through by (−1), we get the net moles of reactant
consumed per mole of reactant fed, or the fractional conversion fCi:
− ​  ∑ ​​​νi​  k​​ ​​ξ ​​k̇ ​​​
​​ṅ ​i​,in​​ − ​​ṅ ​i​,out​​ _____________
_________
​
fC​  i​​ = ​
​ = ​  all k  ​
​​n ​​i̇ ,in​​
​​ṅ ​​i,in​​
Quick Quiz 4.3
If the flow rate of
methane (M) into the
reactor is 100 gmol/h
and the flow rate of
methane out is 20
gmol/h, what is fCM?
Eq. (4.7a)
Rearranging Eq. (4.7a) gives
​​ṅ ​​i,out​​= (1 − ​fC​  i​​)​​​ṅ ​​i,in​​
n˙i,in
Reactor
n˙i,out = (l – fci)n˙i,in
The fractional conversion in a batch reactor is defined in essentially the same
manner:
− ​  ∑ ​​​​νi​  k​​ ​ξk​  ​​
​ni​  ,sys,0​​ − ​ni​  ,sys, f​​ _____________
all k
​
fC​  i​​ = ​ ____________
​
=
​​
​ni​  ,sys,0​​
​ni​  ,sys,0 ​​
​​
Eq. (4.7b)
It is always true that 0 ≤ ​​f​c,i​​ ≤ 1. Percent conversion is simply 100% times
the fractional conversion. There are a few important points to keep in mind.
(a) Conversion is defined only for compounds that are fed to the reactor, never
for products.
(b) If (and only if) there is only one reaction and all reactants are fed to the
reactor at exactly the right stoichiometric ratio, then the fractional conversion of one reactant is the same as the fractional conversion of all other
reactants.
(c) If there is only one reaction and the reactants are fed at nonstoichiometric
ratio, then the fractional conversion of the limiting reactant will be greater
than the fractional conversion of all other reactants.
Example 4.12
Fractional Conversion: Ammonia Synthesis
Ammonia (A) is synthesized from nitrogen (N) and hydrogen (H) by the following
stoichiometrically balanced reaction:
N2 + 3 H2 → 2 NH3
Consider three cases:
Case 1.
Nitrogen and hydrogen are fed to an ammonia synthesis reactor.
The nitrogen feed rate is 1000 kgmol/h, the hydrogen feed rate is
3000 kgmol/h, and ​​ξ ​​̇ = 500 kgmol/h. Calculate the fractional conversion
of nitrogen and of hydrogen.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Case 2.Nitrogen and hydrogen are fed at a 1:5 ratio to an ammonia synthesis
reactor. The nitrogen feed rate is 1000 kgmol/h and ξ​​ ​​̇ = 500 kgmol/h.
Identify the limiting reactant. Calculate the fractional conversion of
nitrogen and of hydrogen.
Case 3.Nitrogen and hydrogen are fed at a 1:5 ratio to an ammonia synthesis
reactor. The nitrogen feed rate is 1000 kgmol/h, and ​​ξ ​​̇ = 1000 kgmol/h.
Identify the limiting reactant. Calculate the fractional conversion of
nitrogen and of hydrogen.
Solution
Case 1.
− ​ν​  ​​ ​ξ ​̇ 500
For nitrogen: ​f​ cN​​ = _____
​  N  ​ = _
​
​= 0.50
​​ṅ ​N​ ,in​​
1000
− ​ν​  ​​ ​ξ ​̇ (3)500
For hydrogen: ​f​ cH​​ = _____
​  H  ​ = _
​
​= 0.50
​​ṅ ​​H,in​​
3000
Nitrogen and hydrogen are fed at stoichiometric ratio (1:3), so the
fractional conversion of the 2 reactants is the same.
Case 2.
For nitrogen: ​f​ cN​​ = _
​  500  ​= 0.50
1000
Hydrogen is fed at 5 times the rate of nitrogen, or 5000 kgmol/h.
(3)500
For hydrogen: ​f​ cH​​ = _
​
​= 0.30
5000
Quick Quiz 4.4
100 kgmol C2H4 and
100 kgmol O2 react to
make C2H4O. Which
reactant is limiting?
What is the maximum
possible fractional conversion of the excess
reactant?
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Hydrogen is fed in excess of the stoichiometric ratio. Therefore, hydrogen
is the excess reactant, nitrogen is the limiting reactant, and the fractional conversion of the limiting reactant is greater than that of the
excess reactant.
Case 3.
For nitrogen: ​f​ C,N2​​ = _
​  1000 ​= 1.00
1000
(3)1000
For hydrogen: ​f​ C,​H2​ ​​​​ = _
​
​= 0.60
5000
Hydrogen is the excess reactant and nitrogen is the limiting reactant.
100% of the nitrogen is consumed by reaction, but there is some hydrogen left.
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Section 4.3 Stream Composition and System Performance Specifications for Reactors
Example 4.13
273
Effect of Conversion on Reactor Flow: Ammonia Synthesis
Ammonia is synthesized from nitrogen and hydrogen:
​N2​ ​​ + 3​H2​ ​​ → 2​NH​3​​
Suppose we want to make 1000 kgmol/h NH3. Assume that N2 and H2 are fed to
a steady-state continuous-flow reactor at stoichiometric ratio, and that 30% conversion is achieved in the reactor. Find the flow rates of N2 and H2 into and out of
the reactor. Assume that N2 and H2 are fed at stoichiometric ratio.
Solution
The block flow diagram is
N2
H2
Reactor
N2
H2
NH3
The differential material balance on ammonia is:
Helpful Hint
If, and only if,
reactants are fed
at stoichiometric
ratio and there is
only one reaction,
then the fractional
conversion is the
same for all
reactants.
​​​ṅ ​​A,out​​ = ​​​ṅ ​​A,in​​ + ​​νA​ ​​​​ξ ​​̇
The basis is 1000 kgmol/h ammonia leaving the reactor, or n​​​ ̇ ​​A,out​​= 1000 kgmol/h.
No ammonia is in the feed to the reactor. Substituting into the material balance
equation, we find
1000 = 0 + 2​​ξ ​​ ̇ or ξ​​  ​​̇ = 500 kgmol/ℎ
N2 and H2 are fed at stoichiometric ratios, so their fractional conversion is the
same, or
​​f​CN​​ = ​​f​CH​​ = 0.3.
(Note that fractional conversion is defined only for reactants, not for products.)
From the definition of fractional conversion:
−​ν​ ​ ​ξ ​̇ −(−1)(500)
​​f​CN​​ = 0.3 = _____
​​  N  ​​ = __________
​​
​​
​​ṅ ​​N,in​
​​ṅ ​​N,in​
or
​​​ṅ ​​N,in​​= 1667 kgmol/ℎ
Using a similar procedure, we find
​​​ṅ ​​H,in​​= 5000 kgmol/ℎ
We can now find outlet flow of N2 and H2 from
​​​ṅ ​​N,out​​ = n​​​ ̇ ​​N,in​​ (1 − f​​ ​CN​​) = 1667(1 − 0.3) = 1167 kgmol/ℎ
​​​ṅ ​​H,out​​ = n​​​ ̇ ​​H,in​​ (1 − f​​ ​CN​​) = 5000(1 − 0.3) = 3500 kgmol/ℎ
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
4.3.4 System
Performance Specifications:
Selectivity and Yield
Quick Quiz 4.5
Suppose there are two
reactions: compounds A
and B react to make C,
and compounds C and
B react to make D.
Is this an example of
reactions in parallel or
in series?
In the perfect reactor, a single desired chemical reaction takes place in a single
reactor. But, in many real reactors, the chemistry is not quite so simple, and
more than one reaction may occur. There are two possibilities (Fig. 4.4):
1. Parallel reactions: The reactant takes an alternate pathway to form different product(s).
2. Series reactions: The desired product reacts further to form another
product.
Most of the time, the additional products are undesired, and our design goal is to
minimize their production while maximizing production of the desired product.
Fractional conversion of reactants is not sufficient to fully characterize reactor performance in these cases. For example, specification of the fractional
conversion of a reactant cannot account for further conversion of a desired
product via a series reaction to an undesired product.
We introduce two new means for characterizing reactor performance when
multiple reactions occur: selectivity and yield. In words, they are defined as
moles of reactant A converted to desired product P
_________________________________________
Fractional selectivity = ​​
​​
moles of reactant A consumed
moles of reactant A converted to desired product P
_________________________________________
Fractional yield = ​​
​​
moles of reactant A fed
(You will sometimes see other definitions of yield and selectivity, but these are
the definitions we use in this book.)
We need more mathematical definitions of these two terms. The first question is: Which of the reactants is reactant A? Usually, we will define selectivity and yield on the basis of either the limiting reactant or the most expensive
reactant. To be clear, we should always state selectivity or yield relative to a
specified reactant.
Reactants
Desired products
Undesired products
Undesired
products
Figure 4.4 Parallel and series reactions. Reactors provide process conditions so that reactants
are converted rapidly to the desired products. Reactants may participate in unwanted side
reactions, forming undesired byproducts. Or, the desired product may undergo further chemical
reactions to undesired byproducts. Optimally, reactor design variables are chosen to maximize
the desired reaction while minimizing all undesired reactions. Realistically, there are often
compromises that must be made between high conversion and high selectivity.
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Section 4.3 Stream Composition and System Performance Specifications for Reactors
275
Next, let’s consider a ratio that is almost the same as selectivity and write
this ratio with our usual notation:
​  ∑ ​​​ ​νP​  k​​ ​​ξ ​​k̇ ​​
Moles of desired product P generated ________
________________________________
​
​ = ​  all k
​
Moles of reactant A consumed
− ​  ∑ ​​​ ​νA​  k​​ ​​ξ ​​k̇ ​​
all k
where the summation is taken over all k reactions, as usual. Note that the “moles
of desired product P generated” is the net generation; the definition includes not
only the desired reaction where product P is generated, but also any reactions
that consume P to make undesired byproducts. Now all we need to consider is
how “moles of reactant A converted to desired product P” is related to “moles
of desired product P generated.” There’s almost always just one reaction in
which reactant A is converted to product P. Let’s call that reaction R1. νA1 and
νP1 are the stoichiometric coefficients of A and P, respectively, in R1. Then:
Moles
of reactant A converted to product P by R1
​νA​  1​​
__________________________________________
​
​ = − ​ _
​νP​  1 ​​​
Moles of product P generated by R1
Putting these two ratios together gets us what we want: fractional selectivity
for converting reactant A to product P, denoted as sA→P
​  ∑ ​​​ ​νP​  k​​ ​​ξ ​​k̇ ​​
​νA​  1​​ _______
_________
​sA​  →P​​ = ​  ​ν​   ​ ​ ​​ all k
​
P1 ​  ∑ ​​​ ​ν​  ​​ ​​ξ ​​̇ ​​
all k
Eq. (4.8a)
Ak k
With this definition, ​sA​  →P​​ is always between 0 and 1. Percent selectivity is simply
100% times fractional selectivity.
Similarly, consider the ratio
​  ∑ ​​​ ​νP​  k​​ ​​ξ ​​k̇ ​​
Moles of desired product P generated
________________________________
_______
​
​ = ​  all k  ​
​​n ​A​̇ ,in ​​
Moles of reactant A fed
Combining this ratio with the ratio −​​ν​A1​​∕ν​​ ​P1​​ gives the fractional yield ​yA​  →P​​
​  ∑ ​​​ ​νP​  k​​ ​​ξ ​​k̇ ​​
​νA​  1​​ _______
_________
​
yA​  →P​​ = − ​  ​ν​   ​ ​ ​​ all k  ​
P1
​​ṅ ​​A,in​​
Quick Quiz 4.6
Why is there a negative sign in Eq. (4.9)
but not in Eq. (4.8)?
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Eq. (4.9a)
With this definition, ​yA​  →P​​ is always between 0 and 1. Percent yield is simply
100% times fractional yield.
The above definitions apply for steady state continuous-flow reactors; similar definitions hold for batch reactors:
​  ∑ ​​​​νP​  k​​ ​ξk​  ​​
​ν​  ​​ all k
​
sA​  →P​​ = _
​  ​νA​  1 ​ ​​ ​​ _______
​​
Eq. (4.8b)
P1 ​  ∑ ​​​​νA
​  k​​ ​ξk​  ​​
all k
​  ∑ ​​​​νP​  k​​ ​ξk​  ​​
​νA​  1​​ _______
_
​
yA​  →P​​ = − ​  ​ν​   ​ ​ ​​ all​nk​
​
P1
A,sys,0​​
Eq. (4.9b)
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Finally, compare Eq. (4.7), (4.8) and (4.9) for fractional conversion, yield,
and selectivity. Notice that
​
yA​  →P​​ = ​fC​  ,A​​ × ​sA​  →P​​
Conversion reports on how much of the reactant has been consumed. Selectivity
and yield tell us how effective we’ve been at converting that reactant to the
desired product. Only two of the three terms are independent.
Eqs. (4.8) and (4.9) may look complicated. We illustrate their use in the
following examples.
Example 4.14
Selectivity and Yield Definitions: Acetaldehyde Synthesis
To make acetaldehyde (CH3CHO), ethanol (C2H5OH) is partially oxidized using O2.
​C​2​​​H5​ ​​OH + ​0.5O​2​​ → ​CH​3​​CHO + ​H2​ ​​O
(R1)
Because oxidation reactions are challenging to control, some of the ethanol is completely oxidized to carbon dioxide and water:
​C2​ ​​​H5​ ​​OH + ​3O​2​​ → ​2CO​2​​ + ​3H​2​​O
(R2)
Also, some of the acetaldehyde is partially oxidized to acetic acid, CH3COOH:
​2CH​3​​CHO + ​O2​ ​​ → ​2CH​3​​COOH
(R3)
(These reactions occur both in large chemical plants and when organisms, including
humans, consume ethanol.) Derive expressions for conversion, yield, and selectivity.
Solution
Reaction (R1) plus (R2) is an example of parallel reactions, whereas reaction
(R1) plus (R3) is an example of series reactions. It is convenient to summarize
the net reaction rate for each compound in a table (a kind of generationconsumption table!).
∑ ​νi​  k​​ ​​ξ ​​k̇ ​​
Compound
C2H5OH
− ​​ξ ​​1̇ ​​ − ​​ξ ​​2̇ ​​
O2
−0.5​​ξ ​​1̇ ​​ − 3​​ξ ​​2̇ ​​ − ​ξ3​  ​​
CH3CHO​​ξ ​​1̇ ​​ − 2​​ξ ​​3̇ ​​
H2O​​ξ ​​1̇ ​​ + 3​​ξ ​​2̇ ​​
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CH3COOH
2​​ξ ​​3̇ ​​
CO2
2​​ξ ​​2̇ ​​
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Section 4.3 Stream Composition and System Performance Specifications for Reactors
Quick Quiz 4.7
If you came up with a
way to prevent the
complete oxidation
of ethanol in
Example 4.14, would
conversion, selectivity,
and yield increase,
decrease, or stay
the same?
277
The fractional conversion of ethanol is
− ​  ∑ ​​​ ​νE​  k​​ ​​ξ ​​k̇ ​​
​​ξ ​​1̇ ​​ + ξ​​  ​​2̇ ​​
​fC​  E​​ = _________
​  all k  ​ = ​ ______
​
​​ṅ ​​E,in​​
​​ṅ ​​E,in​​
The fractional selectivity and yield of acetaldehyde, with ethanol as the reactant
of interest, are
​  ∑ ​​​ ​νA​  c,k​​ ​​ξ ​k​̇ ​​
​​ξ ​​̇ ​​ − 2​​ξ ​3​̇ ​​
​_
ν
​  1​​ ________
(−1) (​​ξ ​1​̇ ​​ − 2​​ξ ​3​̇ ​​) ________
E
​sE​  →Ac​​ = ​  ​ν​   ​ ​ ​​ all k
​ = _
​   ​ ​ _________
​ = ​  1
​
Ac1 ​  ∑ ​​​ ​ν​  ​​ ​​ξ ​​̇ ​​
1 (−​​ξ ​​̇ ​​ − ​​ξ ​​̇ ​​)
​​ξ ​​̇ ​​ + ​​ξ ​​̇ ​​
all k
E,k k
1
2
1
2
​  ∑ ​​​ ​νA​  c,k​​ ​​ξ ​​k̇ ​​
[​​ξ ​​̇ ​​ − 2​​ξ ​​3̇ ​​] _______
​​ξ ​​̇ ​​ − 2​​ξ ​​3̇ ​​
​νE​  1​​ ________
(−1) _________
_
​yE​  →Ac​​ = − ​  ​ν​   ​ ​ ​​ all k  ​ = − ​ _
​ ​  1
​ = ​  1
​
Ac1
​​ṅ ​​E,in​​
​​ṅ ​​E,in​​
​​ṅ ​​E,in​​
1
Example 4.15
Using Selectivity in Process Flow Calculations: Acetaldehyde Synthesis
We want to design and build a plant to produce 1200 kgmol acetaldehyde (CH3CHO)
per hour from ethanol and air. Laboratory data indicate that if we use a new catalyst, and adjust the feed ratio to 5.7 moles ethanol per mole oxygen, we can expect
to achieve 25% conversion of ethanol in the reactor and 100% conversion of O2,
with a selectivity for acetaldehyde of 0.6. The only byproducts are acetic acid and
water. Thus, reactions (R1) and (R3) of Example 4.14 are important; reaction (R2)
is suppressed.
Assume pure ethanol and air (79 mol% N2, 21 mol% O2) are the raw materials
available. Determine molar flow rates of all components in and out of the reactor.
Solution
As always, we start with a flow diagram
C2H5OH
O2
N2
Reactor
C2H5OH
CH3CHO
CH3COOH
H2O
N2
A DOF analysis shows that the problem is completely specified: there are 10 variables
(8 stream, 2 reaction) and 10 equations (5 material balances, 1 specified flow, 2 stream
composition specifications, and 2 system performance specifications).
The basis for the design is 1200 kgmol/h acetaldehyde produced, or
kgmol
​​ṅ ​​Ac,out​​ = 1200 ​ _
​
h
The balance equation for acetaldehyde, assuming steady-state, is
​​ṅ ​​Ac,out​​= 1200 = 0 + (​​ξ ​​1̇ ​​ − 2​​ξ ​​3̇ ​​)
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
From the problem statement, the selectivity for acetaldehyde is 0.6:
​​ξ ​​̇ ​​ − 2​​ξ ​​̇ ​​ 1200
​sE​  →Ac​​= 0.6 = _______
​  1  ​3 = ​ _____
​
​​ξ ​1​̇ ​​
​​ξ ​1​̇ ​​
Therefore
kgmol
kgmol
​​ξ ​​1̇ ​​ = _
​  1200 ​ = 2000 ​ _
​ and ​​
ξ ​​3̇ ​​ = 400 ​ _
​
h
h
0.6
From the problem statement, the fractional conversion of ethanol is 0.25. Combining
this information with the definition of fractional conversion and the above equation gives
kgmol
​​ξ ​​̇ ​​
​fC​  E​​= 0.25 = ____
​  1  ​ = _____
​  2000 ​ or ​​
ṅ ​​E,in​​ = 8000 ​ _
​
​​ṅ ​​E,in​​ ​​ṅ ​​E,in​​
h
Now, we use the steady-state balance equation on ethanol to get
kgmol
​​ṅ ​​E,out​​ = ​​ṅ ​​E,in​​ − ​​ξ ​​1̇ ​​= 8000 − 2000 = 6000 ​ _
​
h
The O2 feed rate is
​​ṅ ​​E,in​​ 8000
kgmol
​​ṅ ​​O,in​​ = ____
​   ​ = _
​
​ = 1400 ​ _
​
h
5.7
5.7
Finally, the acetic acid mole balance equation is
kgmol
​​ṅ ​​AA,out​​ = n​​ ̇ ​​AA,in​​ + 2 ​​ξ ​​3̇ ​​= 0 + 2(400) = 800 ​ _
​
h
The results are summarized below in convenient tabular form. All numbers are
given as kgmol/h.
​​​ṅ ​​i,in​​​​∑​ ​​​​​ν​ik​​ ​​​ξ ​​k̇ ​​​​​ṅ ​​i,out​​
mur83973_ch04_231-320.indd 278
C2H5OH
8000
−2000
6000
O2
1400
−1400
0
N2
53005300
CH3CHO
1200
1200
CH3COOH
800
800
H2O
2000
2000
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Section 4.4 Fractional Conversion and Its Effect on Reactor Flowsheet Synthesis
279
Often, there is a trade-off between high selectivity and high conversion. High
selectivity is generally preferred over high conversion. At high selectivity, low
single-pass conversion, and high recycle, we achieve a high overall conversion
to the desired product, and do not waste raw material making undesired products. If undesired products are made, separation and purification steps are more
difficult, complicated, and costly. If the undesired byproducts are toxic or hazardous, we are faced with high waste disposal costs and plant safety concerns.
High conversion might be favored over high selectivity if the byproducts are not
terribly toxic, the raw materials and/or desired product are toxic, and/or the
reactor operates under extreme conditions of temperature or pressure. In these
cases, high conversion reduces or eliminates recycle, thus reducing reactor vessel size and decreasing safety concerns.
4.4
Fractional Conversion and Its Effect on
Reactor Flowsheet Synthesis
The fractional conversion of reactant to product in a reactor can vary between
0 and 1. The following example illustrates the effect of changing fractional
conversion on the design and operation of a reactor.
Example 4.16
Effect of Conversion on Reactor Flows: Ammonia Synthesis
Ammonia is synthesized from nitrogen and hydrogen:
​N2​ ​​ + 3​H2​ ​​ → 2​NH​3​​
Suppose we want to make 1000 kgmol/h ammonia. Explore the effect of adjusting the
fractional conversion of nitrogen on the flow rates in and out of a continuous-flow
steady-state reactor. Assume that N2 and H2 are fed at stoichiometric ratio.
Solution
The block flow diagram is
N2
H2
Reactor
N2
H2
NH3
The basis is 1000 kgmol/h ammonia leaving the reactor. From the steady-state
balance equation on ammonia we find the extent of reaction:
kgmol
​​ṅ ​​A,out​​ = 2 ​ξ ​̇ = 1000 ​ _
​
h
kgmol
​ξ ​̇ = 500 ​ _
​
h
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Since we have specified a desired production rate of ammonia, the extent of reaction remains constant as the feed rate of reactants changes. From the definition
of fractional conversion:
− ​ν​  ​​ ​ξ ​̇ (3)(500) _____
​fC​  H​​ = ______
​  H  ​ = _______
​
​ = ​  1500 ​
​​ṅ ​​H,in​​
​​ṅ ​​H,in​​
​​ṅ ​​H,in​​
− ​ν​  ​​ ξ (1)(500) ____
​fC​  N​​ = _____
​  N  ​ = _______
​
​ = ​  500  ​
​​ṅ ​​N,in​​
​​ṅ ​​N,in​​
​​ṅ ​​N,in​​
Also,
​​ṅ ​​H,out​​ = (1 − ​​ f​CH​​) ​​ṅ ​​H,in​​
​​ṅ ​​N,out​​= (1 − ​fC​  N​​) ​​ṅ ​​N,in​​
Since the reactants are fed at stoichiometric ratio, ​​ f​CN​​ = ​f​ CH​​.
Now let’s use these equations to examine the effect of varying fractional
conversion on reactor flows. The calculations are easily carried out on a spreadsheet.
16000
Hydrogen in
Nitrogen in
Hydrogen out
Nitrogen out
Flow rate, kgmol/h
14000
12000
10000
8000
6000
4000
2000
0
0
0.2
0.4
0.6
0.8
Fractional conversion
1
The flow rates in and out of the reactor increase drastically as the fractional conversion decreases.
4.4.1
Fractional Conversion and Recycle
If the reactor is operated at low fractional conversion, a large fraction of the
reactants passes through the reactor unconverted to products, as we saw in
Example 4.16. This causes a big problem: We’ve paid for the reactants and
we’re not using them! The solution is simple: Recycle! Recycle changes the
reactor flow sheet: A separator unit must be placed downstream of the reactor,
and a recycle stream must be added to return the unused reactants to a mixer
upstream of the reactor.
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Section 4.4 Fractional Conversion and Its Effect on Reactor Flowsheet Synthesis
Recycle
Reactants
Mixer
Reactor
Separator
Products
Single-pass conversion
Overall conversion
“Tear” strategy
“Group” strategy
Figure 4.5 When conversion is low, recycle of unreacted reactants is often economical.
The flow sheet with recycle requires a mixer and separator as well as a reactor. Two different fractional conversions are defined when a reactor flowsheet with recycle is built.
“Single-pass” conversion is the conversion achieved within the reactor and is calculated
from the difference in reactant flow in and out of the reactor. “Overall” conversion is the
conversion achieved in the process, and is calculated from the difference in reactant flow
in and out of the process. Analysis of flow sheets with recycle demands new strategies.
Helpful Hint
The overall conversion is always larger
than the single-pass
conversion, but the
extent of reaction
calculated from
either measure is
the same!
With recycle, there are two conversions to consider. The conversion based
on flows in and out of the reactor is called the single-pass conversion. The fractional conversion based on flows in and out of the process (includes mixer, reactor, separator, and recycle stream) is called the overall conversion. (See Fig. 4.5.)
The overall conversion is always greater than the single-pass conversion.
Completing process flow calculations on flow sheets with recycle can be
challenging. Here are two strategies to consider when you are faced with recycle
(Fig. 4.5).
1. “Tear” the loop at the reactor inlet, and write material balance equations
around each unit in the loop starting at the “tear” and continuing all the
way around until you are back to where you started. Use the equations to
relate the flow in the tear stream to the flow in the feed stream.
2. Choose a system that groups the mixer, reactor, and separator into one block,
complete process flow calculations for fresh feed and product rates from this
system, then change systems and solve for each process unit separately.
These strategies are illustrated in Example 4.17.
Example 4.17
Low Conversion and Recycle: Ammonia Synthesis
We want to produce 1000 kgmol/h ammonia from nitrogen and hydrogen. Derive
an equation that relates the nitrogen flow rate into the reactor to the single-pass
fractional conversion. Assume the reactants are fed at stoichiometric ratio, and that
all unreacted reactants are recycled.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Solution
The flow diagram with recycle is shown. We assume that the separator works
perfectly—that is, that it completely separates nitrogen and hydrogen from the
product ammonia.
N2
H2
R
N2
H2
FF
Mixer
N2
H2
1
Reactor
N2
H2
NH3
2
Separator
P
NH3
Before we begin our detailed analysis, let’s check the degrees of freedom. The process has 10 stream variables and 1 reaction variable, for a total of 11 variables. There
is 1 specified flow (ammonia production rate), 1 stream composition specification
(reactants are fed at stoichiometric ratio) and there are 8 material balance equations
(2 for the mixer, 3 for the reactor, 3 for the separator). Thus, DOF = 11 − 10 = 1.
A system performance specification of the fractional conversion in the reactor would
completely specify the process.
To evaluate, let’s look at nitrogen flow through the loop, starting and ending with stream 1. With loops, the best strategy often is to tear the loop at the
reactor inlet, and follow the material through the loop starting at the tear and
continuing all the way around until you are back to where you started.
The nitrogen material balance equations for the reactor, separator, and mixer
are, respectively:
​​ṅ ​​N2​​ = ​​ṅ ​​N1​​ − ​​ξ​​1̇ ​​= (1 − ​ fC​  N​​)​​ṅ ​​N1​​
​​ṅ ​​NR​​ = ​​ṅ ​​N2​​
​​ṅ ​​N1​​ = ​​ṅ ​​NR​​ + ​​ṅ ​​NFF​​
We can combine these three equations to get a simple relationship between the
fresh nitrogen feed (stream FF) and the nitrogen feed to the reactor (stream 1):
​​ṅ ​​ ​​
​​ṅ ​​N1​​ = ____
​  NFF ​
​fC​  N​​
Now, if we can determine the nitrogen fresh feed, ​​n ​̇​NFF​​, we can work backward
through the loop to calculate all the process flows for a given fractional conversion.
To do this, let’s think “outside the loop,” by changing our choice of system.
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Section 4.4 Fractional Conversion and Its Effect on Reactor Flowsheet Synthesis
283
N2
H2
R
N2
H2
FF
Mixer
N2
H2
1
Reactor
N2
H2
NH3
2
Separator
P
NH3
If we treat everything inside the dashed line as the system, the process is straightforward to analyze.
​​ṅ ​​AP​​= 1000 = ​​ṅ ​​AFF​​ + ν​ A​  ​​ ​ξ ​̇ = 0 + 2 ​ξ ​̇
kgmol
​
ξ ​̇ = 500 ​ _
​
h
The extent of reaction has not changed from the case with no recycle. The ammonia
production rate is still the same, but because no reactants leave the process, the
overall fractional conversion for the process is 1.0!
Now it is simple to complete the process flow calculations for this system:
​​ṅ ​​NP​​= 0 = n​​ ̇ ​​NFF​​ + ​ν​N​  2​ ​​​​ ​ξ ​̇ = ​​ṅ ​​NFF​​ − 500
kgmol
​​ṅ ​​NFF​​ = 500 ​ _
​
h
Through similar calculations, we find n​​  ​̇​HFF​​= 1500 kgmol/h.
Now we can calculate the flow rate into the reactor (stream 1) as a function of
fractional conversion, from the equation derived from analysis of the recycle loop:
​​ṅ ​​N1​​ = _
​  500 ​
​fC​  N​​
This is exactly the same equation that we derived for the reactor inlet flow in the
absence of recycle (Example 4.16)! The flows in and out of the reactor are the same
as those shown in the graph in the solution to Example 4.16. The difference is that
the nitrogen and hydrogen flows into and out of the process are smaller with
recycle than without. The overall fractional conversion is 1.0, even when the singlepass fractional conversion is much lower.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Even with recycle, low single-pass conversion comes with a cost. Low reactor
conversions translate into higher reactor flow rates, which translate into larger
reactor volumes. This has economic consequences, because larger reactors cost
more. Low reactor conversions may have unwanted safety consequences as well,
if the reactions involve hazardous materials, because larger-volume reactors
have the potential for greater damage in case of accident.
Operation at low fractional conversion with recycle is quite common in
the commodity chemicals business, where businesses operate on small profit
margins and separation technologies are well developed. Are there cases
when we have less than 100% conversion but still don’t recycle? Yes. With
recycle, we must have a feasible technology for separating unreacted reactants from the products. If the separation is expensive relative to the value
of the raw material, and the reactants don’t have to otherwise be removed
to sell the product, then recycling may not be justifiable. For example,
polymerization reactors are usually operated without recycle, and as close
to 100% conversion as achievable, because separation of high-molecularweight molecules is difficult.
4.4.2 Fractional
Conversion, Recycle, and Purge
What if the reactor flow sheet includes recycle but there are contaminants in
the raw materials that are not reactants? This might happen, for example, when
air is used as the source of oxygen and the nitrogen in the air is inert. In that
case, the contaminants would enter the process but would not leave; rather, they
would accumulate in the process which violates steady-state operation. One
option is simply to not use recycle when the raw materials contain unreactive
contaminants. In the pharmaceutical business, for example, recycle is often not
used; the risk of buildup of unwanted and potentially toxic impurities is too great
to justify the savings on raw material costs. Another option is to separate the
contaminants from either the reactor feed or product stream. But what if separation is impractical? A third option is a compromise between 0% and 100%
recycle. This option is implemented by installing a splitter to split off part of
the recycle stream and remove it from the process. This modification to the
reactor process flowsheet is called purge (Fig. 4.6).
Performance of a splitter can be specified as the fractional split fSj
where:
​​ṅ ​​​j​ ​​​
moles leaving in stream j ___
_____________________
​
fS​  j​​ = ​
​ = ​  out ​
​​ṅ ​​in​​
moles fed
Eq. (3.23a)
where n​​​ ̇ ​​i,in​​is the flow rate to the splitter and n​​​  ​̇​j,out​​is the flow rate of stream j
leaving the splitter.
Rearranging Eq. (3.23a) gives:
​​ṅ ​​​j​out​​​ = ​fS​  j​​ ​​ṅ ​​in​​
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Section 4.4 Fractional Conversion and Its Effect on Reactor Flowsheet Synthesis
Purge
Inert in feedstream
= inert in purge
Recycle
Reactants
Helpful Hint
The fractional
split can be
defined on the
basis of either
the purge or the
recycle as the
output stream.
Watch carefully!
Mixer
285
Reactor
Splitter
Separator
Product
Figure 4.6 When the feed stream to a reactor contains an unreactive inert, the inert must
be removed from the recycle loop to avoid accumulation. If separation is too expensive, a
splitter with purge stream is chosen. The flow rate of inert into the process must be the same
as the flow rate of inert out the purge.
In Eq. (3.23a), fractional split is defined as a ratio of total flows, but it is
important to recognize that the identical fractional split applies to any component i. This is because the stream composition does not change with a splitter.
In other words, because the composition of the purge and recycle streams are
the same,
​​ṅ ​​j​​ ​​​ ​​n ​̇​i​j​out​​​
fSj = _____
​  out ​ = _____
​   ​
​​ṅ ​​in​​ ​​ṅ ​​i,in​​
Helpful Hint
With a Splitter, the
composition of all
input and output
streams is the
same.
Example 4.18
for all components i.
To tackle analysis of flowsheets with recycle and purge, it is helpful to
group the mixer, reactor, separator, splitter and recycle stream into one system,
as shown in Fig. 4.6. For steady-state operation, notice that the flow of inert in
the feed must equal the flow of inert in the purge. Using this fact is often the
first strategy to use in tackling purge problems.
Recycle with Purge: Ammonia Synthesis
We want to make 1000 kgmol/h ammonia, from nitrogen and hydrogen. We feed
nitrogen and hydrogen to a steady-state continuous-flow process at stoichiometric ratio. To save money, we purchase nitrogen that is contaminated with 2 mol%
argon. Argon is an inert gas, but it’s too expensive to separate argon from nitrogen, although it’s easy to separate argon from ammonia. Nitrogen and hydrogen
are fed to the process at their stoichiometric ratio. The reactor operates at a
single-pass fractional conversion of 0.2. Show why inserting a purge stream into
the flow sheet is required. Derive an equation that relates the fresh nitrogen feed
rate to the fractional split.
Solution
Let’s first try to use the flow sheet of Example 4.17, with the exception that argon
is in the fresh feed.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
N2
Ar
H2
N2
Ar
H2
FF
Mixer
N2
Ar
H2
1
R
Reactor
2
N2
Ar
H2
NH3
Separator
P
NH3
We’ll start by writing the material balance equations for argon, starting with the
reactor and working around the loop:
​​ṅ ​​Ar2​​ = ​​ṅ ​​Ar1​​
Reactor
​​ṅ ​​ArR​​ = n​​ ̇ ​​Ar2​​
Separator
​​ṅ ​​Ar1​​ = n​​ ̇ ​​ArR​​ + ​​ṅ ​​ArFF​​
Mixer
If we combine the first two equations, we get
​​ṅ ​​Ar1​​ = ​​ṅ ​​ArR​​
which contradicts the third equation because n​​ ̇ ​​Ar,FF​​ ≠ 0.
The problem can be clearly seen if we choose our system to be the entire
process. Argon is entering in the fresh feed, it is not consumed by reaction, and
it’s not leaving the process. Therefore, argon must be accumulating in the system.
But that violates the steady-state condition.
It might be possible to separate argon from the recycle stream and send it on
its way. However, the problem statement says that it is too expensive to remove
argon from nitrogen, so it’ll be too expensive to remove argon from a nitrogen/
hydrogen mix.
There is a third possibility. What if we bleed off (purge) part of the recycle
stream, just enough to get rid of the argon?
N2
Ar
H2
N2
Ar
H2
FF
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Mixer
N2
Ar
H2
1
B
Splitter
R
Reactor
N2
Ar
H2
2
N2
Ar
H2
NH3
3
Separator
P
NH3
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Section 4.4 Fractional Conversion and Its Effect on Reactor Flowsheet Synthesis
287
Now, the material balance equations on argon change just a bit. If we work around
the loop:
​​ṅ ​​Ar2​​ = ​​ṅ ​​Ar1​​
Reactor
​​ṅ ​​Ar3​​ = ​​ṅ ​​Ar2​​
Separator
​​ṅ ​​ArR​​ + ​​ṅ ​​ArB​​ = n​​ ̇ ​​Ar3​​
Splitter
​​ṅ ​​Ar1​​ = n​​ ̇ ​​ArR​​ + ​​ṅ ​​ArFF​​
Mixer
If we combine the first three balance equations we get
​​ṅ ​​ArR​​ + ​​ṅ ​​ArB​​ = n​​ ̇ ​​Ar1​​
which, when combined with the last balance equation gives
​​ṅ ​​ArB​​ = ​​ṅ ​​ArFF​​
Or, the argon purge flow rate must match the argon feed rate. (We reach the same
conclusion if we group the mixer, reactor, separator and splitter into a single system.)
Let’s quickly complete the DOF analysis of the flow sheet including purge.
There are 20 stream and 1 reaction variables, for a total of 21 variables. There is
1 specified flow (ammonia production), 14 material balances (3 mixer, 4 reactor,
4 separator, 3 splitter), and 1 system performance specification (fractional conversion in the reactor). The N2:H2 ratio and the Ar:N2 ratio in the fresh feed are both
known, counting for 2 stream composition specifications. Finally, there are 2 splitter restrictions. This gives a total of 20 equations, and DOF = 21 − 20 = 1.
Specifying the fractional split
​​ṅ ​​ ​​ ​​ṅ ​N​ B​​ ___
​​n ​̇​ ​​
​fS​  B​​ = ____
​  ArB ​ = ___
​   ​ = ​  HB ​
​​ṅ ​A​ r3​​ ​​ṅ ​N​ 3​​ ​​ṅ ​​H3​​
mur83973_ch04_231-320.indd 287
3000
1200
2500
960
2000
720
1500
480
1000
240
500
0
0.00
0.80
0.20
0.40
0.60
Fractional split to purge, fSB
Argon to reactor, kgmol/h
Nitrogen fresh feed, kgmol/h
is necessary to completely specify the process.
Let’s look at how the flows change as f​S​  B​​is varied. Derivation of the material
balance equations and performance specifications is left to the reader. The system
of equations was solved by allowing f​S​  B​​to vary between 0.01 and 0.99.
0
1.00
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
At very low fractional split (low purge, where most of the stream into the splitter is recycled) the nitrogen fresh feed approaches 500 kgmol/h, which is what
we observed in the absence of the argon contaminant. However, the argon flow
to the reactor is high—almost 1100 kgmol/h. This means that the reactor has to
be built large enough to handle a large flow rate of an inert material. In other
words, we are paying for extra reactor volume and not doing anything with it!
At very high fractional split (high purge, where very little of the stream into
the splitter is recycled), the argon flow to the reactor is small but we require
almost 2500 kgmol/h fresh nitrogen feed! This is like not having a recycle
stream at all. The optimum purge is set by further economic analysis of reactor
costs versus raw material costs.
Because raw materials are never 100% pure, a purge stream is frequently
required when there is recycle. There are important economic and safety consequences to selecting purge rates: Either we pay more for a larger reactor in
order to conserve raw material, or we “waste” raw material in order to reduce
the costs of the reactor. If we choose to have a small purge and high recycle,
the concentration of contaminants in the reactor feed is high (and the concentration of reactants is low). This could adversely affect the rate of reaction, or
might provide a chance for unwanted reactions to occur. If the contaminant is
hazardous or toxic, an increase in its concentration presents a safety hazard.
If the concentration or type of contaminant in the raw material changes from
day to day, a high recycle rate may cause problems with process control.
Choosing the optimum recycle and purge rates requires a detailed evaluation
of the economic, safety, and environmental impact.
Evolution of a Greener Process
Round-up® is a popular biodegradable non-selective herbicide; glyophosphate
[N-(phosphonomethyl)glycine] is the active ingredient. When sprayed on foliage, glyophosphate is absorbed by the plant and blocks the action of a specific
enzyme, which prevents the plant from making essential amino acids. Plants
wither and die within a week of being sprayed. (There is ongoing controversy
regarding the safety and toxicity of glyophosphate, as well as regarding the
commercial development of glyophosphate-resistant crop seeds and the appearance of weeds that are glyophosphate-resistant. Still the herbicide remains popular with both farmers and home gardeners.)
A key intermediate in the synthesis of glyophosphate is DSIDA (disodium
iminodiacetate), C4H5NO4Na2. In the conventional process, DSIDA is produced
by reaction of formaldehyde, ammonia, and hydrogen cyanide. A new process
is proposed that uses safer and more environmentally benign chemicals. In this
process, diethanolamine (DEA) is synthesized from ethylene, oxygen, and
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Section 4.4 Fractional Conversion and Its Effect on Reactor Flowsheet Synthesis
289
ammonia in a two-step reaction pathway: ethylene (C2H4) is oxidized to ethylene oxide (C2H4O):
​2C​2​​​H4​ ​​ + ​O2​ ​​ → ​2C​2​​​H4​ ​​O
and then ethylene oxide reacts with ammonia (NH3) to DEA:
​2C​2​​​H4​ ​​O + ​NH​3​​ → ​C4​ ​​​H1​ 1​​​O2​ ​​N
H
OH
C
H
H
H
C
C
H
N
H
H
H
C
OH
H
Diethanolamine, C4H11O2N
Finally, DEA reacts with sodium hydroxide to make DSIDA. A very preliminary
economic analysis looks favorable, so we’d like to pursue some ideas for
designing this new process.
In this case study, we’ll focus on the manufacture of DEA. Our goal is
to synthesize a reasonable block flow diagram for production of 105,000 kg/h
(1000 kgmol/h) DEA.
We’ll start by considering the two reactions in the DEA synthesis pathway.
The two reactions require different catalysts and are carried out in separate
reactors. Therefore our block flow diagram must include two reactors.
Ethylene
Oxygen
Reactor 1
Ethylene oxide
Ethylene oxide
Ammonia
Reactor 2
DEA
Let’s add in mixers and connect the mixers and reactors into a single
preliminary block flow diagram.
Ammonia
Ethylene
Mixer 1
Ethylene
Oxygen
Reactor 1
Ethylene
oxide
Mixer 2
Ethylene oxide
Ammonia
Reactor 2
DEA
Oxygen
We chose to introduce NH3 into Mixer 2, following one of our heuristics
for synthesizing block flow diagrams (Sec. 2.6.1). We could feed the ammonia
to Mixer 1 along with C2H4 and O2, but (1) this increases the required volume
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
(and hence cost) of Reactor 1 and (2) there could be other unwanted reactions,
like oxidation of ammonia.
Notice an assumption we made implicitly—that the reaction goes to completion in both Reactor 1 and Reactor 2. After consulting with some chemist
friends, we learn that it is best to carry out the first reaction such that all the
oxygen but only 25% of the ethylene is converted to ethylene oxide. (Designing
the reactor flowsheet with oxygen as the limiting reactant helps to reduce
unwanted side reactions, like complete oxidation of ethylene to CO2.) This
means that the outlet stream from Reactor 1 contains both ethylene and ethylene oxide. Furthermore, our chemist colleagues tell us that the second reaction
must be carried out in the absolute absence of oxygen. Our preliminary block
flow diagram is in need of some modification to account for these new concerns.
Another heuristic suggests removing byproducts as soon as possible. The
unreacted ethylene isn’t a byproduct, but it isn’t necessary further downstream.
Therefore, we insert a Separator right after Reactor 1 to remove ethylene. If
the Separator can also ensure that there is no trace of oxygen in the ethylene
oxide stream, we’re even better off. Now the preliminary block flow diagram
evolves:
Ethylene
Mixer 1
Ethylene
(Trace oxygen?)
Ethylene
oxide
Ethylene
Ethylene
Oxygen
Reactor 1
Separator 1
(Oxygen?)
Oxygen
Ammonia
Mixer 2
Ethylene
oxide
Ammonia
Reactor 2
DEA
Ethylene oxide
Are there alternative block flow diagrams that might be better? For example, could the process units be connected in a different arrangement? Perhaps
the Separator could be placed after Reactor 2? However, this increases the
volume of ethylene that must be processed in Reactor 2, thus increasing the
size (and cost) of the reactor. Plus, we would still need some sort of Separator
to remove any traces of oxygen from the feed to Mixer 2. We’ll stick with the
arrangement we have.
At this stage in the synthesis of the preliminary block flow diagram, we
might think a bit deeper about raw materials. Specifically, what should we use
as a source of O2? Pure oxygen is expensive. Air is much cheaper, but it contains a lot of N2. One idea is to feed air to a Separator placed upstream of
Mixer 1, which would remove N2 from air before feeding O2 to the rest of the
process. A second idea is to feed air to Mixer 1; all of the O2 would be consumed in Reactor 1 and the N2 would pass through. The N2 could then either
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Section 4.4 Fractional Conversion and Its Effect on Reactor Flowsheet Synthesis
291
be separated along with the ethylene from the ethylene oxide in Separator 1,
or it could be separated from DEA at the tail end of the process by adding a
new Separator 2 after Reactor 2.
The first alternative has the extra expense of the O2/N2 separator, but
reduces the volumetric flow through Reactor 1, hence reducing reactor costs.
The second alternative requires no additional separator, but requires a large
flow of N2 through Reactor 1. The third alternative requires an additional
separator, and requires a large flow of N2 through both Reactors. This is the
least attractive option. It is cheaper to separate N2 from ethylene oxide than
from O2, so the best option is the second alternative. The modified block flow
diagram is:
Ethylene
Mixer 1
Nitrogen
Oxygen
Reactor 1
Ethylene oxide
Ethylene
Nitrogen (Oxygen?)
Ethylene
Nitrogen
(Trace oxygen?)
Ammonia
Separator 1
Mixer 2
Reactor 2
DEA
Ethylene oxide
Let’s complete process flow calculations. We’ll choose compounds as
components, number the streams, use as a basis the desired production rate of
105,000 kg DEA/h (1000 kgmol DEA/h), approximate air composition as
21 mol% O2 and 79 mol% N2, and specify 25% conversion of ethylene in
Reactor 1. We’ll leave the details to the reader and summarize the results in
Fig. 4.7.
Now is a good time to review what simplifying approximations have been
made in getting this far. We have:
∙ Assumed the air was only N2 and O2, and neglected argon and other gases
that are present in air.
∙ Assumed the ethylene was pure, and neglected any contaminants that
might be present.
∙ Assumed Separator 2 perfectly separated all ethylene oxide from the other
gases.
∙ Assumed 100% conversion of reactants to products in Reactor 2.
∙ Neglected any side reactions (e.g., oxidation of C2H4 to CO or CO2 in
Reactor 1).
Making these approximations has greatly simplified the calculations. It’s
a good idea to explicitly list all approximations. As the design progresses, more
realistic approximations will be incorporated.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
E
1
Mixer 1
E
N
O
3
Reactor 1
E
N
O
EO
4
E
N
O
5
Separator 1
2
N
O
A
7
Mixer 2
A
EO
8
Reactor 2
D
9
EO
6
Flow (kgmol/h)
Stream
Ethylene (E)
Nitrogen (N)
Oxygen (O)
Ethylene oxide (EO)
Ammonia (A)
Diethanoloamine (D)
Total
1
8000
8000
2
3760
1000
4760
3
8000
3760
1000
4
6000
3760
5
7
8
1000
2000
1000
1000
3000
9
6000
3760
2000
12,760 11,760
6
2000
9760
2000
1000
1000
Figure 4.7 Preliminary block flow diagram for production of DEA. First alternative.
Now let’s look again at Fig. 4.7. Perhaps you have noticed something odd
about the proposed block flow diagram. We are feeding 8000 kgmol ethylene/h
to Reactor 1, then discarding 75% of it! This is very wasteful. We should
recycle this stream back to Mixer 1.
Ethylene
Nitrogen
(Trace oxygen?)
Ethylene
Mixer 1
Air
Reactor 1
Ammonia
Separator 1
Mixer 2
Reactor 2
DEA
Ethylene oxide
Do you see what the problem is with this block flow diagram? Notice that
nitrogen enters the process with the air, but never leaves the process. Since
nitrogen is not consumed by reaction, it would accumulate inside the system.
Let’s put in Separator 2, to separate unreacted ethylene from the N2, then
recycle the unreacted ethylene back to Reactor 1. We’ll assume that the separator works perfectly. The block flow diagram has now evolved to this:
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Section 4.4 Fractional Conversion and Its Effect on Reactor Flowsheet Synthesis
Nitrogen
(Trace oxygen?)
Separator 2
Ethylene
Nitrogen
(Trace oxygen?)
Ethylene
Ethylene
Mixer 1
Ammonia
Reactor 1
Air
Separator 1
Mixer 2
Reactor 2
DEA
Ethylene oxide
We’ll want to calculate process flow calculations for our new design, but
first let’s check whether the problem is completely specified. The DOF analysis is completed for each isolated process unit as well as for the entire process
and is summarized in Table 4.2. The process is completely specified. This gives
us courage to continue with process flow calculations. Further, the DOF analysis
tells us how many of what kinds of equations we need (e.g., four material balance
equations around Reactor 1). Finally, from the DOF analysis we see that the
best way to solve the equations is to start with Reactor 2 balances, because
that is the only individual process unit that is completely specified.
