1
IA
2
II A
3
III B 4
IV B 5
VB 6
Group CAS
IA
VIII B 8
Element name
VI B 7
Relative atomic mass
Symbol
Atomic number
1
Group - IUPAC
9
VII B
10
11
I B 12
II B
13 III A 14 IV A 15
V A 16 VI A 17 VII A
18 VIII A
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Cooper FrontMatter.fm Page i Tuesday, July 1, 2014 3:51 PM
I N T R O D U C T I O N
T O
ENVIRONMENTAL
ENGINEERING
Cooper FrontMatter.fm Page ii Tuesday, July 1, 2014 3:51 PM
Cooper FrontMatter.fm Page iii Tuesday, July 1, 2014 3:51 PM
I N T R O D U C T I O N
T O
ENVIRONMENTAL
ENGINEERING
C . D AV I D C O O P E R
UNIVERSITY OF CENTRAL FLORIDA
WAVELAND
PRESS, INC.
Long Grove, Illinois
Cooper FrontMatter.fm Page iv Tuesday, July 1, 2014 3:51 PM
For information about this book, contact:
Waveland Press, Inc.
4180 IL Route 83, Suite 101
Long Grove, IL 60047-9580
(847) 634-0081
info@waveland.com
www.waveland.com
Cover design by Kelly Cooper Kwoka
Copyright © 2015 by Waveland Press, Inc.
10-digit ISBN 1-4786-1142-1
13-digit ISBN 978-1-4786-1142-4
All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or
transmitted in any form or by any means without permission in writing from the publisher.
Printed in the United States of America
7
6
5
4
3
2
1
Cooper FrontMatter.fm Page v Tuesday, July 1, 2014 3:51 PM
This book is dedicated to my dear wife of 40 years, Margo,
for her patience with, and understanding of, my need to write.
This book is also dedicated to our three grandsons—
Cooper Harold Hofstetter,
Wyatt Washington Kwoka,
and Abel Oliver Jackson
—in the hope that their environment will be
even better in the future.
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Contents
Preface
1
xiii
What Is Environmental Engineering?
1.1 Introduction
1
1
Environmental Engineers—Who We Are and What We Do
Professional Engineers and Engineering Ethics 3
1.2 Thinking like an Engineer
Problem Solving 5
Units and Significant Figures
1
5
6
1.3 Growth of Human Civilization and
Its Environmental Impacts 7
The Mathematics of Population Growth 8
The Implications of Population Growth 10
Resource Consumption and Pollution Generation 12
1.4 The Rise of Environmental Protection
15
An Environmental Ethic 18
Pollution Prevention, Sustainable Development,
and Green Engineering 20
Environmental Laws, Regulations, and Agencies
24
1.5 The History of the Future 29
■ Problems 29 ■ References 30
2
Chemistry Is Important to Our Business
2.1 Introduction
2.2 Solutions
32
32
Ways to Report Concentrations in Solutions 33
Gas Behavior 39
2.3 Stoichiometry of Chemical Reactions
Balancing Simple Chemical Equations 45
Balancing Redox Reactions 48
vii
44
32
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viii
Contents
2.4 Water Chemistry
51
Equilibrium Concepts 51
Acid-Base Chemistry 52
Solubility Product 57
The Carbonate System 61
2.5 Chemical Reaction Kinetics
67
Overview 67
Elementary Reactions 70
Temperature Effects 76
■ Summary 78
3
■ Problems 78 ■ References 82
A Process Engineering Approach to Solving Problems
3.1 What Is Process Engineering?
83
83
3.2 Flow of Materials—Mass Balances
84
The Basic Balance Equation 87
Steady-State Flow Processes without Chemical Reaction 89
Unsteady-State Flow Processes 95
Balances with Chemical Reactions 96
Unsteady-State Processes 100
3.3 Flow of Energy—Energy Balances
103
Various Forms of Energy and Power 103
Energy Balances 105
Real-World Energy Conversion Processes 107
3.4 Combined Material and Energy Balances
3.5 Numerical Methods
108
111
Simulation of Unsteady-State Processes 111
Iterative Solution of Nonlinear Algebraic Equations
116
■ Problems 120
4
Water Resources
4.1 Water Quantity
128
128
Precipitation Quantities 131
Runoff and Stormwater Management
Groundwater Hydrology 138
4.2 Surface Water Quality
134
141
Dissolved Oxygen 141
Water Pollutants 142
4.3 Microbiological Decomposition of
Organic Materials in Water 146
Natural Systems 146
Engineered Systems 147
4.4 BOD Kinetics
149
Dissolved Oxygen in Rivers and Lakes 154
■ Summary 161
■ Problems 161
■ References 165
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Contents
5
Potable Water Treatment
166
5.1 Source Water Characteristics
5.2 Drinking Water Standards
ix
166
168
5.3 Treatment Processes to Produce Potable Water
5.4 Unit Operations of Water Treatment
172
175
Rapid Mix 175
Coagulation/Flocculation 176
Sedimentation 179
Rapid Sand Filtration 183
Disinfection 184
Aeration for Removal of
Hydrogen Sulfide from Groundwater 187
Lime-Soda Softening of Groundwater 188
Membrane Processes 197
Treatment and Disposal of Residuals 200
■ Problems 200
6
■ References 203
Wastewater Treatment
6.1 Introduction
205
205
6.2 Characterization of Wastewaters
208
Municipal Wastewaters 208
Industrial Wastewaters 211
6.3 Standards for Wastewater Treatment
211
National Pollutant Discharge Elimination System (NPDES) 212
Municipal Discharge Standards 212
Industrial Discharge Standards 214
6.4 Unit Operations/Processes of
Domestic Wastewater Treatment
Overview 214
Bar Screen 216
Grit Chamber 216
Equalization Tanks 216
Primary Treatment 221
Activated Sludge Process
214
223
6.5 Material Balance Approach
227
Overall System Balances 228
Reactor Balances 229
Clarifier Balances 231
6.6 WWTP Biological Kinetics
234
Secondary Clarifier 240
Disinfection 240
6.7 Septic Tanks
241
6.8 WWTP Sludge Treatment and Disposal
■ Problems 247 ■ References 252
243
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x
Contents
7
Air Resources
253
7.1 Management of Air Resources
7.2 The Major Air Pollutants
7.3 Global Issues
253
255
257
Acid Deposition 257
Stratospheric Ozone Depletion 259
Global Climate Change 260
7.4 Legislative and Regulatory Standards
265
7.5 Air Pollution Control: Stationary Sources
269
Particulate Matter Control Devices 269
Gaseous Emissions Control Devices 276
7.6 Air Pollution Control: Mobile Sources
281
Mobile Sources—A Global Problem 281
Characteristics of Engines and Fuels 283
Strategies for Pollution Prevention/Control
from Mobile Sources 285
7.7 Meteorology and Atmospheric
Dispersion of Pollutants 286
Meteorology and Atmospheric Stability 286
The Box Model 287
The Gaussian Dispersion Model 290
Residuals from APC Systems 298
■ Problems 298
8
■ References 301
Solid and Hazardous Waste Management
8.1 Municipal Solid Wastes
303
303
Quantities and Composition 303
Recycling 304
Collection 306
8.2 Landfill Disposal of MSW
310
Description of a Sanitary Landfill 310
Preliminary Design of an MSW Landfill 313
8.3 Thermal Destruction of Waste
8.4 Hazardous Wastes
315
320
Introduction 320
Laws and Regulations 321
Generators 323
Treatment, Storage, and Disposal (TSD)
of Hazardous Wastes 325
8.5 Site Remediation (Soil and Groundwater Cleanup)
Introduction—A Case Study 328
CERCLA 330
Remediation Technologies 331
■ Problems 334
■ References 337
328
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Contents
9
Other Important Topics
9.1 Risk Assessment
339
339
Risk Assessment Process 342
Risk Management 348
9.2 Energy Resources (with Emphasis on
Nuclear Power and Radioactive Wastes)
349
Reducing our Energy Use and Improving the Environment
Nuclear Power 353
9.3 Indoor Air Quality
361
Some Pollutants of Concern 361
Ventilation and Infiltration 363
Material Balance Models for Indoor Air Quality 364
Solutions to IAQ Problems 367
9.4 Noise Pollution
370
Loudness 370
Frequency 372
Duration 373
Subjectivity 374
Noise Attenuation 375
Legislation and Regulations
■ Summary 378
383
Appendix B
386
Appendix C
393
Index
397
377
■ Problems 378
Appendix A
xi
■ References 380
351
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Preface
This text is an introduction to the amazing field of environmental engineering. It is intended to provide a solid foundation for all engineering students
who are interested in this field and its many subdisciplines. The author has
devoted 35 years of his life to teaching environmental engineering, and this
course is one of his favorites. He strongly believes that the approach followed
in this book is very effective in helping students to not only learn the subject
matter but also develop a way of thinking. This text presents a fundamental
method for approaching and solving problems that will benefit readers
throughout their careers. Moreover, it introduces in a logical way much of the
knowledge base used by environmental engineers in current practice.
The presentation begins with a broad overview of environmental engineering, including a discussion of the history, professionalism, and ethics of this
field. Basic calculations, population growth, pollution prevention, and sustainable development are also briefly discussed in the first chapter. Chapter 2 offers
a thorough review of some of the most important topics of chemistry, all of
which are essential to success in environmental engineering. The key concept
of this course—the use of material and energy balances to solve environmental
engineering problems—is covered in Chapter 3. The remaining chapters present important details of various subdisciplines of environmental engineering—
water resources, drinking water treatment, wastewater treatment, air pollution
control, solid and hazardous wastes, risk assessment, energy resources (including nuclear), indoor air quality, and noise pollution.
Environmental engineering is a very broad field, and no one book can present all the knowledge needed in the practice of engineering. Indeed, engineering education is a lifelong pursuit. However, it is the author’s sincere belief that
this book presents critical tools for students and is an excellent introduction to
this dynamic field of engineering work.
The author acknowledges the work of Dr. John Dietz and Dr. Debra Reinhart (his co-authors of a previous, similar textbook), and sincerely thanks them
for allowing him to freely use material from that earlier text. The author extends
his thanks to several colleagues at UCF for their reviews of various chapters,
including: Dr. Steve Duranceau, Dr. Andy Randall, Dr. Dingbao Wang, Mr. Ben
xiii
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xiv
Preface
Fries, Ms. Jackie Sullivan, Ms. Stephanie Bolyard, and Dr. John MacDonald. The
author also thanks Neil Rowe, publisher of Waveland Press, for believing in this
project from the outset; Laurie Prossnitz, editor, for her usual excellent work
and attention to detail; and Debi Underwood, for her fine work with the graphics. A very special thank you goes to Dr. Kurt Westerlund for his excellent contributions to the Solutions Manual.
Cooper.book Page 1 Monday, June 23, 2014 9:58 AM
CHAPTER
1
What Is Environmental Engineering?
1.1 Introduction
This text is an introduction to environmental engineering. However, this
book is intended for engineering students from all disciplines who are interested in learning sound fundamentals and obtaining a quantitative understanding of our air and water resources, environmental quality, pollution
control processes, pollution prevention, management and treatment of solid
and hazardous wastes, and restoration of contaminated land areas. As a formal
discipline, environmental engineering has only been taught since the 1960s.
However, the problems now being solved by environmental engineers have
been faced by people in one form or another for centuries—controlling or preventing human pollution of the environment, making the world better for people, and (more recently) preserving and restoring environmental quality for its
own sake. In this sense, environmental engineering is an old and broad field.
Even today, much environmental work is accomplished by civil, chemical,
industrial, and mechanical engineers, as well as by chemists, biologists, and
other scientists, plus many others in a variety of disciplines.
Environmental Engineers—Who We Are and What We Do
Environmental engineering can be defined as the application of engineering
principles and technology to solve existing pollution problems, to prevent new
pollution problems, and to preserve or improve environmental quality. It is a
branch of engineering that is concerned with (1) the protection of human health
from adverse environmental factors, (2) the protection of ecosystems (local and
global) from the potentially harmful effects of human activities, and (3) the
improvement of environmental quality. The broad discipline of environmental
engineering branches out into many subdisciplines to better understand and solve
specific, complex environmental challenges (see Figure 1.1 on the following page).
The common theme of environmental engineering is a basic understanding
of the interactions of human activities and the environment, how emissions
from human activities can damage the environment, the hazards pollution
presents to people, and how both people and the environment can be protected
1
Cooper.book Page 2 Monday, June 23, 2014 9:58 AM
2
Chapter One
from such effects. Environmental engineers not only design, operate, maintain,
and manage facilities and systems for environmental protection, but also measure environmental quality and continually seek ways to improve it at a reasonable cost. They deal with atmospheric, aquatic, and terrestrial environments, as
well as interactions among these environments, and generate process designs,
feasibility studies, environmental assessments and compliance reports, among
other things. Modern practitioners of environmental engineering try to stay
knowledgeable about all these subdisciplines, though most specialize in one
branch such as air pollution control, water or wastewater treatment, or solid
and hazardous waste management.
Figure 1.1
The many branches of environmental engineering.
Climate
Change
Renewables
Nuclear
Air Pollution
Control
Fossil Fuels
Dispersion
Air Quality
Modeling
Remediation
Assessment
Reclamation
Discharge
Treatment
AIR
WA S T E WATE R
ENERGY RISK
Chemical
Landfills
Drinking
Water
Groundwater
Storm
Water
Incineration
NOISE
ECOLOGY
Surface
Water
SOLID
WASTE
WATER
E
N
V
I
R
O
N
M
E
N
T
A
L
E
N
G
I
N
E
E
R
I
N
G
Hazardous
Waste
Recycling
Disposal
Cooper.book Page 3 Monday, June 23, 2014 9:58 AM
What Is Environmental Engineering?
3
Environmental engineers work in consulting firms; industrial corporations;
local, state, and federal governments; private research organizations; academic
institutions; and, in small but increasing numbers, with public-interest groups.
In recent years, many companies in corporate America have combined the
functions of environmental protection with those of employee health and
safety. Environmental health and safety (EHS) managers (who often have an
engineering degree) are now common in large companies. They lead their companies’ efforts to comply with regulations, prevent or reduce pollution, advocate for sustainability, and protect their workers’ safety and occupational health
(National Association for Environmental Management 2010). About one-fourth
of today’s environmental engineering students go on to get a master’s degree,
specializing in one branch or another, and propelled by a commitment to lifetime learning, they gain a high level of expertise within their chosen subfield of
environmental engineering. In addition, some environmental engineers obtain
doctoral degrees and join university faculties, although many doctoral-level
environmental engineers are employed in all of the above-mentioned sectors.
Regardless of where they work, most environmental engineers have a high
level of satisfaction, feel their work is challenging, and believe that their efforts
benefit society and our quality of life.
Professional Engineers and Engineering Ethics
Becoming a professional engineer is very important, especially for environmental and civil engineers. Engineers who work in a field dealing directly with
public health and safety have a heightened responsibility. We are held to a high
standard; if we design something that fails (a bridge, a drinking water treatment plant, a hazardous waste treatment and disposal facility), then people can
die. Therefore, we must have knowledge, experience, and training that provide
us with the tools we need to do the jobs right. Much like doctors or lawyers, we
must also have credentials that certify our abilities and that are acknowledged
by the public and by regulators. The most widely recognized credential is the
Professional Engineering (PE) license. PE licensure is regulated by the individual states, and is required before one is allowed to supervise certain types of
engineering design jobs or to operate one’s own engineering business. The PE
has proved very useful to many engineers, and obtaining it while in the
employ of others usually results in an immediate raise in salary. Other credentials exist, for example, the Qualified Environmental Professional (QEP), and
these can further one’s career as well.
To become a registered PE in a state, one must first graduate from an
ABET-accredited engineering program. ABET stands for the Accreditation
Board for Engineering and Technology, and is a group organized to promote
and advance the education of engineers. ABET utilizes professional engineers
and educators who conduct on-site visits at colleges and universities seeking
accreditation. They review the school’s engineering courses, curriculum, faculty, and inspect labs to determine if those programs should be accredited.
During their time in an ABET-accredited engineering school, or after graduation, students may take the Fundamentals of Engineering (FE) exam, a mul-
Cooper.book Page 4 Monday, June 23, 2014 9:58 AM
4
Chapter One
tiple-choice, six-hour exam that tests their knowledge of basic math, science,
and engineering topics and with some focus on the discipline (National Council of Examiners for Engineering and Surveying 2013). Those who pass that
exam must work for four years to gain experience. Then the Professional Engineering exam may be taken. The PE exam deals solely with one specific engineering discipline (e.g., environmental, civil, mechanical, chemical). When
passed, the state governing body bestows the license to practice engineering in
that state. Many states have reciprocity and will recognize a license from
another state as sufficient to grant a license in that state.
Part of the PE exam deals with engineering ethics. Regardless of one’s PE
status, an engineer needs to have a solid set of engineering ethics. The dictionary defines ethics as a theory or system of moral values. In other words, ethics
is a sense of knowing what is right and wrong and having a commitment to do
the right thing. As an environmental engineer, you must exhibit honesty, trustworthiness, responsibility, respect for others and for the environment, a sense
of justice, and fairness. Your ethical obligations are to protect public health and
safety, to represent yourself honestly, and to do your best for your clients/
employers in all assignments. Like other professionals, environmental engineers must strive to maintain a good balance among sometimes competing
interests. In our profession, public health and safety are the top priority. Costs
must always be considered, but there are many other factors that enter into our
engineering decision making (see Figure 1.2).
Figure 1.2
Maintaining balance is the key.
Public Health,
Technology,
Regulations,
Experience
Safety, Cost,
Education,
Feasibility,
Ethics
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What Is Environmental Engineering?
5
EXAMPLE 1.1
This example is suggested for class discussion. . . . You are a pollution control
agency inspector visiting a small “one-industry” town. You discover a significant spill near the industrial plant that might contaminate the groundwater
and thus may threaten nearby homeowners’ water wells. You know that this
plant is struggling financially due to the recession. The monetary fine for this
spill would surely be large enough to force the plant to close, and that would
leave the contamination for someone else to clean up. It also would result in
hundreds of workers losing their jobs. The plant owner offers to work on the
cleanup over time if you do not report this violation. Considering your obligations to the public health, to yourself, and to your agency, what action
should you take? (adapted from Taback 2004)
1.2 Thinking like an Engineer
This may be your first engineering course. As such, you probably have not
yet started thinking like an engineer, but don’t worry—you will soon enough!
Engineers are like scientists in that they think logically and precisely, and they
don’t jump to hasty conclusions. But engineers differ from scientists in that we
are always on the lookout for practical solutions to real-world problems, solutions that work and don’t run over budget. In fact, engineers are known as
problem solvers—that’s our claim to fame.
Problem Solving
Engineers are trained to identify a problem, to analyze it into “bite-size”
pieces, solve those pieces, and then to synthesize the solutions back into one
overall cost-effective solution. We often make reasonable assumptions and use
approximations when necessary to get the job done. One simple (but effective)
way to think about problem solving is as a three-step process:
• find out where you are (“here”)
• decide on where you want to go (“there”)
• figure out the best way to get from “here” to “there”
Finding out your starting point is the first part of identifying the problem.
You might get an assignment from a supervisor, be asked to investigate a complaint, or may simply observe an improper or inefficient operation. Determining your goal is the second step of problem identification, and the first part of
the analysis phase. It involves researching and defining acceptable outcomes;
for pollution control projects, these outcomes may be meeting all applicable
emission standards or regulations, or perhaps satisfying internal company policies. Step three includes making engineering design calculations, evaluating
alternatives, and choosing the best solution.
Cooper.book Page 6 Monday, June 23, 2014 9:58 AM
6
Chapter One
As an example, consider the situation where an engineer working for a petrochemical company is called in one day by the plant manager. She tells the
engineer that she has received a complaint from a local resident about a visible
plume of “smoke” from the plant, and to “solve the problem.” First, the engineer must find out exactly what the problem is: which stack was emitting, what
was being emitted, why (i.e., normal or upset operations), when did it happen,
and for how long. Next he must determine if there are any state regulations for
these stack emissions, and determine what steps the plant must take to clean
up the emissions to meet regulations or to satisfy the resident’s complaint.
Then, and only then, can he begin to design the best solution to bring the operation from where it is to where it should be.
Units and Significant Figures
All engineers must be numerate—that is, literate with numbers. Before progressing very far in their chosen curriculum, engineering students must be able
to perform calculations involving many kinds of units and be able to convert
answers from one set of units to another. One of the basic skills of engineers is
to make calculations and keep track of units. This skill can be improved with
practice. It is strongly recommended that when stringing calculations together,
that students use the long line technique, as illustrated in the following example. This technique ensures that you pay attention to units and that the calculations are set up so that units are converted appropriately. It also is easy to check
that your answer is in the requested units. Some common conversion factors
are tabulated in Appendix A.
Students must also pay attention to significant figures. Students should
note that the answers to the questions in the following example problem are
reported to only two, three, or four significant figures. It is common in engineering to use three or four significant figures for all answers, except when
more are needed and are fully justified by precise measurements of the data or
exact values of conversion factors. Even though your hand calculator will give
an answer to part (a) of 6.281286 meters/second, it should be obvious that the
answer cannot be this precise.
EXAMPLE 1.2
(a) A track star runs 1 mile in 4 minutes, 16.2 seconds (256.2 s). What is her
average speed in m/s?
SOLUTION
1 mile 5, 280 ft
1m
¥
¥
= 6.281 286 m/s
256.2 s 1 mile 3.281 ft
(b) A NASCAR driver is going about 200 mph. How fast is this in m/s?
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What Is Environmental Engineering?
7
SOLUTION
200 mi 5, 280 ft
1m
1 hr
¥
¥
¥
= 89 .4036371 m/s (about 90 m/s)
3.281 ft 3, 600 s
hr
mi
(c) The concentration of H2S in water is 4.0 mg/L. How many pounds of H2S
are contained in 10.0 gallons of water?
SOLUTION
10.0 gal ¥
4.0 mg 3.785 L
1g
1 lb
-4
¥
¥
¥
= 3.33 4802 (10 ) lb
1 gal
1, 000 mg 454 g
L
(d) An industrial plant discharges a wastewater containing 0.510 mg/L of a
pollutant. The wastewater flow averages 30.0 L/s. How many lb of pollutant are discharged in a year?
SOLUTION
1g
30.0 L 0.510 mg 86, 400 s 365 day
1 lb
1, 060 lb
¥
¥
¥
¥
¥
=
s
L
1 day
1 yr
1, 000 mg 454 g
yr
(e) A power plant discharges 0.60 lb of SO2 per million Btu of heat input. It
burns 1.20 tons/minute of coal with a heating value of 25,000 kJ/kg. How
much SO2 is discharged to the air, lb/day?
SOLUTION
7
Heat in =
2.58 (10 ) Btu
1 kg
1 Btu
1.20 tons 2, 000 lbs
25, 000 kJ
¥
=
¥
¥
¥
kg
1.055 kJ
min
min
1 ton
2.205 lb
7
SO 2 out =
2.58 (10 ) Btu
m in
¥
0.60 lb SO 2
6
(10)
Btu
¥
1, 440 min
= 22, 300 lb SO 2 /day
1 day
1.3 Growth of Human Civilization and Its Environmental Impacts
Humans now number over 7 billion people on planet Earth. These billions
of humans have a tremendous capacity to consume resources and to produce
wastes. In modern societies, people consume large amounts of food, and use
fertilizers, steel, aluminum, glass, plastics, and other materials, resulting in
large quantities of wastes. They use energy in almost every aspect of their
lives—from transportation to cooking to heating and cooling their homes.
Everyone produces a small amount of personal waste (including excreta, inedible parts of food, worn out clothing, and so forth). In the scattered nomadic or
agrarian societies of the distant past, these small amounts of organic materials
were easily assimilated back into nature. When populations were small and
dispersed, the impacts of humans on the environment were small. With the
advance of civilization and the growth of urban societies, the types and
Cooper.book Page 8 Monday, June 23, 2014 9:58 AM
8
Chapter One
amounts of wastes increased. During the 1700s and 1800s it was rare to find a
city that did not have severe local pollution problems (with serious consequences for public health) due to improper disposal of wastes. Today, modern
societies are producing hazardous and radioactive wastes, “new” types of
wastes (e.g., pharmaceuticals, petrochemicals, pesticides), as well as ever
increasing quantities of “traditional” wastes. Developed countries (DCs) produce much more waste per capita than less developed countries (LDCs), and
thus can have a great impact on the environment. Starting in the 1990s, China
and India and several other LDCs have grown their economies so rapidly that
they now rival or surpass the DCs with regard to their pollutant emissions. The
consequences of these practices and trends are potentially global in scale and
may continue to be felt for centuries to come.
The Mathematics of Population Growth
Percent
Population, billions
Adverse effects of pollution are related both to the numbers of people, to
their crowding into urban areas, and to their use of energy. Also, as the economic well-being of people improves, it seems as though their
Figure 1.3 World population growth and
habits become more wasteful. Huurbanization in the last thousand years.
man population growth (and the
trend towards urbanization) from
World Population Growth
the year 1000 AD to present is depicted
in Figure 1.3, and shows
8
7 billion
that
world
population has been
7
rising
very
rapidly
during the past
6
two
hundred
years
or so. Also,
5
note
from
the
figure
the very re4
cent
and
very
significant
shift of
3
people
from
farms
and
rural
com2
munities
into
urban
centers.
1
The growth curve depicted in
0
Figure 1.3 can be modeled as
1000
1200
1400
1600
1800 2000 2013
exponential growth. Exponential
Year
growth is deceiving in that it
appears slow for a long time, and
Population Percent in Urban Areas
then appears to “speed up” at the
100
end. It is different from linear
growth in that something which is
75
growing linearly changes by the
50 percent
same amount each year, while
50
exponential growth changes by
the same percentage each year.
25
That is, the quantity grows in proportion to how much is already
0
there, as shown in Equation (1.1).
1000
1200
1400
1600
1800 2000 2013
Year
Cooper.book Page 9 Monday, June 23, 2014 9:58 AM
What Is Environmental Engineering?
dN
= rN
dt
9
(1.1)
where:
N = quantity present at time t
t
= time, in arbitrary units (days, years, etc.)
r
= growth rate, expressed as a fraction per unit time
Equation (1.1) has the solution
N = N0 ert
(1.2)
where:
N0 = the amount present initially, at the start of the exponential growth
period
EXAMPLE 1.3
In May of 2013, 300 trees in a stand of 10,000 trees (or 3%) were observed to be
covered with kudzu (a fast-growing vine that completely covers trees, eventually killing them). The previous month only 280 trees had been covered; the
increase was 20 trees newly covered. If growth continues at the same rate, how
long will it take until 30% of the trees will be covered with kudzu? Make two
estimates—one using linear growth and one using exponential growth.
SOLUTION
For the linear growth estimate, it is assumed that each month 20 trees will be
newly covered. To reach 30% or 3,000 trees, the number of months is:
3, 000 - 300 trees
= 135 months or about 11 years
20 trees/month
Using exponential growth, first calculate r based on the observed increase during the month:
r=
20
= 0.0714
280
Next, rearrange Equation (1.2) and then take the natural log of both sides.
Ê N ˆ
= rt
ln Á
Ë N 0 ˜¯
Set
t=
0.30
N
to
and solve for t
0.03
N0
ln (0.30 / 0.03 ) 2.303
=
= 32.2 months (less than 3 years)
0.0714
0.0714
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10
Chapter One
The Implications of Population Growth
Of course, it should be understood that no biological population can continue to grow exponentially forever. Some limit—perhaps food supply, water
supply, space available, or something else—will eventually stop the exponential growth. At that time either the population will stabilize at the sustainable
limit (the carrying capacity), or will crash back down to a much smaller level.
As an example, consider the reindeer population in the following case
study (Miller 1996). In 1910, 26 reindeer were placed onto an island off the coast
of Alaska. Food supplies were plentiful, and there were no wolves or other
predators, so the reindeer population mushroomed to 2,000 by 1935. Then
came several very harsh winters, which, combined with the overgrazing that
had occurred, caused a population crash. By 1950, the herd had plummeted to
only 8 individuals; the reindeer population was reduced by more than 99%!
Of course, people are not reindeer; they have intelligence and free will and
can consciously choose to limit population and/or take other actions to avoid a
disastrous population crash. However, despite efforts of governments and other
groups to educate and convince societies to reduce population growth rates, the
system has a lot of inertia, and progress has been slow. The pressures of population growth continue through the present day. Not only is population growing,
but so too is the use of energy and material resources, which contributes to environmental pollution. This will be addressed further in the next section.
The world population was about 1.0 billion people in 1815, 2.0 billion in
1928, and 3.0 billion in 1960. It reached 6.0 billion in 1999 and 7.0 billion in 2012.
As of May 2013, world population was estimated at 7.1 billion people, with an
overall growth rate of about 1.1% per year—which adds about 78 million people (equal to another Iran) every year. Although this rate of growth is down
from 1.3% per year in 2000, and 2.0% in the 1960s, the world’s population grew
by about 3.5 billion people during the last 45 years. That is, it took from the
beginning of human history until about 1968 to add 3.5 billion people to the
planet, but it took only 45 more years to add another 3.5 billion!
If we investigate growth dynamics by regions, we find that the highest
growth rates are in the least developed countries (in Africa, Latin America, the
Middle East, and Asia)—the ones that can least afford it. Also, we note that the
population distribution in the fastest-growing countries is heavily weighted
toward young people. This over-weighting of youth gives these countries a
growth momentum that is difficult to appreciate and extremely hard to stop.
That is, as children attain child-bearing age, they then have more children. If
each couple has only two children they simply “replace” themselves. This is
called replacement fertility. (Actually, due to accidental deaths of children and
other reasons, replacement fertility rate averages about 2.1 children per couple
in the more developed nations, and somewhat higher in the less developed
countries.) But in many countries the fertility rate is much higher than replacement, so their populations continue to grow.
By the time this book is published, the world population will be about 7.3
billion. Because of the youthful distribution of population, even if child-bear-
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What Is Environmental Engineering?
11
ing-age couples all over the world were to immediately achieve replacement
fertility rates, the momentum of the youthful distribution of population implies
that the total population will continue to grow to about 10 billion before stabilizing by about 2085. Of course, people who grow up in large families tend to
want to have large families. If it takes another 40 years to change people’s
behavior and attain the goal of replacement fertility worldwide, then the stabilized population level will be significantly higher, perhaps about 14 billion. A
world population twice as big as it is today would put enormous strains on the
global environment, and may well be beyond the earth’s carrying capacity for
human beings. Thus it is vital to change people’s behavior and attitudes now.
Such change of human behavior is difficult but not impossible. China now
has 1.3 billion people, or about 18% of the world’s population, with only 7% of
the world’s arable land. For obvious reasons, the Chinese government had
strong incentives to control its country’s population. Beginning in about 1970,
the government of China implemented strict policies to limit birth rates. In
urban areas each couple is allowed only one child; in rural areas the limit is two
children per couple. Through education, political pressures, and economic and
other penalties, the behavior and attitudes of people are changing. China’s
present population growth rate is less than 0.5%, slightly lower than France
and Britain, and slightly more than Italy or Norway. Many Chinese women see
this as an opportunity to overcome the traditional view of the role of women in
China, and to allow them to more fully participate in work, societal, and political institutions.
Right now, human population growth is still growing exponentially, and
more people means more consumption of energy and other resources. We have
been very successful in increasing food supplies, in curing many illnesses, and
in harnessing vast amounts of energy to help sustain our lives. But our successes have serious implications for the future. How long can such growth continue? How long can we continue to use up fossil fuels and mineral resources
to sustain such growth? How long can we discard wastes into our environment? We know that exponential growth cannot continue forever—some limit
will be reached. What is the limit for human population on earth—water supply? food? air quality? What will be the quality of life for the majority of human
beings living on earth when the world population reaches 10 billion or 14 billion? Within any economic system, there has always been (and will always be)
wealthy and poor people, and wealthy and poor countries. With modern communications highlighting these disparities for all to see, there will be increasing
pressures for more immigration from poorer to richer countries, and/or other
forms of resource redistribution. What effects will such pressures have on the
relationships between the “have” and “have not” countries?
These are exceedingly difficult questions, but there is an urgent need to
address them. If people are not successful in controlling world population and
economic disparities very soon, the societal and environmental consequences
may be disastrous. More efforts are needed by the “have” countries to help the
“have nots” with agricultural and industrial development, as well as environmental preservation. The developed countries must also put much more effort
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12
Chapter One
into population control, pollution prevention, and reducing their consumptive
use of energy and other resources.
Fortunately, world leaders are beginning to recognize and address some of
the difficult societal and cultural problems identified in the preceding paragraphs. Meanwhile, as engineers we must also continue to address the technical problems of trying to increase efficiencies, reduce pollution, improve the
distribution and utilization of resources, improve quality, and reduce costs in
order to support and improve our global living conditions.
Resource Consumption and Pollution Generation
Population growth has imposed large demands on the earth’s abilities to
sustain large societies; indeed, it might well be considered the single most
important cause of environmental problems over the past 200 years. A close
second, however, has been the huge increase in the use of energy. Because of
the industrial revolution, people are able to harness fossil fuels and other forms
of energy to do vast amounts of work and to alter their environment. Table 1.1
illustrates the rapid rise and projected growth in energy consumption; Table 1.2
displays some statistics about world population, food production, and mineral
resource consumption in the last 25 years.
Human discards have expanded far beyond personal wastes to include
industrial, mining, petrochemical, and agricultural wastes as well. The evergrowing use of energy and materials to support modern lifestyles produces
large quantities of gaseous, liquid, and solid wastes, which can pollute our air,
water, and land resources. As an example, air pollution is closely and directly
linked to the use of energy (combustion of fossil fuels). Even though we may
try to use materials and energy efficiently, it is inevitable that something will
end up as a waste. As humans gain economic wealth, they want more goods
and services, which in turn requires the use of more materials and more energy.
No other creatures on earth can leverage the use of energy and machinery to
move, change, use, and discard such massive amounts of materials.
Table 1.1
World Energy Consumption—Past, Present, and Future
Energy Use, Quads/year
Fuel
1990
2010
2030
Oil
Natural Gas
Coal
Nuclear
Hydro
Biomass
Solar, wind, geothermal
Total
137
75
89
20
8
24
1
354
161
108
142
28
12
50
4
505
216
162
195
47
14
72
14
722
Note: 1 quad = 1 quadrillion Btu = 1015 Btu.
Data from various sources.
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What Is Environmental Engineering?
Table 1.2
13
World Population Growth, Food Production,
and Resource Consumption Rates
Annual Production or Consumption, units/year
Commodity
Units
People (cumulative)
Energy
Wheat
Rice
Fertilizers (nutrients)
Iron Ore (Fe content)
Copper Ore (Cu content)
Passenger Cars
9
10
Quads
106 MT
106 MT
106 MT
106 MT
106 MT
106
1990
2000
2010
5.25
354
560
519
155
570
8.8
33
6.09
420
586
599
160
1,010
12.6
41
6.87
505
651
672
170
2,400
16.2
77
Data from several sources.
EXAMPLE 1.4
From the data in Table 1.2, calculate average growth rates from 1990 to 2010 for
population and energy use. Assume two models for the growth—linear and
exponential. Express your answers as % per year using 1990 as the base year.
SOLUTION
Population:
9
Linear : =
9
6.87 (10 ) - 5.25 (10 ) people
= 8.1 107 persons/
/yr
20 years
( )
7
=
8.1 (10 )
9
5.25 (10 )
¥ 100% = 1.54% / year
Exponential : First, solve Eq. 1.2 for r
Ê N ˆ
ln Á
Ë N0 ˜¯
r=
t
Ê 6.87 ˆ
ln Á
Ë 5.25 ˜¯
r=
= 0.0134 = 1.34%/year
20
Cooper 01.fm Page 14 Thursday, March 3, 2016 9:42 AM
14
Chapter One
Energy use:
Linear : =
=
Exponential:
505 - 354 quads
= 7.55 quads/yr
20 years
7.55
¥ 100% = 2.13% / year
354
r=
(
)
ln 505 354
= 0.0178 = 1.78%/year
20
As seen in Example 1.4, while recent population growth has averaged less
than 1.5% per year, energy growth has been about 30% higher than that. It is
sobering to consider that we are now capable, as a species, of permanently
changing our environment. And we seem to be doing so in more and more
places around the world.
Finally, it is noted that a waste need not be toxic, hazardous, or even
unpleasant to cause concern. An overly large generation rate of any substance
can create a serious problem. For example, consider the greenhouse gas (GHG)
carbon dioxide (CO2). Most scientists believe that the enormous amount of CO2
being emitted today is changing our world’s climate. Extremes of weather have
been observed in recent years (Derevianko and Balentine 2013), and even more
extreme weather events are projected due to climate change (Gao et al. 2012).
As examples of recent extreme weather, consider the massive hurricanes
Katrina in 2005 and Sandy in 2012. More details about global warming will be
given in Chapter 7, but it seems obvious to many people that there are now
more extremes of heat and drought and massive storms than there were 40
years ago.
Each person, in the simple act of breathing, emits about one-half kilogram
of CO2 into the atmosphere each day, or about 0.2 metric tons (tonnes) per year.
If exhaled CO2 were the only anthropogenic source of CO2, the ecological balance between plants and animals could be maintained. Excess CO2 emissions
come mainly from burning fossil fuels at enormous rates. For instance, one gasoline-fueled car emits about 5 metric tons (tonnes) of CO2 in a year (more than
the weight of the car itself), and an average size coal-fired power plant emits
about 10,000 tonnes of CO2 per day!
EXAMPLE 1.5
One measure of the economic wealth of a society is the prevalence of vehicles.
In 2010, in the United States there were about 310 million people and about
240 million highway vehicles. In China, in 2010, there were about 1.3 billion
people, but only 78 million vehicles. How much CO2 was emitted by vehicles
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What Is Environmental Engineering?
15
operating in each country in 2010? If China eventually reaches the same ratio
of vehicles to people as in the United States, how much CO2 would be emitted
from Chinese vehicles?
SOLUTION
In the United States:
6
240 (10 ) vehicles ¥
5 tonnes CO 2
9
= 1.2 (10 ) tonnes CO 2 / year
veh-year
In China:
6
78 (10 ) veh ¥
5 tonnes CO 2
8
= 3.9 (10 ) tonnes CO 2 / year
veh-year
For the “what if?” scenario for China,
Ê 240 million veh ˆ
9
9
1.3 (10 ) people ¥ Á
˜ = 1.1 (10 ) vehicles
310
million
people
Ë
¯
9
1.1 (10 ) veh ¥
5 tonnes CO 2
9
= 5.5 (10 ) tonnes CO 2 / year
veh-year
From the results above, it can be calculated that US vehicles emit about four
times as much CO2 as Chinese vehicles, but if the ratio of vehicles to people in
China becomes equal to that in the United States, then Chinese vehicles will
emit about five times more than US vehicles. It seems clear that China (and
other developing countries) must not strive to obtain as many vehicles per person as the US has now, and/or Americans must change their driving habits,
and/or the world must find a better way to move people from place to place!
1.4 The Rise of Environmental Protection
Even though the formal discipline of environmental engineering is not very
old, there are numerous examples of environmental engineering in history. Public
water supplies and waste disposal facilities have existed since the days of ancient
Rome, which was supplied fresh water by nine aqueducts. Some of these aqueducts were up to 80 kilometers long and up to 15 meters across. There is a bridge
across the Tiber River in Rome that today carries automobile traffic, yet was built
1900 years ago! The ancient structures in Bath, England, are another example of
Roman construction of public works. In the city of Pompeii, buried by a volcanic
eruption in 79 AD, some homes had running water (albeit transported through
lead pipes), but the sewage was carried away via the stone-paved streets.
In the Middle Ages, local city-states had to defend themselves against warlike neighbors. People who built machines of war became known as engineers,
so the term took on a military context. In the late 1700s, John Smeaton, a nonmilitary builder of roads, buildings, and canals in England, decided he should
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16
Chapter One
more properly be called a civilian (or civil) engineer (Vesilind et al. 1994). The
title became widely adopted by other engineers around the world who
designed and built these kinds of public works.
Prior to the early 1800s, it was common practice to discharge human
wastes into the street where they might seep into the ground or flow slowly
into a nearby ditch and then into a stream, polluting local groundwater wells
and small streams and rivers. In addition to the aesthetic offense, such practices contaminated local drinking-water supplies with pathogens, causing frequent outbreaks of deadly waterborne diseases. As cities grew, public water
supplies often became grossly polluted, and private water supplies became
extremely expensive.
Although it may seem obvious to us today that drinking polluted water
can make you sick, it was not until the mid-1800s that the transmission of disease was linked conclusively to contaminated drinking water. John Snow
traced the 1849 cholera epidemic in London to one drinking-water well located
on Broad Street, and curbed the epidemic by removing the handle from the
pump. Through his work and that of others, water filtration for large public
water supplies came into common practice by the late 1800s. However, it took
considerable time for more advanced technology to come into widespread use.
By the early 1900s, the idea of treating drinking water to kill bacteria became
more widely accepted, and today we have modern facilities that can produce
clean drinking water from almost any raw water source.
Early engineers proposed sewerage systems to separate and transport the
waste discharges away from living areas. The first sewers emptied directly into
the nearest river, lake, or ocean, without treatment. Wastewater treatment in
the early 1900s attempted only to remove the offensive solids (primary treatment). Later, secondary treatment processes were developed to better protect
human health. In the 1950s and 1960s, the US government contributed billions
to the construction of new municipal wastewater treatment plants. However, it
was not until 1972 that federal requirements were established for sewage treatment. Around the same time, the idea of reducing damage to the environment
itself started to gain traction as worthwhile and necessary. Even after that, there
were numerous cases of water pollution from industrial discharges and agricultural wastes in rainfall runoff into lakes and rivers. Fish kills—cases where
thousands of fish were found dead and floating in a suburban waterway that
had no apparent source of pollution—were ultimately traced to rainfall that
washed pollutants off local streets and/or backyards into the lake. (A fish kill is
not a pretty sight—see Figure 1.4.) Again, this is evidence that as we crowd
more and more people into smaller and smaller areas, the effects of “normal”
human activity can be harmful to the environment.
A basic concept of science is that matter cannot be created or destroyed.
When this simple concept is applied to the environmental field, it tells us that
pollution does not simply “go away” when we discharge it to the air, water, or
land. Unfortunately, it took decades, even centuries, of pollution to make people understand that this concept makes no exceptions, and that the earth itself
is not a big enough place for us to just keep throwing things away.
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What Is Environmental Engineering?
17
Figure 1.4 A fish kill in Lake County, Florida, in 1997. (Julie Fletcher/The Orlando
Sentinel; used with permission.)
During the industrial revolution, people lost sight of this basic concept
because the world seemed so vast that when wastes were discharged into the
air or water, they seemed to disappear. Today, we are more aware than ever
that material is cycled through the environment, sometimes with grave consequences. Matter can be changed in form through chemical reactions, and it can
be stored for long periods of time. But it never just goes away.
For example, phosphorus is a key element in sustaining life. Phosphate ore
(the skeletal remains of prehistoric marine life) is mined in central Florida to
produce fertilizer; when the phosphate fertilizers are applied to crops, the
phosphorus that was stored millions of years ago is absorbed into plant fiber
and food, and finally returns to the environment. However, excessive applications of phosphate fertilizers to the lawns and gardens of homes and businesses near lakes can result in excess phosphorus being carried by rainwater
runoff into the lakes. This can produce rapid growth of algae and water weeds,
and may result in a fish kill such as that pictured in Figure 1.4.
Another example involves a case in which a pollutant adversely affected
human health—namely the mercury poisoning that happened to people living
around Minamata Bay in Japan in the 1950s (Harada 1995). Mercury was discharged by nearby industries into the waters of the bay. That mercury accumulated in bottom sediments, was methylated (transformed into methyl mercury
which is soluble) by aquatic microorganisms, and subsequently transferred up
the food chain to fish and then to people. Thousands of people became ill and
many died due to high mercury levels that accumulated in their systems.
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Chapter One
An Environmental Ethic
The history of environmental awareness within the United States is a long
story. Previously, we saw how the rudimentary science and common sense of
the 1700s and 1800s developed into the engineering practices needed to ensure
the protection of public water supplies from contamination with human sewage. Early concepts of nature clearly influenced society’s values about the environment. During the 1600s and 1700s as pioneers struggled for survival, nature
was viewed as dangerous and capricious. It was filled with beasts, droughts,
plagues, blizzards, and other phenomena that caused hardship and death.
Nature was something to be fought and tamed. During the 1800s, as the country expanded westward, nature was viewed as a commodity. It was still dangerous, but it was rich with resources (forests, animals, minerals) that could
and should be exploited.
With the harnessing of energy and the development of machinery, people
were able to extract and utilize many natural resources at rates never before
imagined. These natural resource industries fueled the development of the
United States into a great economic power by the early 1900s. Understandably,
the resource-development philosophy that was so successful in the 1800s persisted well into the twentieth century. However, there were visionaries even in
the 1800s who foresaw that the domination and exploitation of nature was not
right and would end up being self-defeating.
Early environmentalists like George Marsh in the 1870s and Gifford Pinchot
in the 1890s spoke up for the value of the environment for its own sake. With
Pinchot as an advisor, President Teddy Roosevelt helped establish many national
parks in the early 1900s. President Franklin D. Roosevelt, in response to the
Great Depression of the 1930s, created the Soil Conservation Service, the Civilian
Conservation Corps, and other programs that benefited the environment.
In 1949, Aldo Leopold published A Sand County Almanac. This book introduced a new way of viewing the environment and was a radical way of thinking in 1949. Leopold advocated a “land ethic” that espoused reverence for the
land, and which obligated people to respect and care for nature. In other
words, people should be custodians or stewards of the land rather than owners
(this effectively would limit the rights of property owners). During the 1950s
and 1960s, various consequences of severe environmental pollution began to
become apparent and to be widely publicized. It was then that a wide crosssection of society began to realize that we could not continue to dispose of
untreated or improperly treated municipal, industrial, and hazardous wastes
without dire consequences for the environment and for all of us.
In 1962, Rachel Carson authored a book called Silent Spring that may have
had more to do with the general public becoming involved with and more
aware of the environment that any other single event of the 1950s and 1960s.
The title of her book refers to a future springtime when neighborhoods, meadows, and forests are silent because all the birds have died, having been poisoned by environmental pollution. Interestingly, in a 1997 article entitled “Fifty
Years of Progress,” one chemical engineer reminisced about his first job in 1941,
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What Is Environmental Engineering?
19
working at a cyanide plant in New Jersey. One of his recollections was that
“when the HCN in the stack gas became strong enough, birds innocently flying
overhead would fall onto the ground dead at one’s feet. Similarly, poisoned
rats would run out from under buildings and drop dead” (Connolly 1997).
In the 1960s and 1970s, many people became outraged at the state of affairs
of the environment. The idea of protecting the environment for its own sake
became more widely accepted as worthwhile and necessary. People joined
forces in environmental groups such as the Sierra Club or Greenpeace, and
demanded that their legislators do something about protecting the environment.
Their vocal advocacy was what was needed at the time to provoke government
and industry into action. The following excerpts from an Environmental Protection Agency report (US EPA 1980) are provided to help students understand
what prompted the public to push Congress to pursue the passage of environmental laws and regulations.
During the 1950s and 1960s, toxic wastes were placed into drums and discarded in a landfill near the towns of Toone and Teague, Tennessee. When
the landfill closed in 1972, the site held some 350,000 drums, many of them
leaking pesticide wastes. The towns’ water supply was made unusable in
1978 when water leaching from the landfill reached the drinking water
zone; it was contaminated with a number of organic compounds. The towns
no longer had access to uncontaminated ground water, and had to pump
water in from other locations.
Another underground water supply near Denver was contaminated from
disposal of pesticide waste in unlined disposal ponds. The wastes were produced from manufacturing activities of the US Army and a chemical company during the 1943–1957 period.
In 1978, a truck driver was killed as he discharged waste from his truck into
one of the four open pits at a disposal site in Iberville Parish, Louisiana. He
was asphyxiated by hydrogen sulfide produced when the liquid wastes
mixed and reacted in the open pit.
In one of the most publicized environmental disasters of the 20th century,
the health of residents of Love Canal, near Niagara Falls, was seriously damaged by chemical wastes buried in the 1950s. As drums holding the wastes corroded, their contents percolated through the soil into yards and basements,
forcing evacuation of more than 200 families in 1978 and 1979.
By the late 1960s there were calls for action from many groups, and a number of federal legislative initiatives were begun. The 1970s became known as the
“decade of the environment.” As mentioned earlier, the thinking had evolved to
the point where many people believed that the correct approach to the environment was in a custodial or stewardship role. That is we, as custodians, have a
moral obligation to preserve and care for the land. This philosophy was often
justified by enumerating the benefits to humankind of such custodial actions
(increased fishing yields, or better enjoyment of forests, for example). There
were calls for stopping all further land development and growth, but, in the face
of growing populations, this course of action was unacceptable to most people.
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Chapter One
Some individuals even advocate the inalienable rights of nature to exist, even
if no observable benefits accrue to people. This includes not only the rights of
plants and animals to live and enjoy their habitats, but also the rights of the habitat
itself (including the rocks, rivers, and soil). This may be the most enlightened view
of all, but is certainly the most difficult to put into practice. In a society with evergrowing consumption of food, energy, living space, and other materials, engineers
must strive to meet those needs with minimal impacts on the environment.
Pollution Prevention, Sustainable Development, and Green Engineering
Initially, once the need to protect the environment from human pollution
was recognized, the accepted approach to a pollution problem was to design,
build, and operate “end-of-the-pipe” treatment processes. In the last twenty
years or so, however, people have begun to realize that preventing environmental degradation makes much more sense than cleaning up pollution damage after it occurs. In many cases, it is smarter and more cost effective to not
create the pollution during the manufacturing process than it is to put pollution
control devices at the tail end of that process. Pollution prevention is one of the
strategies of the US EPA and of many industrial companies.
One good example of the pollution prevention philosophy is the decision
by EPA in the mid-1970s to mandate the removal of lead from gasoline. (Lead is
an octane booster in gasoline but is a toxic metal in the environment.) By the
early 1990s, motor vehicle emissions of lead to the atmosphere had reduced
dramatically (by more than 98%).
Another example deals with chlorofluorocarbons (CFCs). In the mid-1970s it
was postulated (and later demonstrated in 1985) that CFCs released into the
atmosphere (from leaky automotive air conditioners, Styrofoam manufacture,
metals-degreasing operations, and many other industrial processes) were contributing to the destruction of the stratospheric ozone layer that protects life on
earth from the sun’s ultraviolet rays. Some 46 countries worked together to
develop the Montreal Protocol in 1987 to limit and reduce the manufacture and
use of CFCs. Industry has since diligently sought ways to reduce its dependence
on CFCs. Local environmental agencies implemented regulations to control how
automotive air conditioners could be recharged. Stores stopped selling individual
cans of Freon. Many industrial cleaning processes were changed (for example,
solvent degreasing of parts in the metals-plating industry was replaced with alkaline aqueous detergent washing), and many product substitutions occurred. The
original Montreal Protocol was strengthened and extended to all nations, and as
of 2014, 197 countries have signed the agreement (United Nations Environment
Programme 2014). Most countries have outright bans on the manufacture of
CFCs, and all have pledged to reduce their use. Hydrochlorofluorocarbons
(HCFCs) were developed to replace CFCs in air-conditioning uses (HCFCs are
less damaging than CFCs). At the time of their peak use in the early 1980s, worldwide production of CFCs was over 1 million tons/year. Today, there is essentially
no manufacture or use of CFCs, and work is progressing to replace the HCFCs.
The US EPA has endorsed a hierarchical approach to solving pollution
problems (see Figure 1.5). At the base is pollution prevention/waste minimiza-
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What Is Environmental Engineering?
21
Disposal
Treatment
Recycle—Reuse
Figure 1.5
Hierarchy of solving
pollution problems.
Pollution Prevention—Waste Minimization
tion, which is the most preferred approach. Next comes recycling and reuse of
waste materials. Third comes treatment, and fourth is disposal. These principles are demonstrated with the following example.
EXAMPLE 1.6
A company is discharging 100,000 gallons per day of a wastewater that contains small amounts of dissolved organic compounds as well as dissolved
chromium metal ions. The organics come from a solvent-cleaning process,
and the chromium originates from a 2,000 gallon-per-day chrome-plating
operation. Because of the chromium content, the entire wastewater stream
has been declared a hazardous waste—treatment and disposal of which will
cost a small fortune. You as an engineer are asked to recommend an approach
to solving this problem. What are your thoughts?
SOLUTION
First, investigate whether the solvent-cleaning process can be changed to
clean the parts to be plated with a simple jet wash using water only, or perhaps soap and water. If we can eliminate all use of the organic solvent, we
will be well ahead of the game.
Second, analyze the chrome-plating process to maximize the placement of
the chrome onto the product and to reduce the chrome losses to the greatest
extent possible.
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Chapter One
Third, try to reduce the volume of chrome wastewater and separate it from
the ordinary wash waters that come from other parts of the plant. Perhaps we
can isolate the chromium wastewater to its original 2,000 gallons per day.
Fourth, after the volume of the wastewater containing chromium has been
reduced as much as possible and separated from the rest of the wastewater, try
to use ion exchange or some other process to further separate and recover chromic acid to the extent possible. Finally, chemically treat the remaining small
volume of chrome wastewater to remove the chrome as a solid precipitate, and
dispose of only the chrome precipitate as hazardous waste. The other 98,000
gallons per day of “ordinary” wastewater now can be treated by ordinary
means, saving a considerable amount of money for the company, while simultaneously reducing the threat of chrome contamination of the environment.
Sustainable development is defined as development that meets the needs
of the present without compromising the needs of the future. A simple example: rather than harvesting old-growth forests, we can selectively cut and replant “timber farms.” Under this concept, growth and development are not
halted but economic growth must include recycling and restructuring. Technological progress must consider not only efficiency and profit, but also resource
and energy conservation, and must adapt to a changing world. Many countries
around the world, both developed and developing, recognize the need to
encourage and embrace sustainable development.
Sustainable development recognizes the right of all countries to continue
to develop and grow economically, but also recognizes the rights of the global
environment and the rights of future generations of people. This term was
introduced to the world in 1987 in the United Nations report “Our Common
Future,” and was popularized in the 1992 Earth Summit in Rio de Janeiro, Brazil. At the Earth Summit, more than 100 heads of state and more than 1,400
other leaders, scientists, and planners from 178 nations met to develop plans
and policies for addressing global environmental issues. The major strategies
that resulted from that summit are as follows:
• reduce population growth
• reduce poverty
• reduce the wasting of resources—both matter and energy
• emphasize pollution prevention
• make things that last longer and are easier to repair and/or recycle
• protect habitats and preserve biodiversity
• use renewable resources at their natural rates of renewal
• use locally renewable energy resources such as the sun, the wind, flowing water, and biomass
Ways to achieve sustainable development have been given serious attention by planners and leaders from many countries. Sustainable development is
an especially important concept with regard to current concerns about global
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What Is Environmental Engineering?
23
climate change. For the past three hundred years humans have been clearing
forests and burning the trees, which contributed net carbon dioxide to the
atmosphere. Sustainable forestry is an important step toward restoring a CO2
balance in this arena.
An example of a sustainable industry is the sugarcane industry. Sugarcane
is grown in many countries throughout the world, including the United States
where it grows in Florida, Louisiana, Texas, and Hawaii. In south Florida, it is
grown in large flat fields located north of the Everglades. As it grows, sugarcane very efficiently absorbs CO2 from the air. The cane is harvested and
brought to a mill. There the cane is cut and milled to obtain its sugar juice
(sucrose in water). Using steam generated in boilers at the mill, the sugar juice
is evaporated to produce sugar crystals.
The steam needed for evaporation is produced at the sugarcane factory by
burning the pressed cane fiber (called bagasse), thus avoiding the need to burn
fossil fuels. When the bagasse is burned, CO2 is generated of course, but since
the carbon in the cane fiber came from CO2 that was already present in the
atmosphere, there is little or no net increase for the cycle. Bagasse is the primary fuel source for the mills, and supplies over 95% of all fuel needs (fossil
fuels are often used for the initial start-up of the boilers). The steam is also used
to turn turbines which run the milling machines and generate electrical power
needed to run pumps, blowers, and other equipment in the factory. Sometimes
excess electricity is sold to the grid. This particular industry is sustainable: it is
a fully integrated co-generation operation—fossil fuels are not used, electricity
is not purchased, solid wastes are not produced, and the net addition of CO2 to
the atmosphere is near zero. However, as sustainable as it is, there are still environmental impacts. Large quantities of water are consumed in growing the
cane and producing the sugar. Water that drains off the fields (runoff) has a
high phosphorus content; it can eventually make its way into the Everglades
where it could stimulate plant growth and have significant ecological impacts.
Green engineering goes hand in hand with sustainable development.