Table 4.2
Summary of DOF Analysis
Mixer
1
Reactor Separator Separator
1
1
2
Mixer
2
Reactor
2
Process
No. of variables
Stream variables
766443
18
Chemical reactions
010001 2
No. of equations
Flows 0 0
0
0
0 1 1
Stream
compositions 1 0
0
0
0 0 1
System performance 0 1
0
0
0 0 1
Material balances 343 2 2317
DOF 3
23220
0
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Now we proceed to apply the 10 Easy Steps as we calculate all the process
flows. All flows are in kgmol/h. The equations for the block flow diagram of
Figure 4.8 are:
Specified flow (basis): n​​  ​̇​D9​= 1000
Specified stream composition (air): ​​ṅ ​​O1​∕n​​ ̇ ​​N1​= 21∕79
Specified reactor performance (Reactor 1): ​​f​CE1​​= 0.25 = −​​ν​E1​​ ​​​ξ ​​1̇ ​​∕n​​ ̇ ​​E3​
Material balance equations:
Mixer:​​n ​̇​E3​ = ​​ṅ ​​E1​ + n​​ ̇ ​​E10​
​​ṅ ​O​ 3​ = n​​ ̇ ​​O2​
​​ṅ ​​N3​ = n​​ ̇ ​​N2​
Reactor 1:​​ṅ ​​E4​ =
0=
​​ṅ ​​N4​ =
​​ṅ ​​EO4​=
​​ṅ ​​E3​+ ​νE​ 1​ ​​ξ ​​1̇ ​
​​ṅ ​​O3​+ ν​ O​ 1​ ​​ξ ​​1̇ ​
n​​ ̇ ​N​ 3​
​νE​ O1​ ​​ξ ​​1̇ ​
Separator 1:​​ṅ ​​E5​ =
​​ṅ ​N5
​ ​=
​​ṅ ​E​ O6​ =
Separator 2:​​ṅ ​​E10​ =
​​ṅ ​​N11​ =
Mixer 2:​​ṅ ​​EO8​ =
​​ṅ ​​A8​ =
Reactor 2:
0=
0=
​​ṅ ​​D9​=
​​ṅ ​​E4​
​​ṅ ​N​ 4​
​​ṅ ​​EO4​
n​​ ̇ ​E5
​ ​
n​​ ̇ ​​N5​
​​ṅ ​​EO6​
​​ṅ ​A​ 7​
​​ṅ ​E​ O8​+ ​νE​ O2​ ​​​ξ ​​2̇ ​​
​​ṅ ​A​ 8​+ ν​ A​ 2​ ​​​ξ ​​2̇ ​​
​νD​ 2​ ​​​ξ ​​2̇ ​​
We’ve systematically come up with 20 equations describing the block flow
diagram of Fig. 4.8, and we have 18 stream variables plus 2 reaction variables.
We start with the Reactor 2 balances and work our way systematically through
the remaining equations (or we use equation-solving software). Results are
summarized along with the flow diagram in Fig. 4.8.
Compare stream 1 in Fig. 4.8 with stream 1 in Fig. 4.7. With the new and
improved process, we make much better use of our raw material! By recycling
the unreacted ethylene from Reactor 1 back to Mixer 1, we are ensuring that all
of the ethylene is eventually consumed by reaction, and none leaves the process.
Although the single-pass conversion is only 25%, the overall conversion of ethylene is 100%.
Our block flow diagram is taking shape. At this point the diagram acts as
a springboard for further questions. We might ask, for example, how easy is
it to separate ethylene from nitrogen? We do a little investigating and find out
that separating ethylene from nitrogen isn’t cheap. Once again, we head back
to the drawing board. We need to get the nitrogen out of the process, while
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Section 4.4 Fractional Conversion and Its Effect on Reactor Flowsheet Synthesis
11
E
E
N
E
5
3
E
N
O
Mixer 1
Flow (kgmol/h)
Separator 2
10
1
N
O
N
2
Stream
Ethylene (E)
Nitrogen (N)
Oxygen (O)
Ethylene oxide (EO)
Ammonia (A)
Diethanolamine (D)
Total
Reactor 1
1
2000
2000
2
3760
1000
4760
4
E
N
EO
7 A
Separator 1
3
8000
3760
1000
Mixer 2
6
EO
4
6000
3760
5
7
8
Reactor 2
9
6000
3760
9760
10
6000
2000
2000
12,760 11,760
6
8
EO
A
2000
1000
2000
1000
1000
3000
1000
1000
6000
9
D
11
3760
3760
Figure 4.8 Preliminary block flow diagram for production of DEA. Second alternative.
still recycling the ethylene. Yet, we don’t want to pay for separating the
nitrogen from the oxygen (as discussed before) or for separating the nitrogen
from the ethylene.
Here’s a compromise solution: Bleed off part of the nitrogen/ethylene
stream, and recycle the rest. We do this by replacing Separator 2 with Splitter 1.
Some ethylene leaves with the purge stream, but some is recycled.
To proceed with process flow calculations for this latest design, we need
to first specify the fraction of the ethylene/nitrogen stream fed to the splitter
that is recycled back to Mixer 1. Let’s provide a system performance specification for Splitter 1: that 80% of the feed to Splitter 1 is recycled to Mixer 1.
We could then proceed to complete process flow calculations, using a strategy
similar to that illustrated earlier. We’ll skip the details, and just summarize the
results in Fig. 4.9.
Compare Fig. 4.9 to Fig. 4.8. By installing Splitter 1, we’ve saved the cost
of building and operating an expensive Separator 2. (A splitter can be as simple as a three-way control valve.) But this has come at a price—we are throwing away 1200 kgmol ethylene/h, a raw material that we’ve paid for. And
we’ve increased the flow through Reactor 1, which will increase its size and
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Figure 4.9 Preliminary block flow diagram for production of DEA. Third alternative.
hence its cost. Which alternative is better? We don’t know until we’ve completed a more detailed analysis of equipment and raw material costs. But we’ve
made a good start at laying out the best alternatives.
Summary
∙ Chemical reactors are at the heart of most chemical processes, and engineers must select reaction pathways, choose reactor design parameters, and
design reactor flowsheets to maximize the safety and efficiency of the
process while ensuring that the product meets quality standards.
∙ The extent of reaction (ξ or ​​ξ ​​)̇ concept is very useful in reactor material
balance equations when reaction stoichiometry is known. When the reactants are a complex mixture of materials, or the molecular formula is
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Summary
297
unknown, reactor balances can be completed by using elements as components or by using mass reaction rates.
∙ Continuous-flow, batch, and semibatch reactors all find service in chemical
processes, and their performance is analyzed by the appropriate form of the
material balance equation. For continuous-flow steady-state reactors when
reaction stoichiometry is known, the material balance equation simplifies to
​​ṅ ​​i,out​ = ​​ṅ ​​i,in​ +​  ∑ ​​​ ​ν​ik​ ​​ξ ​​k̇ ​
all k
For batch reactors when reaction stoichiometry is known, the material balance equation simplifies to
​n​i,sys, f​= ​n​i,sys,0​ + ​  ∑ ​​​ ​ν​ik​​ξ​k​
all k
∙ Three useful measures of reactor system performance are conversion,
selectivity, and yield:
of reactant consumed
Fractional conversion = ________________________
​ moles
​
moles of reactant fed
−​  ∑ ​​​νi​  k​​ ​​ξ ​k​̇ ​​​
̇ ​​i,in​​ − n​​ ̇ ​​i,out​​ ____________
​​_________
n
​fC​  i​​ = ​​
​​ = ​  all k  ​
​​n ​i​̇ ,in​​
​​ṅ ​​i,in​​
moles of reactant A converted to desired product P
_________________________________________
Fractional selectivity = ​​
​​
moles of reactant A consumed
​  ∑ ​​​νP​  k​​ ​​ξ ​​k̇ ​​​
​νA​  1​​ ____________
_
​sA​  →P​​ = ​  ​ν​   ​ ​ ​​ all k
​
P1 ​  ∑ ​​​ν​  ​​ ​​ξ ​​̇ ​​​
all k
Ak k
moles of reactant A converted to desired product P
_________________________________________
Fractional yield = ​
​
moles of reactant A fed
​  ∑ ​​​νP​  k​​ ​​ξ ​​k̇ ​​​
​νA​  1​​ ____________
_
​yA​  →P​​ = − ​  ​ν​   ​ ​ ​​ all k  ​
P1 ​​ṅ ​​A,in​​
∙ If fractional conversion is low, then reactants are recycled. This requires
addition of a separator to the flow sheet. With recycle, the overall conversion of raw material to product in the process may be very high even if
the single-pass conversion is low. If any inerts are present in the raw material and recycle is used, then purge may be required. With purge, a splitter is added after the separator.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
ChemiStory: Quit Bugging Me!
Not many people like bugs. They are, for most of us, nuisances. More than
that, about 1% of the known 1 million species of insects are real pests.
Because insects are voracious feeders and disease spreaders, some bugs
threaten human life and health. For example, shortly after the Bolshevik
Revolution, 25 to 30 million Russians contracted typhus, which is spread by
lice; about 3 million died. During World War I, neutral Switzerland suffered
severe food shortages because insects consumed much of their grain.
It was in Switzerland that Paul Hermann
Müller was born, in 1899. Paul was a mediocre student who dropped out of school at the
age of 17. But he loved chemistry, and kept a
lab in his family’s home. Paul eventually
returned to school, completed his Ph.D. in
chemistry in 1925, and went to work for
J. R. Geigy. (Geigy later became Ciba-Geigy,
which later became the pharmaceutical giant
Novartis.) During the early part of Müller’s
career, the company discovered a mothproofing
compound—a chlorinated hydrocarbon that
was a stomach poison for moths. This discovery initiated Geigy’s expansion into the insecJim Gathany/Centers for
ticide business. Prior to this discovery, chemical
Disease Control and Prevention
insect control had been limited to natural compounds like nicotine and rotenone—tropical-plant-based compounds that were
expensive and unstable—and arsenic compounds—cheap stomach poisons
that worked well against chewing insects, but were also highly toxic to humans
and other warm-blooded animals.
In 1935, Müller was assigned the job of finding a better insecticide. He
set several criteria for his “ideal” insecticide. It would: kill by contact rather
than requiring ingestion; be a broad spectrum pesticide, killing many different
kinds of insects; be harmless to fish, plants, and warm-blooded animals; have
no odor; be inexpensive and chemically stable. Müller got to work and started
synthesizing and testing compounds. He started with a few chemical structures;
if a compound looked promising he would make several related compounds.
After 4 years, he had laboriously worked his way through 349 compounds, one
by one. (Today, large “libraries” of compounds are synthesized in parallel by
combinatorial chemistry, and tested in parallel using high-throughput screening. Hundreds if not thousands of chemicals can be synthesized and tested in
weeks.) For his 350th compound, Müller reacted chloral and chlorobenzene
with sulfuric acid as a catalyst, and made dichlorodiphenyl trichloroethane, or
DDT. He reported that DDT, sprayed in glass cages, killed flies. Even better,
the cages remained toxic weeks later. DDT was not degraded by light or
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Summary
299
oxidation, and had a very low vapor pressure, so it would persist after application.
It killed many other insects such as the
Colorado potato beetle larvae. It was
insoluble in water, suggesting that it would
not contaminate the water supplies. Muller
had hit his target—he had discovered a
cheap, powerful, broad-spectrum contact
poison that was very stable.
Geigy patented the use of DDT as
an insecticide. It was an instant hit. In
1942, Geigy sold one pound of DDTlaced insecticide per person in
Switzerland, and rescued the potato
crop. Since Switzerland was a neutral
country, Geigy reported their invention
to both Allied and Axis countries.
swim ink 2 llc/Corbis/Getty Image
Germany was not interested, as they had
their own insecticide development program underway. The United States, however, was very interested. After just
a few months of testing, DDT was declared safe and effective, and its use
skyrocketed. In Naples, Italy, 1.3 million refugees were sprayed with DDT
after Allied Forces recaptured that city, thereby preventing a sure outbreak
of deadly typhus. Pacific islands were sprayed with DDT to kill mosquitoes
before Allied Forces fought to reoccupy the lands. (Earlier in the war,
malaria had incapacitated up to 2 out of every 3 soldiers in the mosquitoinfested islands.)
In August, 1945, DDT was released for civilian use, and was it ever
used! Dairy farmers, apple growers, cattle ranchers and housewives all used
DDT to kill any and all insects. In the 1950s, the World Health Organization
aggressively pushed for DDT spraying to eradicate malaria. This effort was
extraordinarily effective; for example, in Sri Lanka the number of malaria
cases dropped from 2.8 million in 1948 to just 17 in 1963! By this time,
the United States was using over 150 million pounds per year of DDT, and
a nearly equal quantity of other chlorinated hydrocarbon bug killers.
Among all the enthusiasm for this miracle chemical came worrisome bits
of news. Some scientists expressed concern that long-term exposure effects
had never been tested. A few stories emerged about fish kills and the development of DDT-resistant insects. As early as 1948, when Paul Müller was
awarded the Nobel Prize in Medicine for his discovery, he spoke about his
concerns of the effect of DDT on ecosystems, especially with overuse. DDT
started showing up in sites distant from where it had been sprayed—in fisheating birds and cow’s milk. In some places, insect-eating birds died for
lack of food. In other places, DDT killed more beneficial than damaging
(continued)
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
insects, upsetting predator-prey relationships. Baby food makers were unable
to get fruits and vegetables that were free of DDT residues.
DDT’s best features turned out to be its worst. Its broad-spectrum activity meant that entire insect ecosystems were wiped out. Its chemical inertness meant that DDT remained in the environment for months and years
after its initial application. Its insolubility in water meant that it was soluble
in fats and oils, and accumulated in fatty tissue in fish, birds, and mammals.
In 1962, Rachel Carson published her book, Silent Spring.
Analyzing studies on DDT published
over the previous 17 years, Carson
claimed that DDT overusage was
leading inexorably toward a silent
spring, bereft of insects and birds.
This extraordinary book moved the
debate about DDT from the scientists’ labs and conference halls to the
public’s living rooms and town halls.
And in 1964, when evidence surfaced that peregrine falcons and bald
eagles were dying off because of
DDT-caused eggshell thinning, the
tide dramatically turned. The public
outcry against the once-popular
insecticide grew to a loud roar. As
Rachel Carson in her laboratory.
one of its first acts, the Environmental
George Rinhart/Corbis/Getty Images
Protection Agency (EPA), established in the United States in 1971, banned DDT. Anti-DDT feelings remain
strong in the general public: recently a shipment of Zimbabwean tobacco
was blocked from entering the United States because it contained traces of
DDT. (One wonders whether tobacco or DDT poses the greater risk.)
Thirty years after the ban on DDT, the bald eagles and falcons have
returned. Still, because of its incredibly slow degradation rates, it is estimated
that about 1 billion pounds of DDT remain in the environment. Farmers have
switched to other pesticides. But there have been adverse consequences brought
on by the ban on DDT. Some pesticides developed as DDT replacements, such
as parathion, have proved to be acutely toxic to humans. The ban on DDT
spraying, along with the widespread and rapid development of insecticideresistant bugs, has led to a rebound in mosquito populations and a stunning
increase in mosquito-borne disease. Worldwide, 300 to 500 million people
suffer from malaria every year and 1 to 2 million people die from the disease,
mostly small children in Africa. Weakness and fevers brought on by the disease
cripple the fragile economies of entire villages. The burgeoning mosquito
population may be partly responsible for the spread of other pathogens such
as West Nile virus. New chemical, biological, and ecological strategies to
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Chapter 4 Problems
301
control the number of mosquitos without upsetting fragile ecosystems are under
study. These include use of natural insecticides such as bacillus thurngiensis
(BT) to selectively kill larvae, installation of pheromone lure traps, and techniques to encourage mosquito-eating predators such as bats.
Quick Quiz Answers
4.1
4.2
4.3
4.4
4.5
4.6
4.7
1.4 gmol ibuprofen.
Hydrogen is limiting, 20% excess.
​fC​ M​ = 0.80.
C2H4, 50%.
Both!
Because in Eq. (4.10), both ​ν​A1​and Σ​ν​Ak​ ​​ξ ​​k̇ ​ are negative, so selectivity
is positive. In (4.11), ​νA​ 1​is negative, needs minus sign to make yield
positive.
​ξ ​2​̇ ​= 0, so conversion decreases, selectivity increases, yield stays the
same.
References and Recommended Readings
1. The new synthesis route to ibuprofen won several awards, including the
1993 Kirkpatrick Chemical Engineering Achievement Award and the 1997
Presidential Green Chemistry Challenge Award. The example is described
more fully in Real World Cases in Green Chemistry, by M. C. Cann and
M. E. Connelly (2000). Published by the American Chemical Society.
2. Prometheans in the Lab: Chemistry and the Making of the Modern World,
by Sharon Bertsch McGrayne (McGraw Hill, 2001), has more detail on
Paul Müller, Rachel Carson, and the invention of DDT.
Chapter 4 Problems
Warm-Ups
Section 4.1
P4.1 List four chemical reactions you used today. Describe which ones, if
any, belong to one of the categories of reactions listed in Section 4.1.1.
P4.2 Example 4.1 compared traditional and modern synthesis schemes for
ibuprofen manufacture. Evaluate these two schemes in light of the heuristics given in Section 4.1.2. Which heuristics are violated by the traditional scheme and not by the modern pathway?
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Section 4.2
P4.3 Consider the reaction 2A + B → C + 0.5D. If ​​r ​​Ȧ ​= −10 gmol/h, find
​​r ​​Ḃ ​, ​​r ​​Ċ ​, r​​  ​​Ḋ ​and ξ​ ​.̇
P4.4 Sketch out flow diagrams that illustrate the difference between continuousflow, batch and semibatch reactors. Then, explain which type of reactor
you might choose to manufacture: (a) 45,000 metric tons caprolactam
every year, (b) pyrolysis of plastic solid wastes into valuable liquids
and gases, and (c) 12 dozen cupcakes per day.
P4.5 A burner is fed with 100 gmole/s of methane (CH4) and 400 gmoles/s
of O2, where the methane is burned to CO2 and H2O:
CH4 + 2 O2 → CO2 + 2 H2O
gmol
The burner operates at steady state, and it is known that ​​ξ ​​̇ = 100 ​​ _____
s ​​.
With the burner as the system, write the differential component mole
balance equations for CH4, O2, CO2, and H2O.
Section 4.3
P4.6 Propane (C3H8) is mixed with oxygen at a 1:10 C3H8:O2 molar ratio. If
the reaction produces CO2 and H2O, which is the limiting reactant?
What is the percent excess of the other reactant?
P4.7 100 kg of butane (C4H10) is mixed with 1600 kg air and combusted.
Which is the limiting reactant, butane or air? Assume air is 79 mol%
N2 and 21 mol% O2.
P4.8 Consider the reaction A + 2B → C + D. (a) If 100 gmol A and 200 gmol
B are fed to a reactor, and fCA = 0.5, what is fCB? (b) If 100 gmol each
of A and B are fed to a reactor, and fCA = 0.5, what is fCB?
P4.9 Consider the reactions A + 2B → C + D and A + C → E + F. C is
the desired product, and A and B are fed at stoichiometric ratio for
the desired reaction. If fCA = 0.6 and fCB = 0.5, what is yA→C? What
is sA→C?
P4.10 Consider the reactions A + 2B → C + D and A + C → E + F. C is the
desired product, and A and B are fed at stoichiometric ratio for the
desired reaction. If fCA = 0.6 and fCB = 0.5, what are ξ1 and ξ2?
Section 4.4
P4.11 Consider the reaction A + 2B → C + D. 50 gmol/h A and 100 gmol/h
B are fed to a process that includes a mixer, reactor, separator and
recycle stream. The single-pass conversion in the reactor fCA = 0.25,
and ​ξ ​̇ = 50 gmol/h. What is the flow rate of A to the reactor? What is
the overall conversion of A?
P4.12 Explain why some reactor flow sheets must include purge.
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Drills and Skills
Section 4.2
P4.13 Acrylonitrile (C3H3N, used to make carbon fiber, acrylic fibers, nylons,
fumigants, and synthetic rubber) is synthesized by catalytic ammoxidation of propylene (C3H6):
2C3H6 + 2NH3 + 3O2 → 2C3H3N + 6H2O
Propylene, ammonia, and air (79 mol% N2, 21 mol% O2) are mixed
and then fed to the reactor, where the mixture reacts over a catalyst to
make acrylonitrile. The reactor operates at steady state. You are the
process engineer in charge of monitoring the performance of the reactor. One day you determine that the gas flow rate out of the reactor is
7095 gmol/min, and that the gas contains 28.19 mol% water and 1.88 mol%
ammonia, along with N2, propylene, and acrylonitrile, but no O2.
Draw a flow diagram, and use a DOF analysis to show that the
problem is correctly specified. Write the correct form of the material
balance equations for all compounds in this system. Calculate: (a) The
extent of reaction ξ​​  ​​,̇ (b) the flow rate (gmol/min) of acrylonitrile leaving
the reactor, and (c) the flow rates (gmol/min) of propylene, ammonia,
and air fed to the reactor.
P4.14 Dimethyl carbonate (DMC, C3H6O3) can be synthesized by a process
called oxidative carbonylation of methanol:
2CH3OH + CO + __
​​ 1 ​​ O2 → C3H6O3 + H2O
2
A gas containing 80 mol% CH3OH and 20 mol% CO at 2000 gmol/h
is mixed with air (79 mol% N2 and 21 mol% O2) and then fed to a
reactor operating at steady state, where the reaction takes place. The
flow rate of the stream leaving the reactor is 2264 gmol/h, and this
stream contains no O2. Draw a flow diagram and write the correct form
of the material balance equations for all compounds in this system.
Determine the flow rate of air to the reactor, the extent of reaction ξ​ ​,̇
and the composition (mol%) of the reactor effluent.
P4.15 1,3-propanediol (C3H8O2) is a building block in the synthesis of polymers that are used to make fabrics or plastic bottles. 1,3-propanediol is
made commercially by both chemical and biological routes. One chemical route is called hydroformylation, starting from ethylene oxide
(C2H4O):
C2H4O + CO + 2H2 → C3H8O2
A gas stream containing 30 mol% C2H4O, 30 mol% CO, and 40 mol%
H2 is fed to a reactor at 900 gmol/min, where the mixture reacts over
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P4.16
P4.17
P4.18
P4.19
P4.20
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a catalyst to make C3H8O2. The reactor operates at steady state. You
are the process engineer in charge of monitoring the performance of the
reactor. One day you sample the gas stream leaving the reactor and
determine that it contains 36 mol% C3H8O2. Draw a flow diagram and
complete a DOF analysis. Then calculate the total flow rate (gmol/min)
and molar composition of the reactor effluent.
Miscellaneous gas waste streams from a solvent recycling facility are
combined and burned, because purifying the compounds is prohibitively
expensive. A typical combined waste stream contains 5 mol% acetone
(C3H6O), 35 mol% methanol (CH3OH), 20 mol% acetic anhydride
(C4H6O3), 10 mol% benzene (C6H6), along with 15 mol% N2, 5 mol%
O2, and 10 mol% H2O. Calculate the minimum quantity of air needed
to completely combust 100 gmol waste gas stream, and calculate the
composition (mol%) of the combustion gas.
Redo Example 4.3, but with natural gas from Well TX. Compare flue
gas composition from burning Well TX versus Well NM natural gas
and note any substantial differences. You can choose to solve this problem either using elements as components or by writing a set of balanced
chemical reactions and using compounds as components.
Aspirin (acetylsalicylic acid, C9H8O4) is made by combining acetic
anhydride (C4H6O3) with salicylic acid (C7H6O3); acetic acid (CH3COOH)
is the byproduct of the reaction. (Chewing on willow leaves was known
as a way to control pain as far back as 400 bc; the active ingredient,
salicylic acid, is plentiful in the roots, bark, and leaves of willow and
other trees.) 150 g salicylic acid is mixed with 150 g acetic anhydride
in a batch reactor. At the end of the reaction time, 100 g aspirin has
been made. Find rA and RA where “A” is aspirin. Use integral material
balances to find the quantities (g) of all other compounds in the reactor
at the end of the reaction time.
The owner of a local brewery would like to test whether she can use spent
grains as a fuel for its boiler, rather than sending the grains to the landfill
or using as animal feed. Chemical analysis of the grains gives the following elemental composition: 14.4 wt% C, 6.2 wt% H, 78.8 wt% O, and
0.6 wt% S. How much air is needed, per kg of spent grains, to completely
combust the grains? If the maximum allowable SO2 in gas released
to the atmosphere is 0.1 mol%, will you need to install a scrubber to
remove SO2?
You place 1.00 kg spent brewery grains (14.4 wt% C, 6.2 wt% H,
78.8 wt% O, and 0.6 wt% S) in a laboratory reactor. You evacuate all
air in the reactor, seal it and then heat the reactor to initiate pyrolysis.
Gases generated from pyrolysis of the grains are continuously vented
at a constant rate until no material is left in the reactor. The gases
are sampled and found to contain CO, CO2, H2O, and SO2. It takes
30 minutes for the grains to be completely pyrolyzed. Write material
balance equations that model this semibatch reactor. Calculate the
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flow rate of gas (in standard cubic feet per minute), the total amount
of vented gas (standard cubic feet), and the composition (mol%) of the
vented gas.
P4.21 One tablet of Alka-Seltzer contains 324 mg aspirin (C9H8O4), 1904 mg
sodium bicarbonate (NaHCO3), and 1000 mg citric acid (C2H3O (COOH)3).
A reaction occurs when the tablet is dropped into a glass of water. What’s
the reaction between the last two ingredients that produces the fizz? (Hint:
This reaction also produces a weak base.) What volume of gas is produced from one tablet of Alka-Seltzer dissolved in a glass of water? To
solve this problem, first draw the process as a semibatch reactor and then
apply the integral material balance equation.
P4.22 Blue light-emitting diodes can be manufactured from gallium nitride
(GaN) by a process called metalorganic chemical vapor deposition
(MOCVD). A 1 cm × 1 cm chip of Al2O3 is placed inside a laboratoryscale MOCVD reactor to serve as an inert substrate. Trimethyl gallium
[(CH3)3Ga] and ammonia (NH3) are pumped continuously into the reactor, where the following reaction occurs:
(CH3)3Ga (g) + NH3 (g) → GaN (s) + 3 CH4 (g)
The solid GaN deposits on the Al2O3 chip in an even layer while the
methane gas flows continuously out of the reactor.
Ammonia and trimethylgallium are fed to the reactor at a steady
flow rate of 18 and 10 μmol/h, respectively. The exit gas leaving the
reactor flows at 37 μmole/h and contains trimethylgallium, ammonia,
and methane. Calculate the composition (mol%) of the exit gas. If GaN
has a density of 6.1 g/cm3 and a molar mass of 84 g/gmol, how long
will it take to build up a GaN layer 1 μm thick?
P4.23 You are a newly hired process engineer working for the drug company
Nomorepain, Inc. Nomorepain, Inc., is interested in building a plant to
make the painkiller ibuprofen, using the new catalytic reaction scheme
described in Example 4.1. Your job is to measure the reaction rate with
different catalysts as a first step in designing a full-scale reactor. In one
experiment, you mix 134 g isobutylbenzene (IBB) with 134 g acetic
anhydride (AAn) in a laboratory-scale batch reactor, adjust the temperature, add some catalyst, and wait 1 hour. At the end of the hour
you stop the reaction, collect all the material in the pot, and send it for
chemical analysis. The chemist, who’s been working for Nomorepain, Inc.
for many years, reports back that the pot contains 27 g IBB, 52 g AAn,
121 g isbutylacetophenone (IBA), and 68 g acetic acid (AAc). When
you see these results you tell the chemist that the analysis is wrong.
He protests. Who’s right? Write a brief, polite, but convincing memo
to the chemist explaining your reasoning (or write an apology note,
explaining how you could possibly have made the mistake of questioning
his results).
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P4.24 Deepwater injection of CO2 is proposed as one way to reduce atmospheric CO2 levels. However, since CO2 is acidic, injection into the
ocean lowers the pH with adverse effects on marine life. One proposed
solution is to build CO2 sequestration reactors at power plants located
near the ocean. Stack gases rich in CO2 (~10 mol%) are pumped across
a bed of porous limestone (CaCO3) that is continuously sprayed with
water. The reaction produces calcium bicarbonate (Ca(HCO3)2), which
is alkaline and highly soluble in water. This would be pumped out
continuously into the ocean. You want to design a semibatch reactor to
treat 1 ton of CO2 per day, assuming that the reactor would be refilled
once per day. How much limestone must the reactor hold? How much
stack gas (MMSCFD) could be treated with this reactor?
P4.25 Engine exhaust contains a hot mix of gases including N2, O2, NO, NO2,
H2, H2O, NH3, HNO3, CO, CO2, and CH4. As the gas mixture cools and
is run through a catalytic converter, some reactions take place. Determine
a set of independent chemical reactions that involve these (and only
these) species.
P4.26 Is the following set of reactions linearly independent?
H2O + CO → CO2 + H2
2H2 + O2 → 2H2O
4H2 + 2CO → 2CH4 + O2
P4.27 When methanol (CH3OH) and oxygen react, several compounds are
produced: formaldehyde (HCHO), formic acid (HCOOH), CO, CO2,
and H2O. How many independent chemical reactions are there in this
system? Find a set of stoichiometrically balanced independent reactions.
Section 4.3
P4.28 20 lbmol/h propylene, 10 lbmol/h ammonia, and 100 lbmol/h air
(21 mol% oxygen, 79 mol% nitrogen) are fed to a reactor, where they
react to make acrylonitrile and water (as a byproduct). The balanced
reaction is
2C3H6 + 2NH3 + 3O2 → 2C3H3N + 6H2O
90% of the ammonia is converted to product. Calculate the fractional
conversions of propylene and oxygen. Also calculate the flow rate
(lbmol/h) and composition (mol%) of the reactor outlet stream.
P4.29 In the first step in manufacture of the pain medication ibuprofen, isobutylbenzene (C10H14) and acetic anhydride (C4H6O3) react over a catalyst to make isobuylacetophenone (C12H16O), with acetic acid as a
byproduct. 1200 gmol/h of each of the reactants is fed to a reactor
operating at steady state. The exit gas is analyzed and found to contain
42 mol% acetic acid. Write the balanced chemical reaction and draw a
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flow diagram. What is the extent of reaction ξ​ ​̇ (gmol/h) in the reactor?
What is the fractional conversion of each of the reactants?
P4.30 Citral (C10H16O) is extracted from lemongrass oil and is popular in
many consumer products, from dish detergents to ice creams, for its
pleasant lemon-lime fragrance. Alternatively, citral can be made synthetically from butene (C4H8), formaldehyde (CH2O), and oxygen; the
net reaction is:
2C4H8 + 2CH2O + 0.5O2 → C10H16O + 2H2O
A gas stream (1200 gmol/h) containing 35 mol% butene and 65 mol%
formaldehyde is mixed with air (79 mol% N2, 21 mol% O2) at an 8:3
gas:air molar ratio. The mixture is fed to a reactor, where the mixture
reacts over a catalyst to make citral. The fractional conversion of O2 is
0.9. The reactor operates at steady state. Calculate (a) the flow rate of
all components in the reactor outlet stream, (b) mol% citral in the reactor outlet stream, and (c) fractional conversion of butene and formaldehyde. Also specify which reactant is limiting.
P4.31 Waste vapor produced during fiber manufacturing contains 20 mol% CH4,
20 mol% CS2, 10 mol% SO2, and 50 mol% H2O. The plant manager wants
to install a combustion system to burn the vapors and release the combustion gases to the atmosphere. Local environmental regulations limit the
released gases to a maximum of 0.5 mol% SO2, which the manager proposes can be reached by diluting the combustion gases with excess air.
Calculate the air (lbmol) required per 100 lbmol vapor in order to meet
constraint on SO2 content. What is the percent excess air needed?
P4.32 Typically, natural gas contains about 97 mol% methane (CH4) and
3 mol% ethane (C2H6). The gas is combusted with excess air in a poorly
maintained home furnace, and the combustion gases contain some CO as
well as CO2, at a 1:5 CO:CO2 molar ratio. 100% of the natural gas is
combusted, and 90% of the oxygen is converted to products. Find the
composition (mol%) of the product gas, calculate the moles air fed per
mole natural gas, and determine the percent excess air fed to the furnace.
P4.33 Ethylene oxide is produced by partial oxidation of ethylene:
2 C2H4 + O2 → 2 C2H4O
Complete oxidation occurs as an undesired side reaction:
C2H4 + 3O2 → 2CO2 + 2H2O
Ethylene, oxygen, and nitrogen are placed in a batch reactor, with an
initial composition of 10 mol% C2H4, 12 mol% O2, and 78 mol% N2.
The reactor is heated. After some time, the reactor is cooled so that
essentially all the water is condensed. The reactor is opened and the
gases are sampled. All the oxygen is gone, and the ethylene concentration
is 5.1 mol%. What is the fractional conversion of ethylene, and the yield
and selectivity of ethylene oxide from ethylene?
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P4.34 Ethylene glycol (HOCH2CH2OH), used as an antifreeze, is produced by
reacting ethylene oxide with water. A side reaction produces an undesireable dimer, DEG:
C2H4O + H2O → HOCH2CH2OH
HOCH2CH2OH + C2H4O → HOCH2CH2OCH2CH2OH
The reactor feed is 10 gmol/min ethylene oxide and 30 gmol/min water.
If the fractional conversion of ethylene oxide is 0.92 and the selectivity
is 0.85, what is the reactor outlet composition and the yield?
P4.35 In the case study, we made diethanolamine from ethylene, oxygen, and
ammonia. Now, let’s consider just the first reactor, where ethylene oxide
is produced. The reaction is
2 C2H4 + O2 → 2 C2H4O
(R1)
There is an unwanted side reaction—combustion of ethylene to CO2 and
water:
C2H4 + 3 O2 → 2 CO2 + 2 H2O
(R2)
Air (79 mol% N2, 21 mol% O2) is fed as the source of oxygen. The
C2H4:O2 molar ratio in the feed is 3:1. The fractional conversion of
ethylene is 0.20, while all of the oxygen is consumed.
We want to produce 500 kgmol/h ethylene oxide. What’s the reactor
feed rate and feed composition? What is the selectivity and yield of ethylene oxide from ethylene?
P4.36 100 gmol/min of a solution of 70 mol% ethanol/30 mol% water is fed to
a reactor operating at steady state, along with 80 gmol/min of air (79 mol%
N2, 21 mol% O2). Ethanol (C2H5OH) reacts with oxygen to make acetaldehyde (CH3CHO). Acetaldehyde is further oxidized to acetic acid
(CH3COOH). Write the two stoichiometrically balanced chemical equations. What is the byproduct of the reactions? What is the limiting reactant? If there is 100% conversion of the limiting reactant and the
production rate of acetaldehyde is 25 gmol/min, calculate the fractional
conversion of the excess reactant, the yield of acetaldehyde from ethanol, and the composition and flow rate of the reactor effluent stream.
P4.37 Your neighbors are concerned about the operation of their gas furnace.
A handyman came to their door with an offer to check the performance
of their furnace at no expense. The handyman explained that if the CO2
content of the gas leaving the chimney is above 15%, the situation is
dangerous to their health, violates city codes, and can cause chimney rot.
He carefully took a sample of the gas leaving the chimney and reported
that it contained 30% CO2 on a dry basis (with water not included). He
has offered to arrange for the purchase and installation of a safe new highefficiency furnace at a bargain price. Your neighbors are burning natural
gas. What is your estimate of the mole percentage of CO2 in the chimney
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gas (dry basis)? What will you tell your neighbors regarding the handyman
proposal?
P4.38 BUG is a brand-new and exciting chemical, possessing remarkable
insecticide properties, yet biodegradable and nontoxic to birds, fish, or
mammals. Your company is interested in manufacturing BUG, using a
newly discovered synthesis scheme involving six reaction steps. The
scheme is top secret because the patent hasn’t been issued yet; all you
are told is that it requires several reactants, which are labeled A, B, etc.
Each reactor runs under different conditions, so six reactors in series
are required. The reaction scheme, along with fractional conversions
achievable, is given below.
Reaction
Fractional conversion
A + B → C + H2O
0.92
C + HCN → D + CO2
0.95
D + 2NaOH → E + 2H2O
0.97
E + 2HNO3 → G + H2O
0.95
G + 2 F → J + 2NaNO3
0.97
J + K → BUG + 2H2O
0.92
You are in charge of evaluating waste production for this synthesis
scheme. Sketch out the flow diagram. Per 100 kgmol BUG produced,
calculate the moles of each reactant required and each waste product
generated. Assume all reactants are fed at stoichiometric ratio. Compare
your results to what could be achieved if the fractional conversion of
all six steps were 1.0. If you could improve conversion in only one
reaction step, which would you choose and why?
Section 4.4
P4.39 A mixture of CO and H2 is fed to a methanol (CH3OH) synthesis reactor
operating at steady state, where a 50% single-pass conversion of CO is
achieved. The reactor effluent is sent to a separator. All of the methanol
is recovered as product, and the unreacted CO and H2 are recycled. The
methanol production rate is 100 kgmol/h. The overall conversion of both
CO and H2 is 100%. What is the feed rate of CO and H2 to the process,
the flow rate to the reactor, and the flow rate and composition of the
recycle stream?
P4.40 A fresh feed of 110 kgmol/h CO, 230 kgmol/h H2, and 20 kgmol/h N2
is fed to a methanol synthesis process. The fresh feed is mixed with a
recycle stream and sent to a reactor. The reactor effluent is separated,
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where methanol is recovered as pure product and the unreacted CO, H2,
and N2 are recycled. The methanol production rate is 100 kgmol/h.
A splitter is placed on the gas stream to purge 30%, while the remainder is recycled. Draw the flow diagram and complete a DOF analysis.
Then calculate the single-pass and overall conversion of CO and H2,
and the percent N2 in the reactor feed.
P4.41 Hydrogen reacts with iron oxide (Fe2O3) to produce metallic iron
(Fe), with water vapor as a byproduct. 100% conversion of Fe2O3 is
achieved, and the metallic iron is easily separated from the hydrogenwater vapor mixture. The water is condensed, and the hydrogen is
recycled. The hydrogen source is contaminated with 1 mol% CO. The
recycle:fresh feed ratio is 4:1, and the maximum allowable CO in the
gas fed to the reactor is 2.5 mol%. Draw a flow diagram, and complete a DOF analysis. Then calculate the single-pass and overall conversion of H2 as well as all process flows, for a production rate of
1 ton/day metallic iron. (Hint: Consider changing basis to the gas
feed.)
P4.42 In the infamous Flixborough accident in 1974, a nylon manufacturing
plant exploded, killing 28 workers. Fires burned for more than a week.
The accident occurred in a process where air and cyclohexane (C6H12)
reacted to make cyclohexanone (C6H10O). Part of the cause of the explosion was traced to the huge inventory of cyclohexane in the process,
which was necessary because the single-pass conversion in the reactor
was only 6%. Cyclohexanone production was about 5400 kg/h. Assume
that all of the oxygen fed to the reactor was consumed and that the
overall conversion of cyclohexane was 100%. Calculate the fresh feed
rate of air and of cyclohexane to the process, as well as the flow rate
into the reactor.
Scrimmage
P4.43 American settlers made soap by boiling potash (a mix of sodium and
potassium hydroxides and carbonates left over after burning brush) with
animal fat in a pan. Modern large-scale continuous soap-making processes
use a very similar chemistry: sodium hydroxide (NaOH) reacts with
fatty acids produced from fats (beef tallow and coconut oil are the most
common fat sources) in a process called saponification.
Fatty acids have the general molecular formula RCOOH, where R is a
long hydrocarbon chain. The saponification reaction is
RCOOH + NaOH → RCOONa + H2O
Beef tallow produces a mix of the following fatty acids: 32 wt%
palmitic (R = C15H31), 26 wt% stearic (R = C17H35), and 42 wt% oleic
(R = C17H33).
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P4.44
P4.45
P4.46
P4.47
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Inexpensive bars of soap are produced by mixing a 24 wt% NaOH aqueous solution with tallow-derived fatty acids. Inside the saponifier, all the
fatty acid is converted to its sodium salt, and the water content is adjusted
by adding or removing water. The soap product contains 12 wt% water.
If we want to make one metric ton of soap per day, how many kilograms
of tallow fatty acid do we need? How much water (kg/day) needs to be
added or removed?
In baking bread, yeast is added to flour and water to make a bread
dough. The yeast feed on the starches, sugars, and proteins in the flour,
and produce CO2 which causes the bread to rise, as well as water and
more yeast. By elemental analysis, yeast contains 50 wt% C, 6.94 wt%
H, 9.72 wt% N, and 33.33 wt% O. Assume that the flour provides
glucose (C6H12O6) and ammonia (NH3, from protein) for the yeast
growth and metabolism, and that 2 grams of CO2 are produced per
gram of yeast produced. Suppose you start with 150 in3 bread dough
and place it on the kitchen counter to rise for 1.5 hours, after which
the volume has doubled in size. What is the average rate of CO2 production? How many grams of glucose were consumed, and what is the
selectivity for converting glucose to CO2? You may assume that all the
CO2 is trapped in the bread dough and that its volume can be calculated
from the ideal gas law.
Immunotoxins are protein drugs made from antibodies that are designed
to specifically kill cancer cells. These drugs must be endocytosed
(brought inside the cancer cell) to be effective, but once inside the cell
some of the immunotoxin is degraded by proteases (enzymes that cut
up proteins), which destroys their ability to kill the cell. In addition, the
cells recycle (exocytose) some of the internalized drug back to the external environment. You are evaluating the potential of a newly developed
immunotoxin, code-named Hermab. For an immunotoxin to kill a cancer cell, there must be an accumulation of at least 30,000 molecules
inside the cell after 8 hours. Laboratory data indicate that Hermab is
endocytosed at a rate of 62,000 molecules per hour per cell, and exocytosed at a rate of 57,000 molecules per hour per cell. Hermab inside
the cell is degraded at a rate of 2700(1 − e−0.3t) molecules per hour,
where t is the time in hours since the cell was initially exposed to the
drug. Is Hermab likely to be effective?
In Example 4.9, we examined a problem in controlled drug release.
Solve the problem again, except add in one complication: The drug
inside the controlled release device degrades into an inactive form at a
constant rate of 1.1 micrograms/h. Calculate the mass of drug inside the
device at any time, and calculate the total fraction of drug initially loaded
into the capsule that is released over 24 hours.
Hydrodealkylation is a process in which side chain alkyl groups
(like methyl) are removed from aromatics by reaction with hydrogen.
It is an important process in refining of crude oil to higher value fuels
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or into other valuable chemicals. The following reactions take place in
the reactor:
C6H5CH3 + H2 → C6H6 + CH4
toluene
benzene
(R1)
C6H4(CH3)2 + H2 → C6H5CH3 + CH4
xylene
toluene
(R2)
C6H3(CH3)3 + H2 → C6H4(CH3)2 + CH4
pseudocumene
xylene
(R3)
In addition, an unwanted side reaction occurs, in which two benzenes
react to form diphenyl:
2C6H6 → C6H5C6H5 + H2
benzene diphenyl
(R4)
A process stream containing 10 mol% benzene (C6H6), 20 mol% toluene
(C6H5CH3), 30 mol% xylene [C6H4(CH3)2], and 40 mol% pseudocumene
[C6H3(CH3)3] is fed at a rate of 100 gmol to a hydrodealkylation plant.
This process stream is mixed with H2 at 5-molar excess H2 before being
fed to the reactor. Hydrogen and methane are separated from the remaining compounds (lumped together as aromatics). The aromatics are analyzed and found to contain 28 mol% pseudocumene, 1 mol% diphenyl,
19% benzene, and 1% toluene. Calculate the extents of reaction and
fractional conversions of all reactants, the methane production rate, and
the mol% methane in the gas stream.
P4.48 Catalyst performance is usually tested in laboratory or pilot-plant reactors. In one particular run, a supported platinum/tin catalyst was tested
for its effectiveness at dehydrogenating light hydrocarbons, particularly
isobutane. Data from one day were taken as follows:
Reactor temperature: 602°C
Reactor pressure: 768 torr
Inlet gas flow rate: 51.38 SCCM (standard cubic centimeters per minute)
Outlet gas flow rate: 60.59 SCCM
Inlet gas composition (mole percent): 0.086% propane, 32.9% isobutane, 0.068% n-butane, and 66.9% hydrogen
Outlet gas composition (mole percent): 0.35% methane, 0.037% ethane,
0.034% propane, 13.5% isobutane, 13.8% isobutene, 0.20% n-butane,
0.0375% cis-butene, 0.045% trans-butene, and 71.6% hydrogen.
Complete balances on carbon and hydrogen to check the reliability of
the data. Calculate the fractional conversion of isobutane to products.
If the desired product is butenes, what is the selectivity for converting
butanes to butenes? What is the yield?
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P4.49 Catechol is used to make artificial flavorings such as vanillin as well
as pharmaceuticals such as L-Dopa, used to treat Parkinson’s disease.
Synthesis of catechol requires 4 reactions:
Benzene (C6H6) and propylene (C3H6) react to make cumene (C9H12)
Cumene reacts with oxygen (O2) to generate cumene hydroperoxide
(C9H12O2)
Cumene hydroperoxide is unstable and rapidly breaks down to phenol (C6H6O) and acetone (C3H6O)
Phenol reacts with hydrogen peroxide (H2O2) to produce catechol
(C6H6O2), with water as a byproduct.
You work as a process research engineer at a pilot plant (a small-scale
process facility) and your job is to determine operating conditions for
a full-scale operation. In one experiment, you feed a mixture of 1065 g/h
benzene, 745 g/h propylene, 394 g/h O2, and 354 g/h H2O2 to the pilot
plant. The output stream is analyzed and found to contain catechol,
benzene, propylene, phenol, acetone, and water. Complete a DOF analysis to show that this is enough information to completely specify the
process, then calculate the composition (wt% of each compound) and
flow rate (g/h) in the output. Also calculate the fractional conversion of
each reactant fed to the process.
P4.50 Protein C is a blood protein required for clotting. Patients with hemophilia do not make sufficient Protein C. A researcher is developing a
new controlled-release device in which 90 units of Protein C is encapsulated in a polymeric bead. Protein C is released slowly from the bead
into the bloodstream at a rate of 13.5e−0.15t, where the rate is in units/h,
and t is in hours. Protein C in the bead also undergoes an irreversible
degradation reaction, with a degradation reaction rate of 2.3 units
degraded/h. How many units of Protein C are left in the bead after 8 hours?
At this time, what fraction of the Protein C initially in the bead has been
released into the bloodstream, and what fraction has been degraded?
P4.51 A manufacturer of fine cosmetics has an oversupply of glycerol (C3H8O3),
and would like to design a process to convert the glycerol to hydrogen
for use in fuel cells. You have discovered a new catalyst for the reaction
of glycerol with water to form hydrogen, with CO2 as a byproduct. In
a pilot plant test, you feed glycerol and water (1.63 gmol/h glycerol,
5:1 water:glycerol molar ratio) to a reactor, and measure that the reactor
exit contains 35 mol% H2. Find the fractional conversion of glycerol and
water. Then sketch how you would design a large-scale process, using
recycle, to obtain higher overall conversion of glycerol. Maintain reactor
feed and conversion at pilot-plant conditions. Calculate all flows assuming
a fresh feed rate of 550 kg/day glycerol.
P4.52 You are in charge of designing a chemical process that converts reactant
A to product B. Unfortunately, the available source of A is contaminated
with 5 mol% of inert I. Only 15% conversion of A to B is achieved in
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
a single pass through the reactor, so recycle is needed to get better
utilization of A. While it is straightforward to separate A from B, it is
prohibitively expensive to separate I from A, so purge is needed. Your
job is to determine the optimum recycle:purge ratio. Using as a basis a
production rate of B of 100 gmol/h, calculate the fresh feed rate, the
mol%A in the purge stream, the reactor feed rate, and the mol%I in the
reactor feed, first at a high recycle:purge ratio of 4:1 and second at a
low recycle:purge ratio of 1:1. Based on your calculations, comment on
the relationship between recycle:purge ratio in terms of reactor volume
(and hence equipment cost) and raw materials costs.
P4.53 Connie Chemist has developed a new catalyst for making acetaldehyde
from ethanol and air. Besides the desired reaction, some ethanol is oxidized completely to CO2 and H2O. Connie analyzed a laboratory scale
continuous-flow reactor. The feed to the reactor is set at 5.7 moles
ethanol per mole oxygen. Under these conditions, the single-pass conversion of ethanol is 25%, with selectivity to acetaldehyde of 0.92.
Sketch a block flow diagram showing how you would design this process at a large scale, to produce 1200 kgmol/h acetaldehyde. Include
recycle and purge streams as you see fit. Assume that nitrogen, oxygen
and carbon dioxide cannot be economically separated, and that ethanol,
water, and acetaldehyde can all be readily separated from each other.
Complete a DOF analysis of the process. Calculate all flows.
P4.54 Butanal C4H8O is made by reaction of propylene (C3H6) with CO
and H2:
C3H6 + CO + H2 → C4H8O
In an existing process, 180 kgmol/h of C3H6 (P) is mixed with 420
kgmol/h of a mixture containing 50% CO and 50% H2 and with a recycle stream containing propylene. This mix is then fed to a reactor, where
a single-pass conversion of propylene of 30% is achieved. The desired
product butanal is removed in one stream, unreacted CO and H2 are
removed in a second stream, and unreacted C3H6 is recovered and recycled. Draw the flow diagram and complete a DOF analysis. Calculate
the butanal production rate as well as the flow rate of the recycle stream.