According to EPA (2014), “green engineering is the design, commercialization,
and use of processes and products that are feasible and economical while (1)
reducing the generation of pollution at the source and (2) minimizing the risk
to human health and the environment.” This definition holds great promise for
the engineering profession because it encourages us to practice in a way that
protects environmental qualities and at the same time recognizes that we must
produce feasible and economical designs. Some of the key principles of green
engineering are as follows:
• conserve and improve natural ecosystems while protecting human
health and well-being
• ensure that all material and energy inputs and outputs are as inherently
safe and benign as possible
• minimize depletion of natural resources; strive to prevent waste
• create engineering solutions beyond current or dominant technologies;
improve, innovate, and invent (technologies) to achieve sustainability
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24
Chapter One
• actively engage communities and stakeholders in development of engineering solutions
One example of green engineering is the trend toward green roofs. A green
roof is a flat roof of a city building that is covered with shallow-root plants,
planted either in boxes or directly on the roof itself (after a water barrier is
installed). A green roof has been shown to reduce the energy demands of a building (both in summer and winter), while reducing significantly the volume of rooftop runoff of polluted water into nearby ponds, lakes, or rivers. Data on a side-byside comparison of a conventional flat membrane roof and a green roof with
plants about 1–2 feet tall, showed that in both the summer and winter of 2006, the
heat flux through the green roof was about 44–50% less than that of the conventional roof (Cummings et al. 2007). The plants and soil may also absorb various air
pollutants. Figure 1.6 shows a green roof at the University of Central Florida that
was built and is being maintained for research purposes, but which has resulted
in significantly reduced air conditioning and heating bills for the building.
Environmental Laws, Regulations, and Agencies
Environmental protection in this country is supported by a complex web of
laws and regulations, which are created, implemented, and enforced by
numerous individuals, agencies, lawmakers, and the courts.
Figure 1.6 The green roof on the student union at the University of Central Florida. (Courtesy of Mike Hardin, UCF.)
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What Is Environmental Engineering?
25
Environmental laws (whether federal, state, or local) establish broad goals
for environmental quality; set standards of behavior for individuals, industry,
and governments; create agencies to monitor the environment and to oversee
the actions of other groups; and establish penalties for failure to adhere to the
laws. For example, the National Environmental Policy Act (signed into law by
then-President Nixon on January 1, 1970) was a six-page document that set a
new policy for the United States to help ensure that people would consider all
the environmental impacts of any major actions (such as building a new highway). It also led to the creation later that year of the Environmental Protection
Agency (EPA), the main federal agency charged with monitoring environmental quality; creating specific regulations to protect the quality of our air, water,
and land resources; and enforcing those regulations and laws. From the mid1960s to about 1990, the growth in environmental legislation was enormous, as
seen in Figure 1.7 on the following page. Table 1.3 (on pp. 27–28) summarizes
some of the more important laws that were passed between 1970 and 1990.
Environmental agencies are governmental units charged with monitoring
and protecting the environment. There are federal, state, and local environmental agencies, and all must interact, each respecting the (sometimes differing)
opinions, priorities, authorities, and responsibilities of the others in order to get
things done for the ultimate benefit of the environment. In most states, the state
environmental agencies have been given significant powers by the EPA. The
states have significant flexibility in interpreting and enforcing federal mandates and guidelines. An example of the federal/state/local agency organization in Florida is the EPA, the state of Florida Department of Environmental
Protection (FDEP), and the Orange County Environmental Protection Department (OCEPD).
Environmental agencies usually are organized into specific media-oriented
groups (such as air quality or water quality divisions in the EPA) or mission-oriented groups (such as the regulation enforcement division of the EPA). Their
staffs include technical, administrative, legal, and support personnel. Even
non-environmental agencies often have large environmental sections (such as
the Department of Transportation, the Department of Energy, and the US
Army). Of course, we cannot forget about the United States judicial system. In
this, the most litigious country in the world, the courts have played a major
role in directing the course of environmental protection.
Environmental regulations are not laws but are specific rules that have the
power of law. Regulations are created by environmental agencies to protect
existing environmental quality; to limit present or future discharges of pollutants from industry, municipalities, and individuals; and to provide for the legal
means and authority to monitor the actions of other groups and enforce compliance with the regulations. The regulations that have been created by the US
EPA alone take up tens of thousands of printed pages, and can be very confusing to read and understand.
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Chapter One
Figure 1.7
A graphical chronology of environmental laws in the United States.
35
CAAA-90
RCRAA
CWA
25
RCRA
SARA
CERCLA
30
Number of Laws
26
CAAA-77
TSCA
SDWA
20
WPCA
15
CAAA-70
10
WRPA
5
SWDA
FWCA
TGA
0
1880
NEPA
RHA
1900
1920
1940
1960
1980
2000
Year
Key
RHA
TGA
FWCA
SWDA
WRPA
NEPA
CAAA-70
WPCA
SDWA
TSCA
RCRA
CAAA-77
CWA
CERCLA
RCRAA
SARA
CAAA-90
Rivers and Harbors Act, 1899
Taylor Grazing Act, 1934
Fish and Wildlife Coordination Act, 1958
Solid Waste Disposal Act, 1965
Water Resources Planning Act, 1965
National Environmental Policy Act, 1970
Clean Air Act Amendments, 1970
Water Pollution Control Act, 1972
Safe Drinking Water Act, 1974
Toxic Substances Control Act, 1976
Resource Conservation and Recovery Act, 1976
Clean Air Act Amendments, 1977
Clean Water Act, 1977
“Superfund” Act, 1980
RCRA Amendments, 1984
Superfund Amendments and Reorganization Act, 1986
Clean Air Act Amendments, 1990
Note: Symbols without names represent other federal environmental laws in the United States; not all
laws that were promulgated are named in this chart.
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What Is Environmental Engineering?
Table 1.3
27
Major Federal Environmental Legislation, 1970 to 1990
Acronym
Name
Date Enacted
Comments
—Created the Council on Environmental
Quality, and led to creation of Environmental Protection Agency
—Established Environmental Impact Statement process
NEPA
National
Environmental
Policy Act
January 1, 1970
WEQA
Water and
Environmental
Quality Act of 1970
April 3, 1970
—Concentrated on oil pollution and other
discharges by vessels
—Directed the president to designate “hazardous” water pollutants
CAAA-70
Clean Air Act
Amendments of 1970
December 31, 1970
—Established emissions standards for many
sources
—Led to ambient air quality standards
—Technology-forcing legislation led to automobile emissions controls
WPCA
Water Pollution
Control Act
Amendments of 1972
October 18, 1972
—Set a goal of eliminating all discharges of
pollutants into waterways by 1985
—Prohibited the discharge of toxic pollutants
in toxic amounts
—Required regional waste treatment planning and processing
FIFRA
Federal Insecticide,
Fungicide, and
Rodenticide Act
October 21, 1972
—Required that pesticides be registered
—Some pesticides were restricted
—Commercial users must be certified to use
restricted pesticides
—Required that EPA establish procedures and
regulations for disposal or storage of certain
pesticides
ESECA
Energy Supply and
Environmental
Coordination Act
June 22, 1974
—Recognized the interdependence of energy
and the environment
—Allowed some delays in meeting emissions
standards but basically said environment
would not be sacrificed for energy
SDWA
Safe Drinking
Water Act
December 16, 1974
—Provided for federal primary and secondary
drinking water quality standards for public
water supplies
—Provided for regulation of public water
systems
—Provided for protection of underground
drinking water sources
—Regulated all underground injection
TSCA
Toxic Substances
Control Act
October 11, 1976
—Provided pre-marketing review of all new
chemicals
—Provided for direct regulation of manufacture, use, and disposition of all chemicals
—Required extensive testing of chemicals for
hazardous effects
(continued)
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28
Chapter One
Comments
—Mandated comprehensive regulation for
disposal of all wastes (both by-products and
consumer products)
—Defined hazardous wastes; listed a number
of specific hazardous wastes
—Concentrated on “land” pollution rather
than air or water pollution
—Regulated interstate transportation of hazardous wastes as well as disposal, treatment
and/or storage; a “cradle to grave” approach
—Required detailed permit process for everyone handling hazardous waste
Acronym
Name
Date Enacted
RCRA
Resource
Conservation and
Recovery Act of 1976
October 21, 1976
CAAA-77
Clean Air Act
Amendments of 1977
August 7, 1977
—Provided for prevention of significant deterioration of “clean” air regions
—Designated regulations for nonattainment
areas
—Delayed attainment of emissions standards
for automobiles
CERCLA
Comprehensive
Environmental
Response
Compensation and
Liability Act
[Superfund]
December 17, 1980
—Created a special tax on industrial chemicals that goes to a trust fund, commonly
called Superfund
—Directed EPA to perform site cleanups or
take legal action to force responsible parties
to perform the cleanups
NWPA
Nuclear Waste
Policy Act
January 7, 1983
—Authorized the design and construction of
disposal facilities for spent nuclear fuels and
high-level radioactive wastes
—Formed the Office of Civilian Radioactive
Waste Management
HSWA
Hazardous and Solid
Waste Amendments
November 9, 1984
—Established minimum technical requirements for land disposal of wastes
—Modified permitting process for treatment,
storage, and disposal facilities
—Established controls for underground storage tanks
SARA
Superfund
Amendments and
Reauthorization Act
October 17, 1986
—Increased Superfund financial coverage
amounts
—Established emergency planning and public right-to-know procedures
—Required industry to report toxic release
inventories
CAAA-90
Clean Air Act
Amendments of 1990
November 15, 1990
—Addressed urban air quality, especially ozone
—Dealt with mobile sources comprehensively
—Regulated hazardous air pollutants
—Established a program for operating permits for air polluters
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What Is Environmental Engineering?
29
1.5 The History of the Future
Our actions today are making the history of the future! One hundred years
from now, when people look back to the beginning of the 21st century, will they
see continued commitment to environmental protection? Will they see that
population control was successful? How will global warming issues be
resolved? How will the engineers who worked to solve these difficult problems
be judged?
Concern for the environment has become an important part of human society. There are still debates about the best ways to implement environmental
protection policies, and how to balance human economic interests and environmental interests, but there remains little doubt in the minds of millions of people around the world that protection of our environment is critical to the longterm well being of humanity. This environmental ethic is spreading, and many
developing countries are concerned not only about how to develop their economic wealth but how to do so while protecting and preserving their environment. Hopefully, they can learn from some of the mistakes made by the United
States and Europe on their path toward sustainable development. As environmental engineers, we must strive to foster this ethic and to implement sound
solutions to environmental problems. In the truest sense of the phrase, we are
history makers; let’s work to make it a history of which we can be proud.
PROBLEMS
1.1
A horse is running at 30.0 mph. How fast is that in ft/s? in m/min? in furlongs/fortnight?
1.2
How much salt (NaCl) is carried by a river flowing at 30.0 m3/s and containing 50.0 mg/L of salt? Give your answer in kg/day.
1.3
A petroleum refinery loses 0.20% of its input mass of crude oil through
hydrocarbon leaks and spills. The refinery processes 300,000 barrels per
day of crude oil with a density of 7.9 lb/gal. A US barrel is 42 US gallons.
How many tons/yr of hydrocarbons are lost from this refinery?
1.4
During water treatment, a large 20-foot-deep rectangular settling basin is
used to remove suspended solids. Knowing that the basin’s length-towidth ratio is 3:1 and that its width-to-depth ratio is 2:1, what is the
basin’s volume in cubic feet?
1.5
Energy use in the United States in 2011 was 101 quads. Calculate that
energy use in these units:
a. gigajoules
b. barrels of oil equivalent (boe) [1 boe = 6.0 million Btu]
c. kilowatt-hours (kwh)
1.6
A chicken farmer owns a 10-acre farm and starts business with 500 chickens. After one year, he has 4,000 chickens. How many will he have after
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30
Chapter One
one more year? Assuming exponential growth at this rate, how many
chickens would he have after a total of eight years? Does this seem reasonable to you? Why or why not?
1.7
A spherical particle has a mass of 2.00 grams. The particle is a pure compound with a density of 1.5 g/cm3. What is the particle diameter (µm)?
The particle breaks up into one million equal-sized spheres. Consider one
of those small spheres; what is its mass (µg), and what is its diameter
(µm)? (Note the symbol µ stands for “micro” and indicates 10–6.)
1.8
If the farmer of Problem 1.6 wants to sell 10,000 chickens per month, what
is the chicken population he needs to maintain?
1.9
If the world population is 7.0 billion in 2012, and the growth rate is constant at 1.4%, calculate the population in 2030. If the growth rate is constant for another 30 years, what will be the population in 2060?
1.10 Find out how much wastewater and solid wastes are generated each day
at your campus. Where are these two wastes sent? What happens to them?
1.11 Consider the farmer of Problem 1.8 who is selling 10,000 chickens per
month. How many chicken wings is he selling per month? If each wing
weighs 32.0 g, how many kg of wings are being sold? If each chicken weighs
1.2 kg, how many kg of chickens does it take to produce 100 kg of wings?
1.12 Discuss in writing the benefits and limitations of solving pollution problems using the hierarchy shown in Figure 1.5.
1.13 Assume that the world will burn 440 quads of fossil fuels (coal, oil, and
gas) next year. For the total mix of fossil fuels, the average energy content
is 15,000 Btu/pound, and the average carbon mass fraction of the fuels is
0.83. How much CO2 is formed per year from fossil fuel combustion if
3.67 pounds of CO2 are produced for every pound of carbon burned?
1.14 Read your local newspaper thoroughly for a few days. Find an article
relating to environmental pollution or environmental restoration in your
geographical area. Research the issues raised by this article, and write a
short paper supporting one side or the other of the controversy.
1.15 Rainfall runoff is flowing off a certain stretch of roadway into a holding
pond. The road is 0.75 miles long and 60 feet wide. The total amount of
water runoff from the road is equal to the volume of water that would be
at a depth of 0.5 inches on a perfectly flat roadway, if all the water stayed
on the roadway. Assume that all that water flows into the pond. How big
does the pond have to be to exactly hold all that water? Give your answer
in cubic feet and in gallons.
REFERENCES
Carson, Rachel. 1962. Silent Spring. Boston: Houghton Mifflin.
Connolly, John. 1997 (January). “Personal Perspectives: Reflecting on the Past Fifty
Years.” Chemical Engineering Progress, 93(1).
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What Is Environmental Engineering?
31
Cummings, J. B., C. R. Withers, J. Sonne, D. Parker, R. K. Vieira, D. Jackson, and D. Norvell. 2007 (May). “UCF Recommissioning, Green Roof Technology, and Building
Science Training; Final Report.” FSEC-CR-1718-07. Florida Solar Energy Center.
Derevianko, G., and H. Balentine. 2013. “The New Climate Normal: Observed Changes
in the Intensity and Frequency of Extreme Weather Events.” Paper presented at the
106th Annual Conference & Exhibition of the A&WMA, Chicago, IL, June 25–28.
Gao, Y., J. S. Fu, J. B. Drake, and J. F. Lamarque. 2012. “Projected Changes of Extreme
Weather Events in the Eastern United States Based on a High Resolution Climate
Modeling System.” Environmental Research Letters, 7.
Harada, M. 1995. “Minamata Disease: Methylmercury Poisoning in Japan Caused by
Environmental Pollution.” Critical Reviews in Toxicology, 25(1): 1–24.
Miller, G. Tyler Jr. 1996. Living in the Environment. 9th ed. Belmont, CA: Wadsworth.
National Association for Environmental Management (NAEM). 2010. “What Is EHS?”
Accessed July 2013. http://www.naem.org/?page=What_is_EHS
National Council of Examiners for Engineering and Surveying (NCEES). 2013. “FE
Exam.” Accessed July 2013. http://ncees.org/exams/fe-exam/
Taback, H. 2004. “Ethical Obligations of Engineering Professionals.” Presentation at the
2004 Annual Meeting of the Air & Waste Management Association, Minneapolis,
MN, June 24.
United Nations Environment Programme (UNEP). 2014. “Status of Ratification.” Ozone
Secretariat. Last updated March 2014. Accessed April 2014.
http://ozone.unep.org/new_site/en/treaty_ratification_status.php
US EPA (Environmental Protection Agency). 1980. Everybody’s Problem: Hazardous Waste.
Publication SW-826.
US EPA. 2014. “Green Engineering for a Sustainable Environment.” Last updated March
2014. Accessed April 2014. http://www.epa.gov/oppt/greenengineering/
Vesilind, P. Aarne, J. Jeffrey Peirce, and Ruth F. Weiner. 1994. Environmental Engineering.
3rd ed. Woburn, MA: Butterworth-Heinemann.
Cooper.book Page 32 Monday, June 23, 2014 9:58 AM
CHAPTER
2
Chemistry Is Important
to Our Business
2.1 Introduction
Chemistry is the science that deals with the composition, structure, and
properties of substances and the changes they undergo. Environmental chemistry applies this body of science to understanding and predicting the reactions,
fate, and transport of chemical substances in nature, and to the engineering
design of systems to reduce or remove pollution. It is essential that both civil
and environmental engineers have a firm understanding of environmental
chemistry. Unfortunately, too many engineers are inadequately trained in
chemistry and often fail to comprehend the complex chemical issues that challenge them in solving civil and environmental engineering problems. This
chapter provides a review of certain fundamental information presented in basic
chemistry courses, with focus on the application of chemistry to environmental
issues. Prior knowledge of chemistry is assumed, but it is recognized that many
students who are taking this course now may have had only one chemistry
course and it may have been years ago. Therefore it is imperative that those students not only pay close attention to the material presented in this chapter, but
also do outside reading to refresh their knowledge of chemistry.
2.2 Solutions
When two substances are combined but do not chemically react they may
form a suspension, a true solution, or a colloidal dispersion. Figure 2.1 provides an indication of the relative sizes of natural materials. In a suspension,
matter is suspended in a gas or liquid and will settle out relatively quickly.
Usually, suspended matter is greater than 1 micrometer (µm) in diameter. A
true solution is composed of matter homogeneously dissolved in a liquid or
gas (for gases, we usually just call it a mixture, but it actually satisfies the definition of a solution). A true solution cannot be mechanically separated into its
32
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Chemistry Is Important to Our Business
33
component phases withFigure 2.1 Sizes of various natural objects.
out changing the temmeters
perature or pressure of
the solution. The dis102
meters
solved matter is referred
Trees
to as the solute while the
101
1020
Galaxy
dissolving material is the
People
100
solvent. Typically, the
Small pets
Solar System
solute is dissolved as
10–1
molecules or ions which
1010
are mixed uniformly
10–2
Insects
throughout the body of
Earth
the solvent. Somewhere
10–3
Grains of sand
between a suspension
and a solution is a colloi100
10–4
Clay particles
dal dispersion comPersonal
10–5
Algae
Environment
posed of material scattered in a liquid or gas. A
10–6
Molecules
Bacteria
colloidal dispersion can–10
10
not be separated by
10–7
Viruses
gravity; however, it can
Atoms
be mechanically sepa10–8
Colloids
rated. The dispersed ma10–9
terial is typically 0.001 to
1 µm in diameter. All
Hydrogen atom
10–10
natural waters contain
complex combinations of
suspended, colloidal, and
dissolved matter. For example, a surface water (e.g., a river) often contains large
particles of soil suspended in the water (which will settle when the water is allowed to remain quiescent), colloidal algae cells and other small particles that
can be removed only by filtration or centrifugation, and salts and gases (solutes)
that are dissolved and physically inseparable from the solvent (water).
Ways to Report Concentrations in Solutions
In order to fully describe many chemical processes, it is necessary to quantify the components of a solution. The concentration of the solute in a solution
is a quantitative measure of how much solute is in a unit amount of solvent.
For example, when 0.8 grams of salt is dissolved in one liter of water, the resulting concentration can be expressed as 800 mg/L. We could say that this water
is “salty” to the taste, but so is ocean water, which has a concentration (of all
salts) in the range of 3.5% or about 35,000 mg/L! “Salty” is not a specific
enough description; numerical values of concentration allow engineers and scientists to be more precise.
An element or pure compound has properties that are characteristic of that
pure material. For example it has a density, expressed in units of mass/volume
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34
Chapter Two
(e.g., lb/ft3 or kg/L). The density can be thought of as the concentration of
total mass per unit of volume. Another term frequently used in dealing with
solutions or pure substances is the specific gravity. The specific gravity is the
ratio of the density of the substance to the density of a reference material. If the
substance is liquid or solid, the reference material is water; if the substance is a
gas, the reference material is air.
For solutions with water as the solvent (aqueous solutions), we often
express concentrations in units of mass or weight fractions (in this text, mass
and weight are used interchangeably), e.g., mass of solute per mass of solution.
Mixtures of gases are considered gaseous solutions, and are often described
based on mole fractions (moles of solute per mole of solution), or volume fractions, but for both liquids and gases, measures in units of mass/volume are
also common. For solids that are mixtures, mass or weight fractions or percents
apply. For more dilute solutions and mixtures, it is usually much more convenient to express mass fractions as parts of solute weight per million parts of
solution weight (ppm), parts per billion (ppb), or even parts per trillion (ppt).
For solids and liquids, ppm and other such relative measures are always on a
mass basis, while for gases ppm is always on a molar or volume basis. Example
2.1 shows why ppm or ppb is much easier than mass fraction or volume fraction for very small concentrations.
EXAMPLE 2.1
The average concentration of gold in ocean water is 13 parts per trillion (ppt).
Express this concentration as a mass fraction.
SOLUTION
13 ppt = 13 g gold for every trillion [(10)12] g of solution, so
13
-11
12
1 (10 )
= 1.3 (10 )
= 0.000000000013 mass fraction
Mass and Mole Concentration Expressions
Frequently, the concentration of a solution is reported using moles. In
chemistry, a mole is defined as the amount of matter that contains an Avogadro’s Number (6.023 × 1023) of entities (molecules or atoms in this case). The
mass or weight of a mole of an element or molecule is equal to its atomic or
molecular weight, in grams, respectively. Atomic numbers and weights of all of
the elements are provided on the front inside cover of this text. By definition,
one mole of carbon weighs exactly 12.0 grams and contains 6.023 × 1023 atoms.
In chemistry, if we say moles, we always mean gram-moles. However, in engineering it is often convenient to work with other kinds of moles (e.g., a poundmole or a ton-mole). The concept is the same, only the mass units change as
shown in Example 2.2.
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Chemistry Is Important to Our Business
35
EXAMPLE 2.2
How many pound-moles, gram-moles, and ton-moles are contained in 50,000
pounds of NaCl?
SOLUTION
The molecular weight of NaCl is 23 + 35.5 = 58.5. The units can be g/gmol,
lb/lbmol, or tons/tonmol. There are 454 g/lb and 2,000 lb/ton (or 454 gmol/
lbmol and 2,000 lbmol/tonmol).
50, 000 lb NaCl ¥
1 lbmol
= 855 lbmols
58.5 lb
50, 000 lb NaCl ¥
1 lbmol 454 gmol
¥
= 388, 000 gmols
58.5 lb
1 lbmol
50, 000 lb NaCl ¥
1 ton
1 tonmol
¥
= 0.427 tonmols
2, 000 lb 58.5 ton
Another relative measure of concentration is mole fraction. The mole fraction is the ratio of the number of moles of any one component in a solution to
the total number of moles present in the solution. This expression is often used
for solutions in which all components are present at high concentrations. For
example, dry air can be approximated as a mixture of gases with an oxygen
mole fraction of 0.21 and a nitrogen mole fraction of 0.79. This means that there
are 21 moles of oxygen and 79 moles of nitrogen for every 100 moles of dry,
clean air.
Mole/Volume and Mass/Volume Expressions
For dilute solutions, it is common to express concentration as molarity
(gmol/liter). Molarity (M) is defined as the number of gmols of solute divided
by the total volume of solution. Alternatively, concentration can be expressed
as the mass of solute per volume of solution in units of g/L, mg/L, or µg/L for
aqueous solutions, or µg/m3 for gaseous mixtures. In many cases, the mass of
contaminants is fairly low, and one liter of a dilute aqueous solution weighs
approximately 1,000 g. Thus, for water, 1 mg/L = 1 ppm, and 1 µg/L = 1 ppb as
shown below:
1 mg
1 mg
1g
=
= 6 = 1 ppm (by weight)
L
1, 000 g 10 g
(2.1)
1 µg
1 µg
1g
=
= 9 = 1 ppb (by weight)
L
1, 000 g 10 g
(2.2)
If the specific gravity of the solution is not equal to one, the conversion is as follows:
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36
Chapter Two
1 mg
= ppm (by weight) ¥ specific gravity
L
(2.3)
Note that this one-to-one conversion from ppm to mg/L only works for aqueous solutions. For gases, the conversion is more complex as will be shown later.
EXAMPLE 2.3
Calculate the NaCl concentration for a 500-mL solution containing 250 mg of
NaCl in units of mg/L, M, and ppm by weight. The solution is sufficiently
dilute to assume that the specific gravity is 1.0.
SOLUTION
First, calculate the concentration in terms of mg/L:
Concentration =
250 mg
= 500 mg / L
0.5 L
(Note: 500 mg/L is the upper limit for acceptable salt content of drinking water.)
Next, calculate the number of moles that dissolved:
250 mg ¥
1g
1 gmol
-3
¥
= 4.27 (10 ) gmol
1, 000 mg 58.5 g
Next, calculate the molarity of the solution:
-3
Molarity =
4.27 (10 ) gmol
0.5 L
-3
= 8.55 (10 )
gmol/L
Next, calculate the concentration in units of ppm:
500 mg / L = 500 ppm by weight [from Eq. (2.1)]
Another expression frequently used is normality, defined as the number of
equivalents of solute per liter of solution. The number of equivalents is equal
to the number of moles divided by a small whole number, n. The value of n is
determined by either the charge on an ion, the number of available protons in
an acid, the number of available hydroxyl groups in a base, the number of electrons transferred in an oxidation-reduction reaction, or the charge on the positively charged ion (cation) in a molecule. The equivalent weight (EW) of a
compound is its molecular weight divided by n.
The advantage of using normality is that it allows us to equate different
masses of substances that have the same reacting capacity. In order for a solution to be electrically neutral, the number of positive equivalents present in a
solution must be equal to the number of negative equivalents.
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37
EXAMPLE 2.4
Drinking water supplies are often obtained from underground porous limestone
aquifers. Limestone is primarily calcium carbonate (CaCO3), so the water contains calcium ions (which contribute to “hardness” in water—see Section 2.4).
Determine the normality and the molarity of water containing 30 mg/L of
Ca2+ (at this level the water is relatively “soft”).
SOLUTION
The MW of Ca2+ = 40 g/gmol
The Ca2+ ion has a charge of 2, therefore n = 2 equivalents/gmol
The EW of Ca 2+ =
MW 40
=
= 20 g / equivalent
n
2
Concentration of equivalents =
0.030 g/L
= 0.0015 eq/L
20 g/equivalent
Normality = 0.0015 N
Molarity
=
0.030 g/L
40 g/gmol
= 0.00075 M
(Notice that the normality equals n times the molarity.)
It should be clear that the mass or mass flow rate of a pure substance can
be calculated from its volume or volumetric flow rate times its density, and the
mass (mass flow rate) of a pollutant or other substance can be calculated from
its volume (volumetric flow rate) times its concentration. We will formalize this
concept in Chapter 3, but for now simply accept it as fact, and use that fact to
solve Example 2.5.
EXAMPLE 2.5
Refer to Example 2.1. If you could invent a device to extract gold from sea
water, what volume of water would you need to process to obtain one Troy
ounce (31.1 g) of gold (which is worth over $1,000)? The density of ocean
water is 1,027 kg/m3. If it costs you a tenth of a cent for each kg of water processed, will your invention make money?
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38
Chapter Two
SOLUTION
First calculate the mass of water, and next its volume.
mass of water =
31.1 g gold
12
= 2.39 (10 ) g of ocean water
Ê 13 g gold ˆ
Á 12
˜
Ë 10 g water ¯
12
volume of water = 2.39 (10 ) g ¥
1 kg
1 m3
6
¥
= 2.33 (10 ) m 3
1, 000 g 1, 027 kg
The cost to obtain one ounce of gold is:
6
Cost = 2.33 (10 ) m 3 ¥
1, 027 kg
m
3
$0.001
6
= $2.39 (10 )
kg
¥
It will cost almost $2.4 million to obtain one ounce of gold that is worth only
about $1,000, so your invention will definitely not make money!
Volume Per Volume Expressions
Concentrations in gaseous mixtures are frequently expressed on the basis of
mole or volume fraction. Volumetric concentration can be calculated by determining the volume that the gas solute would occupy if present by itself at the
same temperature and pressure as the total solution. This fraction can also be
expressed as ppm by volume, and describes the gas solute volume per million
volumetric units of total gas solution. The reason volumes are used will become
obvious in the next section on gas laws. Table 2.1 summarizes the more common
concentration terms and the types of solutions to which they usually apply.