Suppose now a company contacts you and claims that they can
supply you with a cheaper source of propylene that could replace your
current supply. Unfortunately, the cheaper source is contaminated with
propane at a 5:95 ratio (propane:propylene). The cheaper source is economically attractive if the butanal production rate is maintained at the
current rate and if an overall conversion of 0.90 can be achieved. At
your reactor conditions, propane is an inert, and it is too expensive to
separate propane from propylene so you decide to install a purge stream.
Assume the CO + H2 stream remains the same (420 kgmol/h, 50 mol%
CO) as does the single-pass conversion of propylene in the reactor (0.3).
Given this, show how the block flow diagram must be modified to
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315
accommodate the cheaper source of propylene, and calculate the flow
rate of the contaminated propylene stream to the process, the mol% inert
and the total flow rate of the purge stream, and the mol% inert and the
total flow rate (kgmol/h) of the feed to the reactor.
P4.55 Methanol (CH3OH) reacts to form formaldehyde (HCHO) either by
decomposition to formaldehyde and hydrogen (H2) or by oxidation to
form formaldehyde and water (H2O):
CH3OH → HCHO + H2
CH3OH + _​​ 12 ​​ O2 → HCHO + H2O
A mixture containing 99 mol% methanol and 1 mol% of an inert contaminant is available as feed to a formaldehyde plant. 1000 kgmol/h of
this mixture, along with 200 kgmol/h O2, are fed to a process as fresh
feed. The fresh feed is mixed with a recycle stream and fed to a reactor.
The fractional conversion of methanol achieved in the reactor is 25%.
All of the oxygen is consumed in the reactor. The reactor output is sent
to a separation unit, where formaldehyde, water, and hydrogen are
removed, and methanol and the contaminant are recycled to the reactor
feed. To control the contaminant level in the reactor, a purge stream is
taken off the recycle stream. The maximum contaminant allowed in the
recycle stream is 10 moles contaminant/100 moles methanol. Draw a
flow diagram and complete a DOF analysis. Determine (a) the overall
fractional conversion of methanol to products, (b) the production rate
of formaldehyde, (c) the purge:recycle ratio, and (d) the recycle:fresh
methanol feed ratio.
P4.56 Amines are derivatives of ammonia; they are added to shampoo to
make it foam, used as a building block in carpet fibers, and cause the
stench of rotting fish. Industrially, amines are produced from alcohols
and ammonia over solid catalysts, with water as a byproduct. Alcohols
in turn are produced from alkene hydrocarbons. A synthetic route
direct from alkenes to amines would avoid the cost of producing the
alcohol intermediate.
Your boss is very enthusiastic about reports of a new catalyst that
might be able to achieve this synthetic route by catalyzing two reactions
in the same pot. For example, butene (C4H8), CO, H2, and dimethylamine [(CH3)2NH] react to produce an amine with the molecular formula C7H17N, with H2O as the byproduct.
Unfortunately, the reaction produces a mix of linear and branched
amines, and only the linear amines are of commercial interest.
Here are some data from experiments carried out in a laboratoryscale semibatch reactor. The butene and 15 mmol dimethylamine were
initially dissolved in an organic solvent at the mole ratios listed in the
table. CO and H2 flowed continuously through the reactor at great
molar excess.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Moles
butene/mole
dimethylamine
Conversion of
Selectivity,
Temperature, °C
initially
dimethylamine, % linear amine, %
100
1.2:1
54
65
120
1.2:1
82
60
140
1.2:1
99
52
For each case, calculate the millimoles of butene, dimethylamine, total
amines, and linear amines in the reactor at the end of the experiment.
Imagine you need to design a large-scale process for producing linear
amines. Consider how the reactor operating temperature changes the
process flowsheet. Which operating temperature would you choose?
Explain your answer.
P4.57 PET, the polymer used to make plastic 2-L soda bottles, is made by
polymerization of terephthalic acid (TA; HOOC-C6H4-COOH). In a
synthesis of TA carried out in your company, p-xylene (CH3C6H4CH3)
is dissolved in acetic acid (CH3COOH) at 10 wt%, and 20% excess air
(based on desired reaction) is bubbled through the mixture. The reaction
takes place over a proprietary solid catalyst; conversions and selectivities approaching 100% to the desired product is achieved. However,
about 10% of the acetic acid is completely oxidized to CO2 and H2O
and the reaction is very energy-intensive, so the search is on to discover
a new process. One day Connie Chemist drops by your office with
exciting results. She’s discovered a process in which hydrogen peroxide
(H2O2) and p-xylene are dissolved in supercritical water using a very
dilute, very inexpensive salt as the catalyst. The hydrogen peroxide
decomposes to O2 and H2O, and the O2 reacts with the p-xylene to form
TA. Carried out in a minireactor and using reactants at stoichiometric
ratio, Connie produced about 10 g TA per hour, with yields better than
70% and selectivity better than 90%.
Evaluate the new process vis-à-vis the conventional process.
Sketch out simplified block flow diagrams, and complete process flow
calculations for each process, using the available data and a basis of
100 kg/day TA production. Would you pursue Connie’s idea further?
Why or why not?
P4.58 Ammonia (NH3) and methanol (CH3OH) react over a catalyst to produce methylamine (CH3NH2), which is a useful intermediate in the production of some polymers and pharmaceuticals. Methylamine can react
further over the same catalyst to produce dimethylamine (CH3)2NH, for
which there is a limited market as a specialty solvent. Water is a byproduct of both reactions.
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317
You work in a process research lab, and you are testing a new
proprietary catalyst for methylamine production. In one experiment, you
feed into a laboratory-scale reactor 100 gmol methanol/h and 100 gmol
ammonia/h into the reactor. You collect the reactor effluent and send
it off for analysis. The results: the effluent contains 20.0 mol% ammonia, 9.4 mol% methanol, 18.6 mol% methylamine, 10.9 mol% dimethylamine, and the remainder water.
Write down the two stoichiometrically balanced equations: (R1) for
making methylamine from methanol and ammonia and (R2) for making
dimethylamine from methylamine and methanol. Use element balances
to see if the laboratory analysis is reasonable. From the experimental
data, calculate (a) the fractional conversion of methanol, (b) the yield
of methylamine based on methanol, and (c) the selectivity for methylamine based on methanol, (d) the reaction rate ​​r ​​1̇ ​(gmol/h), and (e) the
reaction rate r​​ ​​2̇ ​ (gmol/h).
P4.59 Formaldehyde (HCHO) is produced by partial oxidation of methanol
(CH3OH). Several side reactions also occur, producing formic acid
(HCOOH), CO, CO2, and H2O.
In one working process, air (21 mol% O2, 79 mol% N2) is mixed with
fresh and recycled methanol and fed to a reactor. The reactor outlet is
sent to a separator, which produces three streams: a liquid product
stream, a pure methanol stream that is recycled, and an offgas stream
that is sent offsite. Your job is to evaluate how well this process is
working. You take samples at several points in the process and find that
(a) the reactor inlet stream is 35 mol% CH3OH, (b) the recycled methanol is pure, (c) the offgas contains 10.9 mol% H2, 6.0 mol% CO2,
0.3 mol% CO, 81.9 mol% N2, and 0.9 mol% O2, and (d) the liquid
product stream contains 30.3 mol% HCHO, plus HCOOH and H2O.
Write down a set of independent, stoichiometrically balanced chemical equations that completely describe all the reactions taking place in
the reactor. Use a DOF analysis to determine if you’ve got enough
stream compositions analyzed to completely describe the process
operation. Calculate the production rate for the two valuable products,
formaldehyde and formic acid, per 100 moles methanol fed to the
process. Then calculate all other flowrates. Finally, if formaldehyde is
a more valuable product than formic acid, how might you consider
adjusting the process operation?
P4.60 Methanol (CH3OH) was originally made by distillation of wood. Now,
methanol is produced primarily from methane and water. In the steam
reformer, two reactions occur:
CH4 + H2O → CO + 3H2
(R1)
CH4 + 2H2O → CO2 + 4H2
(R2)
Steam reforming is carried out at 2-fold excess steam to suppress
unwanted side reactions. The relative importance of (R1) and (R2)
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
depends on the reactor temperature: high temperatures (1500–1800°F)
favor (R1) while lower temperatures (600–700°F) favor (R2). Typical
reactor pressure is 300 psig. At these conditions, all the methane is
converted to products.
The products from the steam reformer are sent to a catalytic reactor
which operates at high pressures (4500 psig) and moderate temperatures
(500–600°F). Two reactions take place:
CO + 2H2 → CH3OH(R3)
CO2 + 3H2 → CH3OH + H2O(R4)
About 15% of the CO/CO2 fed to the reactor is converted to methanol.
The methane available is contaminated with 2% N2. Methanol and
water can easily be separated from CO, CO2, and H2 by cooling and
condensing. CO and CO2 can be separated from H2 and N2 by absorption. It is very expensive to compress gases to the high pressures
required for the second reactor.
Your job is to design a process for making 60 million lbs/year
methanol based on this information. Sketch out a block flow diagram
showing your preferred design. Indicate the chemical species in each
stream. Write one or two paragraphs explaining why you think your
flowsheet is superior to other alternative arrangements. Indicate whether
you would choose to run the steam reformer at high or low temperatures. You do not have to calculate process flows.
Game Day
P4.61 Phthalic anhydride (C8H4O3—we’ll call it PA for short) is widely used
to synthesize plasticizers used in vinyl notebooks and auto interiors. The
old process to make it uses partial oxidation of naphthalene (C10H8).
The process operates at 90% conversion of naphthalene, with 85% selectivity to PA. The byproducts are CO2 and water.
You are studying a newer catalyst that facilitates the production of
PA by partial oxidation of o-xylene (C8H10). In laboratory experiments,
75% conversion of xylene with 65% selectivity to PA was achieved. The
undesired byproducts are—you guessed it—CO2 and water.
Your job is to analyze the economic feasibility of converting to the
new process. First, determine the raw material requirements for a reactor producing 2 metric tons/week of PA, using each of the reaction
schemes. Find current prices of naphthalene, o-xylene, and PA, and
calculate the operating profit or loss, based solely on material costs.
Assume air is free and combustion products are worthless. Which process do you recommend? Then, develop new process flow sheets,
assuming that the reaction products can be separated. What does the
optimum process flow sheet look like? How does this change the
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Chapter 4 Problems
319
process economics, and your thinking on the choice of reaction pathway? Are there any other considerations besides economics that might
affect the choice?
P4.62 Polyvinyl chloride (PVC) is produced by the catalytic polymerization
of vinyl chloride and is used extensively to make products like plastic pipe and film. Your assignment is to design a process for making
vinyl chloride (C2H3Cl). A brief survey of the synthetic chemistry
literature unearths the following reactions involving vinyl chloride or
similar molecules:
C2H2 + HCl → C2H3Cl
(1)
C2H4 + Cl2 → C2H4Cl2
(2)
C2H4Cl2 → C2H3Cl + HCl
(3)
_​​ 1 ​​
2 HCl + 2 O2 + C2H4 → C2H4Cl2 + H2O
(4)
C2H4Cl2 + NaOH → C2H3Cl + H2O + NaCl
(5)
Come up with several different reaction pathways for the production
of vinyl chloride by mixing and matching these five reactions. Using
the market values below, analyze which of your pathways look most
promising.
Ethylene:
$0.43/lb
Dichloroethane:
$0.26/lb
Acetylene:
$1.71/lb
Chlorine:
$0.20/lb
Hydrogen chloride: $0.93/lb
Sodium hydroxide: $1.13/lb
Vinyl chloride:
$0.44/lb
For the reactions listed literature data and pilot plant studies have generated the following information:
Reaction (1) proceeds at 120°C and 5 atm over a catalyst. Essentially
100% conversion can be achieved in a single pass through the reactor.
Reaction (2) proceeds at 95°C and 3 atm pressure over a catalyst.
Under these conditions, 90% conversion can be reached.
Reaction (3) proceeds at 400°C and 20 atm over a catalyst. 80%
conversion can be achieved in a single pass through the reactor.
Reaction (4) proceeds at 300°C and 5 atm. 70% conversion can be
achieved in a single pass through the reactor.
Reaction (5) proceeds at 80°C and 4 atm. Essentially 100% conversion is achievable in a single pass through the reactor.
Dichloroethane is a liquid at the conditions of reactions (2) and (5) and
a vapor at the conditions of reactions (3) and (4). Vinyl chloride and
ethylene are vapors at the conditions of all four reactions listed.
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Chapter 4 Synthesis and Analysis of Reactor Flow Sheets
Also, the following restrictions apply:
(a) No impurities are allowed in vinyl chloride product.
(b) Raw material ethylene contains 2% carbon particles.
(c) Only pure Cl2 and C2H4 are allowed in reaction (2) feed.
(d) Only pure C2H4Cl2 is allowed in reaction (3) feed.
(e) No C2H4Cl2 or C2H3Cl should be in the feed to reaction (4).
(f ) No C2H4 or HCl should be present in feed to reaction (5).
Devise two or three alternative processing schemes for production of
vinyl chloride. Identify the key separations required for each scheme.
Consider the tolerance to feed disturbances or plant upsets.
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5
CHAPTER FIVE
Why Reactors Aren’t Perfect:
Reaction Equilibrium and
Reaction Kinetics
In This Chapter
We consider the effect of chemical reaction equilibrium and chemical kinetics
on reactor performance and discuss ways to design and operate reactors for
optimal output. You won’t become an expert, but you will gain deeper insight
into the inner workings of reactors.
The questions we’ll address in this chapter include:
∙ How does chemical reaction thermodynamics and kinetics affect reactor
design?
∙ How do I choose the optimum reactor temperature and pressure?
∙ Why don’t all reactors achieve complete conversion of reactants?
Words to Learn
Watch for these words as you read Chapter Five.
Chemical reaction equilibrium
Chemical kinetics
Catalyst
Gibbs energy of reaction
Enthalpy of reaction
Gibbs energy of formation
Enthalpy of formation
321
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
5.1
Introduction
We’ve explained what the measures of reactor performance are, but we haven’t
explained why. Why can’t we achieve high single-pass conversion? Why can’t
we achieve high selectivity? There are two major factors that limit reactor
performance: chemical reaction equilibria, and chemical reaction kinetics. You
could write books on each of these topics (in fact many people have!). In this
chapter we can give only a taste of each; you will learn much more in the future.
5.1.1
The Chemical Reaction Equilibrium Constant Ka
Consider a simple reaction where reactants A and B are converted to product
P and waste product W:
​2A + B → P + 3W​
But, if A and B can react to form P and W, can P and W react to form A and B?
​P + 3W → 2A + B?​
Theoretically, the answer is Yes, a chemical reaction can proceed in either
direction.
One way to indicate that both forward and reverse reactions are occurring
simultaneously is to use a two-way arrow:
​2A + B ⇄ P + 3W​
As a practical manner, we engineers usually want the forward reaction, from
reactants to desired products, but not the reverse reaction, from products back
to reactants. But all too often the reverse reaction does happen. By evaluating
chemical reaction equilibrium, we can determine the extent to which the
reverse reaction needs to be considered. We can determine the maximum extent
of reaction possible, and we can adjust reactor variables to achieve the best
conversion and selectivity, within the limits placed by chemical reaction equilibrium constraints.
When a reacting system reaches chemical equilibrium, the concentration
of reactants and products do not change with time. The concentrations of
­reactants and products at equilibrium are quantified by a chemical equilibrium
constant Ka:
v​ i​​
​​K​ a​​  = ​ ∏​​​ ​a​ i​
,eq​ ​​
all i
Eq. (5.1)
where ai,eq is the activity of compound i when the system is at equilibrium, νi
is the stoichiometric coefficient for compound i in the reaction, and Π indicates
that the product of the activity of all compounds participating in the reaction
is calculated.
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Section 5.1 Introduction
323
Illustration: For the reaction
​2A + B ⇄ P + 3W​
the chemical equilibrium constant is
​aP​  ,eq​​ ​a​ W3 ,eq​
​Ka​  ​​ = ​​a​ A−2,eq​​ ​​​a​ B−1,eq​​​​a
​ ​P,eq​​ ​​a​ W3 ,eq​​​ = ________
​​  2
​​
​a​ A,eq​ ​​aB​  ,eq​​
You are certainly very familiar with the word activity, but perhaps not in a
chemical sense. The activity of a compound is related to the “chemical potential” of that compound in a multicomponent mixture. The activity is related to
the amount of each compound in a mixture as well as the phase of the mixture.
Up to this point, we have used zi to indicate mole fraction of compound i in a
mixture. With chemical reaction equilibrium, it is critical to keep track of the
phase of the material. We will use the convention that yi is the mole fraction
of compound i in a vapor, xi is the mole fraction of compound i in a liquid,
and xiS is the mole fraction of compound i in a solid. When the phase is
unspecified, or the mixture contains two or more phases and we are interested
in the mole fraction of the entire mixture, we will continue to use zi.
Illustration: A vapor–liquid mixture contains 40 gmol A and 60 gmol B.
The mixture (F ) is fed to a drum where it is separated into a vapor phase and
a liquid phase. The vapor stream (V ) contains 35 gmol A and 5 gmol B. The
liquid stream (L) contains 5 gmol A and 55 gmol B. For this process,
​​z​ AF​​ = 0.4​​​z​ BF​​ = 0.6​
​​y​ AV​​ = 0.875​​​y​ BV​​ = 0.125​
​​x​ AL​​ = 0.083​​​x​ BL​​ = 0.917​
There is a lot more about activity that you will learn in thermodynamics
courses. In this textbook, we will greatly simplify things and say:
​​a​i​​ = y​​ ​i​​P∕1 atm if i is in the vapor phase
Eq. (5.2a)
where P is the total pressure (in atm).
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​​a​i​​ = ​​x​i​​ if i is in the liquid phase
Eq. (5.2b)
​​a​i​​ = 1 if i is in the solid phase
Eq. (5.2c)
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
There are a number of assumptions that are behind these simplifying equations.
For vapor-phase compounds, we are assuming that the vapor behaves as an
ideal gas mixture. For liquid-phase compounds, our approximation is reasonably good (1) in dilute solutions of compounds in water or another solvent and
(2) in liquid mixtures of two or more compounds that are of similar size and
chemical type. For solid-phase compounds, our approximation is good if the
solid phase is pure.
In order to derive an expression for Ka from a chemical reaction equation
using Eq. (5.1), the reaction must be stoichiometrically balanced, and we must
know the phases of the reactants and products. We will indicate the phase of
compounds in a chemical reaction by writing (g) for gas (vapor) phase, (l) for
liquid phase, or (s) for solid phase, just after the molecular formula.
Finally, a word about the dimension of Ka. Since the activity coefficient ai
is dimensionless, Ka is also dimensionless.
Example 5.1
Deriving Equations for Ka: Three Cases
Derive equations for Ka in terms of mole fractions and pressure for the following
cases:
Case 1.
The gas-phase synthesis of ammonia:
​​N2​ ​(g) + 3​H2​ ​(g) ⇄ 2N​H3​ ​(g)​
Case 2.The synthesis of aspirin (acetylsalicylic acid) in a dilute aqueous solution from salicylic acid and acetic acid:
​​C​6​​H4​ ​(OH)COOH(l) + C​H3​ ​COOH(l) ⇄ ​C6​ ​​H4​ ​(OCOC​H3​ ​)COOH(l) + ​H2​ ​O(l)​
Case 3.
The reduction of barium sulfate ore to barium sulfide:
​BaS​O4​ ​(s) + 4CO(g) ⇄ BaS(s) + 4C​O2​ ​(g)​
(g), (l), or (s) indicates that the reactant or product is in the gas, liquid,
or solid phase, respectively.
Solution
Case 1.
​a​ ​N2 ​
​( ​y​N​  H​3​​,eq​​ P)​​  2​
​(​ y​N​  H​3​​,eq​​)​​  2​ _
H​3​​,eq​
_______________
____________
​​Ka​  ​​ = _
​
​
=
​
​
=
​
​ ​  1  ​​
​a​N​  2​ ​​,eq​​ ​a​ ​H3 ​
(​ y​N​  2​ ​​,eq​​ P)​( ​y​H​  2​ ​​,eq​​ P)​​  3​ (​ y​N​  2​ ​​,eq​​)​( ​y​H​  2​ ​​,eq​​)​​  3​​P​​  2​
​2​​,eq​
Notice that the equation for Ka of gas-phase reactions includes a pressure
term if there is a change in total moles as the reaction proceeds.
Case 2.Using A for aspirin, SA for salicylic acid, AA for acetic acid, and W
for water we write
(​xA​  ,eq​​)(​xW
​  ,eq​​)
​​Ka​  ​​ = _____________
​     ​​
(​xS​  A,eq​​)(​xA​  A,eq​​)
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Section 5.1 Introduction
Quick Quiz 5.1
Consider the gas-phase
reaction 2A ⇄ B + C.
Write down an expression for Ka in terms of
the mole fractions of
A, B, and C. Is there a
pressure term in your
equation?
325
This is a little tricky: this reaction takes place in dilute aqueous solution. xW,eq, the mole fraction of water in the system at equilibrium,
includes not just the water produced by reaction but all the water in
the system. Since the components (other than water) are dilute, we
sometimes say that xW,eq ≈ 1 and
(​xA​  ,eq​​)
​​Ka​  ​​ ≈ _____________
​     ​​
(​xS​  A,eq​​)(​xA​  A,eq​​)
Case 3.Barium sulfate and barium sulfide form two separate solid phases. (The
two solids may appear to be mixed on a microscopic scale, but they
are not mixed on a molecular scale.) Therefore, the activity of each of
the solids equals 1 and:
(​aB​  aS,eq​​)​(a
​ ​C​  O​2​​,eq​​)​​  4​ ____________
(1)​( ​y​C​  O​2​​,eq​​P)​​  4​ _
​y​ ​C4 ​
O​ ​​,eq​
________________
​​Ka​  ​​ = ​
​4 = ​
​4 = ​  4 2  ​​
(​aB​  aS,eq​​)​(a
​ C​  O,eq​​)​​  ​ (1)​( ​yC​  O,eq​​P)​​  ​ ​y​ C ​
O,eq​
To simplify notation, we will omit the “eq” subscript throughout the rest of
this chapter. However, it is critically important to remember that the relationship
between Ka and mole fractions is true only when the reaction mixture is at
equilibrium.
5.1.2
Gibbs Energy of Reaction
Think about what you are doing right now. You are probably sitting up, reading, breathing, perhaps drinking a soda. Hopefully you are thinking. You are
not at your lowest energy state. (If you were, you’d be a dead and decayed
heap on the floor.)
The chemical reaction equilibrium constant Ka is related to the lowest
energy state of a reacting system. There are many ways to describe the energy
state of a system. We are interested now in one kind of energy, called the
Gibbs energy, which we denote as G. When a system is at equilibrium, like our
chemically reacting systems, G is at a minimum (but G is not zero) (Fig. 5.1).
If G could become lower by further chemical reaction, then the system would
not be chemically equilibrated. G of a chemically reacting system changes as
the reaction proceeds to equilibrium. The change in G as the reaction proceeds
from reactants to products is called the Gibbs energy of reaction and is
denoted as Δ​​Ĝ ​​r. (The “hat” indicates that this is the Gibbs energy change per
mole of reaction.) A negative value of Δ
​ ​​Ĝ ​​r​​​indicates that products have lower
Gibbs energy than reactants, and so the equilibrium will favor the formation
of products. A positive value of Δ​​Ĝ ​​°r​  ​indicates the opposite. If ​ΔG
​​ ̂ ​​r​​​ << 0, then
the reaction is irreversible for all practical purposes, and equilibrium conversion to products will reach close to 100%. If ​ΔG
​​ ̂ ​​r​​​>> 0, then the reaction will
not “go” to any great extent, and conversion to products will be close to 0%.
As engineers, we prefer to have a negative ​ΔG
​​ ̂ ​​r​​​.
̂
Δ​​G ​​r describes the thermodynamic driving force for converting reactants to
products. This would be convenient if we had access to a table of Δ​​Ĝ ​​r of every
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Gibbs energy of
reaction = G of
products minus G
of reactants
Gibbs energy
326
Mix of reactants and
products at equilibrium
Lowest G at
equilibrium
reactants
extent of reaction
products
Figure 5.1 The Gibbs energy G of a reacting system changes as the reaction proceeds
from reactants to products. Equilibrium is reached when G reaches its lowest point along
the reaction pathway, thus determining the extent of reaction, and the mixture of reactants
and products, at equilibrium. The Gibbs energy of reaction is the difference in G between
products and reactants.
reaction known to humankind. That would be a pretty long table! Fortunately,
there is a way around this problem, by finding Δ​​Ĝ ​​r at the standard state,
typically at 298 K and 1 atm. This value is known as the standard Gibbs
energy of reaction and is written with a small superscript “°” to indicate standard
state: Δ​​​Ĝ ​°​r​  ​​. To calculate Δ​​​Ĝ ​°​r​  ​​, we must know the standard Gibbs energy of
formation Δ​​​Ĝ ​°​i,  ​​
​  of the reactants and products. A table of values of Δ​​​Ĝ ​°​i,  ​​f​  for
f
some common compounds is included in App. B.
Δ​​​Ĝ ​°​i,  ​​
​  is the Gibbs energy change associated with making the compound i
f
from its elements in their natural phase and state of aggregation, at the standard
temperature and pressure. The phrase “in their natural phase and state of aggregation” needs a bit of explanation. This phrase is needed because not all elements naturally exist as monoatomic compounds. For example, the natural
phase and state of aggregation of oxygen at 298 K and 1 atm is O2(g), hydrogen is H2(g), helium is He(g), bromine is Br2(l), carbon is C(s) and sulfur is
S8(s). For elements at 298 K, 1 atm, and in their natural phase and state of
aggregation, Δ​​​Ĝ ​°​i,  ​​f​  = 0.
Δ​​​Ĝ ​°​r​  ​​ is calculated from Δ​​​Ĝ ​°​i,  ​​ f​  :
​Δ​​Ĝ ​​°r​  ​ = ∑ ​νi​  ​​ Δ​​Ĝ ​​°i,  ​​f​
Eq. (5.3)
where νi is the stoichiometric coefficient of species i (negative for reactants,
positive for products). ​ΔG
​​ ̂ ​​°i,  ​​
​  depends on the phase of the compound, so it is
f
important to select the correct value based on the phase of the compound in
the reaction of interest.
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Section 5.1 Introduction
327
Illustration: From App. B,
Compound​​CH​4​​(g)​​​O​2​​(g)​​​CO​2​​(g)​​​H​2​​O​(g)​​​​H​2​​O​(l)​​
​ΔG
​​  ​​̂ ° ​​  kJ/gmol​
−50.49
0
−394.37
−228.59
−237.19
i, f
For the reaction:
​​CH​4​​​(g) + 2O​2​​(g) ⇄ ​CO​2​​​(g) + 2H​2​​O(g)​
​ΔG
​​ ̂ ​​°r​  ​= 50.49 + 0 − 394.37 − 228.59 = −572.47 kJ / gmol​
For the reaction:
​​CH​4​​​(g) + 2O​2​​(g) ⇄ ​CO​2​​​(g) + 2H​2​​O(​ l)​​
​Δ​​Ĝ ​​°r​  ​= 50.49 + 0 − 394.37 − 237.19 = − 581.07 kJ / gmol​
Helpful Hint
The Gibbs energy
change is calculated per mole of
reaction, not per
mole of reactant or
per mole of product.
​Δ​​Ĝ ​​°i,  ​​f​  and ​ΔG
​​ ̂ ​​°r​  ​​can be positive, negative, or zero. It is easy—and dangerous—
to make a sign error! It is important to note that Δ​​​Ĝ ​°​r​  ​​ depends on the stoichiometric coefficients, so the numerical value depends on the way in which the
balanced chemical equation is written!
Illustration: For the reaction
​NO(g) + 0.5O​2​​(g) ⇄ ​NO​2​​(g)
we find from App. B:
Δ​​Ĝ ​​°NO,
​
​ = 86.57 kJ/gmol
f
Δ​​​Ĝ ​​°​O2​ ​, f​​​ = 0 kJ/gmol
Δ​​​Ĝ ​​​N° O​2​, f​​​ = 51.3 kJ/gmol
Δ​​Ĝ ​​°r​  ​= (−1)(86.57) + (−0.5)(0) + (1)(51.3) = −35.27 kJ/gmol
For the reaction
​NO​2​​(g) ⇄ ​0.5O​2​​(g) + NO(g)
Δ​​Ĝ ​​°r​  ​= (1)(86.57) + (0.5)(0) + (−1)(51.3) = +35.27 kJ/gmol
Quick Quiz 5.2
What is ​ΔG
​​ ̂ ​​r°​  ​​ for the
reaction ​​2NO​2​​(g) ⇄ ​
O​2​​(g) + 2NO(g)​?
mur83973_ch05_321-374.indd 327
For the reaction
​2NO(g) + O​2​​(g) ⇄ ​2NO​2​​(g)
Δ​​Ĝ ​​°r​  ​= (−2)(86.57) + (−1)(0) + (2)(51.3) = −70.54 kJ/gmol
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
5.1.3
Calculating Ka
We now have enough information to calculate a numerical value of Ka—but
only at the standard state temperature of 298 K. The equation is:
​​K​ a,298​​ = exp(−Δ​​Ĝ ​​°r​  ​∕RT )​
Eq. (5.4)
A word about units. Δ​​Ĝ ​​r, Δ​​​Ĝ ​​r°​  ​​, Δ​​​Ĝ ​​i,° ​​
​  have units of energy per mole (e.g.,
f
kJ/gmol, cal/gmol, or Btu/lbmol). In Eq. (5.4), R must have compatible units.
Appropriate values of R include 8.3144 J/gmol K, 1.9872 cal/gmol K, or
1.9872 Btu/lbmol °R. Be careful in using Eq. (5.4). Common errors include:
using inconsistent units for ​−Δ​​Ĝ ​​°r​  ​​ and R, forgetting to use an absolute temperature scale for T, making a sign error, or forgetting to take the exponential.
Illustration: For the reaction
​​NO(g) + 0.5O​2​​(g) ⇄ ​NO​2​​(g)​
​ΔG
​​ ̂ ​​°r​  ​= (−1)(86.57) + (−0.5)(0) + (1​)​(51.3) = −35.27 kJ/gmol​
⎜
⎟
⎛
kJ ​ ​​ ​  1000
_J ​ ​⎞
−​ −35.27 ​ _
(
̂
(
gmol )
kJ )
−Δ​​G ​​r°​  ​
_______________________
​​K​ a,298​​ = exp​(______
​
​
​= exp ​ ​
​ ​ = exp(14.23)​
)
RT
J ​ ​(298 K )
​ 8.3144 ​ _
gmol K )
⎝ (
⎠
6
​= 1.52 × ​10​​  ​​
Determining Δ​​​Ĝ ​°​r​  ​​allows us to calculate Ka at 298 K. But most industrial
chemical reactors don’t operate at 298 K. Are we stuck? Lucky for us, this
problem can be solved by introducing another parameter, Δ​​​Ĥ ​​r°​  ​​, the standard
enthalpy of reaction. Enthalpy is just another measure of energy that is different from, but related to, Gibbs free energy. Δ​​​Ĥ ​r°​​  ​​ is calculated in a manner
very similar to Δ​​​Ĝ ​r°​​  ​​:
​Δ​​Ĥ ​​°r​  ​ = ∑ ​νi​  ​​ Δ​​Ĥ ​°​i,  ​​f​
Eq. (5.5)
where Δ​​​Ĥ ​°​i,  ​​
​  is the standard enthalpy of formation of compound i, which we
f
find in the tables in App. B right next to Δ​​​Ĝ ​°​i,  ​​ f​  .
Now we can use Δ​​​Ĝ ​°​r​  ​​ and Δ​​​Ĥ ​°​r​  ​​ to calculate Ka at any temperature T:
−Δ​​Ĝ ​​r°​  ​ ____
Δ​​Ĥ ​​r°​  ​ _
1 ​ ​​
​​K​ a,T​​ = exp​(______
​
​+ ​   ​
​  1  ​ − ​ _
R ( 298 T ))
298R
Eq. (5.6)
(Eq. (5.6) is appropriate when the standard temperature is 298 K and T is given
in Kelvin. The equation would need to be corrected if Rankine rather than
Kelvin absolute temperature scale is used.) Notice that Eq. (5.6) reduces to
Eq. (5.4) if T = 298, as it should.
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Section 5.1 Introduction
329
Illustration: For the reaction
​​NO(g) + 0.5O​2​​(g) ⇄ ​NO​2​​(g)​
we find from App. B:
Δ​​Ĥ ​​°NO,
​
​ = 90.25 kJ/gmol
f
Δ​​​Ĥ ​​°​O2​ ​, f​​​ = 0 kJ/gmol
Δ​​​Ĥ ​​°N​O2​ ​, f​​​ = 33.3 kJ/gmol
​ΔH
​​ ̂ ​​°r​  ​= (−1​)​(90.25) + (−0.5)(0) + (1)(33.3) = −56.95 kJ/gmol​
The equilibrium constant at 1000 K is
⎜
⎟
⎛
⎞
kJ ​ ​​ ​  1000
kJ
1000
_J
_
_J
−​ −35.27 ​ _
( kJ ​)​ ​(−56.95 ​  gmol ​)​​(​  kJ ​)​ 1
(
)
gmol
1
_______________________
​
Ka​  ,T​​ = exp ​ ​
​+ ​ _______________________
​​ _
​
​ − _
​
​ ​ ​
J ​ ​(298 K )
J ​ ​ ( 298 1000 )
​ 8.3144 ​ _
​ 8.3144 ​ _
(
gmol K )
gmol K )
⎝ (
⎠
​​K​ a,T​​= exp(14.23 − 16.14) = 0.148​
In Figure 5.2, which
reaction(s) have a
negative enthalpy of
reaction? If you wished
to make methanol
from CO and H2, what
temperature range
would you choose?
5
–5
CH4 + H2O ↔ CO + 3H2
+
ln Ka
0
CH3OH ↔ HCHO + H2
+
10
+
Quick Quiz 5.3
+
Ka varies over many orders of magnitude for different reactions and at
different temperatures (Fig. 5.2). If Ka >> 1, then the reaction is considered
irreversible, and at equilibrium essentially all the reactants will be converted
to products. If Ka << 1, then the reaction will not “go”; reactions that fall into
this category are rarely of industrial significance. The magnitude of Ka may
shift dramatically with temperature, depending on the size and sign of Δ​​​Ĥ ​°​r​  ​​.
As can be seen from Eq. (5.6), if Δ​​​Ĥ ​r°​​  ​​ < 0, Ka decreases with increasing T,
whereas if Δ​​​Ĥ ​r°​​  ​​ > 0, Ka increases with increasing T. The choice of temperature
for a chemical reactor is a key design variable, and is based in part on the
desire to have favorable equilibrium.
+
+
+
–10
–20
200
+
–15
CO + 2H2 ↔ CH3OH
400
600
800
Temperature, K
1000
1200
Figure 5.2 The chemical equilibrium constant can change dramatically with temperature.
Reactions with ln Ka > 0 are preferred.
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Example 5.2
Calculating Ka: Ethyl Acetate Synthesis
Ethyl acetate is an industrially important solvent. (Acetates are also partially
responsible for the distinctive odor of many fruits.) Ethyl acetate is synthesized on
a commercial scale by reacting ethanol (C2H5OH) and acetic acid (CH3COOH) in
the liquid phase:
​​C2​ ​​H5​ ​OH(l) + C​H3​ ​COOH(l) ⇄ C​H3​ ​COO​C2​ ​​H5​ ​(l) + ​H2​ ​O(l)​
What is Ka for this reaction at 25°C? at 80°C?
Solution
First we look up Δ​​​Ĝ ​°​i,  ​​f​  and Δ​​​Ĥ ​°​i,  ​​f​  for all the reactants and products. These values
are for forming the pure compounds in the liquid phase at 25°C and 1 atm.
Species
νi
Δ​​​Ĝ ​°​i,  ​​ f​, kJ/gmol
Δ​​​Ĥ ​°​i,  ​​ f​, kJ/gmol
C2H5OH(l)
−1
−174.7
−277.6
CH3COOH(l)
−1
−392.5
−486.2
CH3COOC2H5(l)
+1
−318.4
−463.3
H2O(l)
+1
−237.2
−285.8
Next we calculate Δ​​​Ĝ ​°​r​  ​​ and Δ​​​Ĥ ​°​r​  ​​:
̂
̂
Δ​​
​ G ​°​r​  ​ = ∑ ​νi​  ​​ Δ​​G ​​°i,  ​f​  = (−1)(−174.7) + (−1)(−392.5) + (−318.4) + (−237.2)​
​= +11.6 kJ/mol​
̂
̂
Δ​​
​ H ​​°r​  ​ = ∑ ​νi​  ​​ Δ​​H ​​°i,  ​f​  = (−1)(−277.6) + (−1)(−486.2) + (−463.3) + (−285.8)​
​= + 14.7 kJ/mol​
At the standard state temperature, 298 K,
−11,600 J/gmol
−Δ​​Ĝ ​​r​​
_______________________
​ a​  ​​ = exp​(_
K
​
​)​= exp​
​      ​ ​ = 0.00926​
( (8.3144 J/gmol K)(298 K ) )
RT
​
To calculate Ka at 80°C (353 K) we use Eq. (5.6):
14,700 J/gmol
(​ −11,600 J/gmol)​
________________
​ln ​Ka​  ,353​​ = _____________________
​
​ +
​     ​​ _
​  1  ​ − _
​  1  ​ ​= −3.757​
​(8.3144 J/gmol K)(​​ 298 K)​ ​(8.3144 J/gmol K)​( 298 353 )
​​Ka​  ,353K​​ = 0.0233​
Raising the temperature increases Ka, which means greater conversion of reactants
to products at equilibrium. This is a reaction we want to run at as high a temperature as possible. We are limited by wanting to keep the mixture in the liquid
phase.
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Section 5.1 Introduction
5.1.4
331
Equilibrium Considerations in Reaction Pathway Selection
Chemical equilibrium is an important consideration in selecting appropriate
raw materials and reaction pathways. In Chap. 1 you learned to screen alternative chemical reaction pathways by looking at raw material costs, byproducts,
and atom economy. All of these are important criteria, but none consider
whether the chemical reaction will actually occur under realistic conditions.
As a next step in evaluating the feasibility of reaction pathways, we consider
chemical reaction equilibrium. As a rule of thumb, we look for reactions where
Ka ≥ 1. Lower values may be acceptable only if there are other compelling
reasons to prefer a reaction pathway with poor equilibrium.
Equilibrium constants are functions of temperature, so selection of reactor
temperature is a part of the design process. Reactor temperatures close to ambient are preferred for safety and energy-cost reasons. For simple commodity
chemicals, reactor temperatures up to 500°C are quite reasonable. Even higher
temperatures are possible but require special considerations: The upper limit is
around 1200°C, which is about the temperature of a hot flame. Manufacture
of more complex chemicals, such as polymers or pharmaceuticals, is usually
carried out at much lower temperatures (about 25°C to about 100°C), because
the compounds degrade at higher temperatures.
Example 5.3
Chemical Equilibrium Considerations in Selection of Reaction
Pathway: Safer Routes to Dimethyl Carbonate
Polycarbonates are transparent impact-resistant polymers used for everything from
baby bottles to compact discs to contact lenses. The conventional manufacture of
polycarbonates uses phosgene (COCl2) as one of the raw materials, which is a
highly toxic chemical. Furthermore, the chlorine is not incorporated into the final
product, leading to the production of unwanted chlorinated byproducts. A more
environmentally benign approach for manufacturing polycarbonates uses dimethyl
carbonate (DMC) as a starting material. Besides its use in polycarbonate manufacture, dimethyl carbonate is of commercial interest because it can be used as a fuel
additive and as an electrolyte in lithium-ion batteries.
However, there is a catch. DMC itself is currently produced from the reaction
of methanol with phosgene.
Cl
O
O
C
C
Cl
Phosgene, COCl2
H3CO
OCH3
Dimethyl carbonate (DMC), C3H6O3
​COC​l2​ ​ + 2C​H3​ ​OH ⇄ ​C3​ ​​H6​ ​​O3​ ​ + 2HCl​
(R1)
We are in the hunt for alternative reaction pathways for the production of
DMC that don’t require phosgene. Here are some under consideration, all of
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
which use methanol as one of the reactants and all of which occur in the gas
phase:
Oxidative carbonylation of methanol:
​2C​H​3​OH + CO + 0.5​O2​ ​ ⇄ ​C​3​​H6​ ​​O3​ ​ + ​H2​ ​O​
(R2)
Urea methanolysis:
CO​​N2​ ​​​H4​ ​ + 2C​H3​ ​OH ⇄ ​C​3​​H6​ ​​O3​ ​ + 2N​H3​ ​
(R3)
Carbonylation of methanol with CO2:
​2C​H3​ ​OH + C​O2​ ​ ⇄ ​C3​ ​​H6​ ​​O3​ ​ + ​H2​ ​O​
(R4)
Which pathway would you pick on the basis of chemical equilibrium considerations?
Look at operating temperatures of 100°C to 500°C.
Solution
We need information regarding the chemical reaction equilibrium constant for each
of our reactions. From the data in App. B we collect the necessary information
and organize it in a table that includes the stoichiometric coefficients of each
­reaction. Gibbs energy and enthalpy are all in the gas phase and are reported in
units of kJ/gmol, numbers have been rounded to the nearest 0.1 kJ.
Δ​​​Ĝ ​°​f, ​​i​
Δ​​​Ĥ ​°​f, ​​i​
νi1
CH3OH
−162.3
−200.9
−2
COCl2
−206.8
−220.1
−1
CO
−137.3
−110.5 −1
O2
0.0
νi2
−2
νi3
νi4
−2
−2
0.0−0.5
CON2H4
−152.7
−235.5 −1
CO2
−394.4
−393.5 −1
HCl
−95.3
H2O
−228.6
NH3
−16.6
C3H6O3
−452.4
−92.3
+2
−241.8 +1 +1
−46.2 +2
−570.1
+1
+1
+1
+1
Before completing calculations, let’s take a quick look at the information in
the table. All four reactions consume 2 moles of methanol per mole of DMC generated. In comparing the four different reactions, we note that the equilibrium
constant will be more favorable if the reactants have a relatively higher Gibbs
energy (less negative), and the products have a relatively lower Gibbs energy.
(Look back at Fig. 5.1.) We spot right away that reaction R4 is of concern, given
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Section 5.1 Introduction
333
the very low (very negative) Gibbs energy of CO2. On the product end, equilibrium
of R3 might be relatively unfavorable, since the Gibbs energy of ammonia is
relatively high.
Now we proceed to calculate Δ​​​Ĝ ​r°​​  ​​, Δ​​​Ĥ ​r°​​  ​​, and Ka at 100°C (373 K) and 500°C
(773 K). Results are summarized in the table. Given the negative Δ​​​Ĥ ​r°​​  ​​, Ka decreases
as T increases in all cases.
R1
R2
R3
R4
​ΔG
​​ ̂ ​​°r​  ​​
−111.6
−219.1
−8.3
38.0
​ΔH
​​ ̂ ​​°r​  ​​
−132.7
−299.5
−25.2
−16.6
Ka,373
8 × 1014
7 × 1027
3.7
6 × 10−8
Ka,773
1.8 × 105
1.4 × 106
0.055
4 × 10−9
For R1 and R2, Ka >> 1, so R2 is an attractive alternative to the phosgene reaction, from a chemical equilibrium point of view. As we suspected, R3 is less
attractive, but still feasible, especially at lower temperatures. The very negative
Gibbs energy of CO2 makes R4 infeasible: Equilibrium lies far towards the reactants, and the reaction simply won’t “go.” This is unfortunate, because of the great
interest in capturing and reusing CO2.
Now we want to connect the numerical value of Ka, which can be calculated from ​Δ​​Ĝ ​​°r​  ​​ and ​Δ​​Ĥ ​​°r​  ​​(Eq. 5.6), to the equation that relates Ka to the mole
fractions in the reacting system at equilibrium (Eq. 5.1 and 5.2). At equilibrium,
these numerical values should be the same.
Illustration: The reaction of urea (U) and methanol (M) produces dimethyl
carbonate (DMC) and ammonia (A):
​​CON​2​​​H4​ ​​​(g) + 2CH​3​​OH(g) ⇄ ​C3​ ​​​H6​ ​​​O3​ ​​​(g) + 2NH​3​​(g)​
​y​  ​​ ​y​  2​ ​
​​Ka​  ​​ = _
​  DMC 2  ​​A
​yU​  ​​ ​y​ M ​​
A reaction mixture at 373K contains 32 mol% urea, 14 mol% methanol, 18 mol%
DMC, and 36 mol% ammonia.
​y​  ​​ ​y​  2​ ​
_
​  DMC 2  ​A =
​yU​  ​​ ​y​ M ​​
(0.18)​(0.36)​​  2​
___________
​
​= 3.7
(0.32)​(0.14)​​  2​
From Example 5.3, Ka,373 = 3.7, calculated from ​Δ​​Ĝ ​​r°​  ​​ and ​Δ​​Ĥ ​​r°​  ​​. The two
values agree, indicating that the reaction mixture is at equilibrium.
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
5.1.5
Chemical Reaction Equilibrium and Conversion
Many chemical reactors of industrial importance are operated such that the
materials leaving the reactor are at chemical equilibrium. Even if equilibrium is not reached, knowing the concentration of reactants and products at
equilibrium allows us to calculate the best the reactor can do—we cannot
get a ­conversion higher than what is allowed by the equilibrium constraint.
Ka is useful in design and analysis of chemically reacting systems because
it allows us to calculate the fractional conversion at equilibrium. Such calculations help us make appropriate choices of reactor design variables such
as temperature, pressure, and reactant feed ratio. For reactors where the outlet stream reaches equilibrium, chemical reaction equilibrium relationships
coupled with material balance equations allow us to determine the achievable reactor performance. Our task in this section is to demonstrate strategies for calculating fractional conversion at equilibrium, given a numerical
value of Ka.
To solve these problems, we write material balance equations as usual, and
we recognize that the constraint, that the reactor outlet is at equilibrium, provides a system performance specification. The “trick” is to find a way to
connect Eq. (5.1) for Ka with the material balance equations. For steady-state
continuous-flow reactors, this is accomplished by
(1) writing material balance equations for every compound in terms of stream
variables and the extent of reaction:
​​​n ​​i̇ ,out​​ = ​​n ​​i̇ ,in​​ + ​νi​  ​​ ​ξ ​​̇
(2) relating molar flow out n​​​  ​​i̇ ,out​​​to mole fraction out ​​z​ i,out​​​, using equations of
the form:
​​n ​​i̇ ,out​​
_
​​
​= z​ i​  ,out​​​
​​n ​​ȯ ut​​
where ​​n ​​ȯ ut​​ = ∑all i ​​n ​​i̇ ,out​​
(3) writing Ka in terms of z​ ​​i,out​​​using Eq. (5.2) and then relating all z​ ​​i,out​​​ to ​​ξ ​​̇
by substituting in the material balance equation
​​n ​​i̇ ,in​​ + ​νi​  ​​ ​ξ ​̇
​​zi​  ,out​​ = ________
​
​​
​​n ​​ȯ ut​​
This process yields an equation that relates Ka to n​​​  ​​i̇ ,in​​​ and ξ​​ .̇ ​​ If we know the
flows into the reactor, and we calculate a numerical value for Ka from ​Δ​​Ĝ ​​°r​  ​​
and ​Δ​​Ĥ ​​r°​  ​​, we can solve for ​​ξ.̇ ​​ Once we solve for ​​ξ,̇ ​​ we can calculate outlet
flows and fractional conversion.
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Section 5.1 Introduction
Example 5.4
335
Reactor Performance and Ka: Ammonia Synthesis
For the gas-phase synthesis of ammonia, the equilibrium constant is
​y​​NH​  23​​​  1
___
​Ka​  ​​ = ​​ ______
​y​​N2​ ​ ​​y​​H​  32 ​​​​​  ​  ​P​​  2 ​​
Ka = 6.6 × 105 at 298 K, with pressure in units of atm. Suppose 1000 kgmol/h N2
and 3000 kgmol/h H2 are fed to a reactor, the reactor operates at 1 atm and 298 K,
and the gas leaving the reactor is at chemical equilibrium. What is the fractional
conversion of nitrogen to ammonia?
Solution
We always start with a flow diagram:
N2
H2
Reactor
N2
H2
NH3
at equilibrium
The balanced reaction is
​​N2​ ​ + 3​H2​ ​ ⇄ 2N​H3​ ​​
The three material balance equations are
​​n​  ​​​Ṅ H​3​​,out​​ = 2​ξ ​​̇
​​​n ​​​Ṅ 2​ ​​,out​​= 1000 − ​ξ ​​̇
​​​n ​​​Ḣ 2​ ​​,out​​= 3000 − 3​ξ ​​̇
It’s useful to calculate the total molar flow out:
​​n​  ​​ȯ ut​​ = ∑ ​​n ​​i̇ ,out​​ = 2​ξ ̇ ​+ 1000 − ξ​ ̇ ​+ 3000 − 3​ξ ̇ ​= 4000 − 2​ξ ​​̇
The mole fractions of each component in the effluent stream are
​​n ​​​Ṅ H​ ​​,out​​
2​ξ ​̇
​​y​N​  H​3​​,out​​ = _
​  3  ​ = _
​
​​
​​n ​​ȯ ut​​
4000 − 2​ξ ​̇
​​n ​​​Ṅ ​ ​​,out​​
1000 − ​ξ ​̇
​​y​N​  2​ ​​,out​​ = _
​  2  ​ = _
​
​​
​​n ​​ȯ ut​​
4000 − 2​ξ ​̇
​​n ​​​Ḣ ​ ​​,out​​ 3000 − 3​ξ ​̇
​​y​H​  2​ ​​,out​​ = _
​  2  ​ = _
​
​​
​​n ​​ȯ ut​​
4000 − 2​ξ ​̇
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Helpful Hint:
If, but only if, the
reactor outlet is at
equilibrium, then
the composition of
the outlet stream
must satisfy the
Ka constraint.