Table 2.1
Common Concentration Expressions
Concentration Expressions
Units
Solution or Mixture Type
Mole fraction, ni /nT
Percent by weight, mi /mT × 100
Molarity, moles/L solution
Mass/volume, mi /VT
—
%
M
g/L or
mg/L or
µg/L
N
ppm
ppb
µg/m3
ppm
ppb
Gas or liquid, high concentrations
Solid or aqueous, high concentration
Dilute aqueous solution
Dilute aqueous solution
Normality, equivalents/L of solution
Parts per million by mass, (mi /mT ) × 106
Parts per billion by mass, (mi /mT ) × 109
Mass/volume, mi /VT
Parts per million, Vi /VT or ni /nT
Parts per billion, Vi /VT or ni /nT
Note: for gases, ppm is by volume or moles
Where: n = number of moles; m = mass; V = volume;
Dilute aqueous solution
Dilute aqueous solution
Dilute aqueous solution
Dilute gaseous solution
Dilute gaseous solution
Dilute gaseous solution
i = compound i;
T = total.
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39
Gas Behavior
Gases take part in many environmental processes. For example, the transfer of oxygen into water is essential to support aquatic life. Stripping of nuisance gases such as ammonia, carbon dioxide, or hydrogen sulfide involves the
transfer of these gases from water to the atmosphere. The design and operation
of processes involving control and/or mass transfer of various gaseous air pollutants (e.g., SO2, NOx, VOCs) requires knowledge of gas behavior.
Ideal Gas Law
The ideal gas law states that the product of the pressure (P) and the volume
(V) of a given quantity of an ideal gas is directly proportional to the temperature (T). The ideal gas law is expressed universally as shown below:
PV = nRT
(2.4)
where:
P = absolute pressure, atm
V = volume, L
n = number of moles of gas present, gmol
R = the universal gas constant, 0.08206 L-atm/(gmol-K)
T = absolute temperature, K
The units (and the numerical value) of the universal gas constant, R, are a function of the units of the pressure, volume, and temperature terms, and all units
in Eq. (2.4) must be consistent. For example, another value of R is 0.73 atm-ft3/
(lbmol-R) when P, V, n, and T are expressed in those units. Other common values of R are provided in Appendix B. Most gases behave ideally at low pressures and ambient temperatures.
EXAMPLE 2.6
(a) Determine the volume of one gmol of any gas at standard temperature
and pressure (STP: 25 °C, 1 atm). The value of the ideal gas law constant in
a convenient set of units is 0.08206 L-atm/(gmol-K).
(b) Determine the volume of 48.0 pounds of oxygen at 10 psig and 250 °F. For
this part, a convenient value of R is 10.73 psia-ft3/(lbmol-R).
SOLUTION
(a) First rearrange the ideal gas law
V RT
=
n
P
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40
Chapter Two
Now solve, noting that 25 °C must be converted to Kelvin.
V 0.08206 L-atm ( 25 + 273 ) K
=
¥
n
gmol-K
1 atm
= 24.45 L/gmol
(b) First convert pressure and temperature into their absolute scales.
P = 10 + 14.7 = 24.7 psia
and
T = 250 + 460 = 710 Rankine
Next calculate the number of moles of oxygen
48.0 lb
= 1.5 lbmol
32 lb/lbmol
Finally, rearrange the ideal gas law equation and solve for the volume
V=
nRT 1.5 ¥ 10.73 ¥ 710
=
= 463 ft 3
P
24.7
It must be kept in mind that the volume of one mole of gas will vary with the
temperature and pressure of the gas. Earlier we said that converting from ppm
to a mass/volume concentration for gas mixtures was more complex than for
aqueous solutions. We now present the conversion formula as derived from the
ideal gas law to convert ppm (by volume) to µg/m3. We must know the molecular weight of the gas to apply this formula. Also, we stress that for gases, ppm is
always by volume or moles and not by mass. For liquids or solids, ppm is by mass.
C
mg/m 3
=
1, 000 Cppm MW
(2.5)
24.45
where:
MW = numerical value of the molecular weight of the gas, g/gmol
Note the constant in the denominator is valid only at STP; for other temperatures and pressures, the constant must be recalculated.
Partial Pressure and Partial Volume
The partial pressure and partial volume are concepts that apply to mixtures of gases. The partial pressure can be thought of as the pressure that
would be exerted by that one gas in the mixture if all the other gases were
removed, but the total volume remained the same. The sum of the partial pressures of all the gases in the mixture equals the total pressure. Similarly, the partial volume of a gas is the volume that would be occupied by that one gas if all
the other gases were removed, but at the same total pressure of the original
mixture. That is,
PT = P1 + P2 + P3 + ... Pi = Pi
or
VT = Σ Vi
(2.6)
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41
where:
Pi = the partial pressure exerted by gas i (at VT)
Vi = the partial volume occupied by gas i (at PT)
From the ideal gas law:
Pi =
ni RT
VT
or Vi =
PT =
nT RT
VT
or VT =
ni RT
PT
(2.7)
nT RT
PT
(2.8)
Therefore, for gases, the pressure fraction equals the mole fraction as does the
volume fraction:
Pi
n
= i
PT nT
or
Vi
n
= i
VT nT
(2.9)
where:
ni = number of moles of gas i
nT = total number of moles of gas
EXAMPLE 2.7
A hazardous waste incinerator is burning petroleum hydrocarbons, and is emitting an exhaust gas that contains a small concentration of unburned benzene
vapors. The exhaust gas flows out at a rate of 25,000 standard cubic feet (scf) per
minute (STP = 1.0 atm and 77 °F). The benzene concentration is 1.2 ppm.
(a) Calculate the partial volumetric flow rate of benzene gas emitted (scfm).
(b) Calculate the mass of benzene emitted per day (lb). The MW of benzene is
78, and a convenient value of R for these units is 0.73 atm-ft3/ (lbmol-R).
SOLUTION
(a) 25,000 scfm ×
1.2 ppm
(10)6
= 0.030 scfm benzene
(b) First, calculate the molar flow rate
•
n = 1.0 atm ¥
0.030 scfm
0.73 atm-ft 3
¥ 537 R
lbmol-R
-5
=
mol
7.65 (10 ) lbm
min
Now convert to mass flow rate in the desired units.
•
m=
-5
7.65 (10 ) lbmol 78 lb 1, 440 min
¥
¥
= 8.6 lb/day
min
lbmol
day
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42
Chapter Two
Raoult’s Law
The vapor pressure of a liquid is a measure of the tendency of that liquid to
vaporize, and is a strong, nonlinear function of temperature. A volatile liquid
or mixture of liquids may be in equilibrium with the gas phase above it. At
equilibrium, the partial pressure of one component in the gas phase will be
directly proportional to the mole fraction of that component in the liquid mixture as well as to the volatility of that component as measured by its vapor
pressure. Raoult’s Law may be expressed by the following equation. Note that
the pressure units may be expressed in any consistent set.
Pi = xiPV
(2.10)
where:
Pi = partial pressure of component i in the gas phase
xi = mole fraction of i in the liquid mixture
PV = vapor pressure of component i
The vapor pressures of many liquids are commonly reported in chemistry and
physics handbooks. Raoult’s Law has many applications in environmental chemistry. For example, Raoult’s Law can be used to predict the vapor phase concentration of components of gasoline spilled into the subsurface when the gasoline is
present as a floating pool (nonaqueous phase liquid) on the water table.
Henry’s Law
Often, environmental water pollutants are present in very dilute solutions,
and Raoult’s Law is not applicable. For dilute solutions, a variation on Raoult’s
Law called Henry’s Law can be applied. Henry’s Law states that, under equilibrium conditions, the concentration of a gaseous compound dissolved in a
liquid is proportional to its concentration in the gas phase that is in contact
with the liquid. The proportionality constant is called Henry’s Constant and
takes on many different units, depending on the units of the gas and liquid
concentration terms. For example, in the following equation Henry’s Constant
has units of gmol/L-atm because the concentration of the component of interest in the liquid is expressed in gmol/L, and the concentration in the gas is
expressed as a partial pressure in atm.
Caq = KHPg
(2.11)
Alternatively, if the concentration of the compound in both phases is expressed
in units of gmol/L or mg/L, then Henry’s Constant is unitless. Appendix B
provides Henry’s Constants for several environmentally important gases
expressed in two sets of units. Keep in mind that Henry’s Constant is actually
an equilibrium constant, and thus is very dependent on temperature. Henry’s
Law has important environmental application to situations such as determining the solubility of oxygen in water, the partitioning of hydrogen sulfide
between wastewater and the atmosphere, the stripping of volatile organic compounds from groundwater using a stream of air, and others.
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43
EXAMPLE 2.8
A drinking water supply must be treated to control taste and odor due to the
presence of 6.4 mg/L of H2S. It is proposed to remove the H2S from the water
by transferring it to an air stream in a stripping tower. Within the stripping
tower, water flows downward at 10 million gallons/day (MGD) and air flows
upward at 30,000 standard cubic feet per min (scf/min). The temperature is
25 °C and the pressure is 1 atm.
(a) Calculate the air concentration of H2S in µg/m3 and ppm assuming that
the H2S is completely removed from the water.
(b) Calculate the equilibrium concentration of H2S in the air (in ppm) assuming the finished water concentration is 0.5 mg/L. The Henry’s Constant
for H2S is 0.1022 gmol/L-atm at this temperature.
SOLUTION:
(a) First, calculate the H2S mass flow rate in g/sec.
Water flow rate, Q =
(10 MGD ) (106 gal/MG) (3.785 L/gal )
86 , 400 sec/day
Q = 438 L/sec
•
H 2 S mass flow rate, m = Q ¥ C
= 438 L/sec ¥ 0.0064 g/L
= 2.80 g/sec
Next calculate the volumetric flow rate of air.
30 , 000 scf
1 m3
¥
min
35.31 ft 3
Qair , m 3 /sec =
60 sec
min
= 14.16 m 3 / sec
Now calculate the concentration of the H2S in the air, assuming that all
H2S is transferred to the air. The concentration of H2S in the air is simply
the mass flow rate of H2S divided by the volumetric flow rate of air:
•
Cgas =
Cgas =
m
Q
2.80 ¥ 106 µg/sec
14.16 m 3 /sec
= 198, 000 µg/m3
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44
Chapter Two
Rearranging Equation (2.5) and substituting:
Cppm =
198, 000 µ g/m 3 ¥ 24.45
1,000 ¥ 34
= 142 ppm
(b) Using Henry’s Law, calculate the concentration of H2S in the air if it were
in equilibrium with water that contains 0.5 mg/L of H2S. Rearrange Eq.
(2.11) as
Pg =
Pg =
Caq
KH
0.0005 g/L
(34 g/gmol )(0.1022 gmol/L-atm )
= 0.000144 atm
Since the total pressure is 1.0 atm, the pressure fraction is:
(0.000144 atm) / 1 atm = 0.000144
And the concentration in ppm is simply the pressure fraction × 106
Cg = 144 ppm
The equilibrium gas concentration of H2S (144 ppm) is about equal to the
calculated exit gas concentration (142 ppm) achieved by completely stripping
the H2S from the water. In practice, we would use an excess of air to ensure that
we can strip the H2S satisfactorily, and the exit concentration would be well
below the equilibrium value.
2.3 Stoichiometry of Chemical Reactions
In a chemical reaction, atoms combine in ratios of small integers (e.g., 2
moles of H2 react with 1 mole of O2 to yield 2 moles of H2O). A chemical reaction always follows the law of conservation of mass, which states that matter
can neither be created nor destroyed (e.g., 4 g of H2 plus 32 g of O2 produce 36 g
of H2O). The elements or molecules that are reacting are called reactants; the
elements or molecules formed are called products. The quantitative relationship
among the reactants or products is called the reaction stoichiometry. An understanding of the stoichiometric relationships among reacting atoms or molecules
is essential to the control and design of environmental chemical processes.
There are basically four types of reactions. A metathesis reaction involves
the rearranging of atoms into new molecules and often leads to the separation
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45
of one or more products due to a phase change (gas liberation or solid precipitation). An example of this type of reaction is:
H2CO3
→
CO2(gas) + H2O
(2.12)
Acid-base reactions involve the combination of protons and hydroxide ions to
form water. An example of an acid-base reaction is:
→
NaOH + HCl
NaCl + H2O
(2.13)
Reduction-oxidation (redox) reactions involve the transfer of electrons from a
reduced compound to an oxidizing agent. An example of this type of reaction is:
C6H6 + 7.5 O2
→
6 CO2 + 3 H2O
(2.14)
Finally, some reactions lead to electron sharing between two or more reactants.
An example of electron sharing is:
→
NH3 + H+
NH4+
(2.15)
Balancing Simple Chemical Equations
Just as a sentence expresses ideas and thoughts in written form, the chemical
equation describes the behavior of reactants and products. Conventionally, the
equation is written with the reactants on the left and the products on the right. A
chemical equation is of little use unless it is balanced. Balancing involves “discovering” the correct stoichiometry such that the same number of atoms of each type
in the reactants also appears in the products. Note that the total mass of products
must always equal the total mass of reactants, even though the total number of moles may
not balance. The “discovery” process can be accomplished by inspection (a nicer
word than “guessing”) or by following several simple rules. Most people prefer
the rules approach! The following rules are helpful in the inspection process:
• Start with the atoms that appear least frequently, balance them, and then
proceed to other atoms.
• Make sure the number of atoms to the left and right of the arrow are equal.
• Save the H and O atoms until last.
Example 2.9 illustrates this process for two simple reactions.
EXAMPLE 2.9
Balance the following simple chemical reactions:
(a) Ca3(PO4)2 + H2SO4
(b) C8H18 + O2
→
→
CaSO4 + H3PO4
CO2 + H2O
SOLUTION
(a) Ca, P, and S each appear once on the left and once on the right. S already
appears to be balanced, so start with Ca. Note that there are 3 Ca atoms in the
compound on the left, so we put a 3 in front of the Ca compound on the right.
Ca3(PO4)2 + H2SO4
→
3 CaSO4 + H3PO4
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46
Chapter Two
Similarly, for P, we put a 2 in front of the compound that contains the P on
the right.
→
Ca3(PO4)2 + H2SO4
3 CaSO4 + 2 H3PO4
Notice that now S has become unbalanced; we have 3 S atoms on the right,
so place a 3 in front of H2SO4 on the left.
→
Ca3(PO4)2 + 3 H2SO4
3 CaSO4 + 2 H3PO4
Check the H atoms (we have 6 on the left and 6 on the right).
Check the O atoms (we have 20 on the left and 20 on the right); we are done.
(b) This is a simple combustion reaction, and can be balanced with the same
procedure. There are 8 carbons on the left so place an 8 in front of CO2 on
the right.
C8H18 + O2
→
8 CO2 + H2O
There are 18 H atoms on the left, so place a 9 in front of H2O on the right.
C8H18 + O2
→
8 CO2 + 9 H2O
There are now 25 O atoms on the right, so place a 12.5 in front of the O2 on
the left. Notice that it is OK to have fractional stoichiometric coefficients.
→
C8H18 + 12.5 O2
8 CO2 + 9 H2O
Check to be sure that each atom appears equally on each side. We are done.
Now let us examine the balanced reaction in part (b) of the above example.
It is correct to say that 1 mole of octane reacts with 12.5 moles of oxygen to produce 8 moles of carbon dioxide and 9 moles of water. (Note that we could be
using gmol, lbmol, kgmol, tonmol, etc, and this statement is still correct.) However, it is equally correct (and often more convenient) to discuss the stoichiometry in units of mass. See Example 2.10.
EXAMPLE 2.10
How many kg of oxygen are needed to completely combust 10.0 kg of octane?
How many pounds of water are produced?
SOLUTION
Start with the balanced reaction
→
C8H18 + 12.5 O2
8 CO2 + 9 H2O
Next write the MW of each compound beneath the compound and then multiply it by its stoichiometric coefficient.
C8H18
114
+ 12.5 O2
→
(12.5) 32 = 400
8 CO2
+
8 (44) = 352
9 H2O
9 (18) = 162
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47
Note that the total mass on the left (114 + 400 = 514) equals the total mass on
the right (352 + 162 = 514); this always is true for all balanced reactions. Also,
note that the mass ratio of oxygen to octane is 400/114; the water/octane ratio
is 162/114.
The answers to this example are given by:
10.0 kg octane ¥
10.0 kg octane ¥
400 kg oxygen
= 35.09 kg oxygen
114 kg octane
2.205 lb 162 lb water
¥
= 31.33 lb water
1.0 kg 114 lb octane
EXAMPLE 2.11
You have a sample of water that contains the organic compound C7H12ON2 at
a concentration of 50 mg/L. The compound can be oxidized by bacteria to
form carbon dioxide, water, and ammonia. How many mg/L of oxygen is
needed to biodegrade the compound?
SOLUTION
First balance the equation using our previous approach.
→
C7H12ON2 + O2
CO2 + H2O + NH3
There are 7 C atoms on the left so put a 7 in front of the CO2 on the right; also
there are 2 N atoms on the left so place a 2 in front of ammonia on the right.
C7H12ON2 + O2
→
7 CO2 + H2O + 2 NH3
There are 12 H atoms on the left; only 6 appear in the two NH3 molecules on
the right, leaving 6 for the water. So place a 3 in front of H2O on the right.
C7H12ON2 + O2
→
7 CO2 + 3 H2O + 2 NH3
Now there are 17 O atoms on the right; we have one O atom in the organic compound on the left, so we need 16 more on the left. Place an 8 in front of the O2.
C7H12ON2 + 8 O2
→
7 CO2 + 3 H2O + 2 NH3
The reaction is balanced. Note that we need 8 moles of oxygen for each mole
of organic compound; in terms of mass, that ratio is 256 mg of oxygen for each
140 mg of organic compound. Now use the mass stoichiometry to solve for
the requested information.
50 mg / L compound ¥
256 mg oxygen
= 91.4 mg/L oxygen
140 mg compound
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48
Chapter Two
Balancing Redox Reactions
Simple redox reactions involving oxidation by elemental oxygen often can
be balanced by inspection (as in Example 2.10 and Example 2.11), but that
method can be very frustrating when dealing with more complicated redox
reactions. In general, for redox reactions we use the half-reaction method.
There are two half-reactions in each redox reaction—the oxidation half and the
reduction half. Oxidation involves losing electrons, while reduction involves
gaining electrons. One application of this method uses oxidation numbers—
numbers that indicate the relative state of oxidation of each atom in a compound. As atoms are oxidized, their oxidation numbers get more positive, and
conversely, atoms that are reduced see their oxidation numbers grow more
negative. There are a few simple rules to remember for using oxidation numbers to balance redox reactions.
• Oxidation numbers of elements are always zero
• Oxygen is almost always –2
• Hydrogen is almost always +1
• Group IA metals are always +1; Group IIA metals are always +2
• For uncharged compounds, the oxidation numbers of all atoms in the
compound must sum to 0
• For charged ions, the oxidation numbers of all atoms in the ion must sum
to the charge on the ion
Use of these rules is demonstrated in the following example.
EXAMPLE 2.12
Balance the reactions:
(a) Cl2 + NH3
→
N2 + HCl
(b) FeSO4 + K2Cr2O7 + H2SO4 →
Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + H2O
SOLUTION
(a) First assign oxidation numbers to each atom (notice that the H atom stays
at +1 during this reaction)
0
–3 +1
Cl2 + NH3
→
0
+1 –1
N2 + HCl
The nitrogen atom is being oxidized and the chlorine is being reduced.
Now write the half-reactions:
Oxidation:
–3
N
0
–
3 e–
→
N
but the N on the right is combined as N2, so the oxidation reaction really
should be written as:
–3
2N
0
–
6 e–
→
N2
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49
On the left side of the reaction, the minus 6 (due to the two N atoms) is
cancelled by subtracting 6 electrons, and the total “charges” of 0 is
matched by the oxidation number total of 0 on the right (N2 as an element). Similarly for the reduction half-reaction:
Reduction:
0
–1
Cl2
2 e–
+
→
2 Cl–
Now, the number of electrons lost must equal those gained, so we must
multiply the reduction half-reaction by 3 to have the chlorine gain 6 electrons total.
0
–1
3 Cl2
6 e–
+
→
6 Cl–
Finally, put these coefficients on the whole reaction, and check to ensure
all the other atoms are balanced.
→
3 Cl2 + 2 NH3
N2 + 6 HCl
(b) First, assign oxidation numbers to the key atoms. Note that the oxidation
numbers of S, K, O, and H do not change in this reaction, so the key atoms
are Fe and Cr.
+2
+6
+3
FeSO4 + K2Cr2O7 + H2SO4
→
+3
Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + H2O
Notice that the iron is being oxidized and the chromium is being reduced.
Now write the half-reactions (keeping the atoms balanced consistent with
their subscripts in compounds):
Oxidation:
+2
2 Fe
+3
–
2 e–
→
+
6 e–
→
Fe2 (loss of 2 electrons)
Reduction:
+6
Cr2
+3
Cr2 (gain of 6 electrons)
The number of electrons lost must equal those gained, so we must multiply the oxidation half-reaction by three to yield:
+2
6 Fe
+3
–
6 e–
→
3 Fe2
Now put these coefficients on the whole reaction, and check to ensure all
atoms are balanced.
6 FeSO4 + K2Cr2O7 + H2SO4
→
3 Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + H2O
The Fe and Cr are already balanced. Save the O and H until last, so check
the K and S next. The K is balanced, but the S is not. There are 13 S atoms
on the right and 7 on the left. If we change the FeSO4 on the left then we
will unbalance the Fe, so focus on the sulfuric acid. Since we have a coeffi-
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50
Chapter Two
cient of 6 on the FeSO4, we need a 7 on the H2SO4 to make the sulfur balance. Note that since S only appears in the SO4 ions on both sides, we have
also now balanced all the oxygen except for the 7 O atoms in K2Cr2O7. (We
have 13 SO4 ions on the left and 13 on the right, and that accounts for 52 O
atoms on each side).
6 FeSO4 + K2Cr2O7 + 7 H2SO4
→
3 Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + H2O
Put a 7 in front of the H2O on the right. This balances the O atoms. Now
check the H atoms. We have 14 on the left and 14 on the right. The final
balanced reaction is
6 FeSO4 + K2Cr2O7 + 7 H2SO4
→
3 Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 7 H2O
In summary, a balanced chemical equation provides the basis of a quantitative relationship among the reactants and products and allows one to calculate
masses of reactants consumed and products formed. For example, in the balanced reaction of Example 2.12 (a), three moles of chlorine react with two
moles of ammonia to produce one mole of nitrogen and six moles of hydrochloric acid. On a mass basis, the stoichiometry tells us that 213 g of chlorine
react with 34 g of ammonia to produce 28 g of nitrogen and 219 g of hydrochloric acid.
EXAMPLE 2.13
Using the stoichiometric relationships developed in Example 2.12, calculate
the amount of ammonia in grams that will react with 100 g of chlorine to produce nitrogen and hydrochloric acid.
SOLUTION:
The balanced equation is 3 Cl2 + 2 NH3 → N2 + 6 HCl
The question will be answered first using molar stoichiometry.
Cl 2 reacting =
100 g Cl 2
71 g/gmol
= 1.41 gmol CL 2
From the balanced chemical equation above,
NH 3 reacting = 1.41 gmol Cl 2 ¥
2 gmol NH 3
3 gmol CL 2
= 0.94 gmol NH 3
Finally, calculate the weight in g of 0.94 gmol of NH3.
Mass of NH3 = 0.94 gmol × 17 g/gmol
= 16.0 g NH3
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51
Now this same question will be answered using mass stoichiometry. The balanced equation (with masses) is
3 Cl2 + 2 NH3
213
→
34
N2 + 6 HCl
28
219
Mass of NH 3 consumed = 100 g Cl 2 ¥
34 g NH 3
213 g Cl 2
= 16.0 g NH 3
2.4 Water Chemistry
Equilibrium Concepts
Many reactions are reversible; that is, under certain conditions the reactants will produce products, and under other conditions the products will produce reactants. When the reactants are producing products at the same rate as
the products are producing reactants, equilibrium has been achieved. In more
advanced courses, you will learn that thermodynamic equilibrium occurs
when the total free energy of the system is at a minimum. The approach to
equilibrium is a dynamic process during which small changes in temperature,
pressure, or concentrations can cause a shift in the direction of the reaction. The
change in free energy for any reaction (which can be calculated knowing the
conditions of the reaction) can provide an indication of whether the reaction
can be expected to proceed in the direction written. Reaction equilibrium
occurs when there is no further change in free energy.
A reversible reaction can be portrayed by the following general equation:
aA + bB
↔
cC + dD
(2.16a)
When the system is at equilibrium, there is no net change in the concentrations
of all reactants and products. Chemical equilibrium can be quantitatively
described by an equilibrium constant. The constant is determined from the
ratio of the equilibrium molar concentrations (for dilute solutions) of the products and reactants at the reaction temperature and pressure, and is specific to a
particular reaction. The equilibrium constant can be calculated from thermodynamic properties of the reactants and products in their standard states. For the
general chemical reaction of Equation (2.16a), the equilibrium constant is written as follows:
c
ÈC ˘ ÈD
Keq = Î ˚ a Î
È A ˘ ÈB
Î ˚ Î
where: [ ] indicates concentration in gmol/L
˘
˚
d
b
˘
˚
(2.16b)
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52
Chapter Two
The value of the equilibrium constant is a ratio of the equilibrium concentrations of products to concentrations of reactants, each raised to their stoichiometric
coefficients. The equilibrium constant will have a different value if the temperature or pressure changes. A small K value suggests that only a small fraction of
reactants has been converted to products at equilibrium, while a large value of Keq
indicates that most of the reactants have been converted. The value of K, however,
does not provide information regarding the rate at which the reaction approaches
equilibrium. For example, the Keq for the reaction between hydrogen and oxygen
to produce water suggests that water production is highly favored. However,
hydrogen and oxygen can exist together in a reaction vessel for a very long time
without producing water. But if a catalyst, such as finely divided platinum, is
added, or if a spark is provided, the reaction will proceed to equilibrium so rapidly that it is described as an explosion. The presence of the catalyst or spark does
not affect the value of Keq, only the rate at which equilibrium was reached.
The units of Keq are specific to the reaction and to the medium. The concentrations of dissolved solutes are expressed in units of gmol/L, while for gases,
units often are gmol/m3 or partial pressures. The concentration of a pure solid
taking part in a reaction is assumed to remain constant, and is defined to be
one. A pure liquid taking part in the reaction is assumed to have a mole fraction of one. There are several specialized equilibrium constants, namely the
water constant, Kw, the acid constant, Ka, and the solubility product constant,
Ksp. All will be introduced in this chapter.
Acid-Base Chemistry
Acid-base chemistry is very important in environmental engineering.
Aquatic life is very sensitive to the pH of natural waters. Acid rain has enormous impacts around the world; the acid gases that cause acid rain often are
removed at the industrial sources by caustic scrubbers. Acid-base chemistry
demonstrates an important application of equilibrium principles. An acid is
defined simply as a compound that is capable of donating a hydrogen ion or
proton (H+) in an aqueous solution. (A bare proton does not exist in water;
rather, it associates with a water molecule as a hydronium ion H3O+, but we
shall simply denote it as H+ in this text.) A base is a compound that accepts
that proton (often, but not always, a base provides hydroxide ions (OH–) to
accept the proton to form water). Water is a unique compound in that it produces both a proton and hydroxide ion when it dissociates:
H2O
↔
H+ + OH–
(2.17)
Because this equilibrium lies well to the left, the concentration of water can be
considered to remain constant at 55.5 gmol/L and can be incorporated into the
equilibrium constant. The specialized equilibrium constant for water is called
the water ionization constant (Kw):
K w = È H + ˘ ÈOH - ˘
Î
˚Î
˚
In this form, the value of Kw is 1 (10)–14 at 25 °C.
(2.18)
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53
The pH System
Because only a small amount of water dissociates, as indicated by the low
value of Kw, the concentrations of the hydrogen and hydroxide ions are very
small numbers. In fact, in pure water, the ÈH + ˘ and ÈOH - ˘ concentrations are
Î
˚
Î
˚
each equal to 10–7 gmol/liter. To simplify computations dealing with these
ions, the pH system was developed. The pH of an aqueous solution is defined
as the negative of the logarithm of the hydrogen ion concentration (in gmol/L):
(2.19)
pH = - log ÈH + ˘
Î
˚
Similarly, we can define the pOH as the negative log of the OH– ion concentration:
(2.20)
pOH = - log ÈOH - ˘
Î
˚
From the definition of Kw, it can be seen that at 25° C:
ÈH + ˘ ÈOH - ˘ = 10 -14 and thus
Î ˚Î
˚
(2.21)
pH + pOH = 14
This relationship holds in all aqueous
solutions, no matter what other ions are
present. When the concentrations of
hydrogen and hydroxide ions are equal,
the pH and pOH are both 7. When the
solution is acidic, the concentration of
hydrogen ions is greater than the concentration of hydroxide ions and the pH is
less than 7 (but the pH + pOH still sums
to 14). When the solution is basic, the concentration of hydrogen ions is less than
the concentration of hydroxide ions and
the pH is greater than 7. The pHs of some
common substances are illustrated in Figure 2.2. The importance of pH in controlling certain metal concentrations in water
will be illustrated in a subsequent example problem.
Weak and Strong Acids and Bases
When placed in aqueous solution,
acids and bases tend to dissociate to different degrees, depending on their chemical composition. A typical reaction of a
weak acid is shown as follows:
Figure 2.2 Values of pH for
common liquids.
pH Scale
14.0
13.0
Bleach
12.0
11.0
Ammonia
10.0
9.0
8.0
7.0
Baking soda
Seawater
Blood
Distilled water
Milk
6.0
5.0
4.0
3.0
Coffee
Orange juice
Carbonated soft drink
Vinegar
2.0
1.0
0.0
Battery acid
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54
Chapter Two
HA
↔
H+ + A–
(2.22)
The tendency to dissociate is referred to as the strength of the acid or base. A
strong acid or base dissociates nearly completely, and the equilibrium position
is far to the right in Equation (2.22). In other words, the concentration of the ion
A– is much greater than the concentration of the un-ionized acid, HA. A weak
acid or base tends to remain largely un-ionized, with the equilibrium position
far to the left in Equation (2.22). That is, the concentration of HA far exceeds
that of A–. The strength of the weak acid is determined by the magnitude of its
specialized equilibrium constant, KA, called the acid constant, and defined in
Equation (2.23).
ÈH + ˘ È A - ˘
Î
˚Î ˚ = K
A
ÈHA ˘
Î
˚
(2.23)
For a weak acid, the value of KA is quite small, so for convenience, the p-operator system is often used for these acid constants as follows:
pKA = –log KA
(2.24)
Typically, the pK for a weak acid or base is a positive number; for example, the
KA for acetic acid is 1.78(10)–5 (at 25 °C) and the pKA is 4.75. Although rarely
used, a pK can be determined for a strong acid or base; in this case K will be a
large number and pK will be negative. For example, the KA for hydrochloric
acid is 1,000 at 25 °C and the pKA is –3. Mostly, for strong acids, we simply
assume that they are completely ionized. Appendix B provides some data for
commonly encountered weak acids and bases.
The pH of a Strong or Weak Acid
When an acid is mixed with pure water, the pH of the solution will be less
than 7. How much the pH drops is a function of both the concentration of the
acid in solution and the equilibrium constant for the weak acid. At equilibrium, the concentrations of the various species involved (for example, H+,
OH–, A–, and HA, for a typical monoprotic acid solution) are controlled by a
set of four equations: the equilibrium constant for acid or base dissociation,
the ionization constant for water, the mole balance for the chemical species of
interest, and the electroneutrality balance. An electroneutrality balance sums
the equivalents of charged ions and requires that the equivalents of positive
ions be equal to the sum of the equivalents of the negative ions. For a typical
monoprotic acid, HA, the four equations (with two repeated from the earlier
discussion) are as follows.
Water equilibrium
Acid equilibrium
K w = ÈH + ˘ ÈOH - ˘
Î
˚Î
˚
KA
ÈH + ˘ È A - ˘
˚Î ˚
=Î
[HA ]
(2.18)
(2.23)
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CT = ÈHA ˘ + È A - ˘
Î
˚ Î ˚
Mole balance on A
55
(2.25)
where: CT = total concentration of A
ÈH + ˘ = ÈOH - ˘ + È A - ˘
(2.26)
Î
˚ Î
˚ Î ˚
These four equations can be manipulated to solve problems related to acid and
base equilibrium. In some cases (especially for pH between 6.5 and 7.5), it is
necessary to solve all four equations simultaneously to get correct answers. In
many other cases (usually for pH < 6 or > 8), simplifying assumptions can be
made, as illustrated in the following examples.
Charge balance
EXAMPLE 2.14
What is the pH of a 0.015 M solution of hydrochloric acid (HCl)? HCl is a
strong acid.
SOLUTION
Because HCl is a strong acid, we assume that it is completely ionized. Relative
to Eq. (2.25), we can say that essentially all the 0.015 gmol/L of HCl ionizes,
and there is no HCl left in its original form. Thus, the concentration of H+ is
0.015 gmol/L.
Since pH = - log ÈH + ˘
Î
˚
pH = - log (0.015 )
pH = 1.82
Thus, for a solution of a strong acid, the concentration of H+ is equal to the
initial concentration of the strong acid, but this is not true for a weak acid, as
shown in the following example.
EXAMPLE 2.15
What is the pH of a 0.015 M solution of acetic acid (HAc)? HAc is a weak acid
with a pKA of 4.75.
SOLUTION
First, calculate the KA. Solving Eq. (2.24),
K A = 10 - pK A
-5
K A = 10 -4.75 = 1.78 (10 )
This problem can be solved by assuming that the water equilibrium reaction is
insignificant. Set up a reaction table as you learned in freshman chemistry, and
let x represent the gmol/L of the acetic acid that ionizes to reach equilibrium.
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56
Chapter Two
= H+
HAc
+ Ac–
start
0.015
0
0
reaction
–x
+x
+x
equil
0.015–x
x
x
Stoichiometrically, it can be seen that for every mole of HAc that ionizes, one
mole of H+ and one mole of Ac– are produced. Then:
ÈHAc ˘ = 0.015 - x
Î
˚
ÈH + ˘ = x
Î
˚
È Ac - ˘ = x
Î
˚
At equilibrium, the ratio of the product of the ion concentrations to the molecular form of the acid is equal to the KA:
( x )( x )
(0.015 - x )
-5
= 1.78 (10 )
This equation can be solved for x either using the quadratic equation, or by iterating twice. In the first iteration, assume x is negligible compared with 0.015,
and in the second one, use the value of x obtained from the first calculation.
First iteration:
x2 = (0.015 – 0) (1.78 (10)–5)
x = [2.67 (10)–7]1/2 = 5.17 (10)–4
Second iteration: x2 = (0.015 – 0.000517) (1.78 (10)–5)
x = [2.58 (10)–7]1/2 = 5.077 (10)–4
The answer from the second iteration is close enough to the first iteration
answer that no further iterations are needed. If we had used the quadratic
equation (and had not made any mistakes punching in the numbers), the
exact answer would have been x = 5.079(10)–4, to which our second iteration
answer is very close.
Now, since x = concentration of H+, use this value to determine the pH
pH = – log (5.077 (10)–4) = 3.29
This pH is well below 6.5, so there appears to be no need to use the more precise method of simultaneously solving the four equations presented earlier.
But, the simplifying assumptions must be checked. CT of 0.015 M is much
greater than the concentration of H+ [5.077 (10)–4 M], and at a pH of 3.29, the
H+ produced from ionization of water is negligible compared to that contributed by the HAc.
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57
As seen in Example 2.15, x is frequently far less than CT. In some cases it is
ignored, and even two iterations are not needed. The value of x then can be
obtained from the acid constant as follows:
(2.27)
x = K A CT
Since ÈH + ˘ = x, converting to the logarithmic system:
Î
˚
pH = ½(pKA + pCT)
(2.28)
In the case of Example 2.15, use of Eq. (2.28) would have yielded an answer of
pH = 3.287, or 3.29 to three significant figures.
Buffers
Many solutions exist in nature that are capable of withstanding the addition of strong acids and bases with little change in pH. These solutions are
called buffers. They are generally combinations of weak acids and their salts.
For example, a combination of sodium bicarbonate (NaHCO3) and sodium carbonate (Na2CO3) will form a buffered solution with a pH near the pKA of bicarbonate, 6.35. A buffer functions because the salt (carbonate) is able to absorb
any new acids (hydrogen ions) that may be added, and the acid (bicarbonate) is
able to absorb any new bases (hydroxide ions) that may be added so that the
resulting equilibrium concentrations of H+ and OH– do not change dramatically upon the addition of a new acid or base. The ability of a buffered solution
is not infinite, however, and works best within +1 pH unit of the system pKA.
Solubility Product
Another example of the application of the equilibrium concept is the solubility of solids. This is of great interest if the solid releases some toxic ion into
water. For the general case:
MaZb(s)
↔
aM + bZ
(2.29)
where M is a positive ion (cation) and Z is a negative ion (anion). It should be
understood that these ions have their appropriate charges on them. The general equilibrium constant is:
a
Keq
ÈM ˘ ÈZ
˚ Î
=Î
È M a Zb
Î
˘
˚
˘
˚
b
(2.30)
where the square brackets indicate concentrations in gmol/L. Since the concentration of the solid is defined as unity, a specialty equilibrium constant, known
as the solubility product constant, can be written as:
a
Ksp = È M ˘ ÈB ˘
Î
˚ Î ˚
b
(2.31)
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58
Chapter Two
Solubility product constants are presented in Appendix B for some common
compounds. For many compounds of interest, the Ksp is a very small number,
and this property allows us to remove toxic metal ions from water. Example
2.16 shows how to write the Ksp formula for any compound.
EXAMPLE 2.16
Write the formulas for the Ksp for silver chloride—AgCl, lead hydroxide—
Pb(OH)2, and calcium phosphate—Ca3(PO4)2.
SOLUTION
For AgCl, the Ksp is simply È Ag + ˘ ÈCl - ˘
Î
˚ Î
˚
For Pb(OH)2, the Ksp is ÈPb2+ ˘ ÈOH - ˘
Î
˚ Î
˚
For Ca3(PO4)2, the Ksp is ÈCa 2+ ˘
Î
˚
3
2
ÈPO 4 3 - ˘
Î
˚
2
A solution can be described as being saturated, unsaturated, or supersaturated. Mostly in environmental work we deal with unsaturated or saturated
solutions. An unsaturated solution is not at equilibrium and can dissolve more
solid. Only a saturated solution, which cannot dissolve more solid unless the
temperature or pressure is changed, is at equilibrium. The equilibrium state of
any solution can be determined by comparing the ion product, IP (the product
of the ion concentrations raised to their appropriate stoichiometric coefficients)
to the solubility product constant. If the IP is less than the Ksp then the solution
is unsaturated, and if the IP is equal to the Ksp the solution is at equilibrium
and is saturated.
The term “solubility” is not the same as the solubility product constant,
although the two are related. Solubility is a general term describing the tendency of a solid to dissolve. In most cases the solubility will increase with
increasing temperature, but this is not always the case. For example, calcium
carbonate solubility decreases as temperature increases—which can create
problems in water heaters and boilers. The solubility is also affected by the
presence of other ions in solution which may react with one of the dissolving
products. The solubility is particularly affected by the addition of an ion common to one of the ions in the solid, as seen in Example 2.17.
EXAMPLE 2.17
(a) Calculate the concentration (mg/L) of barium in a saturated solution of barium sulfate, BaSO4, in pure water at 25 °C. The Ksp for BaSO4 is 1.0(10)–10.
(b) What is the equilibrium concentration of barium (Ba2+) in water that
already contains 10–3 M sulfate, SO42– ?
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59
SOLUTION
(a) Let x equal the number of gmol/L of BaSO4 which will dissolve in pure
water as follows:
BaSO4(s)
↔
Ba2+ + SO42–
If x moles of BaSO4 dissolve, then x moles of Ba2+ and x moles of SO42– will
enter the solution.
Ksp = 1.0 ¥ 10 -10
= ÈBa 2+ ˘ ÈSO 4 2- ˘
Î
˚ Î
˚
= x2
and:
x = 1.0(10)–5 gmol/L
(b) If the water initially contains 10–3 M SO42–, after y gmol/L of BaSO4 dissolve, the solution will contain (y + 10–3) gmol/L of SO42– and y gmol/L of
Ba2+. Then:
Ksp = 1.0 ¥ 10 -10
= ÈBa 2+ ˘ ÈSO 4 2- ˘
Î
˚ Î
˚
(
= y y + 10 -3
)
which can be solved either via the quadratic equation or by iteration.
Either method gives
y = 1.0(10)–7 gmol/L
Thus, in the presence of a 0.001 M of the common ion, sulfate, the concentration of the heavy metal, barium, is two orders of magnitude lower than
in pure water.
For many compounds, the Ksp is a very small number, indicating that the
solid is only sparingly soluble. The concentration of many heavy metals can be
controlled by precipitation with ions such as sulfide, hydroxide, or carbonate
because their precipitates (solid forms) have extremely small Ksp values (see
Figure 2.3). Note, however, in Figure 2.3 (on the following page), that at
extremely high pH, solubility of these metals can actually increase due to the
fact that soluble charged metal hydroxide complexes can form such as
Fe(OH)4–. Example 2.18 demonstrates how we might take advantage of the low
solubility of certain substances.
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Chapter Two
60
100
Figure 2.3 Solubility of
some heavy metal hydroxides as a function of pH.
(US EPA, 1983.)
Pb
10
Fe+2
Solubility
(mg/L)
1.0
Ag
Zn
0.1
Cd
0.01
Ni
Cu
0.001
Fe+3
0.0001
6
7
8
9
10
11
12
pH
EXAMPLE 2.18
(a) Calculate the concentration of cadmium as the pH of a solution increases
from 8 to 10. Assume that the solubility of cadmium is controlled only by
hydroxide. The Ksp for cadmium hydroxide, Cd(OH)2, at 25 °C is 2.0 (10)–14.
(b) The national groundwater drinking water standard for cadmium is 0.005
mg/L. Calculate the hydroxide ion concentration in, and the pH of, the
solution that just meets the groundwater standard.
SOLUTION
(a) From the Ksp and the concentration of OH–, the concentration of Cd2+ can
be calculated.
Cd(OH)2 (s)
↔
Cd2+ + 2OH–
Ksp = ÈCd 2+ ˘ ÈOH - ˘
Î
˚ Î
˚
2
At pH = 8, from Eqs. (2.20) and (2.21), the concentration of OH– is 10–6 M,
so the concentration of Cd2+ is:
-14
ÈCd 2+ ˘ = 2.0 (10 )
2
Î
˚
10 -6
( )
-2
= 2.0 (10 )
M
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61
At pH = 10, the concentration of OH– is 10–4 M, so the concentration of
Cd2+ is:
-14
ÈCd 2+ ˘ = 2.0 (10 )
2
Î
˚
10 -4
( )
-6
= 2.0 (10 )
M
You can see that as the pH increases (as the hydroxide concentration gets
bigger) the cadmium concentration (a toxic heavy metal) gets much lower.
(b) In order to calculate the concentration of hydroxide when cadmium is
0.005 mg/L, first the concentration must be expressed in units of gmol/L.
-6
ÈCd 2+ ˘ = 5 (10 ) g/L
Î
˚ 112.4 g/gmol
-8
= 4.45 (10 )
M
From the Ksp, the hydroxide concentration can be calculated:
-14 ˆ
Ê
ÈOH - ˘ = Á 2.0 (10 )
Î
˚ Á 4.45 10 -8 ˜˜
( ) ¯
Ë
-4
= 6.7 (10 )
1/2
M
Then,
pOH = –log (6.7(10)–4) = 3.17
and
pH
= 14 – 3.17 = 10.83
So, to control Cd in water, we would need to add a base, like NaOH, to
raise the pH to 10.83 and precipitate out Cd(OH)2.
The Carbonate System
Earlier we mentioned buffers. One of the most important buffering systems in nature is the carbonate system, composed of carbon dioxide (CO2), carbonic acid (H2CO3), bicarbonate ions (HCO3–) and carbonate ions (CO32–). The
aqueous carbonate system develops from both atmospheric carbon dioxide
and the many solid carbonate species in the earth. Carbon dioxide is involved
in both biological respiration (where it is produced) and photosynthesis (where
it is consumed). The precipitation of calcium carbonate (CaCO3) in pipes and
process tanks can lead to clogging and scaling. The carbonate system helps stabilize pH in many lakes and rivers, to provide a constant pH for aquatic life.
Surface waters can be in equilibrium with these various carbonate sources and
sinks (see Figure 2.4).
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62
Chapter Two
Figure 2.4 Carbon dioxide/bicarbonate/carbonate equilibrium in surface water
with calcium present.
CO2 from Air
Limestonecontaining
lake bed
CaCO3 + CO2 + H2O <
——> Ca+2 + 2 HCO3–
sediment
CO2
Let us start consideration of this system with the dissolution of carbon
dioxide gas into water:
↔
CO2 (g)
CO2 (aq)
(2.32)
Once dissolved, the CO2 hydrolyzes to carbonic acid:
CO2 (aq) + H2O
↔
H2CO3
(2.33)
It is not possible to distinguish between aqueous carbon dioxide and carbonic
acid. The carbonic acid dissociates into ions:
H2CO3
↔
H+ + HCO3–
(2.34)
H+ + CO32–
(2.35)
The bicarbonate ion also dissociates:
HCO3–
↔
If there is calcium present, it can react with carbonate ions to form calcium carbonate:
Ca2+ + CO32–
↔
CaCO3
(2.36)
There are various equilibrium constants for the above reactions. The Henry’s Law
constant applies to reaction (2.32), and to reaction (2.33) by extension. That is,
KH =
CH 2CO 3
PCO 2
-2
= 3.39 (10 )
mole/L-atm
(2.37)
The first ionization constant of carbonic acid is
ÈH+ ˘ ÈHCO -3 ˘
˚ = 4.47 10 -7
Ka1 = Î ˚ Î
( )
ÈÎH 2 CO 3 ˘˚
(or 10 )
-6.35
(2.38)
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The second ionization constant is
ÈH+ ˘ ÈCO 32- ˘
˚ = 4.68 10 -11
Ka2 = Î ˚ Î
( )
ÈHCO 3 ˘
Î
˚
(or 10
-10.33
)
(2.39)
Finally the Ksp for calcium carbonate is
-9
Ksp = ÈCa 2+ ˘ ÈCO 32- ˘ = 4.57 (10 )
Î
˚ Î
˚
(or 10 )
-8.34
(2.40)
The reason that the three Ks in Eqs. (2.38) to (2.40) are also shown in the format
10x is to make it very easy to see the value of the pK (e.g., the pK for Eq. (2.38) is
simply 6.35). Quite often, reference books will give these values as pK values
instead of in scientific notation.
The reader should be able to see that as pH is independently adjusted (say
by adding nitric acid or sodium hydroxide), the species in equilibrium will
shift to try to offset the change in pH. That is, if we add some H+ ions (e.g.,
“acid rain”) to a lake whose carbonate system is in equilibrium, then reactions
(2.34) and (2.35) will shift to the left to remove some of that excess H+. If we
spill some strong base (e.g., “lye”) into the water, which supplies OH– ions that
“soak up” H+, then those same reactions will shift to the right. However, the
values of the equilibrium constants do not change, only the relative concentrations of the species. This behavior is typical of a buffered system. Due to the
values of the constants in Eqs. (2.37) to (2.40), at pH values less than 6, almost
all the carbonate species are in the form of H2CO3, and at pH values above 10,
almost all the carbonate is in the form of CO32–. Between pH 6 and 10, the bicarbonate (HCO3–) form predominates.
The ability of a natural body of water to withstand pH changes is measured by its acidity or alkalinity. Acidity is defined as the capacity to neutralize
a strong base and is calculated as follows:
Acidity (eq/L ) = 2 ÈÎH 2 CO 3 ˘˚ + ÈHCO 3 - ˘ + ÈH + ˘ - ÈOH - ˘
Î
˚ Î
˚ Î
˚
(2.41)
Recall that the square brackets, È ˘ , indicate the units of gmol/L.
Î ˚
Alkalinity is defined as the capacity to neutralize a strong acid and is calculated as follows:
Alkalinity (eq/L) = ÈHCO 3 - ˘ + 2 ÈCO 3 2- ˘ + ÈOH - ˘ - ÈH + ˘
Î
˚
Î
˚ Î
˚ Î
˚
(2.42)
Traditionally, the concentrations of certain chemical constituents in water
and wastewater have been expressed on the basis of calcium carbonate
(CaCO3). Expressing different chemicals as one common material makes it easy
to add their concentrations to get their total effect. As an analogy, consider a
wealthy person with bank accounts in several countries. In order to calculate
his or her total wealth, you must express each currency in terms of a common
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64
Chapter Two
unit (such as ounces of gold). Examples of constituents that are often quantified as CaCO3 are divalent metal cations, such as Ca2+ and Mg2+, and the alkalinity species shown in Eq. (2.42). Expressing these species in terms of a single
component allows the individual species to be summed, indicating their equivalent total reacting capacity. Therefore, this expression is based on the number
of equivalents of each species present. For any species X, to convert its concentration from mg/L as itself to mg/L as CaCO3, use the following equation:
mg/L as CaCO 3 =
mg/L as X
¥ EW of CaCO 3
EW of X
(2.43)
The previously introduced terms, acidity and alkalinity, are often
expressed in mg/L as CaCO3 as follows:
mg/L as CaCO3 = (meq/L of acidity or alkalinity × 50 mg CaCO3/meq) (2.44)
Because calcium is so common in many water systems (both natural and
built), and because calcium can precipitate with carbonate to form a nuisance
scale in pipes and hot water heaters, it is important to understand water “hardness.” Hardness is a term that describes the tendency of water to form such
scales. All polyvalent metal ions contribute, but hardness is primarily caused by
calcium and magnesium ions. Interestingly, hardness is often measured as the
equivalent amount of CaCO3, the principle component of the solid scale that is
formed. So no matter which ions are contributing to the hardness, we can calculate them each as an equivalent amount of CaCO3 and then add them to get
total hardness. We previously saw how to use equivalents in Example 2.4; now,
in Example 2.19, we will see how to calculate and report hardness in water.
EXAMPLE 2.19
A sample of water contains 60 mg/L of Na+, 50 mg/L of Ca2+ and 40 mg/L of
Mg2+. Calculate the hardness and report your answer in mg/L as CaCO3.
SOLUTION
First, note that sodium, bearing a +1 charge, will not contribute to hardness.
Recall that the equivalent weights of Ca and Mg are their molecular weights
divided by 2.
EW of Ca = 40/2 = 20 g/eq (or 20 mg/meq)
EW of Mg = 24.3/2 = 12.15 mg/meq
Next, calculate the concentration of each ion in meq/L.
50
1 meq
mg
Ca ¥
= 2.5 meq/L
20 mg
L
40
1 meq
mg
Mg ¥
= 3.29 meq/L
12.15 mg
L
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Convert each concentration to “as CaCO3 units,” noting that the equivalent
weight of CaCO3 is 100/2 = 50 mg/meq.
For Ca: 2.5 meq/L × 50 mg/meq = 125 mg/L as CaCO3
For Mg: 3.29 meq/L × 50 mg/meq = 164.6 mg/L as CaCO3
Finally, add the two numbers to get total hardness
125 + 165 = 290 mg/L as CaCO3
This is a fairly hard water.
The alkalinity of most natural waters is associated primarily with bicarbonate. The presence of alkalinity in natural waters is a buffer, allowing the pH to
remain fairly constant and near neutral. A neutral pH is important because
many aquatic species have stringent pH requirements for survival. In any lake
or river, the water is open to the atmosphere. Thus the concentration of CO2 in
the water remains constant, and either more CO2 dissolves into the water as
more carbonic acid dissociates in response to more demand for H+ ions, or CO2
releases back into the air as more H2CO3 is formed in response to acid addition.
Example 2.20 illustrates this discussion.
EXAMPLE 2.20
A lake with a surface area of 4,000 m2 has an average depth of 1 m. The lake
water has an initial pH of 6.5 buffered by the carbonate system (alkalinity = 20
mg/L as CaCO3). Calculate the final pH of the lake after receiving 5 cm of acid
rain with a nitric acid (a strong acid) concentration of 0.0002 M (molarity).
SOLUTION
First calculate the increase in volume of water in the lake due to the addition
of the acid rain:
0.05 m rain × 4,000 m2 surface area = 200 m3
= 200,000 L of acid rain
+
Next, calculate the H added:
200,000 L × 0.0002 gmol/L
= 40 gmol H+ added
The final lake volume is its surface area times its depth plus the volume of
rain that fell on it:
4,000 m2 × 1 m + 200,000 L
= 4,200 m3
= 4,200,000 L
Thus the concentration of H+ added is:
ÈH + ˘ = 40 gmol/4,200,000 L
Î
˚
= 9.5 (10)–6 M
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Chapter Two
Because the initial pH was 6.5, the concentration of carbonate was negligible
and the buffering is provided by bicarbonate and carbonic acid. Now, calculate the concentration of bicarbonate and carbonic acid before the addition of
+
the acid rain. Using Equation (2.38) and knowing the pH, we know the ÈH ˘
Î
˚
concentration, so we can solve for the ratio of bicarbonate to carbonic acid.
(10 ) ÈÎHCO
-6.5
K1 =
-˘
3˚
ÈÎH 2 CO 3 ˘˚
= 10 -6.35
ÈH CO 3 ˘˚
This yields the ratio of Î 2
= 0.71
ÈHCO 3- ˘
Î
˚
But the alkalinity was reported as 20 mg/L as CaCO3.
Converting to meq/L (using Equation 2.44):
20 mg/L
= 0.4 meq/L = 0.0004 eq/L
50 mg/meq
And, from Equation (2.42):
0.0004 eq/L = ÈHCO 3- ˘ + 2 ÈCO 32- ˘ + ÈOH - ˘ - ÈH + ˘
Î
˚
Î
˚ Î
˚ Î ˚
Neglecting both carbonate and hydroxide concentrations at this relatively low
pH of 6.5, we get
ÈHCO 3 - ˘ = 4.0 (10 )-4 - (10 )-6.5 M = 4.0 (10 )-4 - 3.16 (10 )-7 = 4.0 (10 )-4
Î
˚
and thus:
-4
ÈÎH 2 CO 3 ˘˚ = 0.71 ¥ 4.0 (10 )
-4
= 2.8 (10 )
M
With the addition of 9.5 (10)–6 M H+ into the lake, 9.5 (10)–6 M of bicarbonate is
converted to carbonic acid with the following resulting concentrations:
ÈHCO 3 - ˘ = 4.0 (10 )-4 - 9.5 (10 )-6
Î
˚
-4
= 3.90 (10 )
and:
-4
ÈÎH 2 CO 3 ˘˚ = 2.8 (10 )
-4
= 2.90 (10 )
-6
+ 9.5 (10 )
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Now calculate the new pH from Eq. (2.38):
K1 =
(
ÈH+ ˘ 3.9 (10 )-4
Î ˚
(2.9 (10) )
-4
) = 10
-6.35
ÈH + ˘ = 3.32 (10 )-7 M
Î ˚
pH = 6.48
Note that because of the effective buffer system present, the pH depression
was limited to 0.02 units. Without any buffering capacity, the pH of this lake
would have dropped from 6.5 to 5.0!
2.5 Chemical Reaction Kinetics
Overview
The study of chemical or biological reaction rates is the study of kinetics,
and involves different concepts than chemical equilibrium (thermodynamics)
discussed previously. Equilibrium considerations are useful for determining
the final outcome of a chemical reaction, but those considerations cannot tell us
how fast that reaction will proceed. For this, knowledge of kinetics is required,
and thus kinetics is a major field of study within chemistry.
Complete definition of reaction rates requires knowing the reaction stoichiometry, the reaction order, and having numerical values for all rate constants.
Evaluation of reaction order and estimation of rate constants must be based on
experimental data and cannot be deduced from a balanced chemical reaction.
Laboratory- or pilot-scale testing is often conducted to define reaction kinetics
in conjunction with the design of pollution control facilities. Kinetic relationships are crucial in the design (sizing) of chemical or biological reactors. The
sizing of reactors is explored in more detail in Chapter 3. In this section, we
focus on the fundamentals of kinetics.
It is useful to recall the basics of collision theory from freshman chemistry.