Since the gas leaving the reactor is at equilibrium, yi,eq = yi,out, and we plug these
expressions into the equation for Ka:
2
( ​​y​N​  H​3​​,out​​)​​  2​ _________
​​
​
Ka​  ​​ = ____________
​
​ ​  1 at​2m ​
3
( ​y​N​  2​ ​​,out​​)​( ​y​H​  2​ ​​,out​​)​​  ​ ​P​​  ​
​(2​ξ ​)̇ ​​  2​ ​(4000 − 2​ξ ​)̇ ​​  2​
1 ​atm​​  2​  ​
6.6 × ​10​​  5​ = ____________________
​    3 ​ ​ _
(1000 − ξ​  ​)̇ ​(3000 − 3​ξ ​)̇ ​​  ​ ​(1 atm)​​  2​
Finally, we want to solve for ​​ξ ​​.̇ This is a nonlinear equation, and there is more
than one value of ​​ξ ​​̇ that satisfies the equation mathematically. But there is only
one value of ξ​​ ​​̇ that satisfies reality. ξ​​  ​​̇ must be nonnegative and ξ​​ ​​̇ cannot be any
greater than 1000 (why?). It is good practice to look at an equation before solving
it, and decide what the upper and lower maximum limits of the solution must be.
We use a spreadsheet, an equation-solving program, a programmable calculator,
or trial-and-error to find:
kgmol
​​ξ ​̇ = 970 ​ _
​​
h
The fractional conversion of nitrogen is therefore 0.97. Using the material balance
equations, we calculate the molar flows out of the reactor:
kgmol
​​​n ​​​Ṅ H​3​​,out​​ = 1940 ​ _
​​
h
kgmol
​​​n ​​​Ṅ 2​ ​​,out​​= 1000 − 970 = 30 ​ _
​​
h
kgmol
​​​n ​​​Ḣ 2​ ​​,out​​= 3000 − 2910 = 90 ​ _
​​
h
Notice carefully the strategy we used to solve this problem.
∙ We used mole balance equations to express the molar flow at equilibrium (in
the reactor outlet) as a function of the inlet molar flow, the reaction rate ​​ξ ​​,̇
and the known stoichiometric coefficients.
∙ We defined the mole fractions as molar flow of species at equilibrium divided
by the total molar flow.
∙ We wrote the equilibrium constant in terms of equilibrium mole fractions of
reactants and products.
∙ If the inlet flows and reaction stoichiometry are known, then the only unknown
is ​​ξ ​​,̇ which can be found from the Ka equation and the known numerical value
of Ka.
In Example 5.4, the reactor temperature and pressure were specified and we
calculated the reactor performance. Alternatively, we could specify the desired
reactor performance and find the reactor T and P necessary to achieve that
­specification. Performance of a reactor operating at equilibrium is influenced
by reactor temperature because Ka is a function of temperature. As a general
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Section 5.1 Introduction
337
rule of thumb, Ka decreases with increasing temperature for oxidation, hydrogenation, and hydrolysis reactions. Ka increases with increasing temperature for
dehydrogenation and dehydration reactions, and Ka is insensitive to temperature
for isomerization reactions. Performance of reactors may be influenced by
­reactor pressure. Ka is not a function of pressure, but equilibrium conversion
may be, if the reaction is gas phase and if there is a change in the number of
moles with reaction. If the number of moles decreases as the reaction proceeds,
conversion increases with increasing pressure, and vice versa.
Example 5.5
Equilibrium Conversion as a Function of T and P: Ammonia Synthesis
In Example 5.4, we observed that the fractional conversion at equilibrium was
high at 298 K and 1 atm. Unfortunately, the reaction kinetics are extremely slow
at this temperature; equilibrium might not be reached in this lifetime.
Temperatures of about 350 to 600°C are necessary for this reaction to come to
equilibrium in a reasonable length of time, using modern commercially available
catalysts.
Your job is to choose an appropriate ammonia synthesis reactor temperature
and pressure, given a target performance specification of 50% single-pass conversion. Assume a reactant feed of 1000 kgmol/h N2 and 3000 kgmol/h H2, and
assume that the reactor can be designed to achieve equilibrium conversion at
the outlet.
Solution
The flow diagram is identical to that in the previous example.
N2
H2
Reactor
N2
H2
NH3
at equilibrium
The balanced reaction is
​​N2​ ​ + 3​H2​ ​ ⇄ 2N​H3​ ​​
First we need to find Ka as a function of T.
​ΔH
​​ ̂ ​​°r​  ​ = ∑ ​νi​  ​​ Δ​​Ĥ ​​°i,  ​f​  = (−1)(0) + (−3)(0) + (+2)(−46,150) = −92,300 J/gmol​
​ΔG
​ °​r​  ​ = ∑ ​νi​  ​​ ​ΔG​°i,  ​f​  = (−1)(0) + (−3)(0) + (+2)(−16,600) = −33,200 J/gmol​
11,100
​ln Ka = −23.85 + ______
​   ​
​
T
In Example 5.4, we obtained an expression for Ka as a function of the extent
of reaction that is still valid for this problem:
​(2​ξ ​)̇ ​​  2​​(4000 − 2​ξ ​)̇ ​​  2​ 1 ​atm​​  2​
​​Ka​  ​​ = ____________________
​    3 ​ ______
​
​​
(1000 − ξ​  ​)̇ ​(3000 − 3​ξ ​)̇ ​​  ​ ​P​​  2​
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Our design target is 50% conversion of nitrogen, or fractional conversion fcN = 0.5.
From the material balance equation and the definition of fractional conversion, we
find that ξ​​ ​​̇ = 500 kgmol/h. Substituting in this value gives
2
2
​[2(500)]​​  ​ ​[4000 − 2(500)]​​  ​ 1 _
__________________________
​​Ka​  ​​ =
​     3 ​ _
​  2 ​ = ​  5.333
​​
(1000 − 500)​[3000 − 3(500)]​​  ​ ​P​​  ​
​P​​  2​
Now we have two equations in Ka, involving two unknowns, T and P. We simply
calculate Ka as a function of T from 350 to 600°C (633 to 873 K), then use the calculated Ka to determine P. As an example, we show the calculations at T = 633 K:
​ln ​Ka​  ​​= −23.85 + _
​ 11100 ​= − 6.31​
633
​​Ka​  ​​ = 1.81 × ​10​​  −3​ = _
​  5.333
​, ​P​​  2​
___________
P = ​ ___________
​  5.333 −3 ​ ​= 54 atm​
1.81 × ​10​​  ​
√
These calculations are straightforward to carry out in a spreadsheet. We plot our
results as P versus T; the line indicates the reactor process conditions that produce
the desired fractional conversion of 0.5. Notice that very high pressures are required.
Development of the mechanical equipment necessary to work at such high pressures was a crucial innovation needed for commercialization of ammonia synthesis.
700
Pressure, atm
600
500
400
fc = 0.5
300
200
100
0
600
5.1.6
650
700
750 800
Temperature, K
850
900
Chemical Reaction Equilibrium, Selectivity, and Yield
As we saw in Chap. 4, in many chemically reacting systems, there is more than
one reaction occurring. Typically, one reaction is the desired reaction, and our
job as engineers is to design reactors and choose operating conditions to maximize conversion and selectivity to the desired products. We can gain insight
into how best to achieve our goals by analyzing selectivity assuming the reacting
system is at equilibrium. Chemical reaction equilibrium relationships coupled
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Section 5.1 Introduction
with material balance equations allow us to determine reactor conversion,
selectivity, and yield.
Let’s consider this type of problem from a DOF point of view. We’ll assume
a steady-state continuous-flow reactor in which the outlet stream is at chemical
equilibrium. Suppose there are R reactants fed to the reactor, and P products
generated as a result of K reactions. There are R stream variables from the reactor inlet, R + P stream variables attributed to the reactor outlet, and K reaction
variables, for a total of 2R + P + K variables. The number of material balance
equations equals the number of compounds, or R + P. The statement that
the exiting stream is at equilibrium adds K constraints: For each reaction, we
write one equation for Ka. We now have counted up R + P + K equations. To
have an equal number of variables and equations, we need R additional constraints.
These are usually provided by specifying the inlet flows of all reactants.
Illustration: Reactants A and B are fed at a 1:1 ratio and 100 gmol/min to a
reactor. Within the reactor, two reactions take place:
​A + B ⇆ C​
​A + 2B ⇆ D + E​
The reactor outlet is at equilibrium.
A
B
Reactor
A
B
C
D
E
at equilibrium
Number of stream variables 7 Number of material balances
5
Number of reaction variables 2 Number of specified flows
1
Number of specified stream compositions
1
Number of specified system performances 2
Total variables
9 Total equations
9
The strategy for analyzing reactive systems at equilibrium when there are
multiple reactions to consider is similar to what you have already applied when
there is only one reaction. However, the mathematics gets a little more complicated. The most confusion arises in the derivation of the equations for Ka in
terms of flows and extents of reaction. The point to keep in mind is that all
of the extents of reaction can appear in each of Ka equations. For example,
with two reactions, the equation for Ka1 can contain both ​​​ξ ​​1̇ ​​​ and ​​​ξ ​​2̇ ​​​. Similarly,
the equation for Ka2 can contain both ξ​​​ ​​1̇ ​​​ and ξ​​​  ​​2̇ ​​​.
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
For steady-state continuous-flow reactors, the strategy is to:
(1) write material balance equations for every compound in terms of stream
variables and the extents of reaction:
​​n ​​i̇ ,out​​ = n​​  ​​i̇ ,in​​ + ​  ∑ ​​​ ​νi​  k​​ ​​ξ ​​k̇ ​​
all k
where the summation is taken over all K reactions.
(2) relate molar flow out n​​​  ​​i̇ ,out​​​to mole fraction out ​​z​ i,out​​​, using equations of the
form:
​​n ​​i̇ ,out​​
_
​​
​= z​ i​  ,out​​​
​​n ​​ȯ ut​​
where ​​​n ​​ȯ ut​​ = ∑all,i ​​n ​​i̇ ,out​​​.
(3) write one equation for Kak for each reaction k in terms of ​​z​ i,out​​​ using
Eq. (5.1 and 5.2) and then relate all z​ ​​i,out​​​ to all ξ​​​  ​​k̇ ​​​by substituting in the
material balance equation
​​n ​​i̇ ,in​​ + ​∑ k ​ ​​ ​νi​  k​​ ​​ξ ​​k̇ ​​
____________
​​zi​  ,out​​ =
​
​​
​​n ​​ȯ ut​​
This process yields equations that relate each Ka to ​​​n ​​i̇ ,in​​​ and all ​​​ξ ​​k̇ ​​​. If we know
the flows into the reactor, and we calculate a numerical value for each Ka from
Δ​​​Ĝ ​°​r​  ​​ and ​Δ​​Ĥ ​°​r​  ​​, we can solve for all values of ξ​​​ ​​k̇ ​​​. Once we find those values,
we can calculate outlet flows, conversion, selectivity, and yield.
Example 5.6
Multiple Chemical Equilibria and Reactor T: NOx Formation.
A significant contributor to airborne pollution is formation of NOx (rhymes with
lox, fox, and socks)—compounds like NO, N2O, and NO2. One of the main sources
of NOx is combustion in automobiles, industrial furnaces, and other sources. What
is the best way to design burners and internal combustion engines so they don’t
produce much NOx? To begin to answer this question, we ask: How do chemical
equilibrium considerations enter into design of engines and burners?
Two reactions of importance are
​​N​2​ + ​O2​ ​ ⇄ 2NO​
(R1)
​2NO + ​O2​ ​ ⇄ 2N​O2​ ​​
(R2)
Both reactions occur in the gas phase, and are most important in the hot flame
after the fuel has been burned. The equilibrium constants for (R1) and (R2) are K1
and K2, respectively, and depend on temperature T as
21,700
​ln ​K1​  ​​= + 2.97 − _
​
​​
T(K )
13,700
​ln ​K2​  ​​= − 17.5 + _
​
​​
T(K )
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Section 5.1 Introduction
341
1. Plot the equilibrium constants K1 and K2 as a function of temperature, from 300 K
(room temperature) to 2000 K (about as hot as a flame could get).
2. A burner has a postcombustion gas composition of 6.4 mol% O2, and 73.8 mol%
N2, with the remainder a mix of CO2 and H2O. The high flame temperature
supports further reaction to form NOx. Calculate the equilibrium composition
of NO, NO2, and NOx (NO + NO2) as a function of flame temperature. Assume
that H2O and CO2 are completely inert because they do not participate in the
NOx-forming reactions and the pressure of the flame is 1 atm.
Solution
1. Temperature has opposite effects on the equilibrium constant for reaction R1
versus R2: Ka increases with T for R1 but decreases for R2. R2 is much more
favored than R1 except at very high temperatures. Thus, NO formation is
favored at high temperatures, NO2 at low temperatures. Due to the very strong
effect of temperature and the large temperature range, we plot ln Ka rather
than Ka.
20
R2
ln Ka
0
–20
R1
–40
–60
400 600 800 1000 1200 1400 1600 1800 2000
Temperature, K
2. We start with a flow diagram. The flame itself acts as a “reactor.” We choose
a convenient basis of 1000 gmol/h total flow rate into the flame (any basis is
fine), and we lump together CO2 and H2O as I (for inert).
64 O2
738 N2
198 I
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Flame
O2
N2
NO
NO2
198 I
at equilibrium
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Next we set up the material balance equations, and derive equations for the mole
fraction of each compound in the outlet stream as a function of the extents of
reaction:
Compound
​​n ​​i̇ ,in
νi1 ​​ξ ​​1̇ + νi2 ​​ξ ​​2̇
​​n ​​i̇ ,out
yi,out
O2
64
−​​ξ ​​1̇ − ξ​​  ​​2̇
64 − ​ξ ​1̇ − ξ​  ​2̇
64 − ξ​​  ​​1̇ − ξ​ ​2​​ ___________
​​
1000 − ξ​  ​2̇
N2
738
−​​ξ ​​1̇
738 − ​ξ ​1̇
738 − ​​ξ ​​1̇ ​​ _________
​​
1000 − ​ξ ​2̇
+2​ξ ​1̇ − 2​ξ ​2̇
+2​​ξ ​​1̇ − 2​​ξ ​​2̇ ​​ __________
​​
1000 − ​ξ ​2̇
NO
0
+2​​ξ ​​1̇ − 2​​ξ ​​2̇
NO2
0
+2​​ξ ​​2̇
+2​ξ ​2̇
+2​​ξ ​​2̇ ​​ _________
​​
1000 − ​ξ ​2̇
198
0
198  ​​
198​​ _________
1000 − ​ξ ​2̇
1000
−​​ξ ​​2̇
I
Total
1000 − ​​ξ ​​2̇
1
For this gas-phase reaction, K1 and K2 are evaluated as (try deriving these equations
yourself!):
​( ​yN​  O​​)​​  2​ _____________________
​(2​​ξ ​​1̇ ​​ − 2​​ξ ​​2̇ ​​)​​  2​
​​K1​  ​​ = ​ _
​
=
​
​y​N​  2​ ​​​​ ​y​O​  2​ ​​​​ (738 − ξ​​  ​​̇ ​​)(64 − ξ​​  ​​̇ ​​ − ​​ξ ​​̇ ​​) ​​
1
1
2
​( ​y​N​  O​2​​​​)​​  2​ _
​(2​​ξ ​​2̇ ​​)​​  2​(1000 − ξ​​  ​​2̇ ​​)
1 atm  ​ ​ = _______________________
​​K2​  ​​ = ​ _
​ ​
​
​
​​
​( ​yN​  O​​)​​  2​ ​y​O​  2​ ​​​​ ( P(atm) ) ​(2​​ξ ​​1̇ ​​ − 2​​ξ ​​2̇ ​​)​​  2​(64 − ξ​​  ​​1̇ ​​ − ​​ξ ​​2̇ ​​)
Given a temperature T, we calculate numerical values for K1 and K2. We then have
two equations in two unknowns, ​​ξ ​​1̇ and ​​ξ ​​2̇ . We solve simultaneously, noting that
there are constraints on what values are physically reasonable, e.g., ​​ξ ​​1̇ + ξ​​ ​​2̇ < 64,​​
ξ ​​1̇ > ξ​​ ​​2̇ . Then we use the obtained values to calculate the mole fractions of NO
and NO2 from the material balance equation. For example, at T = 1500 K, K1 =
1.0 × 10−5 and K2 = 2.3 × 10−4. Using these values for K1 and K2 in the equations
above, we find the solution by using a spreadsheet or an equation solver. (Notice
that the equilibrium constants are very low for both reactions. This indicates that
the extents of reaction will also be low—close to zero. If we notice this, we can
simplify the equations further by noting that 738 − ​​ξ ​​1̇ ≈ 738, 64 − ​​ξ ​​1̇ − ​​ξ ​​2̇ ≈ 64,
and 1000 − ​​ξ ​​2̇ ≈ 1000!) We find ​​ξ ​​1̇ = 0.343 and ξ​​ ​​2̇ = 1.3 × 10−3 gmol/h. Using
these values, we calculate mole fractions: yNO = 6.8 × 10−4 and y​
​​ N​O2​ ​​​= 2.6 × 10−6.
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Section 5.1 Introduction
343
By repeating at other temperatures, we produce a graph of yNO and y​
​​ N​O2​ ​​​ versus T:
Mole fraction at equilibrium, yi
0.01
0.001
yNO
0.0001
10–5
10–6
yNO2
10–7
10–8
500
1000
1500
Temperature, K
2000
NOx formation increases dramatically with temperature. Industrial boilers and
incinerators must operate at a low enough flame temperature to minimize NOx
formation while maintaining a high enough temperature to achieve complete combustion. The ratio of NO to NO2 also shifts dramatically with temperature: below
600 K, NO2 is the main source of NOx but at higher temperatures NO becomes
the most important contributor to NOx.
Example 5.7
Multiple Chemical Equilibria and Selectivity: DEE from Waste Ethanol
Your company, which manufactures cellulose acetate and other cellulose-based
polymers from organically sourced cotton, produces a waste stream that is mostly
water but contains about 10 mol% ethanol. Currently you are paying a waste disposal business to truck away this solution. Might there be a more environmentally
sound use for this waste stream? One idea you have is to react the ethanol to make
diethyl ether (DEE), which is a solvent your company uses in manufacturing the
cellulose products. If this idea works, you eliminate one waste disposal problem
while reducing the amount of DEE solvent you need to purchase.
You learn that the dehydration of ethanol to DEE occurs in the vapor phase:
​​2C​2​​​H5​ ​​OH(g) ⇄ (​​ ​C2​ ​​​H5​ ​​)​​2​​O(g)​ +H​2​​O(g)​
(R1)
and that DEE can further undergo an unwanted dehydration reaction to ethylene
​​​(​C2​ ​​​H5​ ​​)​​2​​O(g) ⇄ ​2C​2​​ ​H4​ ​​(g)​ + H​2​​O(g)​
(R2)
As a first step in determining the feasibility of this idea, you decide to calculate
the conversion and selectivity at equilibrium at two different temperatures, 373 K
and 473 K, and two different pressures, 1 atm and 10 atm. You also wonder
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
whether you can feed the waste solution directly into a reactor or should separate
out some of the water first.
Solution
The flow diagram, using Et for ethanol, W for water and E for ethylene, is
Et
W
Reactor
Et
W
DEE
E
at equilibrium
Using data in App. B, we calculate:
​Δ​​G ​​̂ ° ​​  = −15.19 kJ/gmol​​Δ​​Ĥ ​​° ​​  = −24.63 kJ/gmol​
r1
r1
​Δ​​Ĝ ​​°r2​​  = +30.59 kJ/gmol​​Δ​​Ĥ ​​°r2​​  = +115.89 kJ/gmol​
A quick look at these values indicate that the desired reaction will be favored at
lower temperatures, while the undesired reaction will be favored strongly at higher
temperatures, a conclusion borne out when we calculate Ka at 373 K and 473 K:
​
K1​  ,373​​ = 62.2​K​ 1,473​​ = 11.6
​
K2​  ,373​​ = 0.051​K​ 2,473​​ = 134
The material balance equations are:
​​​n ​​Ė t,out​​ = n​​  ​​Ė t,in​​ − 2​​ξ ​​1̇ ​​​
​​​n ​​Ẇ ,out​​ = n​​  ​​Ẇ ,in​​ + ξ​​  ​​1̇ ​​ + ​​ξ ​​2̇ ​​​
​​​n ​​Ḋ EE,out​​ = ​​ξ ​​1̇ ​​ − ξ​​  ​​2̇ ​​​
​​​n ​​Ė ,out​​ = 2​​ξ ​​2̇ ​​​
Choosing as a basis 100 gmol/s feed into the reactor (​​​n ​​Ė t,in​​ = 10 ​and ​​​n ​​Ẇ ,in​​ =
90 gmol/s​) we derive expressions for Ka (details left to the reader):
​(​​ξ ​​1̇ ​​ − ​​ξ ​​2̇ ​​)​​(90 + ​​ξ ​​1̇ ​​ + ​​ξ ​​2̇ ​​)​
​​
​​K​ 1​​ = ____________________
​
​​ 10 − ​2​ξ ​​1̇ )​​ ​​​  2​
(
​(90 + ​​ξ ​​1̇ ​​ + ξ​​  ​​2̇ ​​)​​​(​2​ξ ​​2̇ ​​)​​​  2​​P​​  2​
____________________
​​K​ 2​​ =
​
​​
​ ​​ξ ​​1̇ ​​ − ξ​​  ​​2̇ )​​ (​​​ 100 + ​2​ξ ​​2̇ )​​ ​​​  2​
(
We can now proceed to solve for ​​​ξ ​​1̇ ​​ and ξ​​  ​​2̇ ​​​at 373 K and at 473 K, at both 1
and 10 atm. This is best carried out with a spreadsheet or equation-solving software, and it is useful to first note two limits: 0​​ ≤ ​ξ ​​1̇ ​​ ≤ 5​ and ξ​​​  ​​2̇ ​​ ≤ ξ​​  ​​1̇ ​​​. Fractional
conversion of ethanol, and selectivity for producing DEE from ethanol, were calculated as
​2​ξ ​​̇ ​​
​​fc​  ,Et​​ = ___
​  1 ​​
10
​ ​​ξ ​​1̇ ​​ − ​​ξ ​​2̇ )​​ ​
(
2
​​sE​  t→DEE​​ = ​ _ ​ ​ ________
​
​
1 2​ξ ​1̇
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Section 5.2 Chemical Reaction Kinetics and Reactor Performance
345
Results are tabulated:
T (K)
P (atm)​​​ξ ​​1̇ ​​​ (gmol/s)​​​ξ ​​2̇ ​​​ (gmol/s)​​f​ c,Et​​​​​s​ Et→DEE​​​
373 1
4.07
1.80
0.81
0.55
373
10
3.83
0.22
0.77
0.94
473 1
4.89
4.885
0.98
0.001
473
4.15
3.77
0.83
0.087
10
The highest conversion (473 K and 1 atm) gives by far the poorest selectivity.
It is very common to see a tradeoff between conversion and selectivity. In general,
high selectivity is better than high conversion, because a lower conversion can be
offset by recycle. Based on those criteria, the best operating conditions are 373 K
and 10 atm. Since water is a product of the reaction, removing some of the water
before the reactor would improve conversion and selectivity.
Sometimes the desired chemical reaction has a highly unfavorable equilibrium conversion at any reasonable temperature and pressure. We could live
with a low single-pass conversion, and simply recycle. Are there other ways
to engineer a chemically-reacting system in the face of unfavorable equilibrium?
The answer is yes. Here are some ideas to consider:
Adjust the feed ratio. Reactants do not have to be fed at stoichiometric
ratio. The fractional conversion of a limiting reactant is higher, and the
conversion of an excess reactant is lower, than if reactants are fed at
stoichiometric ratio.
Remove one of the products continuously. If a chemical reaction and product
separation are judiciously combined in a single piece of equipment, then
the equilibrium is driven towards increased conversion.
Couple to a reaction with favorable equilibrium. A reaction that consumes
a byproduct of the desired reaction drives the equilibrium toward
increased conversion.
We further explore some of these ideas in the case study at the end of this chapter.
5.2
Chemical Reaction Kinetics and Reactor
Performance
Even when reaction equilibrium is highly favorable, it may be a practical
impossibility to achieve equilibrium under industrially-relevant conditions. Why?
Because of slow reaction kinetics.
Chemical kinetics is the science of chemical reaction rates. Kineticists
attempt to understand the dependence of the rate of reaction on temperature,
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
pressure, catalyst, and reactant and product concentrations. Reactor engineering is the art and technology of designing reactors with the right temperature and pressure control, flow pattern, and residence time to efficiently
and safely carry out the desired reactions. Both topics are much too large
for us to cover. Here we will just give a small taste of chemical kinetics and
reactor engineering.
Chemical kineticists define a reaction rate (which we will call ​​r​′i​  ​​) as the
change in concentration of compound i per unit time:
d​c​  ​​
​​r​′i​  ​ = _
​  i ​​
dt
where ci is the molar concentration of i, that is, the moles of i per reactor
volume VR:
​n​  ​​
​​c​ i​​ = _
​  i  ​​
​VR​  ​​
​​r′​i​  ​​has dimensions of [moles/volume-time], whereas ​​r​​i̇ (with dot over), which we
use in the differential mole balance equation, has dimensions of [moles/time].
In general, ​​r​′i​  ​​is a complicated function of T, P, catalyst, and reactant and
product concentrations. Unfortunately, there is no general expression relating​​
r​i′​  ​​to these properties; the relationship needs to be determined by detailed
experiments and parameter fitting. In this section, we will describe three specific
cases, for which simple but useful expressions for r​
​​ ′i​  ​​ are available.
Case 1. First-order irreversible reactions. Consider the conversion of reactant A to product P:
A→P
If this reaction is irreversible (e.g., Ka >> 1), then
d​c​  ​​
​​r​′A ​​  = ​ _A ​= −k​cA​  ​​​
dt
Eq. (5.7)
These kinetics are called “first-order irreversible” because the
rate depends on the reactant concentration raised to the first
power, and does not depend on the product concentration. The
term k is called a rate constant and is always positive and has
dimension of [1/time] for first order reactions. The rate constant
generally increases strongly with temperature, as described by
the Arrhenius expression:
​E​  ​​
​
k = k​ ​ 0​​ exp (
​ − ​ _a ​)​​
T
Eq. (5.8)
where k0 and Ea are experimentally determined parameters, unique
to a specific reaction and catalyst, but independent of temperature.
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347
This simple case is surprisingly useful, describing many important reactions such as pasteurization and sterilization.
Case 2. Growth kinetics. Microorganisms such as E. coli or yeast reproduce
rapidly when fed a nutritious diet. Because these organisms reproduce by doubling (1 becomes 2 becomes 4 becomes 8), their rate
of growth increases as their number increases (as long as they don’t
run out of food). In a sense, microorganisms are both reactant and
product! These kinetics are expressed as:
d​c​  ​​
​​r​′M ​​  = _
​  M ​= k​ g​  ​​​cM
​  ​​​
dt
Eq. (5.9)
where kg is the rate constant for growth and has dimension of
[1/time]. The concentration of microorganisms cM is typically
reported as either number per volume, or mass per volume, rather
than moles per volume.
Case 3. C
atalytic reactions. Two important heterogeneous reactions include
combustion of firewood and catalytic conversion of CO in automobile exhaust to CO2. Many of these reactions have rate equations
of the form
k​Ka​  ds​​ ​cA​  ​​
d​c​  ​​
​​r​′A ​​  = ​ _A ​ = − ​ _
​​
1 + ​Ka​  ds​​ ​cA​  ​​
dt
Eq. (5.10)
where Kads is an experimentally determined constant that depends on
the chemical and physical properties of the catalyst as well as the
nature of the reactant and the temperature. Many enzyme-catalyzed
reactions are modeled by a similar equation, in which case this
equation becomes the famous Michaelis-Menten equation.
We still have the task of relating ​​r​′i​  ​​ to ξ​​  ​​i̇ in the material balance equation.
To do this, we need to turn our attention to the reactor engineering issues: the
reactor volume VR and the reactor mixing pattern. Reaction engineers generally consider three types of reactor: the stirred tank batch reactor, the stirredtank continuous-flow reactor, and the plug-flow continuous-flow reactor (refer
back to Fig. 4.2). In the completely mixed reactor, the temperature and concentrations are uniform throughout the reactor. For a batch stirred tank reactor,
the concentration of reactant decreases over time. For a continuous-flow
stirred tank reactor, the reactor is at the same concentration and temperature
as the reactor outlet stream. In the plug-flow reactor, the fluid moves as a
“plug” through a cylindrical reactor. As a cross-sectional slice moves through
the reactor, the concentration and the temperature change as the reaction proceeds. Since the reaction rate is a function of concentration and temperature,
the reaction rate is a function of position in the reactor. As you can see from
these brief explanations, the choice of reactor greatly influences the extent of
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
reaction achieved. You will learn much more about these topics in chemical
kinetics and reactor engineering classes. Here we illustrate two examples to
show how reactor performance is influenced by combining chemical kinetics
with reaction engineering.
5.2.1
Irreversible First-Order Reaction in a Stirred-Tank
Batch Reactor
First we consider the reaction kinetics. For the first-order irreversible reaction
​
A → P​
we have the kinetic expression
d​c​  ​​
​​r​′A ​​  = ​ _A ​= −k​cA​  ​​​
dt
Eq. (5.7)
The concentration cA is the moles of A per volume of reactor VR
​n​  ​​
​​c​ A​​ = _
​  A  ​​
​VR​  ​​
Substituting this into the rate equation we find
​n​  ​​
d​ _
​  A  ​ ​
(
V
​
​  ​​ )
​n​  ​​
R
​​r​′A ​​  = ​ _
​= −k​ _
​  A  ​ ​​
( ​VR​  ​​ )
dt
If we make an important assumption that the reactor volume VR is constant,
this equation can be simplified:
d​nA​  ​​
1  ​ ​ _
1  ​ ​n​  ​​​
​​ _
​= −k ​ _
​VR​  ​​ dt
​VR​  ​​ A
d​n​  ​​
​​ _A ​= −k​nA​  ​​​
dt
Eq. (5.11)
Now let’s consider the reactor design. For a batch reactor, there is no input or
output and the differential mole balance equation is:
d​nA​  ,sys​​
​​ _​ = ​​r ​​Ȧ ​​​
dt
Eq. (5.12)
Compare Eqs. (5.11) and (5.12). Since nA in Eq. (5.11) is the moles of A in
the reacting system, or n​ ​​A,sys​​​, then ​​​r ​​Ȧ ​​ = −k​nA​  ​​ = −k​nA​  ,sys​​​. The mole balance
equation becomes:
d​nA​  ,sys​​
​​ _​= −k​nA​  ,sys​​​
dt
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Section 5.2 Chemical Reaction Kinetics and Reactor Performance
349
This equation can be re-arranged and then integrated from t = 0 (where n​ ​​ A,sys​​​ =​​
n​ A,sys,0​​​) to tf (where ​​n​ A,sys​​​ = ​​nA​  ,sys, f​​​)
d​nA​  ,sys​​
​​ _
​nA​  ,sys ​
​​ = −kdt​
​​∫
​tf​  ​​
d​nA​  ,sys​​
​ ​​ ​ _
​
=
−​
∫
​
​​ kdt​
​n​  ​​
​nA​  ,sys, f​​
​nA​  ,sys,0​​
A,sys
0
n​ A​  ,sys, f​​
​ln ​ _
​nA​  ,sys,0 ​​​ = −k​tf​  ​​​
or
​​n​ A,sys, f​​ = n​ A​  ,sys,0​​ ​e​​  −k​tf​  ​​​​
Eq. (5.13)
We can convert Eq. (5.13) to our familiar form of the integral material balance
equation by subtracting n​ ​​ A,sys,0​​​from both sides and rearranging:
​​n​ A,sys, f​​ − ​nA​  ,sys,0​​ = ​nA​  ,sys,0​​​(e​​  −k​tf​  ​​​− 1)​
Eq. (5.14)
Eq. (5.14) is a general equation for a batch reactor with first-order irreversible
kinetics.
The fractional conversion of A is:
​nA​  ,sys,0​​ − n​ A​  ,sys, f​​
​​f​ CA​​ = ____________
​
​
= 1 − e​ ​​  −k​tf​  ​​​​
​n​
​​
A,sys,0
Eq. (5.15)
This derivation shows that reactor performance is a function of both reaction
kinetics and reactor design.
Example 5.8
Reaction Kinetics and Reactor Performance: Vegetable Processing
We need to sterilize cans of vegetables in a batch sterilizer. Each can contains 250 mL
and an average of 10,000 spores. For safety and shelf stability, we need to reduce
this to an average of 0.1 spores/can, a 99.999% reduction. But, we also want to
keep the vegetables tasty.
Spore killing is a first-order irreversible reaction, with a rate constant
15000 ​ ​​
​k = 9 × ​10​​ 15​ exp​(− ​ _
T )
where T is temperature (K) and k has units of min−1. Loss of flavor is modeled as
a first-order irreversible reaction, too, with a rate constant
5000 ​ ​​
​​k​ f​​ = 9 × ​10​​  5​ exp​(− ​ _
T )
where T is in K and kf has units of min−1. On the basis of consumer taste panels,
we decide that a 25% loss of flavor is acceptable.
Is there a way to produce canned vegetables that are both safe to eat and tasty?
At what temperature would you heat the cans? For how long?
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Solution
The sterilizer is a batch reactor with first-order irreversible kinetics. Before we
solve, let’s plot the rate constants versus temperature.
105
104
Spore killing
k or kf , min–1
103
102
Taste loss
101
100
10–1
10–2
50
100
150
200
Temperature, °C
250
300
At 100°C (373 K), k for spore killing is much lower than kf for flavor loss.
However, the spore killing rate increases faster than the flavor loss rate with
increasing temperature. Around 160°C, the rates become similar. Therefore, we
expect that temperatures higher than 160°C are necessary for sufficient spore killing and minimum flavor loss.
We can use Eq. (5.15) for first-order irreversible kinetics in a batch reactor:
Spore killing ( fCA = 0.99999):
​​fC​  A​​= 1 − ​e​​  −k​tf​  ​​​= 0.99999​
Rearranging and taking the natural logarithm of each side, we get
ln (1 − 0.99999)
15,000
_____________________
​tf​  ​​ = − ​
​ = 1.28 × ​10​​  −15​ exp​(_
​
​​
T )
−
15000
15
_
​[9 × ​10​​  ​ exp​(​
​)​]​
T
Flavor loss ( fCA = 0.25):
ln (1 − 0.25)
​tf​  ​​ = − ​ ___________________
​ = 3.20 × ​10​​  −7​ exp​(_
​  5000 ​)​
T
​[9 × ​10​​  5​ exp​(_
​  − 5000 ​)​]​
T
We have two equations in two unknowns. We solve to get T = 517 K (244°C) and
tf = 5 ms. Sterilization and pasteurization are often carried out at high temperature
for short times to avoid flavor loss and vitamin degradation.
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Section 5.2 Chemical Reaction Kinetics and Reactor Performance
5.2.2
Growth Kinetics in a Stirred-Tank Continuous-Flow Reactor
The kinetics of growth of microorganisms can frequently be expressed as
d​c​  ​​
​​r​′M ​​  = _
​  M ​= ​kg​  ​​ ​cM
​  ​​​
dt
Eq. (5.9)
where kg is the rate constant for growth and the concentration of microorganisms, cM, is usually reported as either number per volume or mass per volume.
Microorganism growth (fermentation) can be carried out in a continuousflow stirred-tank reactor (Fig. 5.3). The feed stream is a liquid called “medium”
that contains all the nutrients that the microorganisms need. The reactor contents are well stirred, and the “bugs” feed on the supplied nutrients and grow
at the rate shown in Eq. (5.9). The output contains medium that is depleted of
nutrients, as well as microorganisms.
The reactor is initially inoculated with a small amount of the microorganism
of interest but no new microorganisms enter with the feed stream. If the reactor
operates at steady state, the material equation on microorganism (M) simplifies to:
​​​n ​​Ṁ ,out​​ = ​​r ​​Ṁ ​​​
where ​​​r ​​Ṁ ​​​is the rate of growth of microorganism (number per unit time) inside
the reactor. In other words, the increase in microorganisms due to growth inside
the reactor is exactly matched to the flow rate of microorganisms out of the
reactor! We have usually worked in mass or molar units, but for this application, it is convenient to describe the total flow in or out of the reactor in
volumetric units, where
​​​n ​​ȯ ut​​ = ρ​​V ​​ȯ ut​​​
and n​​​  ​​ȯ ut​​​is the total molar flow out, ρ
​ ​is the density of the fluid (moles/volume)
and V
​​​  ​​ȯ ut​​​is the volumetric flow out (volume/time). Then
​​​n ​​Ṁ ,out​​ = c​ M
​  ,out​​ ​​V ​​ȯ ut​​​.
The rate of growth of microorganism r​​​ ​​Ṁ ​​​inside the system depends on the
kinetics as well as the reactor volume VR, or ​​​r ​​Ṁ ​​ = k​ g​  ​​ ​cM
​  ,sys​​ ​VR​  ​​​. The material
balance equation becomes:
​​c​ M,out​​ ​​V ​​ȯ ut​​ = ​kg​  ​​ ​cM
​  ,sys​​ ​VR​  ​​​
Feed
CM,sys
Output
CM,out
Figure 5.3 Schematic of a continuous-flow stirred tank reactor. When used for microorganism growth, these reactors are sometimes called “chemostats.”
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Now notice again that the reactor is well-stirred—which means the concentration
is the same everywhere inside the reactor. If this is true, then the concentration
in the stream leaving the reactor must be the same as the concentration in the
reactor. In other words, c​ ​​M,out​​ = c​ M
​  ,sys​​​! This is a critical point to understand,
because it greatly simplifies the growth-kinetics continuous-flow stirred-tank
reactor material balance to:
​​​V ​​ȯ ut​​ = ​kg​  ​​ ​VR​  ​​​
Eq. (5.16)
Example 5.9
Growth Kinetics in a Stirred-Tank Continuous-Flow Reactor:
Microbial Degradation of Toxins in Wastewater
A wastewater stream is contaminated with 0.125 mmol/L pesticide. Your company
is interested in developing environmentally sound processes for biodegrading contaminated wastewater streams, and you have discovered a strain of E. coli that is
safe to humans but can degrade the pesticide into harmless byproducts. The growth
rate constant of this E. coli strain ​kg​  ​​= 0.11 h−1. The kinetics of degradation of the
pesticide are well-described by a first-order irreversible expression, with k = 9.5 h−1.
You are testing this process in a one-liter chemostat (a type of continuous-flow stirred
tank reactor). (a) What volumetric flow rate of medium should you choose? (b) What
fractional conversion of the pesticide should you expect?
Solution
(a) The volumetric flow rate can be calculated from Eq. (5.16):
​​V ​​ȯ ut​​ = ​kg​  ​​​VR​  ​​= (0.11 h​ ​​  −1​) × (1 L) = 0.11 L/h
(b) With the reactor as system, the steady-state material balance equation for the
pesticide P is
​​​n ​​Ṗ ,out​​ = ​​n ​​Ṗ ,in​​ + ​​r ​​Ṗ ​​​
It is convenient to use volumetric units, which is fine as long as the density
is constant
​​​n ​​Ṗ ,out​​ = ​cP​  ,out​​ ​​V ​​ȯ ut​​​, and n​​​  ​​Ṗ ,in​​ = ​cP​  ,in​​ ​​V ​​i̇ n​​​
The reaction rate kinetics are first-order and irreversible, and the concentration
is that of the pesticide in the reactor, or
​​​r ​​Ṗ ​​ = r​ ′​P ​ ​​  VR​  ​​ = −k​cP​  ,sys​​ ​VR​  ​​​
The material balance equation becomes
​​cP​  ,out​​ ​​V ​​ȯ ut​​ = ​cP​  ,in​​ ​​V ​​i̇ n​​ − k​cP​  ,sys​​ ​VR​  ​​​
At steady-state, V
​​​  ​​ȯ ut​​ = V
​​  ​​i̇ n​​​. Since this is a well-stirred reactor, ​​c​ P,out​​ = c​ P​  ,sys​​​.
We plug in known values and simplify to
​​cP​  ,out​​(0.11 L/h) = (0.125 mmol/L​)​(0.11 L/h) − (9.5 h​ ​​  −1​)c​ P​  ,out​​(1 L)​
We then solve to find
​​cP​  ,out​​= 0.00143 mmol/L​
In other words, 98.8% of the pesticide has been destroyed!
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353
Hydrogen and Methanol
In the 1700s, it was whale oil. In the 1800s, coal was king. And the late 1900s
were dominated by liquid petroleum. Some people now forecast that this
­century will usher in the hydrogen economy. Hydrogen is a clean-burning fuel,
producing nothing but water on oxidation. It is the fuel source of choice for
energy-efficient fuel cells that in the future may power everything from portable phones to huge electrical utility plants. Although many have proposed
that automobiles and trucks may some day run on hydrogen-fed fuel cells,
concern has been expressed about the safety of people driving around with
storage tanks of gaseous hydrogen under high pressure. Methanol has been
proposed as one possible liquid fuel alternative, as it is safer to handle and
store in fueling stations. The idea is that liquid methanol could be stored in
the fuel tank, then converted to hydrogen in situ which could then be fed
to the fuel cell. Direct-methanol fuel cells are an attractive alternative which
can run directly with methanol. Besides their potential use in fuel cells, hydrogen
and methanol are important reactants in atom-economical chemical synthesis
routes. But, hydrogen is not a readily available raw material and must be
synthesized.
In this case study, we’ll examine the synthesis of hydrogen and methanol from methane. Natural gas, an abundant but nonrenewable resource, is
a major source of methane. Methane can be produced from biomass, including from waste products like cow manure using aerobic digesters. In this
Case Study we will take a closer look at how equilibrium considerations
enter into the design of reactors that synthesize hydrogen and methanol from
methane.
Methane is converted to hydrogen via the steam reforming reaction:
​C​H4​ ​(g) + ​H2​ ​O(g) ⇄ CO(g) + 3​H2​ ​(g)​
(R1)
Let’s start by considering the chemical equilibrium constant of this reaction.
This will help us identify reactor temperatures and pressures that will give
reasonable equilibrium conversions. (Remember—we can never do better than
equilibrium conversion.) We’ll use M for methane, W for water, CO for carbon
monoxide, and H for H2. The reaction takes place in the vapor phase. The
chemical equilibrium constant is
​yC​  O​​ ​y​ H3 ​​
​
Ka​  ​​ = ​ ∏​​​ ​( ​yi​  ​​ P)​​  ​νi​  ​​​= P
​ ​​  2​ ​​ _____
​yM
​  ​​ ​yW
​   ​​​​
all i
Ka for methane-steam reforming is calculated from the Gibbs energy and
enthalpy of reaction to be
​ln ​Ka​  ​​= 25.8 − _
​ 24800 ​​
T
where P has units of atm and T is in K.
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Now we can figure out how the fractional conversion at equilibrium
depends on temperature and pressure.
Our first goal is to calculate the equilibrium conversion as a function of
reactor T and P. We’ll sketch out the flow diagram:
CH4
H2O
Reactor
CH4
H2O
CO
H2
at chemical equilibrium
We can choose any basis we want. Let’s choose 1.0 gmol/s CH4 fed to the
reactor. Let’s assume for now that the feed ratio of methane:water is 1:1 (the
stoichiometric ratio). The steady-state differential material balance reaction for
methane M is
​​​n ​​Ṁ ,out​​ = ​​n ​​Ṁ ,in​​ − ξ​​  ​​1̇ ​​= 1.0 − ξ​​  ​​1̇ ​​​
We write similar material balance equations for the other components, sum up
the component molar flows out of the reactor to get the total molar flow out,
and calculate the mole fraction by dividing the component molar flow by the
total material flow. It’s convenient to do this in table form:
​​n ​​i̇ ,in
​n ​i̇ ,out
yi,out = _____
​​
​
​n ​ȯ ut ​
1 − ​ξ ​1̇
1 − ​​ξ ​​1̇ ​​ _______
​​
2 + 2​ξ ​1̇
​​n ​​i̇ ,out
CH4
1
H2O
1
CO
​ξ ​1̇
0​​ξ ​​1̇ ​​ _______
​​
2 + 2​ξ ​1̇
H2
0
Total
2
1 − ​ξ ​1̇
1 − ​​ξ ​​1̇ ​​ _______
​​
2 + 2​ξ ​1̇
3​ξ ​1̇
3​​ξ ​​1̇ ​​ _______
​​
2 + 2​ξ ​1̇
2 + 2​​ξ ​​1̇
1
Now we can write Ka in terms of T, P, and ξ​​ ​​1̇ :
24,800
_
​K​  ​​ e​(​​​  25.8 − ​  T )​​ ​ _________________
(​​ξ ​​1̇ ​​)​(3​​ξ ​​1̇ ​​)​​  3​
​​ _a2 ​ = ​ ___________
​
=
​
​​
​P​​  ​
​P​​  2​
​(1 − ξ​​  ​​1̇ ​​)​​  2​​(2 + 2​​ξ ​​1̇ ​​)​​  2​
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Eq. (5.17)
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Section 5.2 Chemical Reaction Kinetics and Reactor Performance
Quick Quiz 5.4
What is the range of
physically reasonable
values for ξ​​ 1​​ ̇ ?
355
For a given value of T and P, we can solve for ​​ξ ​​1̇ at equilibrium. This isn’t
too hard to do with an equation solver. Then, we use the value of ξ​​ ​​1̇ to calcu­
late yi, ​​n ​​i̇ ,out, and fractional conversion of methane fCM at equilibrium.
What range of values is reasonable for T and P? From an examination
of Eq. (5.17), we see that ​​ξ ​​1̇ increases if (1) the pressure decreases or (2)
the temperature increases. From our heuristics, we prefer not to go any lower
than ambient pressure. Temperatures above about 600°C (873 K) require
­specialized materials of construction, but maybe we could push it a bit on
this end if needed. With these thoughts in mind, let’s examine reactor per­
formance at 1 atm and at 400°C to 800°C. Some results of our calculations
are plotted.
Fractional conversion of methane
or mole fraction H2
0.8
0.7
0.6
0.5
0.4
Mole fraction H2
0.3
Fractional conversion
of methane
0.2
0.1
0
400
450
500
550 600 650 700
Temperature, °C
750
800
At 600°C, equilibrium conversion is about 30%. This is probably accept­
able, since much higher temperatures are frowned on because of materials costs
and added safety concerns.
Are there any other ways to change the reactor design to increase equilib­
rium conversion of methane? Here are a couple of ideas:
1. Increase the amount of water in the reactor to help drive the reaction,
making methane the limiting reactant.
2. Install an in situ adsorber: a solid material to which CO selectively sticks,
such that the mole fraction CO in the gas phase is always maintained
at a low level, e.g., 0.01. (In Chap. 6 and 7 we will discuss adsorption in
more detail.)
Let’s explore these ideas some more.
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Idea 1.Increase the water:methane feed ratio above stoichiometric. Let’s keep
T = 600°C, P = 1 atm, and ​​n ​​Ṁ ,in = 1.0 and vary the stoichiometric
ratio from 1:1 to 9:1. Eq. (5.17) changes somewhat. (See if you can
derive the modified equation.) The results are plotted:
Fractional conversion of methane
or mole fraction H2
0.9
600°C
1 atm
0.8
0.7
Fractional conversion
0.6
0.5
0.4
H2 mole fraction
0.3
0.2
1
2
3
4
5
6
7
Water : methane molar ratio
8
9
Increasing the water:methane ratio greatly increases methane conversion! We pay a price, though—we are handling a lot of extra
steam, and the hydrogen in the product stream is more dilute, which
increases separation costs.
Idea 2.Install an adsorber to remove CO as it is made. The flow diagram
changes: our process unit is a combined reactor and separator.
CH4, 1 gmol/s
H2O
Reactor + separator
CH4
H2O
CO
H2
CO adsorbed to solid
For these calculations we’ll assume that the adsorber maintains
yco = 0.01 in the reactor. (In Chap. 6 and 7 you will learn more about
how adsorbers work and how to determine the gas phase composition
in the presence of an adsorber.) We’ll assume a 1:1 methane:water
feed ratio, 600°C, 1 atm. The material balance equations must be
adjusted to account for the extra outlet stream:
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Section 5.2 Chemical Reaction Kinetics and Reactor Performance
​​n ​​i̇ ,in
CH4
1
H2O
1
357
​n ​i̇ ,out
yi,out = _____
​​
​
​n ​ȯ ut ​
1 − ​ξ ​1̇
1 − ξ​​  ​​1̇ ​​ ____________
​​
2 + 2​ξ ​1̇ − n​  ​ȧ ds
​​n ​​i̇ ,out
CO
1 − ​ξ ​1̇
1 − ​​ξ ​​1̇ ​​ ____________
​​
2 + 2​ξ ​1̇ − n​  ​ȧ ds
0​​ξ ​​1̇ − ​​n ​​ȧ ds
0.01
H2
0
Total
2
3​ξ ​1̇
3​​ξ ​​1̇ ​​ ____________
​​
2 + 2​ξ ​1̇ − ​n ​ȧ ds
2 + 2​​ξ ​​1̇ − ​​n ​​ȧ ds
1
where ​​n ​​ȧ ds = flow rate of CO leaving the reactor adsorbed to the
solid. With this situation,
24800
_
​K​  ​​ e​(​​​  25.8 − ​  T ​​)​
(0.01)​(3​​ξ ​​1̇ ​​)​​  3​
______________________
​​ _a2 ​ = ​ ___________
​
=
0.0737
=
​
​​
​P​​  ​
​P​​  2​
​(1 − ξ​​  ​​1̇ ​​)​​  2​​(2 + 2​​ξ ​​1̇ ​​ − ​​n ​​ȧ ds​​)​​  2​
and we have the specification regarding the separation that
​​ξ ​​1̇ ​​ − n​​  ​​ȧ ds​​
​​y​ CO​​= 0.01 = ____________
​
​​
2 + 2​​ξ ​​1̇ ​​ − n​​  ​​ȧ ds​​
Solving these two equations simultaneously, we find that adding the
adsorber increases conversion from 30% to 65%. Furthermore, the
gas leaving the reactor is more enriched in hydrogen ( ​​y​​H2​ ​​​ = 0.73,
versus 0.35 in the base case), because the byproduct CO is continuously removed. Are there any disadvantages you can think of?
Notice that with both of these ideas we achieved greater fractional conversion, not by violating the law of chemical reaction
equilibrium, but by designing around it!
In the analysis so far, we have assumed that only one reaction takes place. In
reality, chemistry plays tricks on us all the time. As (bad) luck would have it,
most of the time these tricks come in the form of unwanted reactions: Either the
reactants convert to other undesired products or the products themselves undergo
further reactions. How can we design a reactor to give good fractional conversion
and good selectivity when there are multiple chemical reactions? We’ve got a
bag of tricks of our own. To illustrate, let’s take a closer look at steam reforming. Recall that we’ve made CO and H2 by steam reforming of methane,
​C​H4​ ​(g) + ​H2​ ​O(g) ⇄ CO(g) + 3​H2​ ​(g)​
(R1)
and will want to feed CO and H2 to another reactor to make methanol,
​CO(g) + 2​H2​ ​(g) ⇄ C​H3​ ​OH(g)​
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
However, in the steam reforming reactor, CO undergoes a further reaction with
steam to make CO2. This reaction is the famous water-gas shift reaction:
​CO(g) + ​H2​ ​O(g) ⇄ C​O2​ ​(g) + ​H2​ ​(g)​
(R2)
The chemical equilibrium constant for the water-gas shift reaction is
​ln ​Ka​  ​​= − 5.1 + _
​ 4950 ​​
T
where T is in K.
How should I operate the steam reformer, given these complications and the
desire to make methanol? The methanol reaction requires 2 moles H2 per mole
CO, and steam reforming produces 3 moles H2 per mole CO. So we already
have a CO deficit, and the water-gas shift reaction just makes it worse. Are
there temperatures and pressures that allow good conversion in steam reforming while suppressing the water-gas shift? Let’s start by looking at some trends.
The equilibrium expression for the water-gas shift reaction is
​y​  ​​ ​y​  ​​
​​K​ a​​ = _
​  ​y​ CD​​ ​y​ H  ​​​​
CO W
where we use CD for carbon dioxide. Notice that there is no pressure term!
Since the equilibrium conversion for the desired steam reforming reaction R1
decreases with increasing pressure, and the extent of reaction of the undesired
reaction R2 is unaffected by pressure, we should design our reactor at as low
a pressure as feasible, to favor the desired reaction. Therefore, let’s keep the
design pressure at 1 atm.
Now, let’s look at the temperature dependence by plotting ln (Ka) for the
two reactions R1 and R2:
4
2
Water-gas shift (R2)
0
ln Ka
–2
–4
–6
Steam reforming (R1)
–8
–10
–12
400
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450
500
550 600 650
Temperature, °C
700
750 800
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Section 5.2 Chemical Reaction Kinetics and Reactor Performance
359
The desired reaction is highly unfavored at low temperature but highly favored
at high temperatures. The undesired reaction has the opposite trend. This is
good news. We want to operate at reactor temperatures where Ka for the undesired reaction is less than 1, and also less than that of the desired reaction. Our
earlier choice of 600°C, made when we considered only one reaction, is too low.
Let’s pick 725°C, a temperature at which Ka of R1 is greater than that of R2.
We’ll assume that the reactants are fed at stoichiometric ratio (for the
desired reaction R1) and use a basis of 1 gmol/s CH4 fed to the reactor.
CH4, 1 gmol/s
H2O
Reactor
CH4
H2O
CO
H2
CO2
at equilibrium
Let’s use our material balance equations to calculate conversion and selectivity.
In this analysis, we add in CO2 as one of the components, and we include the
rates of both reactions.
​​n ​​i̇ ,in
∑ νik ​ξ​  ​​k̇
CH4
1
−​​ξ ​​1̇
H2O
1
−​​ξ ​​1̇ − ​​ξ ​​2̇
CO
0
​​ξ ​​1̇ − ξ​​ ​​2̇
H2
0
3​​ξ ​​1̇ + ​​ξ ​​2̇
CO2
0
​​ξ ​​2̇
Total
2
​n ​i̇ ,out
yi,out = _____
​​
​
​n ​ȯ ut ​
1 − ​ξ ​1̇
1 − ξ​​ ​​1̇ ​​  _______
​​
2 + 2​ξ ​1̇
​​n ​​i̇ ,out
1 − ξ​ ​1̇ − ​ξ ​2̇
1 − ​​ξ ​​1̇ − ξ​​ ​​2̇ ​​  _________
​​
2 + 2​ξ ​1̇
​ξ ​1̇ − ​ξ ​2̇
​​ξ ​​1̇ − ξ​​ ​​2̇ ​​  _______
​​
2 + 2​ξ ​1̇
3​ξ ​1̇ + ​ξ ​2̇
3​​ξ ​​1̇ + ​​ξ ​​2̇ ​​  _______
​​
2 + 2​ξ ​1̇
​ξ ​2̇
​​ξ ​​2̇ ​​ _______
​​
2 + 2​ξ ​1̇
2 + 2​​ξ ​​1̇
1
Now we have two equations relating mole fractions to the two value of Ka,
and we have expressions for mole fraction of all components in terms of ξ​​ ​​1̇
and ​​ξ ​​2̇ .
We now substitute the expressions for mole fractions from the table into
the equations for Ka and solve. We’ll compare to the case where we didn’t
consider the water-gas shift reaction.
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Water-gas not considered
Water-gas considered
yi
​​n ​​i̇ ,out
yi
​​n ​​i̇ ,out
CH4
0.125
0.38
0.12
0.40
H2O
0.125
0.38
0.10
0.32
CO
0.19
0.62
0.16
0.53
H2
0.57
1.86
0.59
1.88
0.024
0.077
CO2
​​ξ ​​1̇ 0.62
​​ξ ​​2̇ 0.60
0.077
A couple of observations: First, the water-gas shift reaction has had a small
effect on the methane conversion. Second, the H2:CO ratio has changed from
3:1 to 3.6:1. This is bad if our goal is to make methanol, since the stoichiometric ratio for the methanol synthesis reaction is 2:1 H2:CO. Third, the selectivity, based on methane converted to CO, is (0.60 − 0.077∕0.60) = 0.87
(compared to 1.0 in the absence of the water-gas shift reaction). Finally, we’re
making a byproduct, CO2, that will have to be separated out and is of little use.
Are there any other changes we could make? Notice that water is a reactant
in both reactions. What happens if we adjust the stoichiometric feed ratio of
methane to water? Let’s evaluate a couple of different feed ratios: twofold
excess water, and twofold excess methane. We’ll keep the basis as 1 gmol
methane/s fed to the reactor.
CH4:H2O feed = 1:1
CH4:H2O feed = 1:2
CH4:H2O feed = 1:0.5
yi
​​n ​​i̇ ,out
yi
​​n ​​i̇ ,out
yi
​​n ​​i̇ ,out
CH4
0.12
0.40
0.041
0.19
0.26
0.60
H2O
0.10
0.32
0.22
0.99
0.034
0.079
CO
0.16
0.53
0.13
0.61
0.165
0.37
H2
0.59
1.89
0.57
2.63
0.53
1.22
CO2
0.024
0.076
0.043
0.20
0.009
0.021
​​ξ ​​1̇ 0.60
0.81
0.40
​​ξ ​​2̇ 0.077
0.20
0.021
This has made a big difference! With excess water, we’ve achieved an amazing
81% conversion of methane. The cost, though, has come in selectivity, which
has dropped to 0.75. The H2:CO ratio of 4.4 is worse, and we’ve generated
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Summary
361
more CO2. The total flow through the reactor has increased, which means a
bigger, more expensive, reactor. In contrast, with methane in excess, the methane conversion dropped way down to 40%, but selectivity is better at 0.95, less
CO2 is produced, and the H2:CO ratio is lower.
Deciding on the best reactor conditions will require further design analysis.
Can unreacted methane be recycled? How difficult is it to remove CO2? Is
there anything useful we could do with it? Are there other uses for the excess
H2? Are there any other reactions that might occur? Finally, does the reactor
operate close to equilibrium, or is conversion and selectivity dominated by
kinetic considerations?
Summary
∙ Chemical reaction equilibrium limits the maximum achievable conversion and may affect selectivity. The chemical reaction equilibrium constant
Ka is
​​Ka​  ​​ = ​ ∏​​​ ​a​ i​ν​  i​  ​​​​
all i
where ai is the activity of species i at equilibrium and νi is the stoichiometric coefficient of i. To a first approximation,
​y​  ​​P
​​a​ i​​ = _
​  i  ​ ​
1 atm
for a gas
​​a​ i​​ = x​ i​  ​​​
for a liquid
​​a​ i​​= 1​
for a solid
Ka is a function of temperature:
−Δ​​Ĝ ​​°r​  ​ ____
Δ​H°​r​  ​ _
1​ ​
ln ​Ka​  ,T​​ = ______
​
​+ ​   ​
​ ​  1  ​ − ​ _
R ( 298 T )
298R
where
​ΔG
​​ ̂ ​​°r​  ​ = ∑ ​νi​  ​​ Δ​​Ĝ ​​°i,  ​ and Δ​​
​
Ĥ ​°​r​  ​ = ∑ ​νi​  ​​ Δ​​Ĥ ​°​i,  ​f​​
f
By adjusting temperature, pressure, and reactant feed ratio, and by modifying the reactor flow sheet, we can design around some of the limitations
imposed by chemical reaction equilibrium.
∙ The assumption of chemical reaction equilibrium at the reactor outlet can
be used as a reactor performance specification. The strategy is to (a) write
material balance equations for every compound in terms of stream variables
and extents of reaction, (b) derive expressions for mole fraction out of each
compound in terms of stream variables and extents of reaction using the
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
results of step a, (c) substitute in the mole fractions from step b into the
​ν​  ​​
equation K
​​ ​  a​​ = ∏​​a​  i​​​​  i​​, and (d) use the numerical value of K​
​​ a​​​ (calculated
from ​ΔG
​​ ̂ ​​°r​  ​​ and ​ΔH
​ °​r​  ​​) to solve for the extents of reaction at equilibrium.
Conversion, yield and selectivity at equilibrium can then be calculated.
∙ Chemical reaction kinetics may limit the conversion and affect selectivity.
By adjusting temperature, pressure, reactant feed ratio, catalyst, and reaction time, we can overcome some of the constraints imposed by chemical
reaction kinetics. The reactor performance is an outcome of both chemical
kinetics and reaction engineering.
Quick Quiz Answers
5.1
5.2
5.3
5.4
Ka = yByC∕​​y​A2 ​; no.
70.54 kJ/gmole.
CO + 2H2 → CH3OH. Choose T below 400 K.
Must be between 0 and 1.
References and Recommended Readings
1. Perry’s Chemical Engineers’ Handbook and Lange’s Handbook of
Chemistry have extensive lists of values for Gibbs energy and enthalpy
of formation. Appendix B of this text has a limited tabulation.
Chapter 5 Problems
Warm-Ups
Section 5.1
P5.1 Write an expression for Ka in terms of mole fractions for the following
reactions:
​A(g) + 2B(g) ⇄ P(g) + D(g)​
​A(​ l)​+ 2B​(g)​⇄ P(l) + D(l)​
​A(l) + 2B(l) ⇄ P(l) + D(l)​
​A(s) + 2B(g) ⇄ P(s) + D(g)​
P5.2 If ​Δ​​Ĝ ​​°f,  ​
​ = −25.0 kJ/gmol​and ​Δ​​Ĝ ​​°f,  ​
​ = 15.0 kJ/gmol​, what is ​Δ​​Ĝ ​​°r​  ​​ if
A
B
the reaction is:
i.​
A ⇄ B​
ii.​2A ⇄ B​
iii.​
A ⇄ 2B​
iv.​2A ⇄ 2B​
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Chapter 5 Problems
363
P5.3 For each of the following cases, state if Ka < 1 or Ka > 1 at (a) 298 K
and (b) 500 K
​i​.​​​Δ​​Ĝ ​​r°​  ​∕R = 1000/K​ and ​ΔH
​​ ̂ ​​°r​  ​∕R = 3000/K​
̂
​i​i​.​​​Δ​​G ​​r°​  ​∕R = 1000/K​ and ​ΔH
​​ ̂ ​​°r​  ​∕R = −3000/K​
̂
​i​i​i​.​​​Δ​​G ​​r°​  ​∕R = −1000/K​ and ​ΔH
​​ ̂ ​​°r​  ​∕R = 3000/K​
̂
​i​v​.​​​Δ​​G ​​r°​  ​∕R = −1000/K​ and ​ΔH
​​ ̂ ​​r°​  ​∕R = −3000/K​
P5.4 You are designing a reactor to accomplish the gas-phase conversion of​
A(g) + B(g) ⇄ P(g).​​ΔG
​​ ̂ ​​r°​  ​= 20.0 kJ/gmol​and ​Δ​​Ĥ ​​r°​  ​= −15.0 kJ/gmol​.
If the reactor outlet is at equilibrium, for maximum conversion would
you choose high or low temperature? High or low pressure?
P5.5 For the reaction ​A(l) + B(l) ⇄ C(l)​, Ka = 5 at the reactor operating
conditions. The reactor outlet composition is sampled and determined
to be 33 mol% A, 33 mol% B, and 34 mol% C. Is the reactor outlet at
equilibrium?
P5.6 100 gmoles compound A is placed in a batch reactor and allowed to
isomerize to compound C by the reaction ​A(l) ⇄ C(l)​. The extent
of reaction at equilibrium ​ξ = 42 gmoles.​What is the numerical value
of Ka?
P5.7 In the manufacture of high-fructose corn syrup (used in enormous quantities in sodas, fruit-flavored drinks, and other beverages), glucose
(C6H12O6) is isomerized to fructose by using an enzyme as catalyst.
Despite the fact that the reaction proceeds fairly rapidly, the maximum
glucose conversion achievable is less than 50%. What does this tell you
about the approximate value of Ka and Δ​​​Ĝ ​r°​​  ​​ for this reaction?
Section 5.2
P5.8 Microbial degradation kinetics of benzene in contaminated wastewater
can be modeled as first-order and irreversible. If the rate constant
k = 2 × 10−6 s−1, what is the time it takes to achieve degradation of
99% of the benzene?
P5.9 Yeast are grown in a 20-liter chemostat (continuous-flow stirred-tank
reactor). If the growth rate constant kg = 0.24 h−1, what is the volumetric flow rate (liters/hour) out of the reactor?
P5.10 Yeast are grown in a 20-liter fermenter that operates as a continuousflow stirred-tank reactor. If the fermenter contains 2 grams of yeast,
what is the yeast concentration (grams/liter) in the output stream?
P5.11 Small batch reactors can be placed in specialized microwave equipment
to heat the reacting mixture and speed up the kinetics. A company
claims that its microwave technology can achieve 90% conversion of a
reactant to a pharmaceutical product in 4 minutes, whereas conventional
technology can achieve only 52% conversion in 24 hours. If the reaction
kinetics are first-order and irreversible in both cases, what is the percent
increase in the reaction rate constant k for the microwave technology
compared to conventional?
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Chapter 5 Why Reactors Aren’t Perfect: Reaction Equilibrium and Reaction Kinetics
Drills and Skills
Section 5.3
P5.12 100 gmol each of A and B are placed in a reactor where the gas-phase
reaction 2A + B → C + D takes place. The reaction reaches equilibrium. Write an expression for Ka in terms of the mole fractions of each
reactant and product and the total pressure. Then simplify so that you
have an expression for Ka in terms of the fractional conversion of A
and the total pressure. Will increasing the reactor pressure increase or
decrease the fractional conversion of A?
P5.13 Animal cells use glucose (C6H12O6) as their predominant energy source.
Aerobic metabolism of glucose leads to complete oxidation of glucose
to carbon dioxide and water. (Aerobic means that the metabolism requires
oxygen from air.)
​​C6​ ​​H1​ 2​​O6​ ​ + 6​O2​ ​ → 6C​O2​ ​ + 6​H2​ ​O​
What is Δ​​​Ĝ °​​r​  ​​ of glucose oxidation?
This reaction is coupled to the synthesis of ATP (adenosine triphosphate) from inorganic phosphate and ADP (adenosine diphosphate).
Δ​​​Ĝ °​​r​  ​​ of this reaction is +7.3 kcal/gmol.
​ADP + ​H3​ ​P​O4​ ​→ ATP​
If 6 moles of ADP are converted to ATP for every mole of oxygen
consumed by aerobic glucose, what is the efficiency of conversion of the
chemical energy of glucose into chemical energy of the phosophamide
bond in ATP?
P5.14 Dimethyl carbonate (DMC) is made with highly toxic phosgene, and
manufacture of DMC generates chlorinated byproducts. Many other reaction pathways have been proposed; some were discussed in Example 5.3
and two more are presented here:
Scheme 1: React methanol, CO2 and ethylene oxide to DMC and
formaldehyde (CH2O)
Scheme 2: React dimethylether (CH3OCH3) with water to make
methanol, then react methanol with CO2 to make DMC and water.
Evaluate each scheme by calculating the standard Gibbs energy of reaction
and then finding numerical values of Ka at 373 K and 773 K. Are either
of these schemes promising compared to those discussed in Example 5.3?
P5.15 A mixture of 30 mol% CO, 65 mol% H2, and 5 mol% N2 is fed to a
methanol (CH3OH) synthesis reactor, where the following reaction occurs:
CO + 2​H2​ ​ ⇆ C​H3​ ​OH
The reactor is at 200°C and 4925 kPa. The stream leaving the reactor
is at equilibrium.
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Chapter 5 Problems
365
If 100 kgmol/h of the feed mixture is fed to the reactor, calculate
the flow rates of all species leaving the reactor.
P5.16 Steel (Fe) sits out in the air and slowly rusts. What is the most likely
product of oxidation of Fe: FeO, Fe3O4, or Fe2O3? (Use chemical equilibrium to determine.)
P5.17 To generate hydrogen for fuel cells, steam reforming of methane is to
be conducted at in a steady-state continuous-flow reactor. For the reaction
C​H4​ ​ + ​H2​ ​O ⇄ CO + 3​H2​ ​
Ka = 5.2 at 600°C, with pressure in atm. Assume that the reactor feed
ratio is 5:1 (moles:moles) H2O:CH4 and that the reactor operates at 2 atm.
Derive an equation that relates Ka to the fractional conversion of methane. Then determine whether the equilibrium conversion of methane is
greater than or less than 90%.
P5.18 A soap manufacturer is considering converting its excess glycerol
byproduct to hydrogen for use in fuel cells, using a new catalytic process.
The process operates the following reaction:
​​C3​ ​​​H8​ ​​​O3​ ​​​ + 3H​2​​O ⇄ 3C​O​2​​ ​ + 7​H​2​ ​​
You are running experiments on the new catalyst in a laboratory-scale
reactor that operates in the gas phase, at 1.2 atm and 200°C, with a feed
ratio of water:glycerine of 5:1 (molar units). The glycerine feed rate in
our pilot plant reactor is 150 g/hr, and an exit stream hydrogen mole
fraction of 0.54 is measured. The equilibrium constant for this reaction
at 200°C Ka = 55, with pressure in atm. Is the reactor achieving equilibrium conversion?
P5.19 Ammonia (NH3) and methanol (CH3OH) react to produce methylamine
(CH3NH2), which is a useful intermediate in the production of pharmaceuticals. Methylamine undergoes an unwanted reaction with methanol
to produce dimethylamine (CH3)2NH. Water is a byproduct of both reactions. You are testing the performance of a new catalyst in a batch
laboratory reactor. The equilibrium constants for these two reactions are
Ka1 = 4.1 and Ka2 = 2.8 at the reactor operating conditions. Both reactions
take place in the liquid phase. You load 10 gmol of methanol and 10 gmol
ammonia into the reactor, wait 15 minutes, then remove the reactor
contents and analyze to find that it contains 20.0 mol% ammonia,
9.4 mol% methanol, 18.6 mol% methylamine, 10.9 mol% dimethylamine, and the remainder water. Had the reaction reached equilibrium
after 15 minutes?
P5.20 Acetaldehyde (CH3CHO) is produced by dehydrogenation of ethanol
(C2H5OH).
​​C2​ ​​H5​ ​OH ⇄ C​H3​ ​CHO + ​H2​ ​​
120 gmol/h ethanol is fed continuously to a reactor operating at steady
state. If the reactor operates at 200°C and 2 atm pressure and the reactor
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outlet is at equilibrium, what is Ka? What is the composition of the
outlet stream, and the fractional conversion of ethanol?
P5.21 You are interested in developing a process for converting glycerol
(a byproduct of biodiesel production) to hydrogen. The gas-phase
­reaction is
​​C3​ ​​​H8​ ​​​O3​ ​​​ + 3H​2​​O ⇄ 3C​O​2​​ ​ + 7​H​2​ ​​
You put 16.3 gmol water and 3.26 gmol glycerol into a batch laboratoryscale reactor, adjust the temperature to 200°C and the pressure to 1.2 atm.
You allow the reaction to reach equilibrium, at which time you determine that the composition of the gas in the reactor is 17.83 mol% H2O,
0.29 mol% glycerol, 24.56 mol% CO2, and 57.31 mol% H2. Use these
results as well as values for ​ΔG
​​ ̂ ​​f°​  ​​ and ​Δ​​Ĥ ​​f°​  ​​from App. B to find ​ΔG
​​ ̂ ​​f°​  ​​
(kJ/gmol) of glycerol.
Section 5.4
P5.22 Atrazine is a heavily used herbicide, particularly in the Midwest U.S.
corn belt. Typically, atrazine is applied to corn fields about 18–30 days
after planting, at 0.75 lbs/acre. The herbicide degrades slowly in soil by
chemical hydrolysis and microbial activity. If atrazine degradation is
first-order and irreversible with a rate constant k = 0.0005 h−1, about
how many hours will it take for the atrazine in the soil to decay to one
tenth of its original value?
P5.23 The enzyme urease degrades urea into ammonia and carbon dioxide by a
first-order irreversible reaction, with k = 0.045 h−1. If I place 20 mmol/liter
urea into a 1-liter batch reactor containing urease, what will be the
concentration of urea after 24 hours?
P5.24 Enzymes are proteins that catalyze specific reactions. Many enzymes
are commercially important; an example is glucoamylase, which degrades
cornstarch into glucose. Glucoamylase can be used either in its soluble
form, or it can be immobilized onto solid beads which are then added to
the aqueous solution of cornstarch. You are interested in comparing the
rate of cornstarch degradation by soluble versus immobilized glucoamylase. You dissolve 150 mg cornstarch into 1 mL water, add the enzyme,
and then monitor the production of dextrose over time. The quantity
(mg) dextrose produced at each time point is summarized below:
15 min
30 min
60 min
120 min
Soluble
31 57 93
127
Immobilized
88
150
124
145
Plot the data and decide whether you think the degradation of cornstarch
to dextrose follows first-order irreversible kinetics. Then use the data to
estimate the rate constant k for soluble and immobilized glucoamylase.
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P5.25 Some antibiotics are produced by fermentation of fungi. One type of
antibiotic-producing fungus is placed in a 1-liter batch fermentor at an
initial concentration of 1.25 g/liter. The growth rate constant kg = 0.1 h−1.
Use the material balance equation for a batch reactor to derive an
expression for the mass of fungus in the reactor at some later time t.
Then calculate the mass of fungus at t = 12 h.
Scrimmage
P5.26 Proteins are polyamides—polymers of amino acids—with the amino
acids arranged in a linear chain. The sequence of amino acids on the
chain differs from protein to protein. Under normal physiological conditions, the protein is folded into a specific structure. If you heat the
protein the protein “unfolds” and looks more like a cooked strand of
spaghetti. This process of unfolding takes the protein from the “native,”
or N state, to the “unfolded,” or U state.
​N ⇔ U​
The fraction of protein in the N or U state can be measured using a
number of spectroscopic tools. Here are some data for percent of the
protein in the N state as a function of temperature, taken with a 1 mg/ml
protein solution.
T (°C) 35
%N
45
55
56
58
60
61
63
65
70
75
80
100 99.9 90.2 85.5 71.0 50.7 40.0 22.2 10.9 1.5 0.2 0.03
Use the data to determine Δ
​ G
​​ ̂ ​​°r​  ​​ and ​Δ​​Ĥ ​​°r​  ​​of the folding-unfolding reaction.
̂
P5.27 We need to calculate ​Δ​​H ​​r°​  ​​for the oxidation of p-xylene (C8H10) to
terephthalic acid (TPA, C8H6O4), which is used in synthesizing the polymer that is in 2-liter plastic soda bottles. The reaction is
​​C8​ ​​H1​ 0​(l) + 3​O2​ ​(g) ⇄ ​C8​ ​​H6​ ​​O4​ ​(s) + 2​H2​ ​O(l)​
However, enthalpies of formation of the compounds are not available.
We do know that ​ΔH
​​ ̂ ​​°r​  ​= −1089.1 kcal/gmol​for the combustion of
p-xylene(l) to CO2(g) and H2O(l) at 298 K and Δ
​ ​​Ĥ ​​r°​  ​= −770.4 kcal/
gmol​for combustion of TPA (s) to CO2(g) and H2O(l) at 298 K.
Use these data to find a numerical value for ​Δ​​Ĥ ​​r°​  ​​of the p-xylene oxidation reaction. Then look up ​Δ​​Ĥ ​​f°​  ​​ for CO2 and H2O and find ​Δ​​Ĥ ​​°f​  ​​ for
p-xylene.
P5.28 The lactic acid byproduct recovered from cheese plants can be used to
make a variety of chemicals. For example, it may be hydrogenated to
produce 1,2-propanediol, which is in turn used as a polymer precursor
and as a food additive:
​C​H3​ ​CHOHCOOH + 2​H2​ ​ ⇄ C​H3​ ​CHOHC​H2​ ​OH + ​H2​ ​O​
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The process feed is a vapor stream containing 1:3:2 lactic acid:H2:inert
(mole ratios), at a total pressure of 5 atm and a flow rate of 300 mol/h.
The reaction is reversible, with an equilibrium constant Ka = 3.2 at the
reactor temperature. What is the fractional conversion of the limiting
reactant at equilibrium, and the reactor outlet composition? Will increasing the inert concentration in the feed increase or decrease the equilibrium fractional conversion? First give your answer by using qualitative
reasoning, then calculate the conversion at a higher inert concentration
and see if your reasoning was right. Would changing the reactor pressure change the conversion? Prove your results by calculating the effect
of an increase in pressure by 2 atm.
P5.29 One step in making synthetic detergents is preparation of butanal (also
called butyraldehyde) from propene (also called propylene), CO, and
hydrogen:
​C​H3​ ​CHC​H2​ ​+ CO + ​H2​ ​ ⇄ C​H3​ ​C​H2​ ​C​H2​ ​CHO​
The reaction takes place in a gas-phase reactor at 5 atm pressure. Feed
flow rate to the reactor is 1200 gmol/h. The reaction is reversible, with
an equilibrium constant Ka = 8 at the operating temperature of 180°C.
First analyze the case where the feed is 20 mol% propylene, 40 mol%
CO, and 40 mol% H2. Determine the fractional conversion of propylene,
the butanal production rate, and the percent butanal in reactor effluent.
Then, repeat these calculations at several propene concentrations from
10 to 80 mol%. In all cases, CO and H2 are the remainder of the feed,
and are fed at 1:1 mole ratio. Plot your results and examine the trends.
What propene feed percent allows for the maximum conversion? What
propene feed percent provides maximum butanal production rate? Do
you think one of these is optimum, or would you choose a different
propene feed percent? Explain your reasoning.
P5.30 The water-gas shift reaction
​CO + ​H2​ ​O ⇄ C​O2​ ​ + ​H2​ ​​
reaches equilibrium quickly. Calculate Ka versus T in the range of 100
to 1000 °C. Assuming that CO and H2O are fed to a reactor at a stoichiometric ratio, calculate the mole fractions of CO and H2 in the reactor outlet as a function of temperature in this range, if the reactor outlet
is at equilibrium. Plot your results (yCO versus T, ​​y​​H2​ ​​​ versus T). What
reactor temperature would you pick if you wanted the H2:CO ratio out
of the reactor to be 2:1?
P5.31 One method of producing ethanol is the vapor-phase hydration of ethylene
​​C2​ ​​H4​ ​ + ​H2​ ​O ⇄ ​C2​ ​​H5​ ​OH​
Assume that ethylene and water are fed to a reactor at equimolar ratio.
Derive a general expression for the moles of ethanol produced as a
function of temperature and pressure at equilibrium. Use this expression
to generate plots of ethanol production from 150 to 300°C at 1 atm
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pressure, and from 1 to 50 atm at 150°C. Comment on your results in
terms of optimizing reactor operating conditions.
P5.32 Butanes (n-butane and isobutane) are often available in excess at a refinery; they cannot be blended into gasoline in high quantities because they
are too volatile. Their worth can be increased significantly by reacting
them to make iso-octane, a very valuable component of gasoline. This is
accomplished in two reactions. Reaction R1 is dehydrogenation of n-butane
to n-butene:
​​nC​4​​​H1​ 0​​ ⇄ n​C4​​ ​​​H​8​ ​​​​ + H​2​​​
Reaction R2 combines n-butene with isobutane to make iso-octane:
​​nC​4​​​H8​ ​​​ + iC​4​​​H1​ 0​​ ⇄ i​C8​​ ​​​H​1​ 8​​​​
Reaction R1 is typically carried out at high temperature (400°C) whereas
reaction R2 must be carried out at low temperature (150°C) to prevent
unwanted polymerization side reactions. For iso-octane, Δ​​Ĝ ​​°f​  ​= 11 kJ/gmol
and Δ​​Ĥ ​​°f​  ​= −224.1 kJ/gmol.
The available excess mixed butane stream is 40 mol% isobutane and
60 mol% n-butane. Assume that the reactor effluents reach equilibrium
in both cases, and that any separator can completely separate one compound from the other. Synthesize a block flow diagram containing both
reactors and as many separators and recycle streams as you see fit.
Using a basis of 100 gmol/h feed, calculate the flows of all streams.
P5.33 Acetaldehyde (CH3CHO) is produced by dehydrogenation of ethanol
(C2H5OH).
​​C​2​​H5​ ​OH → C​H3​ ​CHO + ​H2​ ​​
(R1)
An undesired side reaction produces ethyl acetate (CH3COOC2H5)
​2​C2​ ​​H5​ ​OH → C​H3​ ​COO​C2​ ​​H5​ ​ + 2​H2​ ​​
(R2)
120 gmol ethanol/h is fed to a reactor operating at steady state. The
reactor operates at 400°C and 2 atm pressure. If the reactor outlet is at
equilibrium, what is the fractional conversion of ethanol, and fractional
selectivity for producing acetaldehyde from ethanol? How could you
adjust temperature and pressure to increase conversion? To increase
selectivity? Is it possible to adjust temperature and pressure to increase
both conversion and selectivity?
P5.34 Ammonium nitrate is used as a fertilizer but, mixed with a bit of fuel
oil, it can explode. Explosive decomposition of ammonium nitrate was
the cause of a serious accident in Texas City, Texas, as well as the
Oklahoma City bombing. Several decomposition reactions are proposed
to occur in an explosion:
​N​H4​ ​N​O3​ ​ ⇄ ​N2​ ​O + ​H2​ ​O​
​N​H4​ ​N​O3​ ​ ⇄ ​N2​ ​ + ​H2​ ​O + ​O2​ ​​
​​N2​ ​O ⇄ ​N2​ ​ + ​O2​ ​​
Balance the reactions. Are these three reactions independent?
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Pure ammonium nitrate is a solid below 169.6°C. What is the equilibrium mixture at 100°C? 250°C? Why do you think fuel oil is added
to ammonium nitrate to make it explosive?
P5.35 Salmonella in raw eggs is a major source of food poisoning. The idea
of pasteurizing eggs in the shell to reduce salmonella has been considered, but the trick is to kill the bacteria without cooking the eggs. One
of the key chemical processes that occurs when eggs cook is denaturation and coagulation of proteins. At temperatures below 60°C, egg
proteins do not significantly denature or coagulate, so one group of
researchers proposed that eggs be pasteurized by exposure to hot air at
55°C. (Food Microbiology 1996, vol. 13, pp. 93–101.) To evaluate this
idea, you inoculate eggs with salmonella bacteria, hold the eggs at 55°C
for various time intervals, then measure the number of bacteria remaining in the eggs. (Bacteria are quantified as colony-forming units per
milliliter, or cfu/mL.) Use these data to estimate a rate constant kf,
assuming first-order degradation kinetics.
0 min
2.0 × 106 cfu/ml
30 min
1.4 × 105 cfu/ml
90 min
9.8 × 102 cfu/ml
130 min
10 cfu/min
P5.36 Broth used in fermentation processes must be sterilized before use.
About 10 L of broth are used in each batch. You’ve decided to use heat
sterilization and need to pick the optimum time and temperature. 10 L
of broth contain about 22,000 spores before sterilization.
Bacterial spores are killed by a first-order irreversible reaction, with
a rate constant that is a strong function of temperature (with T in K):
​​kdeath
​  ​​ = ​10​​  39​ ​e​​  −35000∕T​ mi​n​​  −1​​
The medium in the broth contains a very expensive nutrient at an initial
concentration of 5 mg/L. This nutrient is destroyed by heat, again by a
first-order irreversible reaction, with a rate constant
​​kdestroy
​
​​ = ​10​​  4​ ​e​​  −4000∕T​ mi​n​​  −1​​
Your goal is to produce spore death while minimizing nutrient destruction. First plot the rates of death and destruction at different temperatures from 298 to 500 K. Then determine an appropriate sterilization
temperature and time if your goal is 99.999% spore death, and an
acceptable level of destruction of nutrient is (a) 50% or (b) 5%.
Suppose the kind of spores contaminating the broth change and now
​​kdeath
​  ​​ = ​10​​  40​ ​e​​  −40000∕T​ mi​n​​  −1​​
If you were unaware of this change and continued to operate under the
chosen design conditions, what percentage death would you achieve?
Would this be a worry?
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P5.37 To make some vaccines, viruses are grown in cultured cells in a reactor.
(The viruses are either genetically altered from the wild-type infective virus,
or are treated after they are made, to attenuate their lethality while still
generating an immune response.) You are developing a process to grow a
new virus. From experiments, you know that the virus growth rate depends
on the reactor temperature: the growth rate in a 2 mL batch mini-reactor is
2.9 × 106 viruses per day at 31°C and 3.3 × 106 viruses per day at 37°C.
However, the virus also undergoes a degradation (first-order irreversible)
reaction that is higher at the higher temperature: k = 0.76 day−1 at 31°C
and 2.2 day−1 at 37°C. Assume the reactor is initially inoculated with
3 × 104 viruses. After 12 hours, what is the virus content in the reactor
if it operates at 31°C? at 37°C? Which temperature should you choose?
P5.38 When a patient takes a drug orally, there is a sharp rise in the drug
concentration in the body and then a fairly rapid fall. To maintain a more
stable drug concentration over longer time, drugs are encapsulated in
controlled-release devices. For one such device, the encapsulated drug is
released from the device into the patient at a rate of 0.015nC, where nC
is the quantity (micromoles) of drug in the capsule, and the rate is in
micromoles/h. The drug is chemically unstable, so inside the device it
degrades to an inactive form by a first-order irreversible reaction, with
a rate constant k = 0.003 h−1. Using the controlled-release device as your
system, derive a differential balance equation on the drug. If 60 micromoles of active drug are initially loaded into the device, what is the total
quantity of active drug that is released after 1 hour? after 10 hours?
P5.39 Hydrogen peroxide solutions are used routinely to bleach cotton fibers
before dyeing. A specialty cotton fabric manufacturer produces 50,000
liters/day of spent bleaching water contaminated with 0.3 wt% H2O2.
You’d like to be able to re-use the water in the dyeing process, but the
residual hydrogen peroxide is reactive with the plant-derived dyes,
destroying their color. Your job is to design a batch reactor that removes
99% of the hydrogen peroxide so the water can be re-used. You consider
two possibilities: (1) spontaneous decomposition of hydrogen peroxide
to water and oxygen, or (2) catalytic decomposition of hydrogen peroxide
to water and oxygen. From laboratory data, you know that after 24 h at
37°C, 5% of the H2O2 decomposes in the absence of a catalyst by a
first-order reaction. With the catalyst, the rate of decomposition is
0.0005 gmoles H2O2/L-min, and this rate is independent of the H2O2
concentration. How long would you have to wait to decompose 99%
of the hydrogen peroxide with (1) spontaneous decomposition and
(2) catalytic decomposition?
P5.40 Vegetables “breathe,” even after they are cut and stored in the refrigerator. For example the rate of respiration of cut broccoli is estimated to be:
219 ​y​O​  2​  ​​​​
​​​r ​​​Ȯ 2​  ​​​​ = _
​
​​
0.014 + ​y​O​  2​  ​​​​
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where ​​​r​​​Ȯ ​2​​​is the rate of oxygen uptake by the vegetables, in units of mL
O2/kg broccoli/h and y​
​​ ​O​2​​​is the mol fraction oxygen in the gas surrounding the broccoli. If vegetables “breathe” too fast, they spoil quickly.
However, if their oxygen supply is completely cut off, they die, emitting
foul odors and liquefying in the process.
Packaging films are designed to regulate the oxygen content in
packaged fresh vegetables to control respiration rates, thereby increasing
the shelf life of vegetables. These films allow some limited transfer of
oxygen from the air to the package. In one experiment, 137 grams of
cut broccoli are placed in a container (initially containing air) and covered with a low-density polyethylene packaging film. What is the initial
rate of oxygen uptake (mL O2/h)? Some oxygen transfers across the
film; at steady state it is found that the mole fraction O2 in the container
is 0.008. What is the rate of transfer of O2 across the film at steady
state?
Game Day
P5.41 The first step in the manufacture of nitric acid is the synthesis of NO
from ammonia and oxygen:
​4N​H​3​(g) + 5​O2​ ​(g) ⇄ 6​H2​ ​O(g) + 4NO(g)​
(R1)
At least one unwanted side reaction may also occur:
​4N​H​3​(g) + 3​O2​ ​(g) ⇄ 6​H2​ ​O(g) + 2​N2​ ​(g)​
(R2)
(a) Derive expressions for the chemical reaction equilibrium constants
K1 and K2 for reaction R1 and R2 as a function of temperature. Plot
K1 and K2 vs T for temperatures from 25°C to 400°C. Also derive
expressions for K1 and K2 in terms of the mole fractions of the
reactants and products and the pressure.
(b) Now consider the reactor design. The raw materials available are
ammonia and air (79 mol% N2, 21 mol% O2). Assume that the reactor operates at steady state and that the reactor outlet stream is at
equilibrium. Derive material balance equations for the molar flow
rate in the outlet stream for all compounds, in terms of the molar
flow rates fed to the reactor and the two extents of reaction. Then
use these equations along with the results of part (a) to derive
expression for K1 and K2 in terms of molar flow rates fed to the
reactor and the two extents of reaction. Using a basis of 400 gmol/h
ammonia fed to the reactor, examine how fractional conversion of
ammonia and selectivity of converting ammonia to NO vary as a
function of T, P, and NH3:O2 feed ratio, by examining three temperatures (25°C, 150°C, and 400°C), three pressures (1 atm, 5 atm, and
10 atm), and three reactant feed ratios (5:1, 1:1, and 1:5 NH3:O2).
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(c) Given your calculations in part (b), the knowledge that ammonia
and water are easily condensed and separated from the gases (N2,
O2, and NO) but that separation of the gases from each other is
expensive, and the fact that air is essentially free but ammonia must
be purchased, propose operating conditions (T, P, and NH3:O2 feed
ratio) for the reactor. Sketch out a flowsheet that includes the reactor
but also includes any separators, splitters, mixers, recycle streams,
etc., as you see fit. For the operating conditions that you chose,
complete process flow calculations for your flowsheet.
P5.42 Propylene (C3H6) and benzene (C6H6) react to form cumene (C9H12).
Unfortunately, a side reaction also occurs, in which diisopropylbenzene
is produced. The two reactions are
​​C​3​​H6​ ​ + ​​C6​ ​H6​ ​ → ​C9​ ​​H1​ 2​​
(R1)
​​C​3​​H6​ ​ + ​C9​ ​​H1​ 2​ → ​C1​ 2​​H1​ 8​​
(R2)
In an existing process in your plant, propylene contaminated with 5 mol%
propane (C3H8, unreactive under these conditions) is mixed with benzene
and fed to a reactor. The reactor operates at 3075 kPa (30.75 bar) and the
reaction occurs in the vapor phase. The reactor effluent is partially condensed and sent to a vapor-liquid separator, where all of the unreacted
propylene, plus the propane, is separated out as vapor. Part of this
stream is purged; the remainder is recycled back to the reactor inlet.
The liquid from the separator contains benzene, cumene, and diisopropylbenzene. This is sent to a series of two distillation columns. In the
first distillation column, benzene is taken off the top of the column and
recycled back to the reactor inlet. Cumene and diisopropylbenzene are
sent to the second distillation column, where the cumene product is
taken off the top of the column and stored in a tank for sale. The diisopropylbenzene is burned as fuel.
The reactor has a volume of 8 m3 (8000 L). It is a fluidized bed
reactor, containing a solid catalyst that is fluidized by the gas stream
flowing through it. To keep the reactor behaving properly, the volumetric flow through the reactor is limited to no more than 600 m3/h, calculated at the reactor temperature and pressure and the total molar flow
rate at the reactor inlet. For the purposes of this problem, we will assume
that the reactor acts like a well-stirred reactor, meaning that the temperature and concentration inside the reactor are the same everywhere.
Under these conditions the performance of the reactor is defined by:
​fC​  A​​
​V​  ​​
_
​​  R  ​ = ​ ____
​​
​​n ​​Ȧ ,in​​ − ​rA′​  ​​
where VR is the reactor volume, ​​n ​​Ȧ ,in is the molar flow rate of the limiting reactant A into the reactor, fCA is the fractional conversion of A,
and ​​rA′​  ​​​  is the rate of reaction of A. In this kind of reactor, the rate is
calculated at the concentrations in the outlet stream from the reactor.
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You have a fantastic new catalyst that you are putting into the reactor.
The reaction rate expressions for this new catalyst for (R1) and (R2) are
​​r′​c ​​  = ​k1​  ​​​cp​  ​​​cb​  ​​​
​​r′​d ​​  = ​k2​  ​​​cp​  ​​​cc​  ​​​
where cp, cb, cc and cd are the molar concentrations (mol/L) of propylene, benzene, cumene, and diisopropylbenzene, respectively. The
rate constants are
​​k1​  ​​ = 2.8 × ​10​​  7​ ​e​​  −12,530∕T​​
​​k2​  ​​ = 2.3 × ​10​​  9​ ​e​​  −17,650∕T​​
where T is in K and the rate constants have units of L/gmol s.
Assume that benzene costs $0.22/kg and the 95% propylene/5%
propane mix costs $0.209/kg. The selling price of cumene is $0.46/kg.
Diisopropylbenzene is worth $0.20/kg. Assume that the maximum
allowable reactor pressure is 30.75 bar and that the benzene:propylene
molar ratio fed to the reactor is 1:1. Assume that all the other equipment
(distillation columns, pumps, vapor-liquid separator, etc.) can handle
any process changes.
Sketch out the process and do a DOF analysis. Analyze the performance of the reactor and the process with the new catalyst. Optimize
the process by looking at the influence of reactor temperature, recycle,
and purge on overall economics. Calculate the flow rates of all components in all streams. Calculate your earnings in $/day from considering
only the values of raw materials and products. Explore how the block
flow diagram and the economics would be affected by (a) changing the
benzene:propylene molar ratio or (b) replacing the propylene/propane
stream with a pure propylene source that costs $0.26/kg.
Write a brief report describing your results. Discuss the key issues
in optimizing the process design. Attach the final block flow diagram
(with flows shown). Document the work by attaching, as needed, calculations, tables, and/or graphs.
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6
CHAPTER SIX
Selection of Separation
Technologies and Synthesis of
Separation Flow Sheets
In This Chapter
We take a closer look at separations. Virtually all chemical processes require
some separation units. For example, the available and affordable raw materials
might be impure: the chemical process designer must develop methods for
removing contaminants in the raw materials so they can be further processed.
Or, the chemical reactor is imperfect: the designer must develop techniques for
removing undesired byproducts, recovering reactants for recycle, and purifying
the desired products to meet customer requirements. Usually, several separation
units must be put together to accomplish all these goals.
The questions you’ll be able to answer after finishing this chapter include:
∙ What are the major kinds of separation technologies used in chemical
processes?
∙ What criteria are used to select the best separation technology for a given
problem?
∙ How do we specify the performance of a separation unit?
∙ How do we best synthesize a flow sheet containing multiple separations?
Words to Learn
Watch for these words as you read Chapter 6.
Mechanical separations
Rate-based separations
Equilibrium-based separations
Separating agent
Product purity
Component recovery
Key component
Filtration
Centrifugation
Distillation
Crystallization
Extraction
Adsorption
Absorption
375
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Introduction
6.1
Separations are a major part of modern chemical processes, typically accounting for 50% or more of the total capital and operating cost. They provide a
remarkable variety of challenging problems for the process engineer to solve,
requiring both technical know-how and creative spirit. Fortunately, there are
rules that guide the designer in this task. This chapter will discuss some of
these rules for choosing appropriate separation methods, will demonstrate how
to evaluate the performance of some common separation technologies, and will
show how to generate reasonable separation flow sheets. We will often take
approximate approaches that get you close enough to the exact answer, and we
will liberally use heuristics to guide us.
6.1.1
Physical Property Differences: The Basis for All Separations
Let’s say you’ve just gone grocery shopping and are bagging your own groceries. To make for efficient unpacking, you might put all your frozen foods in
one bag, cleaning supplies in another, fruits and vegetables in a third, and
canned goods in a fourth. What you’ve done, of course, is taken the output
from your trip through the aisles and separated according to final use. You’ve
exploited differences in physical properties (appearance, temperature, container
type) to decide which product goes in which bag.
Similarly, in a chemical process, a mixture of compounds must be separated into product streams. This is accomplished by exploiting differences in
physical properties. An enormous diversity of separation methods are available
to choose from. The first step in choosing a separation technology is to gather
information about the physical properties of the components to be separated.
Then, we ask four questions:
∙
∙
∙
∙
How do the components to be separated differ in physical properties?
Is the difference between the components large?
Can the difference be feasibly exploited?
Do the components go to the correct output stream?
By answering these four questions, we can choose the best physical property
differences on which to base our design.
Example 6.1
Physical Property Differences: Separating Salt from Sugar
Go to your kitchen and get a tablespoon of common table salt (NaCl) and another
tablespoon of table sugar (sucrose, C12H22O11). Mix the salt and sugar together.
Now try to separate the mixture into pure salt and pure sugar.
Solution
You observe that salt and sugar have the following physical properties:
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Section 6.1 Introduction
Physical property
Salt
Sugar
Appearance
White, crystalline
White, crystalline
Size
A few micrometers
A few micrometers
Taste
Salty
Sweet
Meltability over a stove-top burner
Won’t melt
Melts
Solubility in water
Dissolves
Dissolves
Clearly, salt and sugar don’t differ much in terms of appearance or size. These
properties, then, would not provide a good basis for separating salt from sugar.
Appearance and size fail the “large-difference” test.
Salt and sugar differ significantly in taste. Could we design a process that
separates salt from sugar on the basis of taste? We would need some method of
sensing and discriminating salt from sweet and then of placing the salt or sugar
into the appropriate location. Taste as a basis for a large-scale separation process
fails the feasibility test.
Salt and sugar do differ in melting point. Since sugar melts at fairly low temperatures but salt does not, you could design a process that heats the salt/sugar
mixture to a low temperature and allows the sugary liquid to drain off.
What about dissolving the salt/sugar mixture in water? Both compounds are
soluble in water, but sugar much more so than salt. At certain concentrations, all
of the sugar but only some of the salt would dissolve. Salt crystals could then be
recovered from the sugar- and salt-containing solution by filtration. This process
would produce a pure salt stream, but not a pure sugar stream. Perhaps another
solvent could be found that dissolves salt much more than sugar.