Chemical reactions come about when molecules collide with enough energy to
break old bonds and form new bonds to create new molecules. Two factors are
important here: the frequency of collisions and the energy of collisions. The frequency is proportional to the concentration of molecules in the reacting space,
and the energy is proportional to the temperature of the reacting chemicals. The
usual way of representing these dependencies for the generic reaction A → B at
any moment in time is with the form:
Rate of reaction = k CAx
(2.45)
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Chapter Two
where:
k
= the rate constant (which is a strong function of temperature)
CA = the concentration of reactants
x
= order of reaction
Note that for all reactions, the rate is always positive or zero. The rate equation (2.45) says that the rate is proportional to the concentration of reactant
raised to some power. The parameters k and x are determined in the laboratory
by running many experiments and observing how the concentration of A
changes with time under controlled conditions. The rate is measured by watching how fast the reactant A disappears or the product B appears. We define the
intrinsic rate of production of a specific substance in a constant volume wellstirred reactor (a batch reactor) as follows:
RA =
d CA
dt
or RB =
d CB
dt
(2.46)
where:
Ri = intrinsic production rate of species i, gmol/L-time
Ci = concentration of species i, gmol/L
t
= time
Note that Ri can be positive or negative. For the reaction A → B, A is being used
up while B is being produced, so RA is negative while RB is positive. We can say
the rate of production of A is negative, but we know that the reaction is still
proceeding at a positive rate. The intrinsic reaction rate can be expressed in
terms of reactants or products, and the relationships for the general reaction aA
+ bB → cC + dD are:
–RA/a = –RB/b = RC/c = RD/d
(2.47)
These relationships are illustrated in Example 2.21. In Example 2.22, we demonstrate how to calculate the rate from measured concentrations. The units on
the reaction rates shown in Eq. (2.46) are mole/volume-time, but we can also
write the equation in terms of mass/volume-time (in this book we will use a
lowercase r to indicate mass rate of production). The two are related by the
molecular weight of the compound:
ri = Ri MWi
(2.48)
EXAMPLE 2.21
Consider the reaction C3H8 + 5 O2 → 3 CO2 + 4 H2O, which is occurring
catalytically in a water-based fuel cell. If propane is disappearing at a rate of
0.05 gmol/L-min, what is the rate of reaction expressed in terms of disappearance of oxygen? In terms of production of carbon dioxide?
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69
SOLUTION
From Eq. (2.47),
- Rpropane
=
- Roxygen
or - Roxygen = -5 Rpropane = 0.25 gmol/L-min
1
5
This makes sense because we can see from the stoichiometry that 5 moles of O2
are used up for every one mole of C3H8, so the disappearance rate of oxygen is
5 times faster or 0.25 gmol/L-min. The disappearance rate is the negative of
the production rate, so rate of reaction is a positive number. Similarly, in terms
of CO2, the production rate of CO2 is 3 × (–Rpropane) or 0.15 gmol/L-min.
EXAMPLE 2.22
A chrome-plating operation generates wastewater with a chromate concentration of 0.010 gmol/L. The hexavalent chrome in the chromate ion is
reduced (using SO2) to trivalent chromium ions at low pH, which are then
precipitated in another vessel by adding sodium hydroxide. The reactions are:
2 H2CrO4 + 3 SO2
→
2 Cr3+ + 6 NaOH
2 Cr3+ + 3 SO42– + 2 H2O
→
2 Cr(OH)3 + 6 Na+
Determine the (average) reaction rates in molar and mass units of chromate
(CrO42–) and sulfur dioxide (SO2) for a short time interval, assuming that the
reaction achieves 5% completion in the first 2 minutes.
SOLUTION
After 2 minutes and at 5% completion, the remaining chromate concentration is
0.0095 gmol/L. The molar intrinsic reaction rate for chromate is determined first:
- Rchromate = -
ÈC
˘ - ÈC
˘
d [ C]
gmol
= - Î final ˚ Î initial ˚ = 0.00025
dt
tfinal - tinitial
L-min
The mass-based reaction rate for chromate is obtained by multiplication by
the molecular weight of chromate (MW = 116):
gmol ˆ Ê
g ˆ
g
Ê
- rchromate = Á 0.00025
116
= 0.029
˜
Á
˜
Ë
¯
L-min Ë
gmol ¯
L-min
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Chapter Two
The molar and mass reaction rates for sulfur dioxide are obtained by application of Eqs. (2.47) and (2.48):
gmol ˆ Ê 3 gmol SO 2 ˆ
Ê
- RSO 2 = Á 0.00025
Ë
L-min ˜¯ ÁË 2 gmol chromate ˜¯
= 0.000375
gmol SO 2
L- min
gmol SO 2 ˆ Ê 64 g ˆ
Ê
- rSO 2 = Á 0.000375
Ë
L-min ˜¯ ÁË gmol ˜¯
= 0.024
g SO 2
L-min
Elementary Reactions
An elementary reaction is a simple, single-step reaction. Most actual chemical or biological processes consist of many individual steps in series and/or
parallel. The reaction mechanism is the compilation of these elementary reactions. In most cases in environmental engineering practice, it is not feasible or
necessary to fully characterize all elementary reactions for a process of interest.
It is more common to rely on empirical kinetic models that adequately describe
an overall reaction. For example, the thermal oxidation of hydrocarbons to carbon dioxide and water involves hundreds of elementary reactions, but waste
hydrocarbon incinerators are designed using simple empirical models based on
the rate of destruction of one or two specific compounds.
For the general reaction A + B → Products, the reaction rate is often
described by an expression of the following form:
n m
- rA = k CA
CB
(2.49)
in which k, n, and m are constants that are specific to the reaction. The reaction
rate constant, k, is really only constant at one constant temperature. The constants n and m (which need not be integers) equal the reaction order with
respect to compound A and compound B, respectively. The overall reaction
order equals the sum of the orders with respect to each reactant (n + m). The
numbers (n or m) do not necessarily equal the stoichiometric coefficients (a or
b), however, for many elementary reactions, they do. In the simplest cases, one
component (say B) will be present in great excess, and these reactions can be
modeled as reactant A goes to product P:
A
→
P
(2.50)
n
- rA = k CA
(2.51)
with
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In this representation, n is often an integer and has the value 0, 1, or 2. In
those cases the reaction is call a zero-order, first-order, or second-order reaction. These models of chemical reactions are used widely throughout environmental engineering. Note that the units on k must work with the order of the
reaction to yield an intrinsic rate, r or R, with units of concentration per time
(mass/vol-time or moles/vol-time). These three elementary reactions are discussed in the next few paragraphs. A graph illustrating how the concentration
of reactant changes with time in a batch reactor for each of these reactions is
presented in Figure 2.5, which has three parts, one for each order reaction. Each
part is linearized so as to allow easy determination of the rate constant.
Zero-Order Reactions
If n is zero in Eq. (2.51), then the rate expression becomes simply
–rA = k
(2.52)
and the reaction proceeds at a constant rate, independent of concentration of A.
In a batch reactor,
- dCA
=k
dt
(2.53)
The mathematics are trivial. Separating variables and integrating from the concentration at time zero, C0, to the concentration at time t, Ct, and solving for Ct, we get:
C t = C0 – k t
(2.54)
The concentration of A decreases linearly with time as shown in Figure 2.5a,
and the slope of the line is the negative of the rate constant. This behavior
Figure 2.5 Linearized plots of the
behavior of concentration vs. time for
zero-order, first-order, and secondorder elementary reactions.
C0
dC
= –k
dt
–k
C
Ct = C0 – kt
t
(a) Zero-order
ln C0
–k
ln C
dC
= –kC
dt
ln Ct = ln C0 – kt
dC
= –k C2
dt
1
C
k
1
C0
(b) First-order
t
(c) Second-order
t
1
1
=
+ kt
C t C0
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72
Chapter Two
makes it easy in the laboratory to deduce if a reaction is following zero-order
kinetics. The units on k for zero-order reactions are simply concentration/time.
First-Order Reactions
The first-order model, n = 1 in Eq. (2.51), yields:
- rA = -
dCA
= k CA
dt
(2.55)
This is another very simple model, but in spite of its simplicity, first-order
kinetics are used with great success to describe many processes of interest in
environmental engineering.
The differential equation in Equation (2.55) can be solved by separation of
variables and integration, as outlined below. Note that the integration step is
completed with definite integrals. The lower integration limits correspond
with the initial conditions of C = C0 and t = 0.
d CA
= - k dt
CA
Ct
Ú
C0
(2.56)
t
d CA
= Ú - k dt
CA
0
(2.57)
ÊC ˆ
ln Á t ˜ = - k t
Ë C0 ¯
(2.58)
Ct = C0 e–kt
(2.59)
Note that in testing if a reaction follows first-order kinetics, we can take the
natural logs of both sides of Eq. (2.59) to get:
ln Ct = ln C0 – kt
(2.60)
which is the equation of a straight line (see Figure 2.5b) with a slope equal to
the negative of the rate constant. So if our laboratory data plot as a straight line
with this kind of treatment, we feel confident that the reaction follows firstorder kinetics. The units on k for a first-order reaction are simply inverse time.
Second-Order Reactions
The second-order model, n = 2 in Eq. (2.51), yields:
–rA = k CA2
(2.61)
dCA
= kCA 2
dt
(2.62)
and
-
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The mathematical integration of Eq. (2.62) ultimately yields the second-order
model:
1
1
=
+ kt
Ct C0
(2.63)
which is the equation of a straight line (see Figure 2.5c) with a slope equal to
the rate constant when the data are plotted as 1/C versus time. So if our laboratory data plot as a straight line with this kind of treatment, we feel confident
that the reaction follows second-order kinetics. The units on k for a secondorder reaction are inverse concentration divided by time (e.g., L/gmol-s, or L/
mg-min). This is necessary so that when k is multiplied by C2 the resulting
intrinsic rate (r) has the correct units.
EXAMPLE 2.23
A batch chemical reactor achieves a reduction in concentration of compound
A from 100 mg/L to 5 mg/L in one hour. If the reaction is known to follow
zero-order kinetics, determine the value of the rate constant with appropriate
units. Repeat the analysis if the reaction is known to follow first-order kinetics.
SOLUTION
Assuming zero-order kinetics,
- rA = Ct
d CA
0
= k CA
=k
dt
t
Ú d CA = Ú - k dt
C0
0
Ct - C0 = - k t
k=
mg
C0 - Ct 100 mg/L - 5 mg/L
= 95
=
1 hr
L-hr
t
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Chapter Two
Assuming first-order kinetics,
- rA = -
d CA
= k C A1 = k C A
dt
Ct
Ú
C0
t
d CA
= Ú - k dt
CA
0
ÊC ˆ
ln Á t ˜ = - k t
Ë C0 ¯
Ê 5 mg/L ˆ
ÊC ˆ
ln Á t ˜ ln Á
Ë C0 ¯
Ë 100 mg/L ˜¯
k=
=
= 3.0 hr -1
-t
-1 hr
EXAMPLE 2.24
A first-order reaction is proceeding in a batch reactor. It is 40% complete after
30 minutes. How long is required to achieve 95% completion? Repeat for
99% completion.
SOLUTION
The rate constant is determined for first-order kinetics. Note that 40% complete means that 60% of the reactant remains.
ÊC ˆ
È (1 - 0.40 ) C0 ˘
ln Á t ˜ ln Í
˙
C0
Ë C0 ¯
˚ = 0.0170 min -1
k=
= Î
-t
-30 min
Using this value of k, and an algebraic manipulation of the same equation, we
can solve for the time to achieve any desired percent conversion.
For 95% conversion:
ÊC ˆ
È (1 - 0.95 ) C0 ˘
ln Á t ˜ ln Í
˙
C0
Ë C0 ¯
˚ = 176 min
t=
= Î
-k
-0.0170 min -1
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For 99% conversion:
ÊC ˆ
È (1 - 0.99 ) C0 ˘
ln Á t ˜ ln Í
˙
C0
Ë C0 ¯
˚ = 270 min
= Î
t=
-k
-0.0170 min -1
The results from the previous example illustrate a common dilemma in
environmental engineering practice: the rate of the removal reaction decreases as
the desired efficiency of removal increases. Treatment time or size of reactor (and
thus costs) are often directly related to the removal efficiency, with the result that
it is usually much more difficult and expensive to achieve additional removal of
contaminants as the efficiency of removal is increased to very high levels.
Variable-Order Reactions
In some cases (especially biological reactions), the rate behavior may vary
significantly over a range of concentrations. Monod (1949) reported empirical
evidence to support use of a variable-order rate expression to describe bacterial
growth and decomposition of wastes. Variable-order kinetic expressions are
commonly identified as Monod kinetics in environmental engineering practice.
This model has great flexibility to describe many applications, including zeroand first-order situations, and is widely used for characterization of biological
wastewater treatment processes.
The reaction appears to be first-order at low concentrations and zero-order
at high concentrations, and a plot of this behavior is shown in Figure 2.6. The
model for a variable-order reaction rate is given by Eq. (2.64).
- rA =
k CA
K A + CA
(2.64)
where:
k
= rate constant, with units of inverse time
KA = half-saturation constant, with units that match CA
Figure 2.6 Behavior of a
variable-order reaction.
Rate
k
k/2
KA
Concentration of A
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76
Chapter Two
Temperature Effects
The rates of chemical and biological reactions are strongly influenced by
temperature, and we often heat reactants to achieve faster reaction rates. Up to
now, we have only said that the temperature effects are represented in the rate
constant term. A practical example using extreme heating is the incineration of
hazardous waste. Another common example is the mild heating of the liquid
during the anaerobic digestion of sludge, a process in which the microorganisms are particularly sensitive to low temperatures. In other cases (such as
municipal wastewater treatment), adding heat to raise the temperature is not
practical due to the large volume of wastewater. Nevertheless, quantification of
the effect of temperature on reaction kinetics is important to account for seasonal or regional variations in performance. Two models are used in environmental engineering to account for temperature effects on the rate constants.
The first model is the familiar Arrhenius equation:
k = Ae - Ea /RT
(2.65)
where:
k
= rate constant (appropriate units)
A = frequency factor (same units as k)
Ea = activation energy (cal/gmol)
R = universal gas law constant (1.987 cal/gmol-K)
T = temperature (K)
To obtain the temperature dependence of k, we would run several experiments
at several different temperatures to obtain values of k at say four or five temperatures. Then we would linearize Eq. (2.65) to yield:
ln k = ln A -
Ea
RT
(2.66)
which plots as a straight line when ln k is plotted versus 1/T. The slope of that
line is (–Ea / R). The Arrhenius relation has been used for chemical reactions
that potentially might occur over a large temperature range.
The other method, which is commonly used in environmental engineering
for water-based biological systems that do not experience wide variations in
temperature, is the temperature correction factor model:
kT = k20Q(
T - 20)
where:
kT = rate constant at temperature T
k20 = rate constant at 20 °C
Q = temperature correction factor
T = temperature in °C
(2.67)
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77
EXAMPLE 2.25
A chemical reaction is reported to have an activation energy of 12,000 cal/
gmol. Determine the ratio of the rate constant at operating temperatures of
30 °C and 20 °C.
SOLUTION
The rate constant is determined with Eq. (2.66), with conversion of temperature to an absolute temperature scale (Kelvin).
k30
È
˘
12, 000 cal
Í
˙
gmol
Í
˙ = 2.06 10 -9 A
= A exp Í ( )
˙
Ê
cal ˆ
Í Á 1.98
˙
K
30
+
273
)
(
K-gmol ˜¯
˙˚
ÎÍ Ë
k20
È
˘
12, 000 cal
Í
˙
gmol
Í
˙ = 1.04 10 -9 A
= A exp Í ( )
˙
Ê
cal ˆ
Í Á 1.98
˙
+
20
273
K
)
(
K-gmol ˜¯
˙˚
ÎÍ Ë
-9
k30 2.06 (10 ) A
=
= 1.98
k20 1.04 (10 )-9 A
For this specific value of activation energy, the rate of the chemical reaction is
observed to double for a 10-degree Celsius increase in temperature.
EXAMPLE 2.26
A biological wastewater treatment process is known to exhibit first-order
kinetics with a temperature correction factor equal to 1.02. Laboratory studies
at 20 °C established a value for the rate constant of 5.0 day–1. Determine the
required reaction time in a batch reactor to achieve 90% conversion for summer conditions in Florida (assume a wastewater temperature of 30 °C).
Repeat for winter conditions in Iowa (assume 5 °C).
SOLUTION
The rate constant is determined with Eq. (2.67) for the specified temperatures:
kT = k20 Q(T – 20) = (5.0 day–1) (1.02)(T – 20)
k30 = (5.0 day–1) (1.02)(30 – 20) = 6.1 day–1
k5 = (5.0 day–1) (1.02)(5 – 20) = 3.7 day–1
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78
Chapter Two
The reaction time is determined for this first-order process as in Example 2.24:
ÊC ˆ
ln Á t ˜
Ë C0 ¯
t=
-k
t30
È (1 - 0.90 ) C0 ˘
ln Í
˙
C0
˚ = 0.38 days
= Î
-1
-6.1 day
È (1 - 0.90 ) C0 ˘
ln Í
˙
C0
˚ = 0.62 days
t5 = Î
-3.7 day -1
The calculated difference in reaction time underscores the need to consider
site-specific factors (such as temperature) that may influence reaction kinetics
during design. Empirical design guidelines established for one region of the
country may be wrong for other climatic regions; engineers cannot simply use
an “off-the-shelf” design in every part of the country.
SUMMARY
Subsequent chapters in this text provide a description of processes that are
used for pollution abatement. A high percentage of these applications rely on
chemical and/or biological reactions. An in-depth knowledge of chemical reaction stoichiometry is essential in environmental engineering practice, as is
knowledge of solubility, acids and bases, and the gas laws. A quantitative
description of these processes requires a numerical expression of reaction rates.
The most commonly used kinetic expressions in environmental engineering
practice are zero-, first-, second-, and variable-order models. Extensive use of
chemistry is made in the design of engineered systems to control pollution
emissions or to remediate contaminated sites. Chemistry is one of the foundations of environmental engineering, and it is crucial that all environmental engineering students master the fundamentals of chemistry. The applications of
chemistry will be demonstrated throughout the remainder of this text.
PROBLEMS
2.1
Balance the following equations:
a. Na2CO3 + HCl
b. Cl2 + KOH
→
→
NaCl + CO2 + H2O
KCl + KClO3 + HCl
c. Fe(OH)2 + O2 + H2O
→
Fe(OH)3
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79
d. FeSO4 + K2Cr2O7 + H2SO4 →
Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + H2O
e. FeS + HCl
→
FeCl2 + H2S
2.2
Forty (40) mg of calcium (Ca2+) and 60 mg of magnesium (Mg2+) are dissolved in 0.50 L of water. What is the total hardness of the solution
expressed in mg/L as CaCO3?
2.3
The compound C7H4N3O6 burns explosively with oxygen to produce
CO2, H2O, and N2. Write a balanced chemical equation for this reaction
and calculate the mass of oxygen required (in grams) to burn 100 g of this
compound.
2.4
One method of removing phosphate from wastewater effluents is to precipitate it with aluminum sulfate (alum). A plausible stoichiometry (but
not exact because aluminum and phosphate can form many different
chemical materials) is:
2 PO43– + Al2(SO4)3
→
2 AlPO4 + 3 SO42–
If the concentration of phosphate (PO43–) is 30 mg/L, how many kg of
alum must be purchased annually to treat 40 L/s of wastewater? How
many kg/yr of solid precipitate will be formed (dry basis) if all of the
phosphate is precipitated as AlPO4?
2.5
If equal moles of magnesium chloride and aluminum chloride are dissolved in a dilute acid solution, will magnesium hydroxide or aluminum
hydroxide precipitate first as the pH of the solution in increased? At 25
°C, the Ksp of Mg(OH)2 is 8.9(10)–12, for Al(OH)3 it is 1.3(10)–33.
2.6
What is the ratio of the undissociated form of acetic acid to the acetate ion
at a pH of 5.2? The pKA for acetic acid is 4.75 at 25 °C.
2.7
What is the pH of the following solutions?
a. 0.01 M NaOH (strong base)
b. 0.01 M HNO3 (strong acid)
c. 0.01 M HClO3 (weak acid, pKA = 7.6)
2.8
How many kg/day of NaOH must be added to neutralize a waste stream
generated by an industry producing 90,800 kg/day of sulfuric acid, if
0.1% of the sulfuric acid produced is lost to the wastewater? The wastewater flow is 750,000 L/day.
2.9
The solubility of carbon dioxide in pure water is given by its Henry’s Constant. Calculate the concentration of CO2 in rainwater in gmol/L at 25 °C,
given that the atmosphere contains 400 ppm CO2. Assume that [CO2]aq =
[H2CO3]. The Henry’s Constant for CO2 is 1,640 atm/mole fraction.
2.10 Once CO2 is absorbed in water, it reacts according to Equations (2.33)
through (2.35). Calculate the pH of natural rainwater described in Problem 2.9 above.
2.11 What is the alkalinity (expressed in eq/L, and in mg/L as CaCO3) of a
solution containing 0.01 M HCO3– and 0.02 M CO32– at a pH of 10.6?
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80
Chapter Two
2.12 Nickel ion (Ni2+) and OH– are in equilibrium with solid Ni(OH)2 in a 1-L
solution that was formed by dissolving pure Ni(OH)2 in water. What is
the equilibrium concentration of Ni2+ ions (in gmol/L and mg/L)? What
is the final concentration of Ni2+ ions in solution (in mg/L) after the pH is
adjusted to 7 using HCl (ignore the small increase in solution volume)?
The Ksp of Ni(OH)2 is 5.5(10)–16 at 25 °C.
2.13 Balance the following equation:
C2H5OH + O2
→
CO2 + H2O
Calculate the volume (in L, at 25 °C and 1 atm) of CO2 produced from the
oxidation of 100 g of ethanol (C2H5OH) with excess O2.
2.14 If 100 g of HCl reacts with excess sodium carbonate at 35 °C and 1 atm,
how many moles of carbon dioxide will be produced and what volume
will the CO2 occupy? (See Problem 2.1a above)
2.15 If the Henry’s Constant for H2S is 9.74 atm-L/mole at 25 °C, calculate the
mole fraction of H2S present in air at 1 atm above a solution containing
H2S at a concentration of 0.01 M. The solution is at a pH of 7.2. You may
assume that the concentration of S2– is negligible compared to the concentrations of H2S and HS–. The pK1 of H2S is 6.96.
2.16 How many kilograms of oxygen and nitrogen are contained in a 200-ha
lake (1 ha = 10,000 m2) with an average depth of 5 m, if the water is in
equilibrium with the atmosphere which contains 79% N2 and 21% O2?
Assume P = 1 atm and T = 25 °C. At 25 °C the KH for N2 is 8.64(10)4 atm/
mole fraction, and the KH for O2 is 4.38(10)4 atm/mole fraction.
2.17 Determine the correct units for the rate constant(s) in each of the following kinetic expressions (all concentrations are in mg/L):
a. - rA =
k CA
K A + CA
2
b. - rA = k CA
c. - rA = k CA
d. - rA = k CA CB
e. - rA = k
2.18 A chemical reaction (A → B) occurs in a batch reactor. The initial concentration of compound A is 100 mg/L. The concentration that remains after
10 minutes is 90 mg/L. The concentration that remains after 20 minutes is
80 mg/L. Determine the reaction order and the rate constant (with appropriate units).
2.19 A second-order reaction of the type described by Eq. (2.61) is occurring in
a batch reactor. It is 40% complete after 30 minutes. How long is required
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81
to achieve 95% completion? Assume the initial concentration of compound A is 100 mg/L.
2.20 A zero-order reaction (A → B) is known to be 40% complete after 30
minutes in a batch reactor. How long is required to achieve 95% completion? Assume the initial concentration of the reactant is 100 mg/L.
2.21 A variable-order reaction described by Eq. (2.64) with a half-saturation
constant of 50 mg/L and a rate constant of 4.0 mg/L-hr occurs in a batch
reactor. If the initial concentration is 500 mg/L, how long is required to
achieve 75% completion? Which is a more reasonable approximation of
this reaction at this initial concentration—first-order or zero-order?
2.22 The same variable-order reaction as in Problem 2.21 above occurs in a
batch reactor. If the initial concentration is 10 mg/L, how long is required
to achieve 75% completion? Which is a more reasonable approximation of
this reaction at this initial concentration—first-order or zero-order?
2.23 A zero-order reaction is known to have a temperature correction factor (Q)
of 1.01. A batch reactor was used in the laboratory (25 °C) with a reaction
time of 3 hours to achieve a reduction in the concentration of a synthetic
organic compound from 3.0 to 0.1 mg/L. How long will the reaction take
under field conditions (12 °C) to achieve the same treatment efficiency?
2.24 Consider the complete combustion of propane (C3H8). How many lbmol
of oxygen are required to burn 100 lbmol of propane? How many pounds?
2.25 A method used by the US EPA for determining the concentration of
ozone (O3) in air is to pass an air sample through a “bubbler” containing
sodium iodide. Ozone is removed during this process according to the
following equation:
O3 + 2 NaI + H2O
→
O2 + I2 + 2 NaOH
How many moles of sodium iodide are needed to remove 4.0(10)–3 moles
of ozone? How many mg of sodium iodide are needed to remove 0.50 mg
of ozone?
2.26 A liter of water was found to contain 10 mg of benzene (C6H6). What is
the concentration of benzene in mg/L, ppm, and molarity?
2.27 A 5-kg block of dry ice vaporizes to gas at room temperature. Determine
the volume (in L) of gas produced at 25 °C and 1 atm.
2.28 The concentration of CO in the exhaust gas from an industrial furnace
stack is 5.0 ppm. Determine the stack CO concentration in units of µg/m3
assuming the temperature is 350 °F and the pressure is 1 atm.
2.29 What is the final concentration of lead ions in an equilibrium solution of
lead hydroxide in which the final pH has been independently adjusted to
9.5 using sodium hydroxide? Report the final concentration in mg/L.
Note that the Ksp for Pb(OH)2 is 1.4(10)–20.
2.30 By mistake, a worker at an electroplating shop mixes a small amount of
sulfuric acid into a large tank of wastewater containing sodium cyanide.
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82
Chapter Two
The reaction produces a toxic gas, hydrogen cyanide, as shown below:
H2SO4 + 2 NaCN
→
Na2SO4 + 2 HCN
If 2.0 L of pure liquid H2SO4 (ρ = 1.35 g/mL) are mixed into the tank, how
much HCN is formed in kg? Suppose the partial volume of HCN is calculated to be 3,000 L, and the room measures 20 m × 35 m × 5 m. What is the
concentration of HCN in the air in this room in ppm?
REFERENCES
Monod, J. 1949. “The Growth of Bacterial Cultures.” Annual Review of Microbiology, 371.
US EPA (Environmental Protection Agency). 1983. Development Document for Effluent
Limitation Guidelines and Standards for the Metal Finishing Point Source Category. EPA
440/1-83/091 (June).
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CHAPTER
3
A Process Engineering Approach
to Solving Problems
3.1 What Is Process Engineering?
Process engineering is a term used in connection with the processing industries—industries that process materials to make products. Examples of such
industries include oil refining, pharmaceuticals manufacturing, fertilizer production, water and wastewater treatment, food-processing plants, and many
others. The commonality among such diverse industries is that raw materials
are brought into the facility, flow through various unit operations or processes,
and are transformed into products. Process industries tend to have large plants
and consume significant amounts of energy (even making electricity in a coalfired power plant consumes energy). Most operate their facilities more-or-less
continuously, never shutting down the whole plant at the same time.
Process engineering is conducted to improve safety for plant personnel, to
ensure continuous “on-spec” operation, to increase efficiency, to solve day-today operating problems, to design new components for expanded future production, to comply with environmental regulations, and for other reasons.
Although the term process engineering has long been in the “sweet spot” of the
chemical engineering discipline, engineers from a variety of backgrounds
(environmental, mechanical, civil, industrial, etc.) can and do function as process engineers. Because safe and efficient operations, pollution control, pollution prevention, and environmental compliance are so important to the big
process industries, environmental engineers often are ideally suited for this
assignment. No matter what the academic background, all process engineers
must have a good understanding of the flow of materials and energy throughout the facility.
Material and energy balances are key analytical tools for any engineer
working with an existing industrial or municipal process. These calculations
are invaluable in analyzing the details of an existing process, and supplement
the information obtained from instruments that measure various process
parameters. Real-time measurements of the important parameters are essential
83
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84
Chapter Three
for good operation and control of the process, but we cannot measure every
flowing stream in the whole unit—it is simply too expensive, and in the case of
fugitive emissions (i.e., leaks) not technically possible. In addition, material
and energy balance calculations are the only way that engineers can predict
how a new process (one that has not yet been built) will behave.
Material and energy balances are used by engineers in many ways: to
check the accuracy (or at least the consistency) of flow meters, to estimate cost
of certain operations, to set prices on products, and to help predict the results
of proposed operational changes in an existing unit. But, most importantly,
material and energy balances are essential in the design of a new process (we
cannot measure flows or other variables in a process that does not yet exist).