Are there other physical or chemical property differences? Salt dissolved in
water splits into ions, but sugar does not. Salt lowers the freezing point of water
much more than sugar does. Sugar can be chemically degraded to CO2 and water,
whereas salt is fairly unreactive. Can you think of any other differences? Can you
think of practical ways to exploit these differences to achieve the desired separation?
6.1.2
Mixtures and Phases
In separation processes, multicomponent mixtures are separated into streams
of differing composition. The phase of the streams plays a large role in how
separation processes actually work. Therefore, to understand separation processes, you must understand phases and mixtures. Here we’ll review a few
concepts and definitions.
According to Webster’s New Collegiate Dictionary, a phase is a “homogeneous, physically distinct, and mechanically separable portion of matter.” (The
emphasis is ours.) Solids, liquids, and vapors are all phases. Supercritical fluids
and plasmas are examples of other phases that are less commonly encountered.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Quick Quiz 6.1
What are the major
components in each
of the two phases of
carbonated soda?
A phase is homogeneous. Within a phase, the chemical composition and physical properties (e.g., density) are uniform. A phase may be a single component, or a multicomponent mixture of chemical species, with the species
distributed uniformly at the molecular level. Sugar dissolved in water is a
multicomponent mixture but a single phase. Emulsions such as oil-andvinegar salad dressing, or bubbly liquids such as carbonated soda, or
suspensions such as muddy water, are not single phases, because the
components in the mixture are not completely and uniformly distributed
at a molecular level.
A phase is physically distinct. Vapors, liquids, and solids differ in some
fundamental ways. Vapors are much less dense than liquids or solids.
Vapors are highly compressible (i.e., their density changes a lot with
pressure), whereas liquids and solids are almost incompressible. This
means that the behavior of vapors is very sensitive to pressure, whereas
that of solids or liquids is relatively independent of pressure. Vapors
and liquids adopt the shape of their containers, whereas solids retain
their shape independent of their container.
Phases are mechanically separable. One phase can be separated from another
by using mechanical forces and mechanical devices. For example, a paper
filter can be used to separate solid coffee grounds from the liquid beverage.
Figure 6.1 shows several examples of systems that are multicomponent
and/or multiphase.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Figure 6.1 Examples of systems that are multicomponent and/or multiphase. See text for
further discussion. (a) Table salt completely dissolved in water is multicomponent and
single phase, (b) salt mixed with water at a concentration that exceeds the salt solubility
is multiphase, with a multicomponent liquid phase and a single-component solid phase.
(c) Ice cubes in water is an example of a single-component multi-phase system. (d) A pot
of boiling water contains a single-component liquid phase and a multicomponent (air and
water) vapor phase. (e) Turkey drippings separate into two multicomponent liquid phases,
with an oil layer on top and an aqueous phase below. (f) Carbonated soda is a multiphase
mixture of CO2 bubbles dispersed in a multicomponent liquid. (g) A bucket of seawater
and sand contains one solid phase as well as a multicomponent solution. (h) Gold particles
in an ore rock is an example of multiple solid phases.
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Section 6.2 Classification of Separation Technologies
Classification of Separation Technologies
6.2
Separation technologies can be divided into three categories based on their operating mechanism: mechanical, rate-based, and equilibrium-based. Table 6.1
summarizes the key differences between mechanical, rate-based, and equilibriumbased separation processes.
6.2.1
Mechanical Separations
In mechanical separation processes (Fig. 6.2), the feed contains two phases
(e.g., suspended solids in liquid, solid particles in gas, or two immiscible
Table 6.1
Technology
Classification of Separation Technologies
Input
Output
Basis for separation
Mechanical
Two phases Two phases Differences in size or
density
Rate-based
One phase One phase
Differences in rate of
transport through a medium
Equilibrium-based One phase Two phases Differences in composition
of two phases at equilibrium
Phase 1
Two phases
Phase 2
(a)
(b)
Figure 6.2 In a mechanical separator, a mixture of two (or more) phases is divided into
streams of different phases. Mechanical separations exploit differences in size or density. For
example, a centrifuge separates blood into a fraction containing blood cells and a fraction
containing plasma fluid. Photo: David Buffington/Stockbyte/Getty Images
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Table 6.2
Some Mechanical Separation Technologies
Physical
Feed
property
Technology
phases
difference
Filtration
Solid and fluid
Size
How it works
Mixture is pumped across
a porous barrier such
as a membrane; solids
are retained while most
fluid passes through
Sedimentation Solid and liquid
Density
Suspended solid particles
are partially separated
from liquid by gravity
settling
Flotation
Solid and liquid
Density
Two liquids
Two solids
Less-dense solid or
liquid droplets collect
and rise to surface
Expression
Solid and liquid
Size
Wet solids are compressed,
allowing liquid to escape
Centrifugation Solid and fluid
Density
Two liquids
Mixture is spun rapidly;
centrifugal force causes
denser phase or solids
to migrate outwards
fluids), and differences in size or density are exploited to separate the two
phases from each other. Some of these processes are energy-intensive, like
centrifugation; others are not, like sedimentation. See Table 6.2 for some examples of mechanical separation technologies.
Example 6.2
Mechanical Separations: Matching the Problem with the Technology
For the following separation problems, what mechanical separation technology
would you choose? Explain your reasoning.
(a) Removal of dirt from rainwater runoff
(b) Separation of fat from milk
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Section 6.2 Classification of Separation Technologies
381
(c) Recovery of sugar cane juice from chopped cane
(d) Removal of yeast from wine
(e) Separation of low-density from high-density polyethylene
Solution
(a) Sedimentation is an inexpensive and feasible choice for separating many solids
from liquids. Dirt particles will readily settle by gravity due to their density;
the large quantity of material that needs to be processed requires a very inexpensive process.
(b) Centrifugation is a good choice. Milk contains fats, sugars, proteins, salts,
and water. The fats form a separate liquid phase from the aqueous liquid
that contains most of the sugar, salt, and protein. The fat phase is less
dense than the aqueous liquid phase. Separation by sedimentation would
be very slow because of the relatively small difference in the densities of
the two fluids.
(c) Since the sugar cane juice is hidden in small pockets of liquid within the
porous solid cane, expression is the best choice.
(d) If solid yeast particles separate readily, then sedimentation might work.
Filtration might be a better choice, producing a better separation and therefore
a clearer wine.
(e) Mixed plastic waste can be separated using flotation by adding water: the
low-density polyethylene floats while the higher density polyethylene sinks.
6.2.2
Helpful Hint
Rate-based separations are sometimes
compared to shoppers in a mall: Some
shoppers make a
beeline to purchase
only one item and
exit quickly, while
other shoppers
sample every store
and spend all day.
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Rate-Based Separations
Rate-based separation technologies rely on differences in the rate of
transport through a medium of the components to be separated. Most often,
the medium is a porous solid, and the feed and product streams are all the
same phase. If you’ve done any thin-layer chromatography or gel electrophoresis experiments in a chemistry or biochemistry laboratory, then you
are familiar with rate-based processes. The compounds are applied to one
side of the paper or gel. They move through the paper or the gel at different rates, and can be collected one at a time as they emerge from the
opposite side. If you waited forever, all the compounds would move all the
way across the paper or gel, and there would be no separation. Rate-based
separation technologies are very important on an analytical scale; commercially they are most commonly employed in the biotechnology industry
where very high purities are required, in water desalination, and in isotope
concentration. A few examples are listed in Table 6.3 and illustrated in
Fig. 6.3.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Table 6.3
A Few Rate-Based Separation Technologies
Physical
property
Technology
Feed
difference
How it works
SizeMacromolecules
Size
exclusion
dissolved in chromatography solvent
Mixture is injected onto a column containing
porous beads, then solvent is pumped
continuously over column; larger molecules
can’t enter the pores and elute quickly while
smaller molecules enter the pores and take
longer to exit from the column
Microfiltration/ Solutes dissolved
Size
Ultrafiltration
in solvent
Solution is pumped at high pressure across
a membrane with micron- to nanometer-sized
pores. Some of the solvent (e.g., water)
passes through the membrane, but all the
solutes are rejected
Reverse
Solutes dissolved
Size
osmosis
in solvent
Solution is pumped at high pressure across a
membrane with very small pores. Some of
the solvent (e.g., water) passes through the
membrane, but all the solutes are rejected
Gel
Macromolecules
Size,
electrodissolved in
charge
phoresis
solvent
Mixture is injected onto the top of a thin gel
slab or tube which is placed in an electric
field; molecules migrate at different velocities
through the gel
Products eluted
at different times
Mixture
injected
(a)
(b)
Figure 6.3 In a rate-based separation, components in a mixture travel at different rates through a medium. For
example, in gel electrophoresis, protein or DNA fragments migrate through a porous polymer gel in response to an
applied electric field, with the smallest molecules travelling the fastest. Photo: Auburn University Photographic Services/
McGraw Hill
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Section 6.2 Classification of Separation Technologies
Example 6.3
383
Rate-Based Separations: Fresh Water from the Sea
Seawater is a solution containing about 35 g NaCl per 1000 g H2O. In arid parts
of the world, seawater is processed into freshwater by separating salt from water
using a reverse osmosis membrane. These membranes work because the water
passes more freely through the membrane than does dissolved salt.
Seawater in
Brine out
Drinking water out
As shown in the schematic, the membrane (diagonal dashes) is placed in a holder.
Seawater is fed into the holder. Both water (thick arrow) and salt (thin arrow) pass
through the membrane, but the flux (flow rate per area of membrane surface) of
water through the membrane is much higher than the flux of salt.
The manufacturer of a commercially available membrane reports that the flux
of water through the membrane is 9.5 g/s-m2 and that of salt is 3.2 × 10−3 g/s-m2.
If drinking water can contain no more than 0.4 g NaCl per 1000 g water, will this
membrane be good enough to process seawater into drinking water? If the seawater flow rate into the device is 50 g/s and the membrane surface area is 2.0 m2,
what is the flow rate of drinking water out, and what is the concentration of salt
(g NaCl/g H2O) in the brine?
Solution
The device is the system, there are two components, water (W ) and salt (S), one
input (in) and two outputs, drinking water (d ) and brine (b). The flow rate through
the membrane equals the flux times the membrane area. Therefore,
9.5 g
​​​ṁ ​​Wd​​ = ​ _
​
​ (​ 2.0 ​m​​  2​) = 19 g/s​
( s-​m​​  2​)
0.0032 g
​​​ṁ ​​Sd​​ = ​ _
​
​(​ 2.0 ​m​​  2​) = 0.0064 g/s​
( s-​m​​  2​ )
The ratio of NaCl to H2O in the drinking water is the ratio of the flow rates, or
0.0064 g NaCl/s _____________
0.00034 g NaCl __________
0.34 g NaCl
​​m ​​̇ ​​ ______________
_
​​  Sd  ​ =
​
​ = ​
​ = ​
​​
​​m ​​̇ Wd​​
19 g ​H2​ ​O/s
g ​H2​ ​O
1000 g ​H2​ ​O
The salt content of the drinking water falls below the maximum allowable, so the
membrane will be acceptable. The drinking water flow rate is just over 19 g/s.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
The mass ratio of salt: water in seawater is 35 g NaCl/1000 g H2O. The weight
fraction of salt in the seawater is:
​​wS​  ,in​​ = _
​  35  ​= 0.0338​
1035
Since the seawater feed rate is 50 g/s,
​​​ṁ ​​S,in​​ = ​wS​  ,in​​ ​​ṁ ​​in​​= 1.69 g/s​
The steady-state material balance equation for salt over the entire device is:
​​​ṁ ​​Sd​​ + m
​​ ̇ ​​Sb​​ = ​​ṁ ​​S,in​​​
Rearranging and inserting known values, we get
​​​ṁ ​​Sb​​ = ​​ṁ ​​S,in​​ − ​​ṁ ​​Sd​​= 1.69 − 0.0064 = 1.684 g/s​
The material balance equation for water is:
​​​ṁ ​​Wd​​ + ​​ṁ ​​Wb​​ = ​​ṁ ​​W,in​​​
Or
​​​ṁ ​​Wb​​ = m
​​ ̇ ​​W,in​​ − ​​ṁ ​​Wd​​= (1 − 0.0338)50 − 19 = 29.31 g/s​
The ratio of salt to water in the brine is the same as the flow rates of the two
components:
0.0575 g NaCl
1.684 ​ =
​​​m ​​̇ Sb​​∕​​m ​​̇ Wb​​ = ​ _
​ ____________
​​
29.31
g ​H2​ ​O
The mass fraction of salt in the brine is
​​wS​  b​​ = _
​  0.0575  ​= 0.0544​
1 + 0.0575
or about 60% saltier than the seawater.
6.2.3
Equilibrium-Based Separations
In equilibrium-based separation processes, the feed is a multicomponent mixture but a single phase (e.g., solid, liquid, or gas). Within the process, a second
phase is generated. The compositions of the two phases are different. The two
phases are the two products. Generation of the second phase does not happen
spontaneously. Rather, it requires an input of a separating agent. The separating agent can be energy, or it can be an added material. Intelligent choice of
the separating agent is crucial for the success of equilibrium-based separations.
Many of the most popular large-scale equilibrium-based separations add or
remove energy (e.g., heat or cool) to produce a change in temperature and
generate a second phase. These are cases of equilibrium-based separations that
use an energy separating agent (Fig. 6.4). Some of these technologies are listed
in Table 6.4.
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Section 6.2 Classification of Separation Technologies
Table 6.4
Some Equilibrium-Based Separation Technologies That Use Energy as a
Separating Agent
Physical
Product
property
Technology
Feed phase
phases
difference
How it works
Evaporation
Liquid
Liquid and Vapor pressure
vapor
(boiling point)
Liquid mixture is heated until
some of the material vaporizes
Condensation
Vapor
Liquid and Vapor pressure
vapor
(boiling point)
Vapor mixture is cooled until
some of the material condenses
Liquid or
Liquid and Vapor pressure
vapor
vapor
(boiling point)
Mixture is fed into a multistage
column, where repeated
evaporation and condensation
occurs
Crystallization Liquid
Solid and
liquid
Solution is cooled until
solubility limit is exceeded
Distillation
Drying
Solubility at low
temperatures
(melting point)
Solution or
Solid and Vapor pressure
suspension
vapor
Energy separating agent
Feed is heated to volatilize
solvent, leaving behind
nonvolatile solid
Phase 1
Single phase
Heating or cooling
Material separating agent
Phase 2
Phase 1
Single phase
Added material
Phase 2
(a)
(b)
Figure 6.4 In an equilibrium-based separation, a separating agent, which could be either energy or a material, is
used to convert a single-phase feed into two phases that differ in composition. For example, in distillation, energy is
used to partially vaporize hydrocarbons and separate them based on differences in volatility. Photo: Reed Kaestner/Corbis/
Getty Images
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Table 6.5
Some Equilibrium-Based Separation Technologies That Use an Added
Material as a Separating Agent
Separating
Feed
agent
Technology phase
phase
Physical
property
difference
How it works
Absorption
Gas
Liquid
Gas solubility
in added solvent
Gas mixture is contacted with solvent;
one of the components in the gas
is more soluble in the solvent
Adsorption
Fluid is contacted with a solid
material; one of the components in
the mixture sticks to the solid
Fluid
Solid
Affinity for
(gas or
solid surface
liquid)
Leaching
Solid
Liquid
Solubility of solid
components in
added solvent
Solid contains both soluble and
insoluble components; soluble
components dissolve in added
solvent
Extraction
Liquid Liquid
Distribution
between two
immiscible fluids
An immiscible solvent is contacted
with the feed; solute in the fluid
preferentially partitions into the
added solvent
Sometimes, changing the temperature is not feasible. For example, some
materials chemically decompose before they are sufficiently hot to vaporize,
so distillation is not an option. Or materials will condense only at exorbitantly
cold temperatures. In these cases, the separating agent is an added material:
The necessary second phase is generated by adding a material that is a different phase from the feed. Figure 6.4 illustrates the difference between an energy
separating agent and a material separating agent. While a material separating
agent might be used to solve one separation problem, a second separation is
usually needed to recover the material separating agent for re-use. Table 6.5
lists some of the most important equilibrium-based separation technologies that
work in this manner.
Illustration: In Chapter 5’s Case Study, you saw the important role of the
water-gas shift reaction in processing organic feedstocks (biomass or fossil
fuels) into desired products. When CO and H2O react, they produce a gas mix
of CO2 and H2. Separation of H2 from CO2 uses two equilibrium-based separations in sequence: absorption and stripping. Absorption of the acidic CO2 into
a basic liquid solution [such as an aqueous solution of diethanolamine (DEA)]
is an excellent way of separating CO2 from H2, because H2 does not dissolve
in DEA. DEA is an example of a material separating agent. The dissolved
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Section 6.2 Classification of Separation Technologies
CO2 is then stripped from the DEA by heating it, thus removing the CO2 and
recovering the DEA for recycle.
H2 gas
CO2 and H2 gas
Absorber
CO2 in DEA
liquid
CO2 gas
Stripper
DEA liquid
6.2.4
Heuristics for Selecting Separation Technologies
Given the plethora of possible technologies available for separation, it is useful
to have a few heuristics to guide initial selection of feasible technologies.
Heuristics are guidelines, not hard-and-fast rules. Experienced engineers use heuristics wisely, to eliminate clearly unworkable schemes, and to quickly generate
a few reasonable choices that can then be evaluated in more detail. Some useful
heuristics are:
1. If the feed is already two phases, use a mechanical separation technology.
2. If the feed is a single phase, first consider equilibrium-based separation
technologies, especially for products manufactured in large quantities.
3. Consider rate-based separation technologies for small-volume, high-valueadded products that demand high purity.
4. For equilibrium-based separations, consider differences in (a) boiling point,
(b) melting point, (c) solubility in common solvents, and (d) binding to solid
surfaces, in that order. Differences of 10°C or less in boiling point can be
effectively exploited. Larger differences in melting point, solubility, or binding
are usually necessary.
5. Operate at temperatures and pressures as close to ambient as possible, but
prefer temperatures and pressures above ambient rather than below.
6. Avoid adding foreign materials if possible. If a foreign material is added,
avoid toxic or hazardous materials and remove it as soon as possible.
7. For recovery of trace quantities, use separation methods where the cost
increases with the quantity of material to be recovered, not the quantity of
the stream to be processed.
8. For removal of small quantities of contaminants that do not need to be
recovered, consider using destructive chemical reactions rather than physical
separations.
These ideas are illustrated in a few examples. The rationale behind many of
these heuristics will become clearer as we delve further into design and analysis
of separation processes.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Example 6.4
Selection of Separation Technology: Separating Benzene from Toluene
Benzene is a useful feedstock for a number of chemical processes. Benzene is
purified from a refinery stream that contains a closely related compound, toluene.
The CRC Handbook of Chemistry and Physics has this information regarding benzene and toluene:
Boiling point,
°C
Benzene
80.1
Toluene
110.6
Melting point,
°C
5.5
Soluble in
Ethanol, diethyl ether, acetone
−95Ethanol, diethyl ether, acetone,
benzene
What might be a good method for separating a liquid mixture of 50% benzene
and 50% toluene into two pure products?
CH3
Benzene (C6H6)
Toluene (C7H8)
Solution
The feed is a single liquid phase, and benzene is a high-volume, relatively low-value
product. Therefore, by heuristics (1), (2), and (3), an equilibrium-based separation
process is the best. Benzene and toluene differ in a number of ways. Boiling points
and melting points are both quite distinct, so either could be exploited (heuristic
(4)). Heuristic (5) would favor exploiting differences in boiling point, to keep the
process just above ambient conditions. There is no need to go to a process requiring a solvent (indeed, these chemically-similar compounds are soluble in the same
solvents), so no foreign material is added (heuristic (6)). Benzene concentration is
not low, so heuristic (7) does not apply. Distillation is the separation method of
choice, because it is an equilibrium-based separation technology that exploits
differences in vapor pressure, which are related to differences in boiling point.
Example 6.5
Selection of Separation Technology: Removing Viruses from
Engineered Antibodies
The ability to produce engineered proteins has revolutionized the biotechnology
and pharmaceutical industry; such proteins show up in everything from laundry
detergent to cheese to some of the world’s most advanced medicinal drugs.
Monoclonal antibodies (MAbs) are large proteins that are engineered for use to
treat specific cancers and autoimmune disorders. MAbs are produced by genetically
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Section 6.2 Classification of Separation Technologies
389
modified cells that are grown in bioreactors; the cells produce and then secrete
MAbs into the cell culture fluid. Once a production run is finished, the fluid is
collected and the MAbs purified. Contamination of the fluid with viruses is possible, and the viruses must be separated from the MAbs before the product can be
sold. What separation technology is the best choice for separating virus from
MAbs?
Solution
MAbs are proteins, and viruses are genetic material (RNA or DNA) with a protein
coat. Proteins are complex molecules that have both positive and negative charges.
Viruses and MAbs are both soluble in aqueous solutions, and both are destroyed
at high temperature and in most organic solvents. These facts rule out separation
technologies such as distillation or solvent extraction. An equilibrium-based separation technology that involves difference in binding to solid surfaces might be
attractive (heuristic (4)). The concentration of MAb and especially of virus in the
fluid is low: MAb are typically 1–10 g/L and virus is much lower than that. The
quantity of material to be purified is relatively low, the product is very valuable,
and the purity must be extremely high. These considerations suggest a rate-based
separation technology (heuristic (3)).
Some physicochemical properties of virus and MAbs are summarized in the
table
Size
Isoelectric point
Adsorption to
protein A?
MAb
10 nanometers
8–9 typically
yes
Virus
20–400 nanometers
6–7 typically
mostly no
The difference in size suggests the use of a rate-based separation process such
as ultrafiltration, where a membrane with pores of ~20 nm diameter would allow
the MAb through but reject most of the virus particles.
The isoelectric point is the pH at which the material has no net charge. At a
pH of 7, MAb are net negatively charged while virus are neutral or net positive
charge. The difference in charge suggests an adsorption process using positively
charged beads: MAb will adsorb (stick) to the beads while the virus flows through.
Then, increasing the pH to 8 or 9 would neutralize the charge on the MAb and
desorb the protein as a pure product from the beads.
MAbs are known to interact very strongly with a protein called Protein A,
whereas viruses usually don’t interact with Protein A. Is there a way to take advantage of this difference? Protein A can be chemically conjugated to a solid material,
and then the fluid containing MAb and viruses can be put in contact with the solid
material. MAb would adsorb while the virus did not. This is another example of
adsorption separation, but using a different kind of adsorber. It was discovered that
dropping the pH to 3 breaks the interaction between MAb and Protein A, providing
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an easy method for desorbing the MAb. As an added bonus, an acidic pH also
destroys any virus! This is an application of heuristic (8).
It is critically important to remove every trace of virus from an MAb that will
be injected into patients. In practice, the biomanufacturer combines two or more
of these different separation technologies to ensure the safety of the product.
6.2.5
Heuristics for Sequencing Separations
What do we do if we have a multicomponent mixture that must be separated
into three, four, or more products? If there are N products desired, then there
may be as many as N − 1 separation units. The designer has to choose not
only the best technology for each individual separator, but also the best
sequence in which to place the separators. This is a difficult task, but a few
heuristics help make it easier. The underlying basis of many of the heuristics
is simple: Separation costs increase as the volume of material to be processed
increases, and as the two components to be separated become more similar to
each other. Other heuristics come from the need to economize on energy
utilization.
Here are some useful, simple heuristics.
1.
2.
3.
4.
5.
6.
Remove hazardous, corrosive materials early.
Separate out the components present in the greatest quantity first.
Save difficult separations for last.
Divide streams into equal parts.
Avoid recombining components that have been separated.
Meet all product specifications, but do not overpurify.
As you develop flow diagrams requiring multiple separation units, keep these
simple rules of thumb in mind.
Example 6.6
Sequencing of Separation Technologies: Aromatics and Acid
Given the process stream described below, devise what you think is the best scheme
for separating it into three essentially pure product streams (toluene, m-xylene, and
p-xylene), and a waste sulfuric acid stream. Indicate the types of separation technologies you would use and the sequence of separation steps.
CH3
CH3
Toluene
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CH3
CH3
p-xylene
O
CH3
m-xylene
HO
S
O
O
H
Sulfuric acid
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Section 6.2 Classification of Separation Technologies
mol% in
feed
Boiling point
(°C)
Toluene
51
110.6
p-xylene
25
138.4
m-xylene
24
139.1
Component
Sulfuric acid
Trace
330
Melting point
(°C)
Soluble in
water?
Soluble in
benzene?
No
Yes
13.2
No
Yes
−47.2
No
Yes
10.5
Yes
No
−95
Solution
Quick Quiz 6.2
In Example 6.6 we
used water as a solvent
to remove sulfuric
acid. Why not use
benzene as a solvent
to remove the
aromatics instead?
First, consider the physical properties. All compounds are liquids at room temperature. Sulfuric acid is chemically quite distinct, differing tremendously from the
other components in boiling point and in solubility. It is also corrosive. Toluene
and the xylenes are all chemically similar. Toluene differs from the xylenes in
boiling and melting points. The boiling point is above room temperature whereas
the melting point is below; by heuristic (5), separations based on boiling point
differences are preferable. This leads to distillation as the technology of choice for
producing the toluene product. The xylene isomers differ very little in boiling point
or solubility, but differ significantly in melting point. We probably have no choice
but to take advantage of the large melting point difference, even though it requires
slightly colder-than-ambient temperatures. Thus, crystallization is the method of
choice for separation of one xylene from the other. Since sulfuric acid is present
in trace quantities only, separation that scales with the amount of acid and not the
amount of the other streams makes sense. We can use liquid-liquid extraction with
water, since the sulfuric acid is soluble in water, but water and toluene/xylene are
mutually insoluble.
Second, consider sequencing. Sulfuric acid is hazardous, so it’s best to remove
that early. Toluene is present in the largest quantity, and a separation between
toluene and the xylenes leads to a pretty even split. It makes sense to complete
this step next. The separation of p-xylene from m-xylene is the most difficult, so this
should come last.
The proposed separation scheme is shown.
Toluene
Extraction
Water
Sulfuric acid
Toluene
Xylenes
Distillation
Feed
Toluene
Xylenes
Sulfuric acid
Water
m-xylene
m-xylene
p-xylene
Crystallization
p-xylene
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6.3
Separator Material Balance Equations
At their simplest, separation units take in one feed and produce two products
that differ in composition. You already have some experience in using material
balances around separation units and in calculating process flows. Our primary
goal in this section is to review aspects of material balance equations that are
particularly important for analysis of separators.
Since by our definition there is no chemical reaction in a separation unit,
the choice of mass versus mole units is simply one of convenience. The choice
is typically based on whether stream composition specifications and/or any
physical property information are given in mass or molar units.
Since the purpose of separation units is to produce two (or more) products
that differ in composition, their performance is often described in terms of the
compositions of the feed and products. Therefore it is often more convenient
to write material balance equations using mole or mass fraction of that component times the total molar or mass flow rate. As a reminder, we will generally use the convention that
zij = mole fraction of i in stream j
wij = mass fraction of i in stream j
When we wish to indicate the phase of a stream, we will use yi, xi, or xiS instead
of zi, for the mole fraction of i in vapor, liquid, or solid streams, respectively.
The molar flow rate of component i in the input stream to a separator is simply
the mole fraction times the total molar flow rate
​​​n ​​i̇ ,in​​ = ​zi​  ,in​​ ​​n ​​i̇ n​​​
Similarly, the mass flow rate of component i in the input stream is the mass
fraction times the total mass flow rate:
​​​ṁ ​​i,in​​ = ​wi​  ,in​​ ​​ṁ ​​in​​​
In their simplest form, separators have a single input stream and two output (or product) streams which differ in composition. Separators may operate
as continuous-flow steady-state process units, or in batch or semibatch modes
(Fig. 6.5). The most appropriate choice of material balance equation depends
on the mode of operation.
6.3.1
Continuous-Flow Steady-State Separators
Continuous-flow steady-state separators are the workhorses of commodity
chemical plants. For a steady-state continuous-flow separator, the differential
form of the material balance equation is the most useful. The differential component mole balance equation for steady-state continuous-flow separators with
a single input and two outputs, 1 and 2, simplifies to
​​​n ​​i̇ 1​​ + ​​n ​​i̇ 2​​ = ​​n ​​i̇ ,in​​​
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Section 6.3 Separator Material Balance Equations
393
Product 1
Feed
Product 2
(a)
Product 1
Feed
Product 2
t0
tf
(b)
Product 1
Feed
Product 2
t 0 < t < tf
t0
tf
(c)
Figure 6.5 Separators may operate in (a) continuous-flow (b) batch, or (c) semibatch modes.
There are several other semibatch modes not shown.
or, written in terms of mole fractions and total molar flows
​​z​ i1​​ ​​n ​​1̇ ​​ + ​zi​  2​​ ​​n ​​2̇ ​​ = ​zi​  ,in​​ ​​n ​​i̇ n​​​
Generalizing to a single input (feed) stream and any number of output (product)
streams, the material balance equation becomes
​​  ∑ ​​​ ​zi​  j​​ ​​n ​​j̇ ​​ = ​zi​  ,in​​ ​​n ​​i̇ n​​​
all j​​out​
Eq. (6.1a)
In mass units, the material balance equation for an arbitrary number of output
streams is
​​  ∑ ​​​ ​wi​  j​​ ​​ṁ ​​j​​ = ​wi​  ,in​​ ​​ṁ ​​in​​​
all j​​out​
Eq. (6.1b)
Of course, for any separation requiring a material separating agent, Eq. (6.1)
must be modified to include the additional input stream.
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Example 6.7
Continuous-Flow Steady State Separators: CO2 Removal from Flue Gas
Organic fuels, such as natural gas, coal, or even municipal solid waste, are combusted in boilers to generate electricity at power plants. The combustion gas (also
called flue gas) contains N2, O2, H2O, and CO2 and is released to the atmosphere
through the furnace stack (see Example 4.3). One idea for reducing CO2 in the
atmosphere is to place a unit in the furnace stack that continuously separates the
CO2 from the rest of the flue gas. Concentrated CO2 could be used, for example,
in cement manufacturing or in making synthetic fuels. The remaining stream, now
depleted in CO2, could be released to the atmosphere.
The flue gas leaving a boiler has a flow rate of 1260 kgmol/h and contains
75.6 mol% N2, 11.8 mol% O2, 8.4 mol% H2O, and 4.2 mol% CO2. You hope to
develop a separator that produces a concentrated CO2 stream that contains at least
60 mol% CO2, while releasing flue gas that contains only 0.04 mol% CO2 (about
the current atmospheric CO2 content). Find the flow rates of the concentrated CO2
stream and of the flue gas.
Solution
The separator placed in the flue stack is our system. There are one feed stream
and two product streams. We will call flue gas feed stream F, the CO2-rich product stream 1 and the CO2-depleted flue gas stream 2. There are four components,
but we are interested only in CO2 (C) now.
CO2-depleted
f lue gas
2
Separator
1
CO2-rich
product
F
Flue gas
Since this is a continuous-flow steady-state separation, the material balance equation on CO2 simplifies to
​​zC​  1​​ ​​n ​​1̇ ​​ + ​zC​  2​​ ​​n ​​2̇ ​​ = ​zC​  F​​ ​​n ​​Ḟ ​​​
We know the mole fraction CO2 in all three streams, as well as the inlet flow rate.
Inserting these numerical values into the material balance equation yields:
​0.60​​n ​​1̇ ​​ + 0.0004​​n ​​2̇ ​​= (0.042)(1260 kgmol/h)​
There are two unknowns, so we add a material balance equation on the total flows
​​​n ​​1̇ ​​ + ​​n ​​2̇ ​​= 1260 kgmol/h​
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Section 6.3 Separator Material Balance Equations
395
We solve this system of equations to find
​​​n ​​1̇ ​​= 87 kgmol/h​
​​​n ​​2̇ ​​= 1173 kgmol/h​
6.3.2
Batch Separators
Batch separators are commonly used at laboratory scale, and are sometimes
used for manufacture of low-volume products especially when condensed
phases (liquid and solid, or two liquid phases such as oil and water) are
involved. They are only infrequently used at large commodity-chemical scale.
Mechanical separations such as sedimentation or flotation, and some equilibriumbased separations such as crystallization, adsorption, leaching, or extraction,
can be adapted to batch mode.
In a batch separator, as shown in Fig. 6.5(b), the mixture is placed in the
separator all at once (at t = 0), the separation is allowed to occur, and then the
products are removed all together at a later time (at t = tf). Since we are interested in a specified time interval, and no material enters or leaves the separator between t = 0 and t = tf, the integral material balance (Table 3.2) would
seem like a good choice. Using mole fractions and total moles, the integral
balance on component i is:
​​z​ i,sys, f​​ ​ns​  ys, f​​ = ​zi​  ,sys,0​​ ​ns​  ys,0​​​
But this form is unhelpful, because it does not distinguish between the two
different products (two phases) that are left in the separator at t = tf.
A more useful form of the integral balance accounts for the two products,
which we will do by using subscripts P1 and P2 for the system at t = tf :
​​z​ i,sys, f​​ ​ns​  ys, f​​ = ​zi​  ,P1​​ ​nP​  1​​ + ​zi​  ,P2​​ ​nP​  2​​ = ​zi​  ,sys,0​​ ​ns​  ys,0​​​
Eq. (6.2a)
where
​​n​ sys, f​​ = ​nP​  1​​ + n​ P​  2​​ = ​ns​  ys,0​​​
Keep in mind that
​​z​ i,sys, f​​ ≠ ​zi​  ,P1​​ ≠ ​zi​  ,P2​​​
In mass units, the integral material balance equation on component i for a batch
separator is
​​w​ i,sys, f​​ ​ms​  ys, f​​ = ​wi​  ,P1​​ ​mP​  1​​ + ​wi​  ,P2​​ ​mP​  2​​ = w
​ i​  ,sys,0​​ ​ms​  ys,0​​​
Eq. (6.2b)
with
​​m​ sys, f​​ = ​mP​  1​​ + m
​ P​  2​​ = ​ms​  ys,0​​​
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Example 6.8
Batch Separators: Caffeine from Coffee Beans
Green (unroasted) coffee Arabica beans contain 35 grams caffeine per kg dry
beans. The caffeine can be leached from the beans using various solvents. The
decaffeinated beans are roasted and used to make a more relaxing cup of coffee,
while the caffeine is separated from the solvent and then mixed into “energy”
drinks.
Jumpin’ Java is a local café which would like to be able to produce its own
decaffeinated beans on site. The owner wishes to build a batch separator that can
handle 20 kg beans and that leaches 97% of the caffeine from the beans. Ethyl
acetate, a compound present in fruits such as bananas and apples, was chosen as
the solvent for caffeine. If the maximum amount of caffeine dissolved in ethyl
acetate is 1.0 gmol caffeine/100.0 gmol ethyl acetate, how much ethyl acetate
should be added to each batch of 20 kg beans?
Solution
As usual, we start by sketching a diagram of the system. Initially, 20 kg green
coffee beans as well as an unknown amount of ethyl acetate are contacted in a
container. When the caffeine has been extracted, the decaffeinated beans are
allowed to settle to the bottom of the container, and the caffeine-containing solvent
can be easily separated from the beans.
t=0
t = tf
We will use three components: the caffeine C, the green coffee beans B
(a composite material, including everything except the caffeine), and the ethyl
acetate solvent S.
Now let’s write the integral balance equation on caffeine, considering the
caffeine-containing solvent as P1 and the decaffeinated beans as P2. We will work
in mass units:
​​wC​  ,P1​​ ​mP​  1​​ + ​wC​  ,P2​​ ​mP​  2​​ = ​wC​  ,sys,0​​ ​ms​  ys,0​​​
The initial charge of caffeine comes entirely from the beans, or
35 g caffeine
​​ ___________
​
​ ​​(20,000 g beans)​= 700 g caffeine = m
​ C​  ,sys,0​​ = w
​ C​  ,sys,0​​ ​ms​  ys,0​​​
( 1000 g beans )
Since 97% of the caffeine must be removed from the beans, the final mass of caffeine
in the decaffeinated bean product is
​(1 − 0.97​)​(700 g caffeine) = 21 g caffeine = m
​ C​  ,P2​​ = ​wC​  ,P2​​ ​mP​  2​​​
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Section 6.3 Separator Material Balance Equations
Now we need to determine the caffeine content in the liquid product P1. First we
convert the solubility of caffeine in ethyl acetate from moles to mass:
1 gmol caffeine
194 g caffeine ________________
gmol ethyl acetate 0.022 g caffeine
__________________
​​     ​ × ____________
​
​ ×
​     ​ = ______________
​     ​​
100 gmol ethyl acetate gmol caffeine
88 g ethyl acetate
g ethyl acetate
The maximum weight fraction of caffeine in the liquid phase at the end of the
process is
0.022 g C
​​wC​  ,P1​​ = _______________
​
​= 0.0215​
1 g S + 0.022 g C
Plugging all known values into the integral material balance, we find:
Solving,
​(0.0215) ​mP​  1​​+ 21 = 700​
​​mP​  1​​= 31,580 g​
Product P1 is the dissolved caffeine plus ethyl acetate, so
​​m​ S,P1​​= (1 − 0.0215​)​(31,580 g) = 30,900 g ethyl acetate​
30.9 kg ethyl acetate must be added to 20 kg green coffee beans in order to achieve
the desired separation.
6.3.3
Semibatch Separators
Semibatch separators combine features of both batch and continuous-flow
separators. They might find utility, for example, for small-throughput processes,
or where one of the components to be separated is present in very low quantities. One example of a semibatch operation is particle filtration: Fluid with
suspended particles is pumped continuously across a filter at a steady rate, but
particles accumulate on the filter. Eventually the system must be shut down
and the particles cleaned out. A laboratory distillation apparatus is another
example of operation in semibatch mode; a larger-scale version of this is common in the pharmaceutical industry. In this apparatus, a vessel is filled initially
with a multicomponent liquid mixture, and heat is applied. As the temperature
slowly increases, the more volatile components evaporate. Vapors rising from
the liquid surface are continuously removed overhead, where they are condensed and collected in another vessel. The volume in the vessel decreases with
time, and nonvolatile materials become concentrated in the vessel.
Semibatch separators, by their nature, are unsteady-state operations. The
material balance equation must include terms for flows in and/or out of the
system, as well as accumulation inside the system. Thus, analysis of semibatch
separators is generally more challenging than analysis of continuous-flow
steady-state or batch separators. Whether the differential or integral form of
the material balance equation is more useful for semibatch separators depends
on the specific problem to be addressed. For analysis at a single point in time,
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
the differential balance is best. The general form of the differential mole balance equation on component i for a semibatch separator is
d​ni​  ,sys​​
​​ _​ = ​  ∑ ​​​ ​​n ​​i̇ j​​ − ​  ∑ ​​​ ​​n ​​i̇ j​​​
dt
all j​​in​
all j​​out​
Eq. (6.3a)
For analysis of system behavior averaged over a specified time interval, the
integral balance is best. Considering the possibility for the presence of multiple
products (phases) inside the system, a general form of the integral mole balance
equation on component i for semibatch separators is:
​​  ∑ ​​​ ​ni​  P,sys, f​​ − ​ni​  ,sys,0​​ = ​  ∑ ​​​ ​ni​  j​​ − ​  ∑ ​​​ ​ni​  j​​​
all P
all j​​in​
all j​​out​
Eq. (6.3b)
Recall from Chap. 3 that the total amount of material crossing a system boundary can be determined by integrating the flow rate over time. Each term on the
right-hand side of Eq. 6.3b can therefore be expressed as:
​tf​  ​​
​​n​ ij​​ = ​∫ ​   ​​ ​​n ​​i̇ j​​​ dt
​t​0​
Of course, similar equations in mass units, or with mole or mass fractions
and total quantities, can also be used.
Example 6.9
Semibatch Mechanical Separation: Filtration of Beer Solids
Raw beer (density = 1.04 g/mL) contains 0.5 wt% solids. The solids must be
removed before the beer is bottled. Filtration is chosen as the separation technology.
Raw beer at 800 L/h is filtered through a basket filter. The process must be shut
down and the basket cleaned out after 100 kg of solids are deposited in the filter.
What is the rate of deposition of solids in the basket? How long can the filtration
system run until the basket must be cleaned out?
Solution
This is a mechanical separation. The feed contains two phases, liquid and solid,
and filtration separates the two phases into a liquid product and a solids by-product.
The system is the basket filter, in which the filtered solids accumulate over time.
The flow diagram is shown.
Raw beer
0.5 wt% solids
Filtered solids
Filtered beer
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Section 6.3 Separator Material Balance Equations
399
We’ll choose two composite materials—beer (B) and solids (S)—as our components. We need to convert from a volumetric to a mass flow rate. Assuming that
the density of both raw and filtered beer is the same, we calculate:
1.04 g _
1 kg
kg
1000  ​
mL × _
L ​ × ​ _
​​​ṁ ​​in​​ = 800 ​ _
​
​ × ​
​ = 832 ​ _ ​​
L
mL
1000 g
h
h
kg
kg
​​​ṁ ​​S,in​​ = 832 ​ _ ​× 0.005 = 4.2 ​ _ ​​
h
h
This is a semibatch operation. The liquid beer is pumped continuously through
the filter, but the solids are pumped in and then accumulate, with maximum accumulation set at 100 kg solids. To determine over what interval of time this quantity
of solids accumulates, we use the integral material balance equation.
​tf​  ​​
​​mS​  ,sys, f​​ − ​mS​  ,sys,0​​ = ​  ∫​  ​​​ṁ ​​S,in ​​ dt​​
​t0​  ​​
​tf​  ​​
kg
​100 kg = ​ ∫​  ​4.2 ​ _ ​ dt​​
h
0
​​tf​  ​​= 24 h​
The filter should be shut down for cleaning about once per day.
Example 6.10
Rate-Based Separation: Membranes for Kidney Dialysis
Our kidneys are separation devices, removing urea and other waste products from
blood. Patients with kidney failure must go on dialysis, unless and until their
kidneys can be repaired or replaced. In dialysis, membranes are used that allow
passage of some low-molecular-weight solutes (like urea) from the patient’s
blood into fluid that can be thrown away, but do not allow passage of highmolecular-weight solutes (like proteins) that should stay with the patient. Dialysis
is a rate-based separation process, because in practice both urea and proteins pass
across the membrane, but the rate of urea passage is much greater than that of
proteins.
Your job is to evaluate the performance of several dozen membranes provided
by various manufacturers, for possible use in a new kidney dialysis machine that
your company is manufacturing. You quickly build a small test device, consisting
of a stirred tank with a membrane holder to hold the test membrane. Above the
membrane flows dialysis fluid. The system is designed so the volume of fluid in
the stirred tank below the membrane does not change. The experiment is simple:
You load a sample of urea-containing plasma (blood with the cells removed and
anti-coagulant added) into the small tank, insert the test membrane in the holder,
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
pump dialysis fluid across the membrane, then measure the concentration of urea
in the stirred tank over time.
Urea
Dialysis f luid
Dialysis f luid
Test membrane
Stirred tank
with plasma
The design goal is for the urea concentration in the plasma sample to drop
to 3% of its original concentration in 3 hours. However, you’ve got a lot of membranes to test, and not much time. You’d like to be able to measure the urea
concentration after 30 minutes, and then predict whether or not the membrane
meets the design goal. You do know one useful thing: The mass flow rate at which
urea (U) passes through the membrane ​​ṁ ​​U, out decreases linearly with a decrease
in the mass of urea in the tank mU, sys:
​​​ṁ ​​U,out​​ = β​mU​  ,sys​​​
where β is a constant that characterizes the membrane’s performance.
What is the maximum percent urea that should be left in the tank after 30 minutes,
to ensure that the design goal is met?
Solution
We’ll choose as our system the tank containing the plasma, and urea as our component. This separation operates in a semibatch mode.
Let’s start with the differential material balance equation, Eq. (6.3a), written
for urea. No urea enters the system; so:
d​mU​  ,sys​​
​​ _​ = − ​​ṁ ​​U,out​​ = −β​mU​  ,sys​​​
dt
Rearranging and integrating from t = 0 to t, we find
t
d​mU​  ,sys​​
​ ​​ ​ _
​
=
−
β ​
∫
​
​dt​​
​  ,sys​​
​mU​  ,sys,0​​ ​mU
0
​​∫
Quick Quiz 6.3
If the design goal of
Example 6.10 changed
and the membrane had
to remove 90% of the
urea in 1 h, what
would be the minimum
β required?
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​mU​  ,sys​​
or
​mU​  ,sys​​
​ln​ ​ _
( ​mU​  ,sys,0 ​​​ )​= − βt​
which is our design equation.
At the end of 3 hours (t = 3 h), the design goal is to drop the urea concentration to 3% of its initial value:
​​mU​  ,sys, f​​= 0.03 × ​mU​  ,sys,0​​​
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Section 6.4 Stream Composition and System Performance Specifications for Separators
401
Inserting these values into our design equation, we solve for the value of β that
meets the design goal:
0.03​mU​  ,sys,0​​
​ln​(​ _
​mU​  ,sys,0 ​
​​ )​= ln(0.03) = −β(3)​
​β = 1.17 ​h​​  −1​​
So, if the membrane has a value of β = 1.17 h−1 (or greater) it will meet design
criteria. What decrease in urea will be achieved with such a membrane after
30 minutes (0.5 h)? We return to our design equation:
​mU​  ,sys,0.5h​​
​ln​ ​ _
​​ )​= −1.17(0.5) = −0.585​
( ​mU​  ,sys,0 ​
​mU​  ,sys,0.5h​​
​​ _
​
​m​
​​ = 0.56​
U,sys,0
To meet the design goal, the urea in the tank must decrease to at least 56% of the
initial quantity in the first 30 minutes. It is interesting to plot how the urea quantity in the tank changes with time.
Mass of urea in tank/initial mass
of urea in tank
1
0.8
0.6
β = 1.17 h−1
0.4
0.2
0
0
0.5
1
2
1.5
Time, h
2.5
3
Because the mass flow rate out decreases with time, almost half of the removal is
accomplished in the first 30 minutes, and it takes another 2.5 h to remove the
remainder.
6.4
Stream Composition and System
Performance Specifications for Separators
In a perfect separator, the products are pure, and each component in the feed
ends up entirely in the appropriate product stream. The perfect separator is rare
indeed. More realistically, the products are not pure, and there is not complete
recovery of the desired component in the appropriate product stream.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
1
Feed
B
C
Separator
2
(a)
1
B
C
Feed
A
B
C
D
E
A
B
C
Separator
2
B
C
D
E
(b)
Figure 6.6 Separation of mixture containing key components B and C. (a) A perfect
­separator has only the two key components in the feed, and the output streams are pure.
(b) A real separator. Component B is preferentially recovered in product 1, while component
C is preferentially recovered in product 2. The nonkey components A, D and E are nondistributing. In some separators, the nonkey components distribute into both product
streams.
Quick Quiz 6.4
Suppose a real separator has only two components, B and C, in
the feed, but B and C
distribute to both product streams. If you
know the flow rate and
composition of the
feed but nothing else,
what is the DOF?
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In Fig. 6.6, a “perfect” and a “real” separator are compared. (We consider
only separators with one input and two outputs.) In the “perfect” separator
(Fig. 6.6a), the feed is a mixture of two components B and C, product 1 is
pure B, and product 2 is pure C. In the “real” separator (Fig. 6.6b), the feed
is a mixture of not only the two components B and C that are to be separated,
but also contaminants A, D, and E. Although B becomes more concentrated
in product 1 than it is in the feed, and C becomes more concentrated in product 2, both products contain both B and C.
To simplify the analysis of a real separator somewhat, we can identify
“key” and “nonkey” components. Since the separator in Fig. 6.6b is designed to
separate B from C, these become the key components. The nonkey components,
A, D, and E, are just along for the ride. We then assume that the key components
distribute into both product streams, albeit with most of B going to product 1
and most C going to product 2. The nonkey components are assumed to be
nondistributing, that is, they exit the separator in either product 1 or product 2,
not both. Which product stream depends on the physical properties of the
nonkey components and the basis for the separation chosen. For example, suppose the basis for separation is a difference in molecular size. If size increases
in the order A < B < C < D < E, then all of A, most of B and a bit of C
exit in product 1, whereas a bit of B, most of C, and all of D and E exit in
product 2.
A comparison of the degrees of freedom of the perfect versus real separators shown in Fig. 6.6 is enlightening. We assume steady-state operation, and
that the flow rate and composition of the feed stream are known. Specification
of the flow rate and composition of the feed stream is sufficient to ensure that
the perfect separator is completely specified. If both the feed rate and feed
composition to the real separator are known, but nothing else, the separator is
underspecified. In this example there are 5 specifications provided: 1 flow rate,
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Section 6.4 Stream Composition and System Performance Specifications for Separators
Table 6.6
403
DOF Analysis of a Perfect Separator and a
Real Separator (Figure 6.6)
Perfect
separator
Stream variables
4
Specified flows
Specified stream composition
Specified system performance
Material balances
Total equations
12
1
1
0
2
4
DOF
0
Real separator,
non-distributing
nonkeys
1
4
0
5
10
2
and 4 stream composition (because there are 5 components in the feed). There
are also 5 material balance equations, one for each component. However, there
are 12 stream variables, so the real separator requires two additional specifications when the nonkeys are nondistributing (Table 6.6). These specifications
are typically provided in one of three ways: product purity, component recovery,
or separation factor. We’ll describe each one in turn.
Product purity is a stream composition specification. Product purity specifications set the acceptable content of a component in a product stream. Most
of the time, the goal is to operate at purities as close as possible to the minimum acceptable to the customer, because higher purity entails higher costs.
We define fractional purity of product stream j as
moles (or mass) of desired component i in product stream j
________________________________________________
​Fractional purity = ​
​​
total moles (or mass) of product stream j
For steady-state continuous-flow separators, a mathematical definition using
molar units is
​​n ​​i̇ j​​
​
​zi​  j​​ = _
​   ​​
Eq. (6.4a)
​​n ​​j̇ ​​
Quick Quiz 6.5
Write the equation for
product purity wij for a
batch separator.
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where zij is the mole fraction of the desired component i in product j. It is
always true that 0 ≤ zij ≤ 1.0 and that
​​ ∑ ​​​ ​zi​  j​​ = 1​
all i
Eq. (6.4b)
For each product stream j, if there are N components we can specify at most
N − 1 product purities. Purity may also be specified in mass units, or for batch
separators. Percent purity is simply fractional purity × 100.
Component recovery is a system performance specification that relates
the output of the process unit to the input. The goal is usually to operate at the
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Quick Quiz 6.6
Write the equation for
fRij in mass units for a
batch separator.
Helpful Hint
Purity calculated in
mole units is numerically different
than purity calculated in mass units.
However, recovery
is numerically the
same for either
mass or mole units.
highest recovery possible, because higher recovery means more product sold
at a given feed rate. We define fractional component recovery as
moles (or mass) of component i in product j
​Fractional recovery = ​ ____________________________________
​​
moles (or mass) of component i in feed
For steady-state continuous-flow separators and using molar units, we define
fractional recovery fRij as
​​n ​​i̇ j​​
​zi​  j​​ ​​n ​​j̇ ​​
​​f​ Rij​​ = _
​   ​ = ​ _ ​​
Eq. (6.5a)
​​n ​​i̇ F​​ ​zi​  F​​ ​​n ​​Ḟ ​​
where the F subscript indicates the feed, or input, stream. (Recovery may also
be specified in mass units, or for batch separators.) Percent recovery is simply
fractional recovery × 100. According to this definition, it is always true that
​0 ≤ ​fR​  ij​​ ≤ 1.0​. Since
​​  ∑ ​​​ ​fR​  ij​​ = 1​
Eq. (6.5b)
all j​​out​
for each component i, if there are only two products then there can be only one
independent recovery specification per component.
Illustration: A mixture containing 50.0 mol% H2 (molar mass 2 g/gmol),
25.0 mol% CH3OH (32 g/gmol) and 25.0 mol% CO2 (44 g/gmol) is fed to a separator at a flow rate of 100 gmol/h. Two products leave the separator. Product 1
flow rate is 72.5 gmol/h (50 gmol/h H2, 2.5 gmol/h CH3OH, 20 gmol/h CO2).
Product 2 flow rate is 27.5 gmol/h (22.5 gmol/h CH3OH, 5 gmol/h CO2). There
are three components, CH3OH (M ) and CO2 (C ) which are keys, and H2 (H )
which is a non-distributing nonkey. In molar units, product purities are:
​​zH​  1​​ = _
​  50  ​ = 0.69​, ​​zM
​  1​​ = _
​  2.5  ​ = 0.034​, ​​zC​  1​​ = _
​  20  ​ = 0.276​,
72.5
72.5
72.5
​​zH​  2​​ = _
​  0  ​ = 0.0​, ​​zM
​  2​​ = _
​  22.5 ​= 0.82, ​zC​  2​​ = _
​  5  ​= 0.18 ​
27.5
27.5
27.5
Purities in mass units are calculated by first converting molar flows to mass
flows using the molar masses. For example, H2 mass flow rate in product 1 is
​​​ṁ ​​H1​​= (50 gmol/h)(2 g/gmol) = 100 g/h​
Similar calculations provide the mass flow rates in all streams, from which
product purities are calculated on a mass basis:
​​wH​  1​​ = _
​  100  ​ = 0.094​, ​​wM
​  1​​ = _
​  80  ​ = 0.075​, ​​wC​  1​​ = _
​  880  ​ = 0.83​,
1060
1060
1060
​​wH​  2​​ = _
​  0  ​ = 0.0​, ​​wM
​  2​​ = _
​  720 ​= 0.77, ​wC​  2​​ = _
​  220 ​= 0.23​
940
940
940
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Section 6.4 Stream Composition and System Performance Specifications for Separators
405
Fractional recoveries are the same whether calculated from molar or mass flow
rates:
100 ​ = 1​, ​​f​  ​​ = _
80  ​ = 0.1​, ​​f​  ​​ = _
880  ​= 0.8​
​​fR​  H1​​ = _
​  50 ​ = ​ _
​  2.5 ​ = ​ _
​  20 ​ = ​ _
RM1
RC1
800
50 100
25
25 1100
​​fR​  H2​​ = 0​, ​​fR​  M2​​ = 0.9​, ​​fR​  C2​​ = 0.2​
Separation factor is a third way to characterize the performance of a
separation unit. The separation factor gives a measure of how well we have
separated the two key components from each other, and can be thought of as
a measure of the selectivity of the separation process. The separation factor is
generally defined only in terms of the key components, not the nonkey components. If we have two components, B and C, to separate, and two products,
1 and 2, then the separation factor αBC is
quantity of B in product 1
quantity of C in product 2
​Separation factor = ​ ​ _______________________
​ ​​​ ​​× ​ ​ _______________________
​ ​​
( quantity of C in product 1 ) ( quantity of B in product 2 )
or, for steady-state continuous-flow separations using mole units:
​​n ​​̇ ​​ ​​n ​​Ċ 2​​
​​α​ BC​​ = _
​  B1 ​ ​ _
​​
​​n ​​Ċ 1​​ ​​n ​​Ḃ 2​​
Quick Quiz 6.7
What is the purity, recovery, and separation
factor of a “perfect”
separator?
Eq. (6.6a)
(The separation factor may also be specified in mass units, or for batch separators.)
The separation factor is related to purity or recovery specifications; equivalent
expressions are derived by combining Eq. (6.6a) with Eq. (6.4) or (6.5):
​z​  ​​ ​zC​  2​​
​​α​ BC​​ = _
​  ​zB​  1 ​ ​ ​​ _
Eq. (6.6b)
​  2 ​​​​
C1 ​zB
​ f​  ​​ ​ f​ RC2​​ _
​f​  ​​
​fR​  C2​​
​​α​ BC​​ = _
​  RB1 ​ _
​
​ = ​  RB1  ​ ​ _
​​
​fR​  B2​​ ​fR​  C1​​ (1 − f​R​  B1​​) (1 − f​R​  C2​​)
Eq. (6.6c)
By convention, we define B and C such that always ​1 ≤ α
​ B​  C​​ < ∞​.
Illustration: For the carbon dioxide-methanol separator described in the previous illustration, the separation factor is
​z​  ​​ ​zM
​  2​​ _
0.276 _
0.82
​​αC​ M​​ = _
​  ​zC​  1 ​ ​ ​​ _
​  1 ​​​ = ​  0.18 ​ × ​  0.034 ​= 37​
C2 ​zM
It can also be calculated from mass fractions or fractional recoveries
​w​  ​​ ​wM
​  2​​ _
0.77  ​= 37​
​​α​CM​​ = _
​  ​wC​  1 ​ ​ ​​ _
​ = ​  0.83 ​ × ​ _
w
​
​
C2 M1​​
0.23 0.075
​ f​  ​​ ​ f​ RM2​​ _
0.9 ​= 36​
​​αC​ M​​ = _
​  RC1 ​ ​ _
​ = ​  0.8 ​ × ​ _
0.2 0.1
​fR​  C2​​ ​fR​  M1​​
(The difference is round-off error in product purity calculations.)
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Let’s return to our real separator of Fig. 6.6b. From Table 6.6, we know we
need two additional independent specifications. Frequently used sets of specifications include (1) key component purities in each product (e.g., xB1 and xC2),
(2) key component purity and recovery in one product (e.g., xB1 and fRB1), and
(3) key component recoveries in each product (e.g., fRB1 and fRC2).
In the next few examples, we clarify the definition of separator performance
specifications. Then we show how separator performance specifications are
coupled with material balance equations and feed composition specifications
to design and analyze separation units.
Example 6.11
Defining Separator Performance Specifications:
Separating Benzene from Toluene
Distillation is a ubiquitous separation technology that is used in everything from
oil refining and industrial chemicals manufacturing to bioprocessing. In a distillation column, the feed enters more or less near the middle of the column and two
product streams are removed. One product stream, the “distillate,” leaves the top
of the column and is enriched in the more volatile (higher vapor pressure, lower
boiling point) material. The other stream, called the “bottoms” (guess where it exits
the column!), is enriched in the less volatile component. In this example we evaluate distillation of two aromatics, benzene and toluene. Since benzene is more
volatile than toluene (see Example 6.4), the distillate is benzene-rich and the bottoms product is toluene-rich.
The feed to a distillation column contains 60 wt% benzene and 40 wt% toluene. The feed rate is 100 g/s, and the column is operated in the steady-state
continuous-flow mode. The distillate stream is 57 g/s benzene and 1.2 g/s toluene.
Calculate (a) purity of distillate, (b) purity of bottoms, (c) fractional benzene
recovery in the distillate, (d) fractional toluene recovery in the bottoms, and (e)
the separation factor.
Solution
We start with a flow diagram:
Feed 100 g/s
60% benzene
40% toluene
Distillate
57 g/s benzene
1.2 g/s toluene
Separator
Bottoms
benzene
toluene
All information is given in units of g/s and mass fraction, so we will stick with
those units. We’ll use b and t subscripts for our components benzene and toluene,
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Section 6.4 Stream Composition and System Performance Specifications for Separators
and F, D, and B subscripts for our feed stream and distillate and bottoms product
streams, respectively.
g benzene
​​ṁ ​​ ​​
​​ṁ ​​bD​​
(a)​
Distillate purity = w
​ b​  D​​ = _
​  bD ​ = _
​
​ = _
​  57  ​ = 0.98 ​ _ ​​
​​ṁ ​​D​​ ​​ṁ ​​bD​​ + ​​ṁ ​​tD​​ 57 + 1.2
g distillate
(b) To calculate the purity of the bottoms, we first need to find the flow rates of
each component in that stream from material balances. For benzene,
​​​ṁ ​​bF​​ = w
​ b​  F​​ ​​ṁ ​​F​​ = ​​ṁ ​​bD​​ + ​​ṁ ​​bB​​​
​(0.6)100 = 57 + m
​​ ̇ ​​bB​​​
​​​ṁ ​​bB​​= 3 g/s​
Similarly, for toluene,
​​​ṁ ​​tF​​ = ​wt​  F​​ ​​ṁ ​​F​​ = ​​ṁ ​​tD​​ + ​​ṁ ​​tB​​​
​(0.4)100 = 1.2 + m
​​ ̇ ​​tB​​​
​​​ṁ ​​tB​​= 38.8 g/s​
Therefore, the bottoms purity is
g toluene
​​ṁ ​​ ​​
​​wt​  B​​ = _
​  tB ​ = _
​  38.8  ​ = 0.93 ​ _ ​​
​​ṁ ​​B​​ 38.8 + 3
g bottoms
The purity is based on the component that is enriched in that product stream.
With the material balances solved, the remaining calculations are straightforward.
(c) Fractional recovery of benzene in the distillate:
​​m ​​̇ ​​ 57 g/s
​​fR​  bD​​ = _
​  bD ​ = _
​
​= 0.95​
​​m ​​̇ bF​​ 60 g/s
(d) Fractional recovery of toluene in the bottoms:
​​m ​​̇ ​​ 38.8 g/s
​​fR​  tB​​ = _
​  tB ​ = _
​
​= 0.97​
​​m ​​̇ tF​​
40 g/s
(e) Separation factor:
​​ṁ ​​ ​​ ​​ṁ ​​ ​​ 57 _
​​αb​  t​​ = _
​  bD tB ​ = _
​   ​ × ​  38.8
​= 614​
​​ṁ ​​tD​​ ​​ṁ ​​bB​​ 1.2
3
Example 6.12
Purity and Recovery Specifications in Process Flow Calculations:
Separating Benzene and Toluene
Feed (100 g/s) containing 60 wt% benzene and 40 wt% toluene is sent to a distillation column. The distillation column recovers 95% of the benzene in the distillate
product, which is 98 wt% pure benzene. Calculate the flow rates and compositions
of the distillate and bottoms products.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Solution
The flow diagram is shown. There are 6 stream variables, 1 specified flow rate,
2 specified stream compositions, 1 performance specification (95% recovery of
benzene in the distillate) and 2 material balance equations. Therefore, DOF =
6 − (1 + 2 + 1 + 2) = 0 and the problem is completely specified.
Distillate
wbD = 0.98
Feed F
wbF = 0.6
wtF = 0.4
Distillation
column
Bottoms
Streams will be designated as F (feed), D (distillate), and B (bottoms). Benzene
and toluene will be designated by subscripts b and t, respectively. All flow rates
are in units of g/s.
The distillate product purity is specified to be 98 wt% benzene.
​​ṁ ​​ ​​
​​wb​  D​​= 0.98 = _
​  bD ​​
​​ṁ ​​D​​
The fractional recovery is specified: 95% of the benzene fed to the distillation
column must be recovered in the distillate product, or
​​ṁ ​​ ​​ ​wb​  D​​ ​​ṁ ​​D​​
​
​​
​0.95 = _
​  bD ​ = _
​​ṁ ​​bF​​ ​wb​  F​​ ​​ṁ ​​F​​
Substituting in the known basis, feed composition specifications, and product
purity specifications, we get:
0.98​​ṁ ​​D​​
​0.95 = _
​
​​
0.6(100)
g
​​​ṁ ​​D​​ = 58 ​ _s ​​
We have two components (benzene and toluene) and can write two material balance
equations. We will write material balance equations on benzene and on total mass.
(Why? Because we have already solved for the total mass flow in the distillate.)
The material balance equation on total mass is:
​​​ṁ ​​F​​ = ​​ṁ ​​D​​ + ​​ṁ ​​B​​​
g
​​​ṁ ​​B​​= 100 − 58 = 42 ​ _s ​​
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Section 6.4 Stream Composition and System Performance Specifications for Separators
409
The material balance equation for benzene is
​​w​ bF​​ ​​ṁ ​​F​​ = ​wb​  D​​ ​​ṁ ​​D​​ + ​wb​  B​​ ​​ṁ ​​B​​​
​0.6(100) = 0.98(58) + x​ ​b,B​(42)​
​​w​ bB​​ = 0.075​
Since the mass fractions in each stream must sum to 1:
​​w​ tD​​ = 0.02​
​​w​ tB​​ = 0.925​
As a check, we write the toluene material balance:
​​w​ tF​​ ​​ṁ ​​F​​ = w
​ t​  D​​ ​​ṁ ​​D​​ + w
​ t​  B​​ ​​ṁ ​​B​​​
​0.4(100) = 0.02(58) + 0.925(42)​
​40 = 40​
The separation factor is
​w​  ​​ ​wt​  B​​ _
​​αb​  t​​ = _
​  ​wb​  D ​​​ _
​   ​ = ​  0.98 ​ _
​  0.925 ​= 604​
tD ​wb​  B​​
0.02 0.075
Example 6.13
Fractional Recovery in Rate-Based Separations:
Membranes for Kidney Dialysis
In Example 6.10, you evaluated a test apparatus for selecting membranes appropriate for kidney dialysis, using as a design criteria 97% removal of urea in the plasma
sample in 3 hours.
Let’s suppose you completed your experiments and identified two membranes
that you want to investigate further. For FlowTru membranes, ​β = 1.1 ​h​​  −1​​, while
for DiaFlo membranes, β
​ = 1.7 ​h​​  −1​​. A secondary design goal is to minimize loss
of protein through the membrane; specifically, at least 95% of the protein in the
plasma should be retained in the tank. You repeat the experiment, but measure
protein concentration in the plasma sample. The flow rate of protein through the
membrane decreases with the mass of protein in the tank (just as for urea), and
you define a new parameter βP, where
​​​ṁ ​​P,out​​ = ​βP​  ​​​mP​  ,sys​​​
where the subscript P refers to protein. From the data you calculate that for FlowTru
membranes, βP = 0.016 h−1, whereas for DiaFlo membranes βP = 0.05 h−1.
Which membrane, FlowTru or DiaFlo, would you choose?
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Solution
This is a rate-based separation. The goal is to separate urea and protein, using
differences in the rate of transport of the two through the membrane. If the
separation worked perfectly, all of the urea and none of the protein would be
recovered in the dialysis fluid product, and none of the urea but all of the protein
would be recovered in the treated plasma product. However, the separation is not
perfect.
Dialysis f luid
Dialysis f luid
with urea and
protein
Urea
Protein
Test membrane
Tank with plasma
In Example 6.10 we derived a design equation that should apply to either urea
or protein:
​mU​  ,sys​​
​ln​ ​ _
m
​
( U​  ,sys,0 ​​​ )​= − βt​
​mP​  ,sys​​
​ln​ ​ _
( ​mP​  ,sys,0 ​​​ )​= − ​β​ P​​t​
If we consider the dialysis fluid exiting the apparatus as the urea-enriched product 1,
then the fractional urea recovery in that product fRU1 at any time t is
​mU​  ,sys​​
​​fR​  U1​​= 1 − _
​ ​m​
​= 1 − ​e​​  −βt​​
U,sys,0​​
The treated plasma is the protein-enriched product 2, so the fractional recovery of
protein in the plasma product fRP2 is
​mP​  ,sys ​​
−​βP​  ​​t
​​fR​  P2​​ = ​ _
​mP​  ,sys,0 ​​​ = ​e​​  ​​
We’ll use these equations to evaluate the membranes.
For FlowTru, after 3 h,
​​fR​  U1​​= 1 − ​e​​  −βt​= 1 − e​ ​​  −(1.1​h​​
−1
​)(3h)
​= 0.963​
or 96.3% of the urea is removed. The protein remaining in the tank is
​​fR​  P2​​ = e​ ​​  −0.016(3)​= 0.953​
For DiaFlo, similar calculations give 99.4% removal of urea but only 86% retention
of protein.
Neither membrane quite meets both specifications. What if we adjust the operating times? Let’s use the equations for fRU1 and fRP2, and plot fractional recoveries
versus time, to get insight into the characteristics of these systems.
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Section 6.5 Recycling in Separation Flow Sheets
1
Protein in plasma
0.8
Fractional component recovery
Fractional component recovery
1
0.6
Urea in dialysis f luid
0.4
0.2
0
FlowTru membranes
0
0.5
1
1.5
Time, h
2
2.5
3
Protein in plasma
0.8
0.6
Urea in dialysis f luid
0.4
0.2
0
DiaFlow membranes
0
0.5
1
1.5
Time, h
2
2.5
3
Increasing the operating time will lead to a greater removal of urea. Let’s fix
the urea removal at the design specification of 97% and calculate the time required
to achieve that. For FlowTru
​​fR​  U1​​= 0.97 = 1 − ​e​​  −1.1t​​
​t = 3.2 h​
At this operating time, protein remaining in plasma is
​​fR​  P2​​ = e​ ​​  −0.016(3.2)​= 0.950​
For DiaFlo, similar calculations show that 97% removal of urea is reached in 2.1 h,
at which point the percent protein remaining in the plasma is 90%.
So neither membrane fulfills all the design requirements. The advantage of a
shorter operating time for DiaFlo to the patient is significant, and may overcome
the concern about protein loss.
6.5
Recycling in Separation Flow Sheets
We used recycling to advantage in chemical reactors, to overcome low reactor
conversion. Is recycle useful for separations? How does recycle affect recovery
and purity?
A separation flow sheet that includes recycle is shown in Fig. 6.7. We need
a splitter in Fig. 6.7 because we can’t recycle all of product 2 from the separator back to the feed. Why? Because the flow sheet would then produce only
product 1, and at steady state product 1 would have to be exactly the same as
the feed, which would have gained us nothing! This mental exercise helps
us to qualitatively see what happens when we join recycle with separation.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Feed
B, C
F
Mixer
M
Separator
1
Product 1
B, C
2
Product 2
B, C
S
Recycle
R
Splitter
Figure 6.7 Separator flow sheet with recycle.
At high recycle, product 1 becomes more and more like the feed in composition. The recovery of component B in product 1 increases as recycle increases,
but the purity decreases. The opposite trends are observed with product 2—the
recovery of component C decreases, but the purity increases.
Let’s put this qualitative reasoning on a more quantitative basis. We
define single-pass separator recoveries: fRB1 and fRCS are the fractional recovery of component B in stream 1 and the fractional recovery of component C
in stream S, respectively, based on the input stream M to the separator. We
define overall fractional recoveries: fRB and fRC are the fractional recoveries
of component B in product 1 and component C in product 2, respectively,
based on the input stream F to the process. Finally, we define a fractional
split fS as the fraction of the splitter feed S that is recycled to the mixer
​( ​fs​  ​​ = ​​n ​​Ṙ ​​∕​​n ​​Ṡ ​​)​.
From these definitions of fractional recoveries and the material balance
equations, we find
​fR​  B1​​
​​n ​​̇ ​​
​​f​ RB​​ = _
​  B1  ​ = ​ _____________
​​
​​n ​​Ḃ F​​ 1 − ​fs​  ​​ (1 − f​ R​  B1​​)
​​n ​​̇ ​​ ​ f​ RCS​​ (1 − f​ s​  ​​)
​​f​ RC​​ = _
​  C2  ​ = ​ _
​​
​​n ​​Ċ F​​
1 − f​S​  ​​ ​fR​  CS​​
The left-hand side of these equations is the overall fractional recovery—based
on feed to the process and products from the process—while the right-hand
side includes only performance specifications for the individual units on the
flow sheet. (Don’t try to memorize these equations—they can be derived easily
from analysis of the flow sheet.)
Product purities are
​fR​  B​​ ​​n ​​Ḃ F​​
​​n ​​Ḃ 1​​
​​z​ B1​​ = _
​
​ = ​ __________________
​​
​​n ​​Ḃ 1​​ + ​​n ​​Ċ 1​​ ​ f​ RB​​ ​​n ​​Ḃ F​​+ (1 − f​R​  C​​)n​​  ​​Ċ F​​
​fR​  C​​ ​​n ​​Ċ F​​
​​n ​​Ċ 2​​
__________________
​​z​ C2​​ = _
​
​ = ​
​​
​​n ​​Ḃ 2​​ + ​​n ​​Ċ 2​​ ​ f​ RC​​ ​​n ​​Ċ F​​+ (1 − f​R​  B​​)n​​  ​​Ḃ F​​
We use these equations to explore how overall recoveries and purities change
with recycle in the following example.
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Section 6.5 Recycling in Separation Flow Sheets
Example 6.14
413
Separation with Recycle: Separating Sugar Isomers
Fructose and glucose are isomers—with the same molecular formula (C6H12O6),
but different molecular structures. They are both simple sugars, but fructose tastes
much sweeter than glucose. Fructose is naturally present in fruit, but is made commercially on a very large scale by hydrolysis of cornstarch to glucose, followed
by enzymatically catalyzed isomerization of glucose to fructose. High-fructose corn
syrup is a mix of mainly glucose and fructose in water that is widely used in sodas,
juice-flavored beverages, and many other sweetened foods.
Because of chemical reaction equilibrium constraints, the fractional conversion
of glucose to fructose in the isomerization reactor is less than 50%. A reactor
produces a mix, at 500 kg/h, containing 8 wt% fructose and 12 wt% glucose in
water. The mix is fed to a separator. 90% of the fructose fed to the separator is
recovered in product I and 90% of the glucose is recovered in product II. The water
distributes such that the total sugar concentration remains 20 wt% in both products.
(a) Calculate the flow rates and purities of the two products.
(b) Fructose is a more valuable product than glucose. Consider ways to use recycle
to adjust recovery of fructose and/or purity of the fructose-enriched product.
Solution
(a) The flow diagram is shown (why is it OK to ignore the water?)
40 kg fructose/h
60 kg glucose/h
Product I
fructose-enriched
Separator
Product II
glucose-enriched
Let’s start by writing material balance equations. This is a steady-state, continuousflow process. We’ll use F for fructose and G for glucose, and we will indicate
the product stream by subscripts I or II.
kg
​​​ṁ ​​F,feed​​ = 40 ​ _ ​ = ​​ṁ ​​F,I​​ + ​​ṁ ​​F,II​​​
h
kg
​​​ṁ ​​G,feed​​ = 60 ​ _ ​ = ​​ṁ ​​G,I​​ + ​​ṁ ​​G,II​​​
h
The separator performance is specified through component recoveries:
​​m
​ ̇ ​​F,I​​ = ​fR​  F,I​​ ​​ṁ ​​F,feed​​= 0.90(40) = 36 kg/h​
​​​ṁ ​​G,II​​ = ​fR​  G,II​​ ​​ṁ ​​G,feed​​= 0.90(60) = 54 kg/h​
Also considering the material balance equations, we find that 86% of sugar in
product 1 is fructose, and the total sugar flow rate is 42 kg/h. In product II
93% of the sugar is glucose, and the sugar flow rate is 58 kg/h.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
(b) Now we want to look at the effect of recycle. One thing we notice is that we
are losing 10% of the fructose in product II. Can we use recycle to increase
recovery of the fructose in product I? Here is the proposed scheme.
40 kg fructose/h
60 kg glucose/h
Mixer
Separator feed
M
S
Recycle
R
Product I
fructose-enriched
Separator
Splitter
feed
Product II
glucose-enriched
Splitter
Now we have more material balance equations:
Mixer:​​​m ​​̇ F,feed​​ + ​​ṁ ​​FR​​= 40 + ​​ṁ ​​FR​​ = ​​ṁ ​​FM​​​
​​​ṁ ​​G,feed​​ + m
​​ ̇ ​​GR​​= 60 + ​​ṁ ​​GR​​ = m
​​ ̇ ​​GM​​​
Separator:​​​m ​​̇ FM​​ = ​​ṁ ​​FS​​ + ​​ṁ ​​F,I​​​
​​​ṁ ​​GM​​ = m
​​ ̇ ​​GS​​ + ​​ṁ ​​G,I​​​
Splitter:​​​m ​​̇ FS​​ = ​​ṁ ​​FR​​ + ​​ṁ ​​F,II​​​
​​​ṁ ​​GS​​ = ​​ṁ ​​GR​​ + ​​ṁ ​​G,II​​​
If the separator performance remains the same, then:
​​m ​​̇ F,I​​
​0.90 = _
​
​​
​​ṁ ​​FM​​
​​ṁ ​​ ​​
​0.90 = _
​  GS  ​​
​​ṁ ​​GM​​
The fractional split is a design variable that has not yet been specified:
​​ṁ ​​ ​​ ​​ṁ ​​GR​​
​​fS​  ​​ = _
​  FR ​ = _
​
​​
​​ṁ ​​FS​​ ​​ṁ ​​GS​​
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1.00
Purity of fructose in product I
Recovery of fructose in product I
Now we can solve the system of equations as a function of fractional split. (It’s
a good idea to solve for fS = 0.0, the base case of no recycle, as a check.) Let’s
plot fractional recovery of fructose and purity of product I as a function of fS.
0.92
0.84
0.76
0.68
0.60
0.00
0.20
0.60
0.40
Fractional split
0.80
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415
Section 6.6 Entrainment: Incomplete Mechanical Separation
As the recovery of fructose in product I increases, the purity decreases. If a
lower-purity product is acceptable, then recycle is a good idea. What would
happen if we changed the splitter position?
Recycle
40 kg fructose/h
60 kg glucose/h
Mixer
Separator feed
Product I
fructose-enriched
Splitter
Splitter
feed
Separator
Product II
glucose-enriched
Briefly, the fructose recovery in product I decreases, but the purity increases.
(We’ll leave the details for you to calculate.)
6.6
Entrainment: Incomplete
Mechanical Separation
Recall that in a mechanical separation, a two-phase feed is separated into two
phases, which are the two products. The phases are separated on the basis of
differences in density or size. In a “perfect” mechanical separator, the separation into two phases is complete. Real separators often suffer from incomplete
mechanical separation of the two phases, or entrainment. Consider, for
example, particles or liquid droplets suspended in a vapor stream. If this
two-phase stream flows rapidly past a plate, the particles or droplets will tend
to collect on the plate. Some of the particles or droplets, however, will remain
“caught up” or entrained in the vapor stream, and the separation will be
incomplete (Fig. 6.8). Entrainment is even more of a problem in liquid-solid
Captured particles
Entrained particles
Figure 6.8 Entrainment in mechanical separation. The particles are carried in the rapidlyflowing fluid stream. As the stream impinges on the plate, the particles tend to collect on
the plate. However a few renegade particles are swept along with the fluid and manage to
escape the collector plate. Separation is incomplete.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Quick Quiz 6.8
If the seawater in the
bucket is 3.5 wt% salt,
what is the mass fraction salt, if any, in the
entrained liquid in the
wet sand?
Example 6.15
separations. Think, for example, of scooping up sand and seawater in a
bucket. The sand rapidly sediments, and the seawater can be easily poured
off as a single phase. Still, the sand is wet—the seawater becomes entrained
in the solid phase.
In equilibrium-based separation technologies such as distillation, solvent
extraction, or adsorption, the two product streams are two different phases.
These two phases need to be mechanically separated. Thus, entrainment can
also affect the performance of equilibrium-based separations.
Entrainment affects both component recovery and product purity.
Analysis of the effect of entrainment on separator performance is simplified
if one reasonable approximation is invoked: The composition of the entrained
material is the same as that of its bulk phase. This idea is illustrated in the
next example.
Accounting for Entrainment: Coffee Making
Roasted coffee beans contain a complex mixture of chemical species, some of
which are soluble in water (e.g., caffeine) and some of which are not. Typically,
beans contain about 60 wt% soluble components and 40 wt% insoluble components. To make coffee, ground roasted coffee beans are contacted with hot water.
Most of the soluble components are leached out of the beans by the hot water,
making an aromatic, addictive brown liquid.
Suppose 65 grams of coffee beans are contacted with 1800 g (about 8 cups)
of hot water. After 10 minutes, about 85% of the soluble components in the beans
are leached into the liquid—the liquid phase contains 33 g solubles plus water
while the ground beans contain 6 g solubles and all the insolubles. (Not all the
solubles in the beans actually dissolved.) The resulting solid-liquid mixture is
poured into a coffee filter and allowed to drain out. Most of the liquid passes
through the filter and goes into the coffeepot. Some liquid is entrained with the
coffee grounds captured by the filter; specifically, 2.5 grams of liquid solution is
entrained per gram of dry solid material. How much coffee is in the pot? What is
the percent solubles in the coffee?
Solution
Coffee brewing is an example of a solvent leaching process, in which a soluble
component is removed from a solid by addition of a solvent. Leaching is very
common in food processing; other examples include tea brewing and extraction of
fish and vegetable oils. Leaching is an equilibrium-based separation: A second
phase (hot water in this example) is added to the solids feed, and soluble components in the solids are transferred to the added phase. After leaching, the solids
and liquids are separated by filtration, a mechanical separation technology.
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Section 6.6 Entrainment: Incomplete Mechanical Separation
417
Hot water, 1800 g
Ground coffee beans B
39 g solubles
26 g insolubles
Coffee liquid
Spent grounds
Wet grounds, G + L
Coffee C
Helpful Hint
When entrainment
occurs in separation
of phases, assume
that the composition
of the entrained
phase is the same
as the composition
of the bulk fluid
phase.
In this problem we focus on the operation of the filter. There are three components: solubles s, insolubles i, and water w. (Solubles and insolubles are composite materials—mixtures that can be treated as single components because they
behave identically in the process unit.)
The feed to the filter is a two-phase mixture. The coffee liquid contains 33 g
solubles plus all the water (1800 g). The spent grounds is 6 g solubles plus 26 g
insolubles. The two product streams after the filter are liquid coffee C and wet
grounds. The wet grounds contain both dry ground solids G and entrained liquid L.
It is advantageous to consider the two phases in the product stream, because the
entrained liquid L has the same composition as the liquid coffee product C. What
is this composition? The coffee product and the entrained liquid in the wet grounds
have the same composition as the liquid phase in the feed!
g solubles
33
​​ws​  C​​ = ​ws​  L​​ = _
​
​ = 0.018 ​ _ ​​
1800 + 33
g solution
By the same reasoning, the water mass fraction in the coffee and in the entrained
liquid is 0.982 (g water/g solution).
The solid phase in the feed to the filter contains 6 g solubles and all the
insolubles originally in the beans, which is (0.4)65 or 26 g insolubles. Since the dry
spent grounds G contain all of this material,
​​mG​  ​​= 6 + 26 = 32 g​
Treating the filter as a batch separator, the integral material balance equation is
used. With all quantities written in grams, the solubles material balance equation is
​39 = ​ws​  C​​​mC​  ​​ + ​ws​  L​​​mL​  ​​ + w
​ s​  G​​ ​mG​  ​​ = 0.018(​mC​  ​​ + ​mL​  ​​) + 6​
​33 = 0.018(​mC​  ​​ + ​mL​  ​​)​
Finally, we know that 2.5 g liquid is entrained per g of dry grounds, or
​​mL​  ​​ = 2.5​mG​  ​​= 2.5(32) = 80 g entrained liquid​
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Quick Quiz 6.9
Calculate the mass
fraction of insolubles
in the coffee grounds
without entrainment
and then with
entrainment.
Inserting this value into the solubles material balance equation yields
​​mC​  ​​= 1753 g coffee​
The coffee contains 0.018(1753) or 31.6 g coffee solubles. 33 g out of the 39 g
solubles in the beans actually dissolved in the water. Thus, in the absence of
entrainment, the recovery of solubles in the liquid is (33/39) or 0.85. The recovery
of solubles in the coffee product has been decreased by entrainment:
​​fR​  sC​​ = _
​  31.6 ​= 0.81​
39
Recovering Proteins from Fermentation Broths
Proteins, produced by fermentation of genetically modified microorganisms
such as bacteria or yeast, are used widely in industry. For example, enzymes
are added to laundry detergent to remove protein stains, or used as supplements
in animal feed to aid digestion, or added to industrial wastewater streams to
reduce pollutant levels.
In this case study, we will examine the purification of a protease that
is used in laundry detergents. The protease cleaves protein stains into smaller
peptides and amino acids which are soluble and can simply be washed away.
Proteases used for this application must remain stable at the alkaline pH
(9–12) and hot water (40–50°C) conditions in the washing machine. Strains
of Bacillus bacteria have been discovered that make enzymes known as
alkaline proteases that are ideal for this application. To produce the enzyme,
Bacillus bacteria are cultured in a complex liquid solution (called “medium”
or “broth”) that contains nutrients such as yeast extract and glucose, as well
as salts and minerals. The particular enzyme we are interested in making,
which will go by the name “ProTase,” has a molar mass of 66,000 g/gmol.
During fermentation in a 5000-liter batch reactor, the bacteria consume
nutrients, grow and reproduce, and secrete proteins (ProTase as well as other
proteins) into the broth. After 48 hours, the contents of the fermentor are
harvested, and the protein must be purified from this complex mixture.
For this case study, the starting materials for the separation flow sheet are
(all in kg):
Cells
Salts
Glucose
Proteins
Total and cell (NaCl and and organic (including
broth debris
others)
acids
ProTase) ProTase Water
5000
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100
310
60
50
(7)
4480
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Section 6.6 Entrainment: Incomplete Mechanical Separation
419
We make the following observations that influence the sequence of separations as well as our choice of separation technology:
1. The broth contains both liquid and solid materials. Since it is a two-phase
mixture, a mechanical separation to separate liquid from solid would be a
good first step. The desired product is soluble in water and so will be
recovered in the liquid phase.
2. ProTase shares physical and chemical properties with the other proteins in
the broth. Purification of the protease from the other proteins may be a
difficult separation and therefore should be saved for the end of the separation sequence.
3. Water is the most abundant component. It might be useful to remove most
of the water relatively early in the sequence, in order to reduce the volume
of material that needs to be processed.
4. Glucose, organic acids, and salts are low molecular weight species (typically
<200 g/gmol) compared to ProTase and the other proteins in the mixture.
We may be able to take advantage of the difference in molecular size.
These observations lead us to propose a preliminary sequence of separation tasks:
Water
Broth
Sep 1
Solids: cells
and cell debris
Sep 2
Sep 3
Sep 4
Glucose, salts,
organic acids
Other
proteins
ProTase
Let’s next consider how to accomplish the first separation task. We know
that the solids are more dense than the fluid. From the table of mechanical
separation technologies (Table 6.2), filtration, sedimentation, and centrifugation are all reasonable choices. After conducting a few experiments in the lab,
we learn that sedimentation is very slow, and that both filtration and centrifugation result in a fair amount of entrained liquid in the solids: the “cake” from
filtration is about 40 wt% solids, while centrifugation produces a cake with
only about 20 wt% solids. We choose filtration for Separator 1. We lose some
entrained liquid in the solid cake, but the fluid phase is completely clear of
solid materials; this is advantageous because solid particles could plug up
downstream processing units. By applying material balances and accounting
for entrainment, we calculate the quantities of materials leaving the filter. The
results are shown in the table, with all quantities in kg; calculations are left to
the reader. (Proteins includes ProTase as well as other proteins. ProTase mass
is shown in parentheses to indicate that it is already included in the total protein mass.)
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Feed to
Component
filter
Filter cake
(solids + entrained
liquid)
Flowthrough
(liquid)
Cells, cell debris
100
100
0
Salts
310
9.5
300.5
Glucose, organic acids
60
1.8
58.2
Proteins
50
1.5
48.5
ProTase
(7)
(0.2)
(6.8)
Water
4480
137
4343
Total
5000
250
4750
The next step in the sequence calls for removing water in order to reduce
the volume. We don’t want to remove all of the water; rather, we want to
increase the protein concentration in the aqueous solution. Water is more volatile than any of the other compounds, which might lead us to think of evaporation or distillation. However, these processes would require high temperatures,
which might damage the proteins. Furthermore, the energy required to evaporate that much water would be significant. What other technologies might
work? ProTase is a “trace” quantity compared to the total quantity of broth to
be processed, and one of our heuristics is “use separation methods where the
cost increases with the quantity of material to be recovered, not the quantity
of the stream to be processed.” Applying this heuristic to our problem, we look
for separation technologies that scale with the quantity of protein rather than
of water. A search of the literature reveals the option of protein precipitation,
an example of an equilibrium-based separation using a material-separating
agent. Specifically, ammonium sulfate causes proteins to become insoluble in
water without damaging the protein. The protein precipitate can be removed
by centrifugation or filtration, and the remaining ammonium sulfate recovered
for re-use. In the lab, however, you find that precipitation of the proteins is
too slow, and the amount of ammonium sulfate required is too high, for this
process to be attractive on a large scale, so this idea is abandoned.
What about rate-based separations? The molecular weight of proteins is
much greater than water. Ultrafiltration might be attractive. (Ultrafiltration is
not the same as standard filtration. In standard filtration, a solid and fluid phase
are separated from each other, whereas in ultrafiltration, dissolved solutes in a
fluid phase are separated based on differences in rate of flow through a sizeselective membrane.) With careful choice of ultrafiltration (UF) membrane,
much of the water (as well as some of the salts, organic acids, and glucose)
are pushed through a membrane while most of the proteins are retained.
Compared to evaporation or distillation, ultrafiltration operates at temperatures
that don’t damage ProTase. The size of the process scales with the amount of
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Section 6.6 Entrainment: Incomplete Mechanical Separation
421
material to be processed, which goes against one of our heuristics. However,
compared to precipitation, there is no need to add secondary steps to recover
the salts. Given all of these considerations, ultrafiltration is chosen as the means
to remove water. Laboratory experiments indicate that it is feasible to increase
the protein concentration to ~20 wt%, that 95% of the protein is retained by the
membrane, and that the mass ratios of salt:water or glucose plus organic
acids:water remains the same in both product streams. Applying material balances as well as the recovery and purity specifications, we obtain the following
quantities.
Feed to UF
membrane
Retentate
Filtrate
Salts
300.5
11.7
288.8
Glucose, organic acids
58.2
2.3
55.9
Proteins
48.5
46
2.5
ProTase
(6.8)
(6.5)
(0.3)
Water
4343
170
4173
Total
4750
230
4520
Component
Ultrafiltration has removed 96% of the water, with the added benefit of
flushing out much of the salts and small organic compounds as well, thus
completing separation tasks 2 and 3. We have accomplished much of what we
wished to do in one step rather than in two!
Separation task 4 may be the most challenging, because we need to separate ProTase from the other proteins. This requires a closer look at the properties of the different proteins in this mixture. You find that the molecular
weights of the proteins range from about 14,000 to about 100,000. ProTase, at
66,000 g/gmol, falls right in the middle, making size-based separation infeasible. Proteins carry many charges, both positive and negative, and the isoelectric point (pH at which the protein has net zero charge) of ProTase is 10.5,
whereas the other proteins all have isoelectric points of 7 or lower. If the pH
is adjusted to 8, ProTase will be positively charged whereas the remaining
proteins will be negatively charged. This property difference provides the basis
for an adsorption technology called ion exchange (IEX). Briefly, the proteincontaining solution is brought into contact with polymeric beads that carry a
negatively charged group (such as carboxymethyl or sulfopropyl groups).
ProTase will adsorb to the beads while the remaining proteins (as well as the
glucose) will not. The unbound proteins are washed away with water, after
which ProTase is eluted (desorbed) by either adjusting the pH to above 11, or
adding a buffer containing a high concentration of salts. Experiments in the
lab establish that ion exchange adsorption recovers 85% of ProTase while
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
adsorbing only a small amount of the other proteins, that the eluted protein is
90 wt% ProTase, and the eluted liquid is 25 wt% protein and 6 wt% salt. This
purity is sufficient for use in laundry detergent; the product can be dried to
remove remaining water if need be.
Our flow sheet, as well as the ProTase quantity, purity, and recovery at
each separation step, is shown. Purity is based on the total of all other materials. Recovery is calculated relative to the initial quantity of ProTase in the
broth.
Broth
Concentrated
Flowthrough
protein
liquid
solution
Filtration
Ultraf iltration
IEX adsorption
Cake:
40% solids,
60% liquid
Water, salts,
glucose,
organic acids,
trace protein
90% ProTase
10% other protein
water
salts
Dryer
Proteins,
ProTase (trace),
water, glucose,
organic acids,
salts
ProTase
product
Water
Broth Post-Filtration Post-UF Post-IEX Post-Dryer
ProTase (kg)
7
6.8
6.5
5.5
5.5
Protein (kg)
50
48.5
46
6.1
6.1
Total (kg)
5000
4750
230
24.4
7.6
Purity (%)
0.14
0.14
2.8
22.5
72.3
97
93
78
78
Recovery (%)
Summary
∙ Separations account for 50% or more of the total capital and operating
costs of a typical chemical process facility. There is enormous diversity in
the choice of separation technologies, but they can be handily classified as
one of three kinds: (a) mechanical, (b) rate-based, and (c) equilibriumbased. Separations work by exploiting differences in physical and/or
chemical properties of the species to be separated. Engineers use heuristics
to guide their choice of technology.
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Summary
∙ Most large-scale separators operate in steady-state continuous-flow mode.
Batch or semibatch modes are chosen sometimes for smaller-scale processes, where solids are handled, or where rate-based technologies are
employed.
∙ Three useful measures of the performance of a separation process are
purity, component recovery, and separation factor:
quantity of desired component i in product stream j
​Fractional purity = ​ ___________________________________________
​​
quantity of product stream j
​​n ​​i̇ j​​
​​ṁ ​​ij​​
​
zi​  j​​ = _
​   ​ or ​
wi​  j​​ = _
​   ​
​​n ​​j̇ ​​
​​ṁ ​​j​​
quantity of desired component i in product j
​Fractional recovery = ​ ____________________________________
​​
quantity of desired component i in feed
​​n ​​i̇ j​​
​​ṁ ​​ij​​
​
fR​  ij​​ = ​ _ ​ = ​ _ ​
​​n ​​i̇ , feed​​ ​​ṁ ​​i, feed​​
​f​  ​​ ​fR​  B2​​
​z​  ​​ ​zB​  2​​ _
​​n ​​̇ ​​ ​​n ​​Ḃ 2​​ _
Separation factor = α
​ A​  B​​ = _
​  ​zA​  1 ​​​ _
​  ​z​   ​​​ = ​  A1 ​ ​ _
​ = ​  RA1 ​ _
​   ​
​​n ​​Ȧ 2​​ ​​n ​​Ḃ 1​​ ​fR​  A2​​ ​fR​  B1​​
A2 B1
∙ Splitters and recycle can be employed in separation flow sheets to improve
component recovery or product purity. In most separation flow sheets,
there is a tradeoff between higher selectivity or higher product purity.
∙ Entrainment occurs when two phases are not completely mechanically
separated. Entrainment affects component recovery and product purity and
must be accounted for in analysis of the performance of separators.
ChemiStory: How Sweet It Is
Sugarcane is a perennial grass, native to tropical southern Asia. After
Christopher Columbus brought sugarcane to the New World, the European
colonial powers Spain, England, and France rapidly established sugarcane
plantations on the tropical Caribbean islands and produced molasses—a
brown, unrefined sugar syrup. Ships transported the molasses to New
England, where it was made into rum, the rum was shipped to slave traders
in Africa, and then the ships returned to the islands with new slaves to work
the plantations.
When slaves revolted on French Caribbean islands, French plantation
owners fled to New Orleans. Louisiana’s rich soils and extensive waterways
proved conducive to establishment of a sugar industry, and by the early
1800s, sugarcane was the dominant crop grown in southern Louisiana.
A great deal of slave labor and fuel was required to produce sugar. The cane
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
was first crushed at a mill to release the
juices, then water was evaporated in the
sugarhouse. Working in the “Jamaica
train,” slaves ladled boiling sugar juice
from one steaming open kettle to another.
The work was hot, dirty, and dangerous.
When the sucrose concentration in the
syrup was finally high enough to crystallize, the juice was cooled, allowing
further crystallization. The juice-crystal
mixture was stored in a barrel with a perforated bottom, allowing the molasses to
run out and leaving behind brown sugar
crystals. This raw sugar was shipped
north for further refining into white sugar
Norbert Rillieux
by re-crystallization.
GRANGER
In 1806, just 3 years after the Louisiana
Purchase, Norbert Rillieux was born in New Orleans. He was the son of a
wealthy white cotton merchant and his mixed-race mistress—a “quadroon,”
or one-fourth Black and three-fourths white. He grew up a free man of
color—educated, well-to-do, with the right to own land and slaves but not
to vote or marry whites. During the 1820s, as sugar became king in
Louisiana, Norbert was sent to Paris for his higher education and became
interested in mechanics and thermodynamics. (Rather ironically, it was more
common for southern free men of color than for whites to be educated at
European universities, which were far more advanced than American schools
Sugar manufacture in Antigua, West Indies. Drawing by William Clark, 1823.
DeAgostini/Getty Images
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Summary
425
at that time.) For a gifted student such as Rillieux, Paris in the 1820s was
the place to be. The Industrial Revolution was underway. Engine efficiency,
and the relationship between heat and work, were hot topics. The Parisian
Sadi Carnot published pioneering studies of steam engines in 1822–1824
and conducted work leading to formulation of the second law of thermodynamics. (Carnot’s work preceded by about 30 years that of James Joule,
whose work led to formulation of the first law of thermodynamics—the
energy balance.) Rillieux became especially interested in latent heat—the
energy required to convert liquid to vapor.
At that time, much of European sugar derived from the sugar beet.
French scientists and engineers had worked extensively to establish the
science underpinning sugar processing and to use scientific reasoning as
the basis for technology development. This was a far cry from the empirical and tradition-bound methods of Louisiana sugarmakers. Evaporation of
water from sugar beet juice required an enormous amount of energy, and
the French were attempting to develop methods to use the energy in the
steam emanating from the boiling sugar juice. Rillieux became interested
in this problem; he was familiar with the Jamaica train method of transferring sugarcane juice from one kettle to the next during the boil-up process.
His idea was to set up a cycle, where steam evaporated from one pot would
give up its latent heat to provide the energy for evaporation from the next
pot. From his study of thermodynamics, Rillieux knew that heat flows only
from hot to cold, but he also knew that the boiling temperature dropped
with a drop in pressure. He figured he could build a series of three enclosed
containers, each operating at greater vacuum than the previous. The syrup
would boil at progressively lower temperatures, and the steam from one
container would be used to heat the next.
The idea was great on paper, but Rillieux needed to prove it would work
by building a prototype. Unfortunately, the French economy was sputtering
by this time, and he could not find funding or a manufacturer to test his idea.
In contrast, on the other side of the Atlantic, the sugar business was undergoing explosive growth. In 1833, Rillieux left Paris and returned to Louisiana,
as the chief engineer at a sugar refinery owned by the wealthy Edmond
Forstall. It was probably a difficult decision, since he was returning to a land
where slavery was still legal. What Rillieux found upon his return was that
Louisiana sugar production technology was still in the dark ages. Sugar produced by the Jamaica train was dark, heavy, and dirty, but federal tariffs
protected the processors from competition. Sugar syrup evaporation took a lot
of energy and labor, and the local swamps were stripped bare of timber.
Forstall hired Rillieux to solve the problems of poor sugar quality, high
energy use, and high labor costs. Rillieux worked for 10 years, first for
Forstall and then alone, perfecting his triple-effect evaporator design, filing
patents, and building prototypes. These failed to work reliably, because the
equipment was home-made. He needed a professional machinery company to
build the equipment to tight specifications, and he needed financial backing.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
Rillieux’s chance came when he met the millionaire planter Judah Benjamin,
the first openly Jewish U.S. senator. Finally, Rillieux was able to get a
professionally built apparatus. The system had 3 stages and was fueled
entirely on discarded dried cane (called bagasse). It was installed on
Benjamin’s plantation in 1843 and worked extraordinarily well. Profits went
up 70%. To top it off, the sugar was a much higher quality, as good as any
produced by secondary refiners in the north. In fact, Benjamin’s beautiful
white sugar crystals were prize winners.