Whether we are assessing the impact on the local environment of a proposed
plant, or calculating the profit or loss expected from a new chemical process, or
sizing equipment to be purchased, material and energy balances are essential.
The use of fundamental conservation (or “balance”) equations (mass,
energy, and momentum) is always an appropriate approach for describing natural or engineered systems. Sometimes, we have existing data or we can generate data to support the development of empirical equations or general design
guidelines. But, public health and safety concerns may preclude us from conducting experiments to generate the data needed (as in the case of transport of
radioactive releases from nuclear power facilities). So, in the absence of data,
these conservation equations may represent the only method available.
The development of material and energy balance equations is emphasized
throughout this text. Such development will reinforce a methodology that is
basic to other engineering courses and disciplines. Conservation of mass underlies the continuity equation in fluid mechanics and mass transfer equations, as
well as reactor design, in both chemical and environmental engineering. Force
and momentum balances are critical in civil and mechanical engineering courses
(statics, dynamics, mechanics of materials). Principles of energy conservation
are incorporated into Bernoulli’s Equation (fluid mechanics), Kirchoff’s Law
(electrical engineering circuits), and heat-work relationships (thermodynamics).
These basic equations (which are used throughout engineering practice) are all
derived from fundamental conservation laws. Mastery of the concepts in Chapter 3 is essential to your development of process-engineering skills.
3.2 Flow of Materials—Mass Balances
All industries that manufacture products have at least two things in common: the flows of raw materials and finished products (into and out of the plant),
and the consumption of energy. Conservation of mass is a fundamental principle of engineering, which simply states that matter can neither be created nor
destroyed. While not strictly true for nuclear reactions, the principle can be
viewed as being exact for ordinary physical and chemical processes. Put another
way, this principle tells us that if one or more streams of material are flowing into
a defined region of space (such as a tank, a reactor, or even an industrial plant),
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A Process Engineering Approach to Solving Problems
85
then material must be either flowing out of that space at the same total mass flow
rate, or else material will accumulate in, or be withdrawn from, that space.
Regardless of whether we measure the volumetric flow rate of a gas, the linear
velocity of a liquid flowing in a pipe, or simply count the number of 50-pound
bags of a dry chemical received per hour, we must convert our measured flow
rates to proper units to make a material balance. The only units that are always correct in a material balance equation are mass/time or moles/time. In certain special
cases, it is permissible (and more convenient) to make a volumetric flow balance,
but the assumptions implicit in this approach must be understood. Specifically, for
volume balances, all streams must have the same density—a condition often
approximated closely by streams of water (whether slightly polluted or not).
These conservation equations require dimensional consistency. Careful attention to units is therefore essential. Identification and diagnosis of errors in calculations begin with an examination of units for dimensional consistency. Calculations
performed in contemporary American practice often use either English or metric
units. The ability to work in both systems is necessary for most engineers.
In Chapter 2 we introduced the concept of mass flow rates by saying the
mass flow of a pure fluid is its volumetric flow rate times its density, and now
we formalize that relationship in Eq. (3.1a):
•
m = Qr
(3.1a)
where:
•
m = mass flow rate, kg/s
Q = volumetric flow rate, m3/s
 = fluid density, kg/m3
Also, recall from Chapter 2 that the mass flow of any particular substance dissolved in the fluid (or being carried by the fluid) is given by the fluid’s volumetric flow rate times the concentration of that substance. Again, we formalize
that concept with Eq. (3.1b):
•
mi = QCi
(3.1b)
where:
Ci = concentration of the substance i in the fluid, kg/ m3
The continuity equation tells us that, under steady conditions, the mass
flow rate of a fluid flowing in a conduit is constant no matter how the area of
the conduit changes. We can easily convert from a fluid linear velocity to a volumetric flow rate provided that the area normal to flow is known. The volumetric flow rate is simply the linear velocity times the cross-sectional area of
flow, as shown in Eq. (3.2):
Q = uA
where:
u = fluid linear velocity, m/s
A = area normal to the flow, m2
(3.2)
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86
Chapter Three
If the density is known, the mass flow rate of the fluid can be related to the
fluid linear velocity—see Eq. (3.3a). Similarly, if the concentration of a particular substance is known, its mass flow rate is given by Eq. (3.3b):
•
m = uAr
•
mi = uACi
(3.3a)
(3.3b)
Note also that the units do not have to be as given above for the equations to be
valid. The units must simply be consistent (e.g., lb/hr, ft3/hr, lb/ft3, etc.).
EXAMPLE 3.1
Water is flowing in a 4-inch, Schedule 40 pipe with a velocity of 3.00 m/s. The
pipe splits into two 2-inch, Schedule 40 pipes with an equal volumetric flow
in each. Calculate the linear velocity and mass flow rate in each of the 2-inch
pipes. The density of water is 1,000 kg/m3 and the inside diameters of the 4inch pipe and the 2-inch pipe are 10.23 cm and 5.250 cm, respectively.
SOLUTION
First calculate the mass flow rate in the 4-inch pipe:
∑
•
m 4 = 1, 000
kg
m3
¥ 3.00
m p
1 m2
2
¥ ¥ (10.23 cm ) ¥
s 4
(100 cm )2
= 24.66 kg/
/s
Exactly half of this flow goes into each 2-inch pipe. The volumetric flow is:
Q2 = 12.33
kg
1
¥
s 1, 000 kg/m 3
= 0.01233 m 3 /s
and the linear velocity is
u2 =
0.01233 m 3
1
¥
2
s
p ( 5.25 cm )
1 m2
¥
4
(100 cm )2
= 5.70 m/s
Notice that even though the total volumetric flow rate in the two 2-inch pipes
is the same as in the one 4-inch pipe, the linear velocity in each of the smaller
pipes is higher than that in the larger pipe.
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A Process Engineering Approach to Solving Problems
87
The Basic Balance Equation
A material balance is simply an accounting of all materials into and out of
an identifiable process area. Consider first an overall mass balance. We want to
account for the total mass flow rates regardless of chemical type. Since mass can
neither be created nor destroyed, we can say with certainty that whatever mass
flows into the process area must either flow out of the area or accumulate within it. In
other words, the rate of accumulation of mass within the designated process
area is equal to the inflow rate minus the outflow rate. Mathematically, we write
dM •
•
= min - mout
dt
(3.4)
where:
M = total mass within the boundaries of our system
Usually there are several different chemical components flowing through a
process. Quite often, a chemical reaction is employed to change a less valuable
raw material into a valuable product. Other times, the process is designed to
use chemical or biological reactions to change harmful waste products into less
harmful ones. An independent material balance equation can be written for
each chemical component (or for all components but one, if an overall balance
is also made). In the general case, a material balance equation has the following
form for component i:
(Accumulation Rate)i = (Input Rate)i – (Output Rate)i + (Generation Rate)i
or
ARi = IRi – ORi + GRi
(3.5)
Usually, we are interested in steady-state operations, which means that nothing
changes with time, thus there is no accumulation. If there are no chemical reactions generating (or destroying) component i, and there is no accumulation, the
balance becomes very simple:
(IR)i = (OR)i
(3.6)
In most cases, the mass flow of a particular substance will be the product of the
volume flow of the stream carrying that substance times the concentration of
that substance in the stream.
Choosing the Basis and the Boundaries
A key step in making a material balance is clearly defining the region in
space that is to be material balanced, including the boundaries across which
mass is flowing. To aid in doing this, process engineers draw simplified flow
diagrams. Blocks represent reactors, tanks, separation columns, and so forth,
and lines with arrows indicate material flow. For example, Figure 3.1 could be
used to represent a process in which a wastewater stream is aerated and the
pollutants converted in the reactor (aeration basin) to sludge, which is separated from the water in the clarifier and then discharged separately from the
treated water.
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88
Chapter Three
Figure 3.1 A simplified process flow diagram.
gases
clarifier
treated
water
wastewater
aeration basin
air
recycle
sludge
waste
sludge
In Figure 3.1, the dotted line represents the boundaries for material balance. In this case, it represents a processing unit material balance, rather than a
reactor balance or a clarifier balance. Often we make several material balances
on a single unit and its individual equipment to better understand the process.
This seemingly simple concept is a very powerful tool, and will help the
engineer solve many process-related problems throughout his or her career.
The problem-solving technique using material balances can be formally organized into five steps.
Step 1. Draw a diagram with boundaries. Label the known streams and
write down given data. Once everything known has been identified on
the diagram, you are ready to solve the problem.
Step 2. Choose a basis. The most general approach is to use the equations
as written and solve in terms of flow rates as the basis. However, often it
is more convenient to pick a specified time interval or amount of material as the basis and solve in terms of mass of material only. That is, we
may choose as a basis either a unit of time (an hour, a day, etc.) or a unit
of material (1,000,000 gallons of wastewater, one metric ton of sludge
produced, etc.).
Step 3. Write the general equation. After drawing a diagram (with boundaries) and deciding on a basis, the next step is to write down the general
unsteady-state equation; that is, repeat Eq. (3.5).
Step 4. Simplify the general equation, eliminating terms that do not apply.
Also, at this time insert specific terms for the general terms. For exam-
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A Process Engineering Approach to Solving Problems
89
ple, if your input stream is a volumetric flow rate times a concentration,
replace IRi with QCi.
Step 5. Substitute numbers and solve.
Steady-State Flow Processes without Chemical Reaction
Steady state is an important condition for design and analysis. By definition, when a process is operating at steady state, nothing changes with time. This
means all flow rates, temperatures, pressures, liquid levels, etc. are constant.
While true steady state is rarely achieved for extended periods of time, it is
often approximated to a reasonable degree. Design is usually based on steadystate conditions, and achieving steady state is a goal of most process operators.
Thus, processes can be analyzed neglecting the accumulation rate term (at
steady state there can be no accumulation—either positive or negative). This is
a big help, because the accumulation rate term usually leads to differential
equations. Without this term, often only algebraic equations remain.
A very common balance is the one performed on a mixing point. A mixing
point is a point in space where two streams come together. Theoretically, a
point has no volume, so there can be no accumulation. Mixing at the point is
assumed to be complete and instantaneous, and there is no time for reaction.
Therefore, the balance simplifies to “input equals output.” That means the
mass flow rates of each substance in and out of the mixing point are equal, but
it does not mean that the inlet concentrations are the same as those in the outlet
stream. Making material balances on a mixing point using the 5-step process is
illustrated in Example 3.2.
EXAMPLE 3.2
A treated wastewater stream flowing at 5,000 L/s contains 75 mg/L of solid
particles. It flows into a river that was flowing at 35,000 L/s and carrying 5
mg/L of solids. What is the concentration of particles in the combined stream
(river plus wastewater)? Assume that the overall density of each stream is
essentially identical.
SOLUTION
Step 1—Draw the diagram. A diagram with boundaries is presented below.
Treated Wastewater
Q = 5,000 L/s
C = 75 mg/L
Combined Stream
Q=?
C=?
Mixing
Point
River
Q = 35,000 L/s
C = 5 mg/L
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90
Chapter Three
Step 2—Choose a basis. The basis we pick is the continuous flow rate through
the mixing point, the point where the two streams join and mix.
Step 3—Write the general equation.
AR = IR – OR + GR
Step 4—Simplify the general equation.
In this problem we have no accumulation in the mixing point, and no generation of particles or water by reaction. We can write:
IR = OR
Note that two mass balances can be written for this mixing point; the first is
on the water, the second is on the particles. In each balance there are two
inputs to the mixing point and one output.
Water balance:
Qwwww + Qriverriver = Qmixmix
Solids balance:
QwwCww + QriverCriver = QmixCmix
We know all variables in the two equations except for Qmix and Cmix. With
two unknowns, two equations are necessary to solve this problem, and we
have them.
Step 5—Substitute known terms and solve the simplified equations:
Water balance:
Qww + Qriver = Qmix
Since the density of each stream is almost identical, density divides out of the
equation and we find that
Qmix = Qriver + Qww
(a so-called volume flow balance)
or
Qmix = 35,000 L/s + 5,000 L/s = 40,000 L/s
Solids balance:
Thus,
5, 000
mg
mg
L
L
¥ 75
+ 35, 000 ¥ 5
= Qmix Cmix
s
L
s
L
Cmix = 13.75 mg/L
Notice that the flow rate of the output stream is the sum of the two input
streams, but that the concentration of solids in the output stream is NOT the
sum of the input concentrations.
In Example 3.2, we saw that two equations were necessary to solve the
problem. The two equations were two mass balances—one on the mass of
water, and one on the mass of solids. Even though the water balance could be
simplified such that the units were m3/s, technically the only units that are correct for a material balance are mass/time or moles/time.
Another simple concept used in materials balance problems is that of a
splitter point. A splitter point is simply a place where one stream divides into
two (or more) streams. At a splitter point, because it is a point, there can be no
accumulation and no chemical reaction. Indeed, at a splitter point, we do noth-
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A Process Engineering Approach to Solving Problems
91
ing to change the composition of the stream, we just split it into two flows. It is
like pouring all the iced tea from a pitcher into two different-sized glasses. All
properties of each of the outlet streams (the tea in the glasses) are the same as
the inlet stream (the tea in the pitcher); only the flow rates (amounts of tea in
the glasses) change.
A splitter point is different from a separator tank. A stream that enters a
separator tank can emerge as two different streams, each with its own composition. For example, consider a child playing at the beach. She scoops up some
ocean water near the shore that has a lot of grains of beach sand suspended in
the turbulent water. A few moments later, she pours most of the water out
(stream 1), but it is mostly clear because the sand particles have settled to the
bottom of the pail. Then she dumps the pail upside down to remove the wet
sand (stream 2). That is an example of a separator that operates at nonsteady
state because the water drops sand during some period of time. If the pail were
built to continuously receive turbulent ocean water and to discharge continuously two separate streams (wet sand and clear water), then it would operate
as a steady-state separator.
Another useful concept is that of recycle. A stream can be recycled inside a
process unit for a variety of reasons. But, if the material balance boundaries are
drawn such that the recycle loop is totally enclosed within a material balance
diagram boundaries, it is as if the recycle stream does not exist. We can ignore the
recycle stream in those material balances because it does not cross the boundaries. (If we want to know some details about the recycle stream, we must draw a
different diagram). In the next example, the material balance approach is
applied to a steady-state flow process in a system that has a recycle loop, a mixing point, and separator tanks.
EXAMPLE 3.3
The following diagram depicts a clarifier-thickener system in which solids are
concentrated and removed from a wastewater stream before the wastewater
stream is discharged.
B
A’
A
D
C
E
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92
Chapter Three
Data for figure:
Stream
Flow, L/s
C, mg/L
A
B
C
D
A’
E
100
90
2,000
25
5,000
150
Determine the flow rate of stream E, the concentrations of solids in stream E,
and the solids concentration in stream D.
SOLUTION
In this problem, four diagrams can be drawn for a material balance analysis,
and two balances can be written for each drawing. When there is recycle, it is
usually best to start with a diagram such that the recycle stream does not
cross the boundaries. The first material balance diagram then represents the
overall balance as shown below.
B
A’
A
D
C
E
From this diagram, we see that stream A flows into the process area and
streams B and E flow out. The two equations we can write are the total mass
balance and the solids balance. Assuming steady state, the balance simplifies
to Inputs = Outputs. The total mass balance becomes a volumetric flow balance if we make the usual assumption of constant density for these three
water streams, and the solids balance is done as usual.
QA
= QB + QE
QA CA = QB CB + QE CE
We have two equations and two unknowns, so we can solve for QE and CE.
QE = 100 – 90 = 10 L/s
and
CE = (QA CA – QB CB ) / QE
CE = (100 × 2,000 – 90 × 25) / 10
= 19,775 mg/L
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A Process Engineering Approach to Solving Problems
93
Let’s try our next balance diagram around the mixing point where streams A
and D meet.
A
B
A’
D
C
E
The balance equations are:
QA + QD = QA′
QA CA + QD CD = QA′ CA′
We know QA and QD so we can solve for QA′ = 100 + 150 = 250 L/s
But even after that we still have two unknowns (CD and CA′) and only 1 equation left. We must pick another place to make a balance.
Let’s pick the thickener (a separator tank).
A
B
A’
D
C
E
The equations are:
QC
= QD + QE
QC CC = QD CD + QE CE
Here we have only two unknowns, so we can solve
QC = 150 + 10 = 160 L/s
and
CD = (QC CC – QE CE ) / QD
= (160 × 5,000 – 10 × 19,775) / 150 = 4,015 mg/L
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94
Chapter Three
In many cases, material balance equations can be solved sequentially, as in the
previous examples. But sometimes, they must be solved simultaneously, as in
Example 3.4.
EXAMPLE 3.4
Reconsider Example 3.2, but change the problem statement to read as follows.
A treated wastewater stream flowing at 5,000 L/s and carrying 75 mg/L of
solids, mixes into a river that is flowing at an unknown flow rate. Prior to
mixing, the river was carrying 10 mg/L of solids. After mixing, a sample from
the combined stream shows a concentration of 22 mg/L solids. What was the
flow rate of the river prior to mixing with the wastewater? Assume all
streams have equal density.
SOLUTION
The diagram is similar:
Treated Wastewater
Q = 5,000 L/s
C = 75 mg/L
Combined Stream
Q = ? L/s
C = 22 mg/L
Mixing
Point
River
Q = ? L/s
C = 10 mg/L
and the initial balance equations are similar,
Qriver + Qww = Qmix
QwwCww + QriverCriver = QmixCmix
but now the two unknowns are Qriver and Qmix. So the equations must be
solved simultaneously.
Substitute Qriver + Qww into the second equation for Qmix and solve for
Qriver. That yields
ÊC
- Cmix ˆ
Qriver = Qww Á ww
Ë Cmix - Criver ˜¯
Ê 75 - 22 ˆ
Qriver = 5, 000 Á
Ë 22 - 10 ˜¯
Qriver = 22, 100
L
s
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A Process Engineering Approach to Solving Problems
95
Unsteady-State Flow Processes
So far we have only worked with steady-state balances, which means that
the accumulation rate term in the general equation has always been zero. Let
us now consider an unsteady-state situation. By definition, unsteady state
means that some things do change with time, and that mass can accumulate or
de-accumulate in any particular region in space. A good example of unsteady
state is the chemical reaction in a batch reactor that was used so often in Chapter 2. Let us first focus on unsteady-state flow processes without chemical reaction. For this case Eq. (3.5) simplifies to
AR = IR – OR
(3.7)
In general, we will write the AR term as a change in mass with time, and for a
particular species, Eq. (3.7) becomes:
dMi •
•
= mi - in - mi -out
dt
(3.8)
or
d (VCi )
dt
= Qin Cin - Qout Cout
(3.9)
Usually, unsteady-state situations will result in a differential equation, but
sometimes they do not, as seen in Example 3.5.
EXAMPLE 3.5
Consider a hilly area in which new homes are being constructed. As rain falls
on the site, without any controls, the runoff water would carry mud and dirt
into a nearby creek, polluting it badly. A detention pond is built to catch the
runoff and allow the mud to accumulate in the bottom of the pond while the
water flows out of the pond and into the creek. For simplicity, assume the rain
is steady, all flows are steady, and ignore any transient effects on the pond
water concentration.
The water runoff generated by the rainstorm is 5.0 m3/min, and has a concentration of particles of 4.00 g/L. The nearby creek flows at 120 m3/min and
normally has a concentration of 15.0 mg/L of particles. The detention pond
will catch and remove most of the mud, leaving the overflow water with a
concentration of 400 mg/L. Assume that 5% of the water that comes into the
pond will seep down through the bottom (filtering out all particles), while the
rest will overflow into the creek. Calculate the mass of mud accumulated in
the detention pond after three hours of this rainfall/runoff operation.
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96
Chapter Three
SOLUTION
1. Draw the diagram.
runoff
to creek
seepage
2. Choose a basis. In this case the problem asks for the amount of mud accumulated in three hours—the basis will be the time of 3 hours.
3. Write the general equation
ARmud = IRmud – ORmud + GRmud
4. Simplify
ARmud = IRmud – ORmud
5. Solve the mass balance for the pond, noting that 5.0 m3/min = 5,000 L/
min, and 4.0 g/L = 4,000 mg/L
ARmud = 5, 000
mg
mg
L
L
¥ 4, 000
- 0.95 ¥ 5, 000
¥ 400
min
L
min
L
7
mg
6 mg
- 1.90 (10 )
min
min
7
mg
kg
1 kg
¥ 6
= 18.1
min 10 mg
min
= 2.00 (10 )
= 1.81 (10 )
(mud is accumulating at the bottom of the pond at the rate of 18.1 kg/min)
The total amount of mud in the pond is calculated by multiplying the rate
by the time basis.
18.1
kg
min
¥ 60
¥ 3 hr = 3, 258 kg
min
hr
Balances with Chemical Reactions
In most processes with a reaction step, a total mass balance by itself does
not answer all our questions. Indeed, a reactor is built specifically to convert at
least one reactant into at least one product. Hence, individual mole balances for
each of the components of interest are usually required. In the general balance
equation, the term “generation rate” refers to the rate of generation of a particular component by chemical reaction within the system boundaries. It is often
represented as an intrinsic reaction rate multiplied by the system volume. Note
that if a component is being used up in the reaction, then its rate of generation
is negative. An intrinsic reaction rate was defined in the previous chapter as
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A Process Engineering Approach to Solving Problems
97
the time rate of change in the concentration (due to reaction only) of a given
component. In the laboratory we can most easily obtain intrinsic reaction rates
from small, constant volume, well-stirred (batch) reactors. In practice, we use
intrinsic reaction rates to design large-scale continuous flow reactors.
Recall that for a simple A  B reaction, a kinetic model to predict RA and
RB is:
n
RA = - kCA
(3.10)
n
RB = kCA
(3.11)
where:
Ri = production rate of species i, moles/volume-time
k = a temperature dependent rate constant
n = an empirically determined order
Note that the production rate of A is negative because A is being used up. Note
also that because 1 mole of A produces one mole of B, the production rate of
the product B is equal to the destruction rate of A (and is expressed as a function of the concentration of reactant A). Again, the generation rate of B often is
just RBV, where V is the volume of the reactor.
Reactor Models
Three common models of ideal chemical reactors are the batch reactor, the
completely mixed flow reactor, better known as the continuous flow stirred
tank reactor (CSTR), and the plug flow reactor (PFR). In practice, there are
many examples of real reactors that behave closely like one or another of these
three ideal models. We must accomplish the material balances for these types
of reactors differently due to their very different characteristics.
The batch reactor is a single tank that is being continuously stirred so that
in all parts of the tank the contents are uniform at any instant in time. There is
no flow into or out of this tank, and the volume remains constant, but the concentrations of reactants and products in the tank change with time. Under
these conditions, the material balance equation becomes:
ARi = 0 – 0 + GRi
(3.12)
V dCi / dt = RiV
(3.13)
or more specifically
Note that V can be divided out of Eq. (3.13), and it would then be identical to
Eq. (2.46). If species i is a reactant, Ri is negative, and if species i is a product, Ri
is positive. The difference between the two equations is that Eq. (3.13) has been
derived from a formal material balance. The term RiV is the generation rate; it
has units of moles/time and depends upon the size of the reactor, V. Assuming
a first-order reaction, after mathematical integration, we end up with Eq. (2.59)
as we saw from Chapter 2.
Ct = C0 e–kt
(2.59)
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98
Chapter Three
The CSTR reactor is that of an overflowing
tank in which the contents are rapidly and continuously mixed. It is similar to a batch reactor in that
there are no differences in concentrations of any
species anywhere in the tank. However, it is vastly
Q
different in that there is a flow into and out of the
CSTR (see Figure 3.2). Because the tank is full and
CAe
overflowing, the volume is constant, and the flow
CBe
rate out of the tank is exactly equal to the flow rate
into the tank. Since the outlet stream is continuously withdrawn from the tank and the contents of
the tank have the same composition everywhere, the concentrations in the outlet stream are identical to those in the tank. These are the concentrations at which
all reactions occur. Another big difference is that whereas the batch reactor is
inherently unsteady state, the CSTR usually operates at steady state. A steadystate material balance on a reacting component (A → B, first order) in a CSTR
results in a simple algebraic expression as shown below:
Figure 3.2 Schematic diagram
of a continuous flow stirred tank
reactor (CSTR).
Q
CA0
CB0
(3.14)
0 = QCA 0 - QCA e - kCA e V
Again, we emphasize that the concentration we use in the generation rate term
is the in-tank concentration, which for a CSTR equals CA , the exit concentration.
e
EXAMPLE 3.6
Calculate the CSTR volume required for 98% conversion of component A. The
kinetics are: RA = –kCA with k = 0.10 s–1. The inflow rate is 75 L/s with CA =
0
0.05 gmol/L.
SOLUTION
Solving Eq. (3.14) for the volume of the reactor and noting that Qe = Q0 = Q,
we get:
VR =
(
Q CA 0 - CA e
kCA e
) = Ê Q ˆ ( 1 - CA
ÁË ˜¯
k
e
/ CA 0
)
CA e / CA 0
Substituting and solving:
Ê 75 L/s ˆ Ê 0.98 ˆ
VR = Á
= 36, 750 liters
Ë 0.1 s -1 ˜¯ ÁË 0.02 ˜¯
The other widely used ideal reactor model is that of a plug flow reactor
(PFR). In this model, the reactor is pictured as a long, narrow tube, through
which fluid is flowing. Refer to Figure 3.3. Flow is assumed to be one-dimensional; velocity, concentration, and temperature are all constant with radial
position, but concentration varies with length. Longitudinal dispersion is
Cooper 03.fm Page 99 Thursday, March 3, 2016 9:46 AM
A Process Engineering Approach to Solving Problems
Figure 3.3
99
Schematic diagram of a plug flow reactor.
Q
Q
CA0
CAL
Δx
L
assumed to be negligible. Often, temperature does not vary much with length,
and the analysis is very much simplified if the reactor can be assumed to be isothermal. We will now use the material balance approach to develop the basic
steady-state design equation for an isothermal plug flow reactor.
Consider the simple reaction A  B with first-order kinetics occurring in a
PFR operating at steady state. Start by making a material balance on component A in a small volume increment of the reactor (V = Sx, where S is the
cross-sectional area) located at an arbitrary position x along the length of the
reactor. The mole balance equation (subject to our assumptions) becomes:
0 = Qx CA x - Qx +Dx CA x +Dx + RA SDx
(3.15)
Assume that S remains constant throughout the reactor and that Q, the volumetric flow rate, does not change with x. Dividing through by Sx and noting
that Q/S = u, the linear velocity, we obtain
0=
u(CA x - Cx +Dx )
Dx
+ RA
(3.16)
Now let x approach zero and take the limit. For a first-order reaction (n =
1), substituting for RA from Eq. (3.10), Eq. (3.16) becomes
0 = -u
u
dCA
- kCA
dx
dCA
= - kCA
dx
(3.17)
(3.18)
which can be solved by separation of variables and integration. We write
CA L
Ú
CA 0
L
dCA - k
=
Ú dx
CA
u 0
(3.19)
which upon integration is
ln
CA L
CA 0
= -k
L
u
(3.20)
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100
Chapter Three
or, in exponential form, is
CA L = CA0 e - kt
(3.21)
where:
 = reactor residence time (= L / u
or
V / Q)
Note the striking similarity between the PFR equation (3.21) and the integrated
batch reactor equation (2.59).
EXAMPLE 3.7
Rework the problem of Example 3.6, except calculate the volume of a PFR
required for the same 98% conversion.
SOLUTION
From Eq. (3.20),
ÏÔ .02 CA
0
ln Ì
C
A0
ÔÓ
¸Ô
-1
˝ = -0.10 s t
Ô˛
Solving for  we get  = 39.1 seconds. But,
 = V/Q
so
V = Q = (75 L/s)(39.1 s)
V = 2,930 L
Note that the PFR volume is much less than the CSTR volume for the same
conversion.
Unsteady-State Processes
Although steady-state operations are desirable, and knowledge of the
steady-state condition is useful for design, the transient or unsteady-state
response of a process is also important. In general, an unsteady-state material
balance is represented by Eq. (3.5), which may have one or more terms equal to
zero, but the AR term is always non-zero, by definition. An in-depth study of
the dynamic behavior of systems is a complete course in itself. However, we
need to be aware of transient processes; therefore some examples of responses
of simple systems to simple forcing functions are presented in Table 3.1. Finally,
the use of the unsteady-state material balance equation is illustrated with the
following example.