But the situation in the South was deteriorating in the years leading up
to the Civil War. Legal rights that free men of color had enjoyed for years
were curtailed. Even as he traveled around the state installing his nowheralded invention, Rillieux was forced to stay in slave quarters. Patent
examiners challenged his legal right to file patents as a nonwhite. Increasingly
frustrated, he moved back to France in the 1860s, married a young French
woman, took up the study of Egyptian hieroglyphs, and abandoned his engineering and science work.
Rillieux’s invention may have had the greatest impact in central Europe,
where farmers eagerly adopted new technology, and both land and labor
were scarce resources. By 1888, about 150 Rillieux evaporators were
installed in Germany, Austria, and Russia. By 1900, German agriculture was
transformed; by building on a strong scientific and technological basis the
country began exporting food products that led to expansion of the German
economy. Rillieux’s multi-stage evaporator is still in use worldwide, adapted
to a slew of energy-intensive processing industries.
Quick Quiz Answers
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
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Liquid: water and sugar. Vapor: CO2
Because the sulfuric acid is present in small quantities but the aromatics
in large; liquid-liquid extraction scales with the quantity of material to
be recovered, so we would need a lot of benzene. Additionally, benzene
is more difficult to separate from the other aromatics and is more expensive than water.
β = 2.3 h−1.
DOF = 2. (6 stream variables, 1 specified flow, 1 specified composition,
2 material balances)
​wi​  j​​ = ​mi​  j​​∕​mj​  ​​​.
​fR​  ij​​ = ​mi​  j​​∕​mi​  F​​ = w
​ i​  j​​ ​mj​  ​​∕​wi​  F​​ ​mF​  ​​​.
Purity = 1.0, recovery = 1.0, separation factor approaches infinity.
3.5 wt%.
0.8125 without entrainment, 0.78 with entrainment.
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Chapter 6 Problems
427
References and Recommended Readings
1. Physical property data useful for initial selection of appropriate separation
technologies are available in reference books such as the CRC Handbook
of Chemistry and Physics, Lange’s Handbook of Chemistry, or Physical
and Thermodynamic Properties of Pure Chemicals (DIPPR database), published by Taylor and Francis. The Knovel Engineering and Scientific
Online Reference is also a useful source.
2. For more on the life and times of Norbert Rillieux, see Prometheans in
the Lab, by S. B. McGrayne.
Chapter 6 Problems
Warm-Ups
Section 6.1
P6.1 List three physical property differences between NaCl and H2O that
might be used as the basis for a separation.
P6.2 Limestone (CaCO3) is essentially insoluble in water, while NaCl can be
dissolved in water up to 360 g/L. If you add 10 g CaCO3 and 400 g
NaCl to a beaker and then fill with water to 1 L, how many phases are
there? What components are in each phase?
P6.3 100 mL hexane, 100 mL water, and 1 mL of an oil-based dye are mixed
gently in a cup, then allowed to sit. Two layers form. Identify the two
layers, and explain whether the dye is in the top or bottom layer.
P6.4 A simple salad dressing is made by mixing vinegar and oil, while mayonnaise is a condiment made by mixing vinegar, oil, and egg yolk (along
with seasonings). Is this salad dressing one phase or two? Is mayonnaise
one phase or two? For both foods, list the major components in each phase.
Section 6.2
P6.5 In the case study, both filtration and ultrafiltration are used. Briefly
explain why filtration is a mechanical separation but ultrafiltration is
rate-based.
P6.6 Match up each separation technology (left) with the physical property
difference exploited (right).
crystallizationdifference
liquids
adsorption
difference
liquid-liquid extraction difference
distillation
difference
membrane filtration
difference
absorption
difference
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in solubility in two immiscible
in
in
in
in
in
freezing point
binding to solid
size
volatility
solubility of gas in a liquid
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
P6.7 Briefly explain the similarities and differences between flash vaporization, condensation, and distillation.
P6.8 Briefly explain the similarities and differences between adsorption, absorption, and solvent extraction. Draw flow sheets to illustrate your explanation.
Section 6.3
P6.9 A feed stream F containing two components A and B is fed to a continuous-flow steady-state separator. Two product streams, P1 and P2,
leave the separator. Both streams contain A and B, but A is concentrated
more in P1 and B is concentrated more in P2. Write in symbols the
material balance equations for A, B, and total, using mass fractions and
mass flow rates.
P6.10 A separator operating in semibatch mode is initially loaded with a feed
F containing two components A and B. One product stream, P1, is
removed continuously from the separator. The other product stream, P2,
is collected all at once from the separator when the process is concluded.
Both P1 and P2 contain A and B, but A is concentrated more in P1 and
B is concentrated more in P2. Write in symbols the material balance
equations for A, B, and total, using mass units.
Section 6.4
P6.11 100 gmol of a gas containing 25 mol% CO2 and 75 mol% N2 is to be
separated into two products, P1 and P2. P1 will contain 20 gmol CO2
(C) and 10 gmol N2 (N). What is zCP1, zNP2, fRCP1, and fRNP2?
P6.12 A feed to a separator contains 4 components: A, B, C, and D. One
product stream contains A, B, and C. The other product stream contains
B, C, and D. Identify the key and nonkey components. If the feed flow
rate and feed composition are given, determine how many additional
specifications are required for DOF = 0.
P6.13 A gas stream containing 25 mol% CO2 and 75 mol% N2 is separated into
two product streams; one is 80 mol% CO2 and the other is 93.3 mol% N2.
Calculate the separation factor α​
​​ C​O2​ ​-​N2​ ​​.​
Section 6.5
P6.14 For the separator flow sheet with recycle (Fig. 6.7), the overall recovery
of component B, fRB, is related to the recovery of component B
in the separator unit, fRB1, and the fractional split fS, as ​​f​ RB​​ = ​fR​  B1​​∕
(1 − ​fS​  ​​ (1 − f​ R​  B1​​))​. Derive the relationship between fRB and fRB1 for two
cases: fS = 0 and fS = 1. Explain briefly why your result makes sense
in each case.
Section 6.6
P6.15 Limestone (CaCO3) is essentially insoluble in water, while NaCl can be
dissolved in water up to 360 g/L. 10 g CaCO3 and 400 g NaCl are
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Chapter 6 Problems
429
placed in a beaker and water is added to the 1-liter mark. After stirring
overnight, the liquid is poured off. Remaining in the beaker is the solid
plus entrained liquid. What is the composition of the entrained liquid?
Drills and Skills
Section 6.2
P6.16 You are faced with solving several different separation problems, as
listed below. For each problem, choose the best separation technology
from this list: distillation, sedimentation, flash vaporization, condensation, absorption, filtration, leaching, crystallization, solvent extraction,
adsorption. Write the name of the chosen technology in the table. State
whether it is a mechanical separation, or an equilibrium-based separation.
Separation problem
Best separation technology
Recovery of antibiotics from
fermentation broth
Removal of isopropanol vapor
from air
Recovery of limestone sludge from
saline solution
Recovery of soybean oil from
soybeans
Removal of colored impurities from
high fructose corn syrup
Separation of methane from digested
manure
Separation of CO2 and H2
Separation of ethylbenzene and styrene
Removal of yeast from beer
Recovery of potassium nitrate from
aqueous solution
P6.17 Consider the following separation technologies: drying of solids, adsorption, distillation, electrophoresis, absorption, reverse osmosis. Make a
table showing, for each of these technologies, (a) whether it is mechanical, rate-based or equilibrium-based, (b) if equilibrium-based, whether
it has an energy separating agent or a material separation agent, (c) the
phase or phases of the feed, (d) the phases of the two products.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
P6.18 In regions of the world where freshwater is scarce, water is obtained by
desalination of seawater. Identify three physical properties that differ
between water and NaCl (the major salt in seawater) and select at least
three separation technologies that could be used to produce pure water
based on these property differences. Briefly explain which of the technologies is the best choice to provide desalinated water to a small city.
P6.19 Given the process stream described in the following table, devise what
you think is the best flow diagram for separating it into four essentially
pure product streams. Indicate the separation technology you would
choose and the sequence of separation steps. Briefly explain your reasoning. A quantitative answer is not required.
Normal
Mol% in
boiling
Compound
feed
point (°C)
Normal
melting
point (°C)
Soluble in
water?
Soluble in
benzene
Naphthalene
12
218
80.2
no
yes
Ethylene glycol
18
197
−11.5
yes
no
Ethylbenzene
32
136.2
−95
no
yes
Styrene (vinylbenzene)
38
145.2
−30.6
no
yes
P6.20 A liquid bromine (Br2) stream contains 2% chlorine (Cl2) and 0.02%
chloroform (CHCl3) as contaminants. The contaminants must be removed
before the bromine can be used for further fine chemical manufacturing.
The problem is that the boiling point of chloroform is very similar to
that of bromine. However, the following reaction takes place at 250°C
over a catalyst:
_​  3 ​ ​Br​
2
2​​
+ ​CHCl​3​​ → ​CHBr​3​​ + _​ 32 ​ ​CI​2​​
Sketch out a process flow sheet for removing the chlorine contaminants
from the liquid bromine. Write a paragraph justifying your design.
Sections 6.3 and 6.4
P6.21 100 gmol/min of a gas stream containing 30 mol% ethane (C2H6) and
70 mol% methane (CH4) is fed to a distillation column, where it is
separated into an overhead product containing 90 mol% methane and a
bottoms product containing 98 mol% ethane. Calculate the overhead and
bottoms flow rates and the fractional recoveries of methane and ethane
in their corresponding product streams.
P6.22 A mixture (18 mol% A, 32 mol% B, 23 mol% C, and 27 mol% D) is
fed at 100 gmol/min to a continuous-flow steady-state separator with
two product streams. Product 2’s flow rate is 81 gmol/min and its
composition is 3.7 mol% A, 34.6 mol% B, 28.4 mol% C, and 33.3 mol% D.
Draw and label a flow diagram. Calculate the flow rate and composition
of Product 1. What is zA1? fRB2? What is αAB?
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Chapter 6 Problems
431
P6.23 A student is interested in separating 10 gmol of a mixture of 30 mol%
ethanol and water in a laboratory distillation apparatus. He loads a
round-bottom flask with a liquid ethanol–water mixture. The flask has
a long neck with a side-arm. He connects one end of a long tube to the
side-arm and then positions the other end of the tube over a beaker. He
wraps ice around the long tube. Then he places the round-bottom flask
on top of a heating mantle and begins the distillation process. As the
solution in the flask is heated, vapor escapes through the tube, condenses,
and collects in the beaker. The vapor flow rate is held constant at
0.25 gmol/min. The mole fraction ethanol in the vapor and the liquid
are related as yE = 1.2 xE. Explain whether this separation is an example
of continuous-flow steady-state, semibatch, or batch. Using the liquid
in the flask as the system, write a differential material balance equation
on ethanol and also on total moles in the system. Calculate the gmol
liquid remaining in the flask as well as the mole fraction ethanol in the
liquid after 20 minutes.
P6.24 A fermentation broth contains 2 wt% antibiotic, which is to be recovered
by extraction with an organic solvent. For simplicity, you can assume
the only other component in the broth is water. 10 kg broth is mixed
with 20 kg solvent. After settling, the mixture forms two liquid phases;
all of the water and 1% of the antibiotic are in the aqueous phase, and
all of the solvent plus the remaining antibiotic are in the “oily” phase.
Calculate the wt% antibiotic in each of the two phases.
P6.25 The major proteins in whey are alpha-lactalbumin (ALA) and beta-lactoglobulin (BLG). ALA is a valuable ingredient in infant formula, while
BLG is useful in foods as a gelling agent. The two proteins differ in
isoelectric point and so can be separated by using a technology called
charged membranes. Briefly, a positive charge is placed on porous membranes such that most of the negatively charged ALA can flow through
the membrane while the positively charged BLG is rejected. 3.0 g ALA
and 7.0 g BLG are dissolved in 1000 g water and processed through the
charged membranes. 3% of the BLG and 87% of the ALA flow through
the membrane (this product is called the filtrate) while the remainder is
rejected (the retentate). Calculate the separation factor αALA-BLG.
P6.26 One way to separate hydrogen sulfide (H2S), a toxic and smelly gas,
from methane (CH4) is to take advantage of a chemical reaction. “Sour”
methane is pumped across a vessel containing pellets of zinc oxide
(ZnO), where H2S reacts with ZnO, leaving behind solid ZnS. The water
leaves as a vapor phase along with methane.
​​H2​ ​​S(g) + ZnO(s) → ZnS(s)​+ H​2​​O(g)​
You need to treat 1 million SCF/day sour methane, which contains
0.75 mol% H2S. The “sweetened” gas must contain no more than
0.01 mol% H2S. You would like to remove and discard the ZnS no more
than once per day. Write material balance equations on H2S and ZnO,
using the pellet-packed vessel as the system operated in semibatch mode.
What is the minimum quantity of ZnO that must be packed into the vessel?
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
P6.27 One proposed solution for removing CO2 from the atmosphere is injection
into the ocean. To achieve this, CO2 must be first captured from the air
and concentrated. CO2 is acidic, so direct injection has adverse effects
on marine life. One solution is to build CO2 sequestration reactors at
power plants located near the ocean. Combustion gases released into the
atmosphere through the stack contain about 10 mol% CO2. These gases
would be pumped across a bed of porous limestone (CaCO3) that is continuously sprayed with water. The CO2 adsorbs onto the limestone and
reacts with it, producing calcium bicarbonate (Ca(HCO3)2), which is alkaline and soluble in water (5 gmoles per 1000 liters). The calcium bicarbonate solution would be pumped continuously into the ocean, while the
remaining combustion gases would be released to the atmosphere. You are
in charge of designing a process to treat 1 ton/day CO2 in combustion
gases, and would like to remove 98% of the CO2 in the combustion gases
and to replace the limestone once per day. How much limestone (kg) is
required each day? What is the flow rate (L/day) of water required?
P6.28 A gas mixture containing 72 mol% CH4, 13 mol% CO2, 12 mol% H2S, and
3 mol% COS is to be purified in an absorber by contacting the gas with
a liquid solvent. The gas is fed at 3200 gmol/h and the gas feed rate/solvent
feed rate ratio is 3:1. The solvent absorbs 97.2% of the H2S in the gas feed
stream. The COS concentration in the exiting gas stream is 0.3 mol%. CH4
and CO2 are not absorbed in the solvent at all, and no solvent leaves with
the gas. First complete a DOF analysis and show that the problem is completely specified. Calculate the flow rate and composition of the exit gas
and the concentration of H2S and COS in the exit liquid (solvent) stream.
Sections 6.5 and 6.6
P6.29 Popcorn is to be dried with hot air as shown in the flow sheet. The gas
stream recycle flow rate is 4 times the hot air feed flow rate. The desired
popcorn production rate is 50 kg/hr. What feed rate of hot air is needed
if the exit air is to be at 15 volume% water? Give your answer as a
volumetric flowrate (liters/h). Model air as an ideal gas. Assume the air
temperature is 80°C and the pressure is 1 atm. Find the moisture content
of the air entering the dryer.
Exit air
Recycle air
Hot air: 80°C
2 vol% moisture
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Corn
25 wt% moisture
Dryer
Corn
10 wt% moisture
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Chapter 6 Problems
433
P6.30 Ceramic particles are ground into a powder in a machine called a ball
mill, a cylindrical rotating device that is filled with steel balls. The
particles are fed into the ball mill, and as the mill rotates, some of the
particles are ground into fine powder. The powder plus the larger particles are sent to a screen, which separates the powder product from the
larger unground particles. 70 percent of the particles fed to the mill are
not ground sufficiently into powder, and most of these are recycled back
to the mill inlet. The ceramic particles fed to the process contain 1 wt%
contaminant particles which are too hard to be ground; to avoid excessive buildup in the mill of these hard particles, 10% of the recycle
stream is split off as a purge stream. The system is designed to produce
10 tons/day ceramic powder. Draw and label a flow sheet. Identify
purity and recovery specifications, and complete a DOF analysis. Then
calculate the flow rate of all streams.
P6.31 A 30 wt% Na2CO3 aqueous solution is fed at 10,000 lb/h to an evaporator, where 40% of the water is removed. This produces a slurry containing crystals of pure Na2CO3 plus an aqueous solution of 17.7 wt% Na2CO3.
This slurry is fed to a filter, which produces a filter cake and a filtrate
solution. The filter cake entrains 1.0 lb solution per 4.5 lb crystals.
Calculate the filter cake production rate, the purity (wt% Na2CO3) of the
filter cake and the fractional recovery of Na2CO3 in the filter cake.
P6.32 100 kg/h of a 30 wt% KNO3 aqueous solution is cooled to 5°C, which
causes some of the KNO3 to precipitate as a solid. The remaining solution is 14 wt% KNO3. The solid and liquid phases are separated by
filtration. The filter cake entrains 1 kg liquid per 9 kg solid. Calculate
the purity (wt% KNO3) of the filter cake, the fractional recovery of
KNO3 in the filter cake, and the separation factor.
P6.33 For the KNO3 separation problem described in P6.32, your supervisor
suggests recycling 50% of the filtrate back to the feed. How would
recycle change product purity, fractional recovery, and separation factor?
P6.34 Sulfur dioxide (SO2) is manufactured by the oxidation of solid sulfur S,
using air (79 mol% N2, 21 mol% O2) as the source of oxygen. S and O2
are fed at stoichiometric ratio and the reaction goes to 100% completion.
The reaction generates a lot of heat, so cool inert gas would be useful to
maintain a cooler reactor temperature. Ideally there would be 5.0 moles
inert gas per mole O2 fed to the reactor. Eddie Engineer proposes to use
nitrogen in the air as the inert gas. In this scheme, air and sulfur are fed
to the reactor, and the sulfur dioxide is separated from nitrogen downstream of the reactor. Eddie estimates that the separator will recover 97%
of the SO2 in a product stream that is 99 mol% SO2. The other product,
which is nitrogen-rich, could be recycled to provide sufficient inert gas
to the reactor inlet. Draw and label a flow sheet that illustrates Eddie’s
idea. Complete a DOF analysis, choosing 100 gmol/min sulfur feed as a
basis. What fraction of the nitrogen-rich stream should be recycled?
Calculate the flow rates and compositions of all streams. (Hint: First solve
the case with 100% recovery of SO2, to obtain an initial estimate.)
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
P6.35 100 gmol/min of a gas stream containing 30 mol% ethane (C2H6) and
70 mol% methane (CH4) is fed to a distillation column, where it is
separated into an overhead product containing 90 mol% methane and a
bottoms product containing 98 mol% ethane. A customer wants to purchase a product that contains 97% methane. To produce this, the overhead from the column described above is sent to a second column. The
overhead from the second column is the desired product, and the bottoms product, which has a flow rate of 100 gmol/min, is recycled back
to the first column. Calculate the flow rate of the final product and the
composition of the recycled stream. Calculate the fractional recoveries
of methane and ethane in their corresponding product streams.
P6.36 The average Wisconsin cheese plant makes 300,000 lb/day of whey as
a byproduct. Dumping the whey into the nearest river is not a great
solution for disposal of this waste product. It’d be much better to
develop ways to make products from whey.
Cheese whey contains about 93.4 wt% water, 0.9 wt% protein, 5 wt%
lactose (C12H22O11, milk sugar), 0.2 wt% lactic acid (CH3CHOHCOOH),
and 0.5 wt% inorganic salts. Here is some information about each of
these components.
Proteins: A mixture of proteins with molecular weights from 15,000
to 150,000. Highly soluble in water. Will precipitate as a gel if
concentrated to 50 to 60 wt%. Adsorbs to ion exchange adsorbents
if low salt concentration. Valuable as animal feed, or as an ingredient in processed food.
Lactose: Molecular weight = 342. Soluble in water to about 200 g/L.
In large concentrations has an unwelcome laxative effect in mammals. Degrades at high temperatures. Does not adsorb to ion
exchange adsorbents. Of some value as animal feed, in fermentation broth, or as a raw material for polymer production.
Lactic acid: Molecular weight = 90. Very soluble in water. Fairly
high mobility in electric field. A colorless viscous liquid that is
highly soluble in ethanol and ether.
Mineral salts: primarily calcium and sodium salts, with molecular
weight of 50 to 100. Highly soluble in water as ions. High mobility in an electric field. Salts limit palatability of feedstuffs.
Your job is to develop a preliminary flow diagram for making two
valuable products—a high-protein, low-salt dry solid and a high-­lactose,
low-salt dry solid—from cheese whey. Sketch out what you think is
the best block flow diagram. Propose a specific technology for each
separation. Write one or two paragraphs describing your design and
justifying why you think your design is best.
P6.37 Benzene is chlorinated to chlorobenzene
​​C6​ ​​ ​H6​ ​​ + ​Cl​2​​ → ​C6​ ​​ ​H5​ ​​ Cl + HCl​
The Cl2 concentration in the feed is kept below 10 mol% to prevent
unwanted additional chlorination of the monochlorobenzene. Essentially
all of the chlorine reacts under the reactor conditions.
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Chapter 6 Problems
In a second reactor, the byproduct HCl is converted back to Cl2 by
oxidation:
​4HCl + ​O2​ ​​ → 2​Cl​2​​ + 2​H2​ ​​O​
The reaction is reversible; at the reactor conditions 60% of the reactants
are converted to products.
You are given the following information:
No O2 or H2O is allowed in the feed to the first reactor.
Cl2 dissolved in water will preferentially partition into carbon tetrachloride, with a distribution coefficient (moles Cl2 per liter carbon
tetrachloride/moles Cl2 per liter water) equal to 5.0.
The solubility of Cl2 in water is 3.5 g/liter.
The solubility of HCl in water is 720 g/liter.
Sketch out a process flow sheet for production of chlorobenzene. Show
the components in each stream on your flow sheet. Indicate the basis
for separation in all cases (for example, difference in volatility or solubility). You do not have to do any calculations.
P6.38 Propylene (C3H6) and chlorine (Cl2) react to produce allyl chloride
(C3H5Cl) with hydrogen chloride (HCl) as a byproduct. Several unwanted
reactions can also occur, of which the major unwanted byproduct is 1,3
dichloropropane (C3H6Cl2).
Given the reactor effluent information described below, devise what
you think is the best process flowsheet for producing pure allyl chloride,
while recycling unreacted propylene and chlorine back to the reactor
feed. Indicate the types of separation technologies you would use and
the sequence of separation steps. Briefly explain your reasoning.
A quantitative answer is not required.
Component
Relative amount,
weight basis
Normal boiling
point, °C
Solubility in
water, wt%
1,3 dichloropropane
1.8
112
Insoluble
Acrolein chloride
0.2
84
Insoluble
Allyl chloride
9.3
50
0.33
Chlorine
3
−34
0.35
105
−48
0.89
93
−85
72
Propylene
Hydrogen chloride
P6.39 Acetaldehyde (C2H4O) is produced by partial oxidation of ethane (C2H6)
over a catalyst:
​​C2​ ​​​H6​ ​​ + ​O2​ ​​ → ​C2​ ​​​H4​ ​​O + ​H2​ ​​O​
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
A number of side reactions also occur, the most important of which are:
​​C​2​​​H6​ ​​ + ​3.5O​2​​ → ​2CO​2​​ + ​3H​2​​O​
​​C2​ ​​​H6​ ​​ + ​1.5O​2​​ → ​CH​3​​OH + CO + ​H2​ ​​O​
In a process to produce acetaldehyde, ethane at 6000 gmol/h is mixed
with 30,952 gmol/h air. The fresh feed is mixed with a recycle stream,
then fed to a reactor. The ethane:oxygen ratio in the reactor feed is
maintained at 6:1. The reactor outlet stream is fed to gas-liquid
Separator 1, where N2, CO, CO2, and C2H6 are taken off the top and
recycled. Part of the recycle stream is split off and sent to a flare to
be burned. This purge stream is analyzed for composition: It contains
10% C2H6, no O2, and the CO2:CO ratio is 2:1. The bottoms stream
from Separator 1 is sent to Distillation Column 2, where acetaldehyde
and methanol (CH3OH) are separated from water. 100% of the acetaldehyde and 95% of methanol is recovered in the overhead of
Distillation Column 2. Acetaldehyde is further separated from methanol in Distillation Column 3. 98% of acetaldehyde is recovered in the
overhead, while 95% of methanol is recovered in the bottoms of Column
3. A simplified process flow diagram is shown. You may assume that
air is 79 mol% N2, 21 mol% O2, and that the products from the distillation columns are essentially pure, with only trace contaminants.
Identify the key components in Separator 1, Distillation Column 2
and Distillation Column 3. Given the heuristics you learned, do you think
this is the best sequence of separation tasks? Explain why or why not.
Complete a DOF analysis, and show that the process is correctly
specified. Calculate the composition and flow rate of the purge gas,
and the fractional yield, selectivity, and conversion of acetaldehyde
from ethane for the overall process. Calculate the overall purity (mol%
acetaldehyde) in the product stream leaving Distillation Column 3, and
the percent recovery of acetaldehyde leaving the reactor that is recovered in the final product.
Splitter
C2H6
O2
N2
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Mixer
Reactor
To f lare
N2
CO2
CO
C2H6
Separator
1
Distillation
column
2
Distillation
column
3
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Chapter 6 Problems
437
P6.40 Consider the case study where the enzyme ProTase was purified from
a fermentation broth. Re-analyze the process, but with the following
differences: (a) centrifugation (producing a 20 wt% solids cake) was
used instead of filtration for the first step and (b) ultrafiltration to only
15 wt% (rather than 20 wt%) protein content is achievable. Determine
the purity of the final product as well as the overall fractional recovery
of ProTase. Compare to the case study and comment on whether these
changes have a minor or major impact.
P6.41 High-purity silicon, used in making electronic devices and solar cells,
is produced from two inexpensive raw materials, sand (SiO2) and coke
(C). Typically, about 500 kg sand and 200 kg coke are placed in a retort.
Sand and coke are then heated to about 400°F, which produces solid
silicon and carbon monoxide gas:
​​SiO​2​​(s) + 2C(s) → Si(s) + 2CO(g)​
Si and CO are easily separated into solid and gas phases. Because
of impurities in the sand, the solid Si phase is about 98.5 wt% pure.
This is not sufficiently pure, so two additional reactions are employed.
First Si reacts with chlorine gas to make tetrachlorosilane:
​Si(s​)​​ + 2Cl​2​​(g) →​ SiCl​4​​(g)​
Tetrachlorosilane reacts with magnesium:
​​SiCl​4​​(g) + 2Mg(s) →​MgCl​2​​(s) + Si(s)​
MgCl2 is soluble in water (54.2 g per 1000 g water), whereas Si is
not. Therefore, sufficient water is added to dissolve the MgCl2. Solid
and liquid are separated by filtration. Si is recovered as a filtrate cake
along with some entrained liquid (0.1 lb liquid per lb solid). The cake
is sent to a dryer to remove the entrained liquid.
Draw and label a flow diagram. Assuming 100% conversion of the
Si-containing compounds in each of the reactions, calculate the quantities of all raw materials, the quantity of dried Si product, and the purity
of the final product.
P6.42 A light alkane mixture is one of the products made when crude oil is
distilled at a petroleum refinery. The mixture must be further separated
into its components, which are valuable feed stocks for a variety of
industrially important reactions. Distillation is the separation technology
of choice, and each distillation unit produces two products: “overhead”
and “bottoms.”
In one plant, the light alkane stream contains 10 mol% methane,
30 mol% ethane, 15 mol% propane, 30 mol% butane, and 15 mol%
pentane. This stream (1000 kgmol/day) is sent to Separator 1. 100% of
methane and ethane, and 44.6% of propane is recovered in the overhead,
while all the remainder is in the bottoms. The overhead from Separator 1
is mixed with the overhead from Separator 4 in Mixer 1, and the output
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
from Mixer 1 is sent to Separator 2. 99.5% of methane fed to Separator 2
is recovered in its overhead; 99.83% of ethane and all propane fed to
Separator 2 is recovered in its bottoms. The bottoms from Separator 2
is fed to Separator 3. 100% of methane and 99.5% of ethane fed to
Separator 3 is recovered as overhead; 98.5% of propane fed to Separator 3
is recovered in the bottoms. The bottoms from Separator 1 is fed to
Separator 4. 96.4% of propane fed to Separator 4 is recovered as overhead, which is sent to Mixer 1 as mentioned previously. 100% of butane
and pentane fed to Separator 4 is recovered in its bottoms, which is fed
to Separator 5. 100% of propane and 99% of butane fed to Separator 5
is recovered as overhead, while 100% of pentane fed to Separator 5 is
recovered as bottoms.
Draw and label a flow diagram of the process. Identify the key
components in each separator. Complete a DOF analysis to determine
if the entire process is correctly specified. Then solve for all flows and
compositions.
P6.43 In a process for making cellulose acetate, an aqueous acetic acid waste
stream (30% acetic acid, 0.2% sulfuric acid, and water) is produced. (All
compositions are mass percent.) A solvent extraction process, using
ether as the solvent, was developed to purify and concentrate the acetic
acid.
The process is described as follows. The aqueous acetic acid waste
stream is fed to an extraction column, along with the solvent diethyl
ether, which is contaminated with a small bit of water. The ether-rich
phase leaving the top of the extraction column contains 24% acetic acid,
water, and ether. This is fed to a solvent recovery distillation column.
The overhead from this column contains 98.8% ether and 1.2% water
and is recycled back to the extraction column. The bottoms from the
solvent recovery column contains 60% acetic acid and 40% water, and
is fed to an acid finishing distillation column. The bottoms from the
acid finishing column contains 99% acetic acid with the remainder
water; this concentrated acetic acid stream is the desired product. 67.5%
of the acetic acid fed to the acid finishing column is recovered as product. The overhead from the acid finishing column, which is dilute acetic acid in water, is recycled back and mixed with the fresh feed. There
is 1 lb acetic acid recycled for every 2.3 lb acetic acid in the fresh feed.
The water-rich stream leaving the bottom of the extraction column
contains 7% ether, acetic acid, water, and sulfuric acid. This stream is
fed to an ether-stripping distillation column. The overhead from the
ether-stripping column contains 98.8% ether and 1.2% water; this is
recycled back to the extraction column. The bottoms contains 0.1%
ether, acetic acid, water, and sulfuric acid and is discarded. To make up
for the loss of ether in this bottoms column, fresh ether solvent (contaminated with 1% water) is mixed with the other recycled ether streams
and fed to the extraction column.
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439
(a) Evaluate the choice of separation technologies and the sequence of
separations, in light of the heuristics. You may want to look up
relevant physical properties of the components in this process. Why
is solvent extraction used as the first step rather than distillation?
(b) Draw and label a process flow diagram. For each of the four separators identify the two key components that are being separated.
Complete a DOF analysis and show that the process is correctly
specified, except for the choice of a basis. Calculate the flows and
compositions of all streams, assuming that the feed rate of the aqueous acetic acid waste stream is 1000 lb/h. Summarize the flows and
compositions on a table accompanying your flow sheet. What fraction of the acetic acid fed to the process is recovered in the concentrated product?
P6.44 The following feed stream is to be separated by a series of distillation
columns into four products:
Product
Product
Product
Product
Species
Flow rate, gmol/h
Pentane
4000
Benzene
1000
Toluene
1000
Orthoxylene
6000
1:
2:
3:
4:
98%
90%
90%
99%
pentane, no orthoxylene or toluene
benzene, 4% toluene, no orthoxylene
toluene, 2% benzene, no pentane
orthoxylene, no pentane or benzene
Your supervisor proposes the following design. Do you think this is the
best sequence? To analyze, find the boiling points of the four compounds, and review the heuristics on sequencing of separations. If not,
propose an alternative design. For the design you choose, calculate the
product flow rates and purities and the fractional recoveries of each
species in the appropriate product.
Product 2
Product 1
Product 3
Product 4
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
P6.45 A batch thickener is simply a cylindrical tank with an opening at the
bottom, used sometimes for sedimentation operations. Initially the bottom opening is closed, and the tank is filled with a slurry. (A slurry is
just a mixture of liquid solution and suspended solid particles.) The
material is allowed to settle, and then after some time clear liquid is
drawn off the top of the tank and a thickened sludge is pulled out the
bottom opening.
One liter of a slurry containing 2 g NaCl and 230 g limestone
(mostly CaCO3) is poured into a small cylindrical glass tank. The tank
is equipped with a bottoms drawoff. The slurry comes 36 cm up the
side of the tank. After 8 h settling time, an interface between sludge
and clear liquid is apparent, 10 cm up the side of the tank. (The top
of the clear liquid is still 36 cm up.) The sludge is carefully drained
out the bottom and then the clear liquid is removed from the top. What
is the quantity (in total liters and in g NaCl and limestone) and composition (in g/L NaCl and limestone) of the two product streams?
The solubility of limestone in water is 0.015 g/L, the solubility of
NaCl in water is 360 g/L, and the density of limestone can be taken
as 2.7 g/cm3.
There are three components in the feed stream and two product
streams. Is this process designed to separate limestone from NaCl, limestone from water, NaCl from water, or . . . ? What is the product purity
and product recovery? What is the separation factor?
P6.46 In a process to synthesize the pain reliever acetylsalicylic acid (ASA,
also known as aspirin), the effluent stream from the final reactor contains 11 wt% ASA and 2 wt% sodium acetate in water. A dried powder
is the desired final product. The product purity, measured as weight
percent ASA in the final dry powder, is a key value.
(a) The simplest method to achieve a dried product is to evaporate the
water. If evaporation is chosen, what product purity results?
(b) Sodium acetate is highly soluble in water, so an alternative
process is suggested. Part of the water is removed by evaporation, then the concentrated solution is cooled to crystallize
some pure ASA. The filter cake, which contains 1 kg entrained
solution for each 4 kg of ASA crystals, is then dried. At the
crystallizer/filter temperature, the solubility of ASA in water is
35 wt%. If enough water is evaporated so that 50% of the ASA
fed is recovered as dry product, what is the ASA product final
purity?
(c) Obtain a general equation that relates product purity to fractional
recovery. Plot your expression. What purity do you predict at 90%
ASA recovery? At 10% ASA recovery?
(d) Suggest changes to this process that could increase recovery, purity,
or both.
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Chapter 6 Problems
441
P6.47 A 30 wt% Na2CO3 solution is fed at 10,000 lb/h to an evaporator/
crystallizer system shown in the figure. The filter cake contains 3.5 lb
crystals per lb entrained solution, and the entrained and recycled solution both contain 17.7 lb Na2CO3 per 100 lb of solution. What is the
production rate of crystals? How much water is removed in the evaporator? If 40% of the water fed to the evaporator is evaporated, what
is the ratio of recycled solution to fresh feed? What is the purity
(% Na2CO3) of the product stream?
Water
10,000 lb/h
30% sodium
carbonate
Evaporator
Crystallizer/
filter
Crystals and
entrained
solution
Recycled solution
You’d like to improve the performance of the process. The evaporator
has a maximum capacity of 7500 lb water evaporated per hour. The
concentration of Na2CO3 is fixed at its solubility limit of 17.7 lb/100 lb
solution. (a) Assuming that the entrainment remains the same, will
changing the recycle ratio improve product purity and/or recovery?
Explain. (b) Suggest at least one other process modification and analyze
how your proposal will affect product purity and/or recovery.
P6.48 The xylene isomer p-xylene is a starting material for the production
of polyester fibers. m-xylene is blended into gasoline, and is less valuable than p-xylene. The volatilities of the two xylenes are quite similar, but their melting points are different, so crystallization is proposed
as a means to separate them.
A mixture of 30 wt% p-xylene and 70 wt% m-xylene is fed at a rate
of 100 kg/h to a heat exchanger, where the stream is cooled to −45°C.
At this temperature, some p-xylene crystallizes as a solid, and the
remaining solution is 16.5 wt% p-xylene. The stream is next sent to a
crystallizer/filter unit, where crystals are separated from the liquid solution. Some solution is entrained with the crystals at 1 kg solution per
15 kg crystals. What is the production rate and purity of the crystals
leaving the process (including the entrained liquid)? What is the composition and flow rate of the solution leaving the filter? What is the
fractional recovery of p-xylene in the p-xylene-rich product?
A colleague proposes the following idea: Why not send the liquid
solution to an isomerization reactor, where some of the m-xylene fed to
the reactor is converted to p-xylene. (See diagram.) The product from
the reactor, which is 30 wt% p-xylene, is then mixed with the fresh feed.
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
What is the fractional recovery and purity of p-xylene in the p-xylenerich product? What do you think about this idea?
100 kg/h
30% p-xylene
70% m-xylene
30% p-xylene
70% m-xylene
Cooler
Isomerization
reactor
T = −45°C
Crystallizer/
filter
Crystals
Entrained solution
Solution
P6.49 The main active ingredients in dishwashing detergents are surfactants,
or surface-active molecules, which alter oil-water interfaces. These molecules have a part that is oil soluble, which dissolves oil and grease,
and a part that is water soluble, which partitions into the water phase,
thereby solubilizing the oil and grease. Detergents differ from soaps
because detergents are synthesized from hydrocarbons, while soaps are
made from natural fatty acids.
In one process to manufacture a surfactant, a straight-chain hydrocarbon C10H22 (decane) is chlorinated in a photocatalytic step to produce
monochlorodecane (C10H21Cl, MCD):
C10H22 + Cl2 → C10H21Cl + HCl
An undesired reaction also occurs, where MCD is chlorinated to dichlorodecane (C10H20Cl2, DCD):
C10H21Cl + Cl2 → C10H20Cl2 + HCl
The product from the reactor is fed to a high-pressure separator, where
chlorine and HCl are taken off as gases, and liquid products are sent to
a distillation unit. The liquid product contains 40 mol% decane, 50 mol%
MCD, and 10 mol% DCD, and is at a flow rate of 1000 gmol/min.
A separation flow sheet is to be designed to recover 99% of the MCD
in this stream in an MCD-rich product stream that is 97 mol% pure.
The DCD-rich product should contain no decane. The decane product
is recycled back to the reactor; to avoid excessive chlorination the
decane product should contain no DCD and no more than 2 mol%
MCD. Since there are 3 products, the separation train requires two distillation columns.
Propose two different configurations for the two columns in the
distillation unit, keeping in mind that the normal boiling points of decane,
MCD, and DCD are 174°C, 215°C, and 241°C, respectively. Identify
the key and nonkey components in each. Assume that 97% of the
decane fed to the distillation unit is recovered for recycle. Use a
DOF analysis to determine whether the problem is completely specified.
Calculate the flow rates and compositions of streams (as many as possible)
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Chapter 6 Problems
443
for the two different configurations and summarize your results in table
form along with the two block flow diagrams. Is one configuration
preferable? If so, why?
Game Day
P6.50 Ethylene glycol [C2H4(OH)2, an antifreeze] is produced in two steps.
The simplified process flow diagram is sketched.
In the first reactor R-1, ethylene (C2H4) is mixed with air (79% N2,
21% O2) and oxidized to ethylene oxide (C2H4O) in the gas phase over
a silver catalyst:
C2H4(g) + _​​ 12 ​​ O2(g) → C2H4O(g)
Complete combustion of the ethylene to water and carbon dioxide can
also occur, as an undesired side reaction:
C2H4(g) + 3O2(g) → 2CO2 + 2H2O(g)
All of the O2 fed to the process is consumed. CO2, H2O, and C2H4O
are separated from unreacted C2H4 and N2 by absorption in column A-1
into cold water. C2H4, O2, and N2 are recycled back to the reactor inlet,
with a purge stream taken off. C2H4O and CO2 are then separated from
the water in distillation column D-1. The water is discarded, and CO2
is then removed from C2H4O by absorption in column A-2 into triethanolamine (TEA). C2H4O is sent to reactor R-2.
In the second reaction, ethylene oxide is mixed with liquid water,
in which it is very soluble. Ethylene oxide reacts with the water to
produce ethylene glycol:
C2H4O(g) + H2O(l) → C2H4(OH)2(l)
Ethylene glycol can react with ethylene oxide to produce diglycol in an
unwanted side reaction:
C2H4O(g) + C2H4(OH)2(l) → (C2H4OH)2O(l)
Ethylene oxide is very reactive, and 100% conversion of ethylene oxide
to products is achieved. Water and diglycol are separated from ethylene
glycol in two distillation columns D-2 and D-3.
Identify each separation unit. Specify the main purpose of the separation unit and the separation technology used. Explain why that separation technology is or is not the best choice for that separation problem.
Simplify the flow diagram to show what it would look like if the
unwanted reactions did not occur.
The fresh ethylene feed rate to the process is 1000 gmol/h. The
C2H4/O2 ratio in the fresh feed is 2:1. 50 gmol/h CO2 is removed in
the absorber A-2. The fractional conversion of C2H4 to products in
R-1 is 0.25. Calculate the production rate of C2H4O from R-1, the
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Chapter 6 Selection of Separation Technologies and Synthesis of Separation Flow Sheets
composition and flow rate of the purge gas, and the flow rate of the
recycle stream to R-1.
Freshwater is fed to R-2 at a water:C2H4O ratio of 5:1. For every
10 mol of ethylene glycol produced, 1 mol of diglycol is produced.
Calculate the production rate of ethylene glycol. Calculate the overall
conversion of ethylene to ethylene glycol for the entire process.
The process has been operating successfully for about 10 years.
Suddenly, the price of ethylene doubles. Propose two process modifications that would improve ethylene utilization. Identify the effects of
these proposed modifications on all parts of the plant (i.e., increases/
decreases in flows or compositions). Explain your reasoning.
Purge
H2O
Ethylene
R-1
A-1
Air
Ethylene
glycol
H 2O
D-3
C2H4O
CO2
C2H4
N2
D-2
C2H4O
H2O
CO2
D-1
H2O
CO2
A-2
C2H4O
R-2
TEA
TEA
H2O
Diglycol
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CHAPTER SEVEN
7
Equilibrium-Based
Separation Technologies
In This Chapter
We take a closer look at equilibrium-based separations. These types of separation technologies are probably the most prevalent in the chemical process
industry. You will learn how to find the relationship between temperature,
pressure, and composition when two phases are in equilibrium with each other.
This relationship is critical information in the design and analysis of equilibriumbased separations, because the constraint of phase equilibrium places limits on
component recovery and product purity, somewhat akin to the way in which
chemical reaction equilibrium places limits on conversion and yield in reactors.
The questions you’ll be able to answer after finishing this chapter include:
∙ Why is phase equilibrium so important in the design of some separation
processes?
∙ How do we know the composition of two phases in equilibrium?
∙ How do we combine phase equilibrium information with material balances
to design and analyze separation processes?
∙ How do we choose the optimum temperature and pressure?
Words to Learn
Watch for these words as you read Chapter 7.
Equilibrium-based separations
Separating agent
Key component
Phase equilibrium
Equilibrium stage
Gibbs phase rule
Distillation
Crystallization
Extraction
Adsorption
Absorption
445
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Chapter 7 Equilibrium-Based Separation Technologies
7.1
Introduction
Equilibrium-based separations are probably the most common types of separation technologies. This category includes technologies that are used in a wide
variety of industries, from food processing to pharmaceutical manufacturing to
oil refining. Distillation, crystallization, extraction, absorption, adsorption—
these and related technologies are ubiquitous and essential in making the
­everyday products of modern life.
Equilibrium-based separations, just like all separations, work by exploiting
differences in physical properties of the components to be separated. For
equilibrium-based separations, these physical property differences result in
differences in the way the components distribute into two phases. In each
section of this chapter, you will learn first how to determine the distribution
of components between the two phases, and second how to use that knowledge
to evaluate specific types of separations.
7.1.1
Phases: A Brief Review
In equilibrium-based separation processes, multicomponent mixtures are separated into two phases of differing composition. Therefore, to understand separation processes, you must understand phases.
Recall from the previous chapter that a phase is a “homogeneous, physically distinct, and mechanically separable portion of matter.” Solids, liquids,
vapors, and supercritical fluids are all phases.
A phase is homogeneous. Within a phase, the chemical composition and physical properties (e.g., density, viscosity) are uniform. A phase may be a
single component, or a phase may be a multicomponent mixture of chemical species, with the species distributed uniformly at the molecular level.
A phase is physically distinct. Vapors, liquids, and solids differ in some
fundamental ways. Vapors are much less dense than liquids. Vapors are
highly compressible (i.e., their density changes a lot with pressure),
whereas liquids and solids are almost incompressible. This means that
the behavior of vapors is very sensitive to pressure, whereas that of
solids or liquids is relatively independent of pressure. Vapors and liquids
adopt the shape of their containers, whereas solids retain their shape independent of their container.
Phases are mechanically separable. One phase can be separated from
another by using mechanical forces and mechanical devices.
Terminology. A gaseous phase will be defined as a “vapor” for compounds
that, when pure, exist as a condensed phase (liquid or solid) at or near
ambient conditions. The word “gas” will typically describe compounds
that remain in the gaseous phase at temperatures and pressures near
ambient. Thus, steam is a vapor, but air is a gas. A liquid will be defined
as a “mixture” if all the components are normally liquid when pure, but
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Section 7.1 Introduction
as a “solution” when one or more of the components is normally solid
when pure. A “fluid” is either a gas or liquid; a “condensed phase” is
either a liquid or solid.
Single-Component Phase Equilibrium
7.1.2
106
104
Solid
10−2
10−4
nce
1
S
Va
por ublim
-so
atio
lid
coe n
xis
te
Pressure, mmHg
102
Melting
Liquid-solid coexistence
Let’s consider first a single-component system: water. Its phase depends on the
temperature T and pressure P. You probably already know, for example, that
water changes from liquid to vapor at 100°C and 760 mmHg. We call this the
normal boiling point Tb of water. You probably also know that water changes
from solid to liquid at 0°C and 760 mmHg—the normal melting point Tm of
water. You may know that water coexists as vapor, liquid, and solid at a single
temperature and pressure, 0.01°C and 4.58 mmHg, which is the triple point
of water. There is one other point you should know about—the critical point.
The critical point for water is 1.67 × 105 mmHg (critical pressure Pc) and
374°C (critical temperature Tc). If the temperature and pressure are both
above the critical point, then the material becomes a supercritical fluid which
is neither liquid nor vapor. All of this information is very succinctly presented
on a P-T phase diagram, such as that is shown in Fig. 7.1.
10−6
−100
Liquid
e
ling
tenc
Boi
exis Critical point
o
c
quid
TC = 374°C
or-li
Vap
PC = 1.67 × 105 mmHg
Triple point
T = 0.01°C
P = 4.58 mmHg
Vapor
0
200
100
Temperature, °C
300
400
Figure 7.1 Pressure-temperature diagram for pure H2O. Based on data from Perry’s
Chemical Engineers’ Handbook, 6th edition, McGraw Hill.
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Chapter 7 Equilibrium-Based Separation Technologies
Quick Quiz 7.1
When vapor and liquid
water are both present
and the temperature is
100°C, what must the
pressure be?
Quick Quiz 7.2
What changes more as
the pressure changes,
the boiling point temperature or the melting
point temperature?
Quick Quiz 7.3
If H2O is at 10,000
mmHg and 120°C,
what phase is it?
Let’s examine Fig. 7.1 closely. First find the triple point, which by definition is the point where three phases exist simultaneously. Once we know
we are at the triple point of water, we know we must be at 0.01°C and 4.58
mmHg; we have no flexibility about picking the temperature or the pressure!
Next, identify the solid lines, which are called coexistence curves. They
show all the combinations of P and T at which two phases (liquid and vapor,
liquid and solid, or solid and vapor) may coexist. If two phases are present
simultaneously, then there is only one degree of freedom, e.g., if P is specified then T is known from the coexistence curve, or alternatively if T is
specified then P is known.
The pressure corresponding to a specific T on the liquid-vapor or solid-vapor
coexistence curve is called the saturation pressure P sat. Vapor at T and P lying
on the coexistence curve is called saturated vapor. Vapor at T and P lying
below the vapor-liquid coexistence curve of Fig. 7.1 is called superheated. Liquid
at T and P lying on the coexistence curve is called saturated liquid. Liquid
above the vapor-liquid coexistence curve in Fig. 7.1 is called subcooled. As an
example, at 100°C and 760 mmHg, water is either a saturated liquid or a
­saturated vapor (or a mixture of both). At 100°C and 10,000 mmHg, water is a
subcooled liquid. At 100°C and 0.01 mmHg, water is a superheated vapor. H2O
at −50°C and 10,000 mmHg is a solid. These points are marked on Fig. 7.2.
The diagram also shows how to evaluate changes in phase with changing P at
constant T, or with changing T at constant P.
Suppose I have saturated steam at 120°C.
What pressure must
it be?
104
Pressure, mmHg
Suppose I have pure
H2O at 100°C and 760
mmHg. Is it a vapor, a
liquid, both, neither, or
unknown?
106
Solid
Saturated liquid
Saturated solid
Decrease T
at constant P
Increase P at constant T
448
102
1
10−2
10−4
10−6
−100
Subcooled liquid
Saturated vapor
Saturated liquid
Superheated vapor
Saturated vapor
Saturated solid
0
100
200
Temperature, °C
300
400
Figure 7.2 How to read a pressure-temperature diagram.
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Section 7.2 Multicomponent Phase Equilibrium and the Equilibrium Stage Concept
Helpful Hint
For a pure compone