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A Process Engineering Approach to Solving Problems
Table 3.1
Transient Responses of Some Simple Systems
to Some Simple Forcing Functions
System
Forcing function
CSTR
(A → B)
Step increase in feed flow rate
Forcing Graph
Q
Step increase in concentration
of reactant A
t
0
t
0
t
0
t
0
t
0
t
CAe
t
Q
CAe
t
CA0
CAe
0
Spike increase in concentration
of reactant A
t
CA0
0
Step increase in concentration
of reactant A
0
CAe
0
Step increase in feed flow rate
t
CA0
0
Spike increase in concentration
of reactant A
Response Graph
CAe
0
PFR
(A → B)
101
t
CA0
CAe
0
t
EXAMPLE 3.8
A company has been discharging its nonreactive waste into a holding pit for a
long time. For the last several years the pit has been full and overflowing into
a local river. The waste concentration has been nearly constant at 10 mg/L of
the pollutant of interest (as has the pit overflow stream). At this concentration, there have been no adverse effects on the river. Suddenly there is a process change and the company’s waste stream concentration goes up to 100
mg/L. Given that the waste stream flow rate is 100,000 L/day and the holding pit volume is one million liters, calculate the concentration of the pit’s
overflow stream after 10 days. Assume the pit is well mixed.
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102
Chapter Three
SOLUTION
AR = IR – OR + GR
The pollutant balance on the pit is:
d ( Ce V )
dt
= QCi - QCe + 0
where Ce is the concentration in the exit stream at any time.
dCe Q
QCi
+ Ce =
dt V
V
The hydraulic residence time of the pit, , is defined as
V
, so the above equaQ
tion becomes
dCe 1
1
+ Ce = Ci
t
dt t
Since both  and Ci are constant with time, the solution to this linear, firstorder differential equation is
Ce (t) = C0 + (Ci – C0) (1 – e–t/ )
or
Ce (t) = C0e–t/τ + Ci (1 – e–t/τ )
in which C0 is the concentration in the pit outlet overflow stream just prior to
when the inlet stream jumps to 100 mg/L, and Ci is the new inlet concentration of 100 mg/L. This solution checks with our intuitive understanding of
the situation: At t = 0, Ce = C0, and at t = , Ce will equal Ci .
Noting that τ has a value of 10 days, and substituting numbers into the
equation above, we get
Ce = 10 mg/L + (100 – 10) mg/L (1–e–10/10 )
Ce = 10 + 90 (0.632)
Ce = 67 mg/L after 10 days
It certainly is possible that the situation described by Example 3.8 could
have involved a waste material that was also reacting in the pit. In that case,
none of the terms in the general material balance equation would be zero.
When there is reaction in an unsteady-state problem, the math can become
much more complicated because the reaction term may be nonlinear. So let us
save the solution to that problem for a bit later in this chapter after we have
learned a simple numerical technique for solving differential equations.
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A Process Engineering Approach to Solving Problems
103
3.3 Flow of Energy—Energy Balances
Energy can flow through a process area and be balanced just the same as
mass. An energy balance can be a very useful tool for all kinds of engineers in
all kinds of industry. In this section we will review several forms of energy and
power, and learn how to make energy balances.
Various Forms of Energy and Power
Energy is often defined as work or the capacity to do useful work (however, we also classify waste heat as energy). Power, on the other hand, is the
rate of doing work or the rate of expending energy. On a global scale, we have
a continuous flow of energy into and through the environment. Unlike matter,
the earth is not a closed system with respect to energy. We depend on a continuous flow of high-quality energy (in the form of solar radiation) into the biosphere, just as we must have the flow of low-quality thermal radiation away
from the earth. Not only does the sun drive our hydrologic cycle, power our
winds, and drive other physical processes, but through green plants it is the
basis for energy flow up the food chain.
To do work requires energy; but no process is 100% efficient in converting
energy into useful work. There will always be waste heat produced in any natural or engineered energy conversion process. However, units can be mathematically converted among the many forms of energy without incurring “losses,”
just as mass can be converted from pounds to grams. Also, energy and power
can be related mathematically by dividing or multiplying by a unit of time.
Energy has many forms (thermal, electrical, chemical potential, etc.), and
many units of measure. Power may also take on many different units, although the
most common deal with electrical power (kilowatts, megawatts) or with mechanical power (horsepower). Various conversion factors for energy and power are presented in Appendix A. Because fuel combustion is so important in today’s way of
life, Table 3.2 presents some typical values of the energy content of various fuels.
Table 3.2
Nominal Energy Content of Various Fuels*
Form
Higher heating values of fuel*
1,000 std cubic feet (MSCF) of natural gas
1 barrel (B) of crude oil (42 gallons)
1 gallon of gasoline
1 gallon of ethanol (anhydrous)
1 pound of coal
—Pennsylvania anthracite
—Illinois bituminous
—N. Dakota lignite
BTU
1.03 (10)6
6.0 (10)6
1.26 (10)5
83,800
13,500
11,500
7,200
kwh
302
1,760
36.9
24.6
3.96
3.37
2.11
kJ
1.08 (10)6
6.33 (10)6
1.33 (10)5
88,500
14,200
12,100
7,600
* Energy content of fuels (especially coals) may vary widely depending on the composition of the fuel.
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104
Chapter Three
EXAMPLE 3.9
Let’s suppose that a family of four uses 1,550 kwh of electrical energy on average every month. How many gallons of gasoline is this equivalent to? Where
does this family use more energy in a year, operating their home or operating
two cars (say about 20,000 miles at 20 miles per gallon)? If gasoline is $4.25/gallon and electricity is 15 cents/kwh, where does the family spend more money?
SOLUTION
Operating their home:
1, 550
1 gal gasoline 12 months 504 gal gasoline (equiv.)
kwh
¥
¥
=
year
month
36.9 kwh
1 year
Driving their cars:
20, 000 miles 1 gal gasoline 1, 000 gal gasoline
¥
=
year
20 miles
year
Therefore, the family expends more energy operating their cars each year.
Money spent per year:
1, 550 kwh 12 months $0.15
¥
¥
= $2, 790 year (for electricity)
month
year
kwh
vs.
1, 000 gal $4.25
¥
= $4, 250 year (for gasoline)
year
gal
This family spends more money on gasoline for their cars than they do for
operating their home.
EXAMPLE 3.10
(a) What is the annual average electrical power delivery to the family in Example 3.9? Actually, most of the electricity used is delivered to homes in relatively short periods of time—the “peak” hours are 6–9 AM and 4–8 PM on
week days. Assume that peak power delivery to a home is 15 kw. Calculate the electrical energy used if the 15 kw load is sustained (b) for one 24hour day, or (c) for one whole year.
SOLUTION
(a) 1, 550 kwh ¥ 12 months ¥ 1 year ¥ 1 day = 2.12 kwh = 2.12 kw
hr
month
year
365 days 24 hours
Cooper.book Page 105 Monday, June 23, 2014 9:58 AM
A Process Engineering Approach to Solving Problems
(b) 15 kw ¥
24 hours
¥ 1 day = 360 kwh in one day
day
(c) 15 kw ¥
24 hours 365 days
¥
¥ 1 year = 131, 000 kwh in one year
day
year
105
Energy Balances
Just like mass, energy is conserved. In the study of thermodynamics, this
simple statement is called the first law of thermodynamics. Regardless of what
it is called, in the analysis of any process all energy inputs and outputs as well
as energy accumulation must be taken into account. The form of energy can
change through an energy conversion process (e.g., when natural gas is
burned, its chemical potential energy is converted into thermal energy), but
basically, the energy that goes into a process unit must come out or accumulate
in the process area. (Accumulation of energy is often indicated by a rise in temperature.) Thus, an energy balance is very similar to a mass balance. However,
it must be recognized that, unlike mass, energy can be transferred by radiation
and/or by conduction through the walls of containers and pipelines, in addition to being transferred with material by bulk flow. We will make use of mass
and energy balances throughout this course, just as engineers make use of
them throughout their careers.
A flowing stream of material carries with it an associated energy flow. If we
consider only thermal energy for the moment, enthalpy is a more useful property to engineers than energy. Enthalpy is a thermodynamic property of material which depends on temperature, pressure, and composition. Enthalpy has a
precise mathematical thermodynamic definition, but here we will use the layman’s definition: “heat content.” The change in enthalpy that occurs when
material passes from one set of conditions to another often is more useful to us
than the absolute value of enthalpy. In many real-world situations, the change
in enthalpy of an amount of material can be related to its change in temperature:
H = m Cp T
(3.22)
where:
H = enthalpy change, cal or Btu
m
= mass, g or lb
Cp = specific heat at constant pressure, cal/(g-°C) or Btu/lb °F
T = temperature change, °C or °F
This equation can easily be extended to matter that is flowing from one place to
another as well as to matter that is stationary, as long as a constant flowing mass
is considered. The assumptions inherent in Eq. (3.22) are that Cp is constant over
the range of temperatures, that the effect of pressure is negligible (or pressure is
Cooper.book Page 106 Monday, June 23, 2014 9:58 AM
Chapter Three
constant), that any change in composition has negligible effect on enthalpy, and
that there has been no change of phase during the process. Phase changes of a
pure compound can be accounted for easily by the following equation:
H = m
(3.23)
where:
 = the heat of phase change (i.e. vaporization, liquefaction, or sublimation),
cal/g or Btu/lb
Again, Eq. (3.23) can just as easily be applied to a mass flow rate, in which case
the enthalpy would be an enthalpy flow rate. Figure 3.4 traces the change in
temperature and enthalpy as heat is added to a pure solid (ice) until it is converted to pure vapor, and Example 3.11 illustrates some typical calculations
with enthalpy.
Figure 3.4
Temperature-enthalpy diagram for water.
steam
Enthalpy per unit mass
(cal / g)
106
Cp vapor
ΔHvaporization
liquid
Cp liquid
ΔHfusion
ice
0
T (°C)
100
EXAMPLE 3.11
A 30,000 L/day stream of wastewater must be heated from 15 °C to 40 °C for a
treatment process to work properly. At what rate must we put heat into the
stream? Assume the stream has thermal properties similar to water. Use the
specific heat of water as 4.18 kJ/kg-°C for this problem.
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A Process Engineering Approach to Solving Problems
107
SOLUTION
The heating process can be pictured as shown below.
heater
stream
(H2)
stream
(H1)
heat (Q)
The general form of the energy balance is the same as that for a material balance.
AR = IR – OR + GR
which for this problem simplifies to:
•
•
0 = H1 + Q – H2 + 0
(Note, Q is often used as the symbol for heat flow rate as well as for volumetric flow rate.)
•
•
•
Thus, Q = H 2 - H1 = D H
•
•
But, D H = mC p DT , so
•
D H = 30 , 000
•
kg 4.18 kJ
L
¥1
¥
¥ ( 40 - 15) ∞C
day
L
kg-∞C
6
D H = 3.14 (10 ) kJ/day
6
Thus, Q = 3.14 (10 ) kJ/day
NOTE: The way this problem was worded (i.e., calculate the heat put into the
stream), there are no heat losses (heat transfer inefficiencies) to consider. In a
real situation, in order to transfer that much heat into the stream, a device
called a heat exchanger is used. In this device, as in all real-world devices,
there are inefficiencies—this is the topic of the next section!
Real-World Energy Conversion Processes
Based on common sense, we can accept the idea that there is no real-world
process that is 100% efficient in using energy to do useful work, or in transferring energy from one useful form to another useful form. As an example, if we
use electrical energy to lift an elevator full of people, not all of the electrical
energy will be used to do the lifting. Some portion of the electrical energy will
be “lost” as heat in the motor wiring, as friction in the gears, etc. In the study of
thermodynamics, this common-sense idea is known as the second law of thermodynamics. Note that “losing energy” is not a violation of the first law: the
total amount of energy is conserved. But the second law requires that some of
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108
Chapter Three
the energy input to this process must be dissipated in a useless form (such as
low-temperature heat). The following example demonstrates this concept.
EXAMPLE 3.12
Regarding Example 3.11, assume that the heater uses steam to heat the wastewater.
(a) Calculate the heat that must be contained in the steam in order to transfer
the required amount of heat into the wastewater stream. Assume that this
old heater has heat losses to the surroundings of 15%. Now assume that
the steam is produced in a furnace burning a fuel oil that is similar in heat
content to crude oil. Assume that the production of steam from the chemical potential energy in the oil has a thermal efficiency of only 60%; the
remaining heat goes out the stack with the hot gases, or is lost through the
hot walls of the pipes or through the furnace walls, or other places.
(b) Calculate the hourly feed rate of oil, in gal/hour.
SOLUTION
(a) From the previous example, the heat required by the wastewater was 3.14
(10)6 kJ/day. This is one “output” of the heater. The other output is the
“heat losses.” At steady state, the energy flowing into the heater (the heat
contained in the steam) must be the sum of these two outputs, or:
Input = 3.14 (10)6 + losses = 3.14 (10)6 + 0.15 × input
Therefore, the input (heat contained in the steam) is:
3.14(10)6 kJ/day
= 3.69(10)6 kJ/day
0.85
(b) The heat contained in the steam comes from burning oil, but only 60% of the
heat released by burning the oil actually ends up in the steam. Therefore,
6
Heat released by oil =
3.69 (10 ) kJ/day
0.6
6
= 6.15 (10 ) kJ/day
6
The oil feed rate =
6.15 (10 ) kJ/day
6
6.33 (10 ) kJ/B
¥
42 gal 1 day
¥
= 1.70 gal/hr
1B
24 hr
3.4 Combined Material and Energy Balances
We do nothing new when we combine material and energy balances. In fact,
the two complement each other and assist the engineer in truly understanding
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A Process Engineering Approach to Solving Problems
109
the process. The previous example problem showed how calculating an energy
balance can help determine a required firing rate of a fuel. In addition, combined
energy and material balances are quite useful in gaining a preliminary understanding of the major environmental effects of large electricity-generating plants.
Fossil fuel-fired (primarily coal and natural gas) power plants are one of the
mainstays of the US electricity-generating infrastructure. In 2011, the electric
power industry generated a net 3.9 (10)12 kwh of electricity from all sources (fossil fuels, nuclear, hydro, wind, biomass, etc.), and together, coal and gas provided
the primary energy to produce about two-thirds of it. Figure 3.5 will help the
reader understand the flow of energy and materials (fuel, air, ash, exhaust gases,
pollutants) through a coal-fired power plant as described in the next paragraph.
In a typical steam-electric power plant, fuel (usually coal or natural gas) is
burned in a boiler where the heat released by the combustion is utilized to boil
high-pressure water into steam. The steam gets superheated by the hot combustion gases, and then flows through a turbine-generator. Some of the thermal
energy of the high-pressure steam is converted into mechanical energy by turning the turbine at high speed; the steam exits the turbine at low pressure. The
turbine is coupled to a generator that transforms the mechanical energy into
electrical energy. Meanwhile, the exhaust gases from the combustion of fuel in
air flow through heat exchange equipment to try to capture as much heat as
possible, transferring the heat first to boiler feedwater and then to the incoming
air. The exhaust gases then flow to air pollution control (APC) equipment (more
APC is needed at coal-fired plants than at gas-fired plants) to control the pollutants that were generated or released during the combustion. The low-pressure
steam is condensed by a separate stream of cooling water to be able to re-use the
high-purity boiler feedwater. This condensation of low-pressure steam removes
Figure 3.5
Simplified process flow diagram for a coal-fired power plant.
steam
electricity
turbine
generator
furnace / boiler
evaporation
boiler
feed
water
coal
condenser
cooling
water
flue gas
crusher
air
bottom ash
air
air
preheater
cooling
tower
make-up
water
flue gas to air pollution control system
(contains fly ash and gaseous pollutants)
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110
Chapter Three
even more thermal energy as waste heat than was used to generate the electricity. In all modern plants, the waste heat removed by the cooling water is transferred to the atmosphere in a cooling tower (in older plants, sometimes rivers or
lakes were used as sources of “once-through” cooling water). This large amount
of heat is released to the environment in the cooling tower by evaporation of a
small percentage of the circulating cooling water. The use of combined energy
and materials balances is demonstrated in the following example.
EXAMPLE 3.13
A 1,000 MW coal-fired power plant is burning anthracite coal from Pennsylvania which has 6% ash and 2.5% sulfur. The plant has a thermal efficiency of
40%. Assume that the overall removal efficiencies for ash and SO2 are 99.5%
and 88% respectively. Calculate:
(a) the rate of heat emitted to the environment (kJ/s)
(b) the rate of coal input to the furnace (kg/day)
(c) the rate of ash emissions to the atmosphere (kg/day)
(d) the rate of SO2 emissions to the atmosphere (kg/day)
SOLUTION
First, do an energy balance on the plant (recall that power is a rate of energy).
Ein (fuel)
Eout (electricity)
1,000 MW
Eout (heat)
With a thermal efficiency of 40%, only 40% of the input energy is converted to
useful electricity, so
Ein =
1, 000 MW
= 2, 500 MW
0.40
(a) The heat emitted to the environment is:
Heat = (1 - 0.40 ) ¥ 2, 500 MW = 1, 500 MW or
Heat = 1, 500 MW ¥
1, 000 kW 1 kJ/s
6
¥
= 1.5 (10 ) kJ/s
1 kW
1 MW
(b) To calculate the coal input rate, we need the energy content. From Table
3.2, one pound of this coal contains 14,200 kJ or 3.96 kwh.
Coal input = 2,500 MW ¥
6
1 kg
1, 000 kW 24 hr
1 lb
¥
¥
¥
1 MW
1 day 3.96 kwh 2.2 lb
= 6.89 (10 ) kg/day
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A Process Engineering Approach to Solving Problems
111
(c) To get ash emissions, first calculate ash input,
6
Ash input = 6.89 (10 ) kg/day ¥
0.06 kg ash
1 kg coal
5
= 4.13 (10 ) kg/day
Then multiply by (1.0 – removal efficiency)
5
Ash emissions = 4.13 (10 ) kg / day ¥ (1.00 - 0.995)
= 2, 065 kg / day
(d) Assume that all the incoming sulfur is oxidized to SO2, i.e.
S + O2
→
SO2
The incoming sulfur is just:
Sin = 6.89 (10)6 × 0.025 = 1.72 (10)5 kg/day
Since the mass ratio of SO2 to S is 64 to 32, and since the problem specifies
88% removal efficiency, SO2 emissions are:
5
SO 2 = 1.72 (10 )
kg 64 kg SO 2
4 kg
¥
¥ (1 - .88 ) = 4.13 (10 )
day
day
32 kg S
3.5 Numerical Methods
Engineers use numerical methods in many, many applications in their jobs.
With powerful computers on each desk, engineers today can solve problems
that were impossible thirty years ago. Spreadsheets are used all the time for
ordinary calculations (e.g., addition, subtraction, multiplication, division), but
also can be used to solve differential equations and sets of nonlinear algebraic
equations via iterative techniques. This section provides a basic introduction to
numerical techniques that are easy to learn, yet powerful enough to be useful in
many types of problems where analytical solutions are difficult or impossible.
Simulation of Unsteady-State Processes
In general, the unsteady state is modeled using Eq. (3.5) with the accumulation rate term not being zero. In many cases, the resulting equation is simple
enough to be integrated analytically, but in many other cases it is too complicated for easy analytical solution. Although there are several sophisticated
numerical methods for solving such an equation, the finite difference
approach, based on Euler’s Method, is a simple technique for solving a differential equation. The finite difference method uses the first term of a Taylor
series expansion of the first derivative of a function to estimate future values of
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112
Chapter Three
the function. It is valid for linear or nonlinear equations. Examples of nonlinear
equations include when a variable is raised to a power, appears in an exponential or logarithmic term, or when two variables are multiplied or divided.
The finite difference
method involves using the
Figure 3.6 Finite difference method for “predicting”
first derivative or “slope” of
function values.
the function to “look ahead”
a small increment of time
(the time step, ∆t), and to use
y (t1)
t2 = t1 + Δt
that slope along with the cury (t2) actual
rent value of the function to
estimate (numerically) the
y (t2) numerical
“new” value of the function
at the “new” time (t + ∆t). We
y
march forward through time,
one step at a time, to eventually get a set of values for the
function over the time domain of interest. This is
shown schematically in Figure 3.6.
0 t1 t2 Δt
In order to use the finite
t
difference method to numerically integrate an equation,
it is necessary to have an equation that “solves for” the first derivative of a variable as a function of other variables. Let’s say that we want to predict the concentration of a certain substance over time as it reacts in a batch reactor. As a
result of making a material balance, we have an equation that equates the time
derivative of the concentration (dC/dt) with some function of the concentration. We start with a known value of C at time zero (call this C0), and calculate
the value of the slope (dC/dt) at time zero. Now we convert the slope (dC/dt) to
a finite difference over a small increment of time (∆t), and calculate C at the
new time (Cnew) from the old value of C, the old value of the slope, and the size
of the time step. Then we replace the old value of C with the new value, step
forward to the next time, and repeat the process.
As an example consider a first-order chemical reaction, A → B, occurring in
a batch reactor with a rate expression rA = –kC. (To keep the symbols simpler,
the concentration of A will be represented without the subscript A in this
development.) In the batch reactor, dC/dt = –kC, so we have an analytical
expression for the rate of change of concentration with time. Then we make the
finite difference approximation
dC DC
@
Dt
dt
(3.24)
∆C = Cnew – Cold
(3.25)
But,
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A Process Engineering Approach to Solving Problems
113
Substituting Eq. (3.25) into Eq. (3.24), we get
(Cnew - Cold ) = dC
(3.26)
Ê dC ˆ
Cnew = Cold + Dt Á
Ë dt ˜¯
(3.27)
Dt
dt
Solving for Cnew
We can easily solve Eq. (3.27) if we have an analytical expression for dC/dt. In
this case, we do: dC/dt = –kC. Start at time zero with C = C0 and calculate the
initial value of dC/dt. We solve for the value of dC/dt at time zero using the initial (or old) value of C. Next, we solve for the new value of C (the value that we
project at time t + ∆t) using the old value of C and the “old” value of dC/dt. In
this case it is simply –kCold, but in general we can represent the dC/dt part of the
equation as some function of Cold and time.
Cnew = Cold + ∆t × f (Cold, t)
(3.28)
Finally, Cold is replaced by Cnew , time is incremented, and the process is repeated.
Cold = Cnew
(3.29)
tnew = told + ∆t
(3.30)
This process is repeated for the desired time of the simulation.
Notice how, with the ∆t shown in Figure 3.6, the prediction of the new yvalue is in error compared with the actual or analytical solution. The error is
greater near the left side of the curve due to the steeper curve (greater rate of
change of y) in the beginning. The error could be very large if (a) the time steps
are very large or (b) if the function is not as smooth and “well-behaved” as
shown in Figure 3.6.
The accuracy of the simulation can be improved by using smaller increments for the time step (see Figure 3.7). Note that in Figure 3.7a, we see a very
large error between the actual value of y(t2) and the numerical value for y(t2). In
Figure 3.7b, the time step is one-third the size of that used in Figure 3.7a. The
elapsed time indicated by time t2 in Figure 3.7a is exactly equal to the elapsed
time indicated by time t4 in Figure 3.7b, but it takes three sets of calculations to
get from time t1 to time t4 in the (b) diagram, whereas it only took one set to get
to same time (t2) in the (a) diagram. However, the large error indicated in the
(a) diagram between the actual and numerical solutions has become very small
in the (b) diagram.
Reducing the step size increases the accuracy, but also increases the number of computations to get to a particular end time, which would certainly be a
problem if the calculations were performed by hand. For spreadsheet applications, however, reducing the step size may not noticeably increase the length of
computer time necessary for completion of the simulation. It is noted that use
of extremely small time increments may introduce significant numerical errors
inside the computer as well as result in excessive computer time.
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Chapter Three
114
Figure 3.7 Effect of time step size on simulation accuracy: (a) large time steps, (b) small
time steps.
y (t1)
y (t4) actual
y (t2) actual
y (t1)
y (t3)
y
y
y (t4) numerical
y (t2) numerical
Δt
0
t1
Δt
t2
t1 t2 t3 t4
0
t
(a) large time steps
t
(b) small time steps
The finite difference technique is demonstrated in the next two examples.
Example 3.14 uses a very simple chemical reaction (for which we already have
an analytical solution); this example not only illustrates the technique, but also
shows how the accuracy of the numerical answer improves with smaller time
steps. Example 3.15 demonstrates the major advantage of a numerical technique; it solves a problem for a reaction with complex kinetics, which is not
easy to integrate analytically.
EXAMPLE 3.14
A first-order reaction (A → B) is conducted in a constant volume batch reactor. The rate constant is 0.1 min–1 and the initial concentration is 100 mg/L.
Determine the concentration of A that remains after 30 minutes of reacting.
Use both analytical and numerical solution methods. Compare the analytical
answer to the numerical answers for various time step sizes.
SOLUTION
Analytical method: The mass balance equation for reactant A in a batch reactor is:
ARA = IRA – ORA + GRA
d (V CA )
dt
V
= 0 - 0 + rA V
d CA
= - k CA V
dt
d CA
= - k dt
CA
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A Process Engineering Approach to Solving Problems
115
As presented in Chapter 2, the analytical solution for this case, derived by
separation of variables and integration, was:
Ct = C0e–kt
After 30 minutes of reaction, the concentration of A remaining is:
C30 = (100 mg/L) e-(0.1) (30) = 4.9787 mg/L
Numerical Method: For this problem, the differential equation is
dC
= - kC
dt
and the finite difference equation, solved for Cnew is
Cnew = Cold + ∆t × (–kCold)
The numerical procedures are presented below for the full 30 minutes using a
time step of 5 minutes. Then the first six iterations of a solution using 1-minute
time steps are shown. Finally, the 30-minute concentrations are compared for
time steps of several different sizes. As will be seen, reducing the time step size
from 5 minutes to 0.01 minute greatly improves the accuracy of the final answer.
It is strongly suggested that a table be set up like the one shown below to
help keep track of the calculations in this procedure. Once the student
becomes comfortable with the procedure, the table can be set up and the calculations can be performed entirely within a spreadsheet.
For ∆t = 5 minute
Start time
min
Cold
mg/L
–kCold
mg/L-min
∆t
min
Cnew
mg/L
End time
min
0
5
10
15
20
25
100
50
25
12.5
6.25
3.12
–0.1(100) = –10
–0.1(50) = –5
–0.1(25) = –2.5
–0.1(12.5) = –1.25
–0.1(6.25) = –0.625
–0.1(3.125) = –0.3125
5
5
5
5
5
5
100 + 5(–10) = 50
50 + 5(–5) = 25
25 + 5(–2.5) = 12.5
12.5 + 5(–1.25) = 6.25
6.25 + 5(–0.625) = 3.125
3.125 + 5(–0.3125) = 1.5625
5
10
15
20
25
30
For ∆t = 1 minute
Start time
min
Cold
mg/L
–kCold
mg/L-min
∆t
min
Cnew
mg/L
End time
min
0
1
2
3
4
5
100
90
81
72.9
65.61
59.05
–0.1(100) = –10
–0.1(90) = –9
–0.1(81) = –8.1
–0.1(72.9) = –7.29
–0.1(65.61) = –6.561
–0.1(59.05) = –5.905
1
1
1
1
1
1
100 + 1(–10) = 90
90 + 1(–9) = 81
81 + 1(–8.1) = 72.9
72.9 + 1(–7.29) = 65.61
65.61 + 1(–6.561) = 59.05
50.05 + 1(–5.905) = 53.14
1
2
3
4
5
6
This process continues until the end time is 30 minutes, at which the final concentration is 4.2391 mg/L. A similar process is followed for time steps of 0.1
Cooper.book Page 116 Monday, June 23, 2014 9:58 AM
116
Chapter Three
minutes and 0.01 minutes. Note that for a ∆t of 0.01 minutes, it takes 3,000
rows of calculations to reach 30 minutes, but with the ability of a spreadsheet
to copy whole rows of formulas down by using the mouse, 3,000 rows of calculations is easily done. The following table displays the final results and
clearly shows how the errors are reduced with smaller and smaller time steps.
Simulation Results (analytical solution = 4.9787 mg/L)
∆t
min
C @ 30 min
mg/L
|Error|
mg/L
Percent Error
%
5
1
0.1
0.01
1.5625
4.2391
4.9041
4.9712
3.4162
0.7396
0.0746
0.0075
68.62
14.86
1.50
0.15
Iterative Solution of Nonlinear Algebraic Equations
The complete description of a system may require many mass and/or
energy balances. Independent equations may be obtained by making balances
using different boundaries or different conservative properties. The resulting
system of equations often must be solved simultaneously. Obtaining an analytic solution may be very difficult if there are many unknowns and/or if nonlinearities exist in the equations. Systems of linear algebraic equations can be
solved using matrix methods, but even one nonlinear equation may require
numerical solution methods.
The solution methodology presented in this section relies exclusively on
iterative calculations. This unsophisticated, brute-force approach requires very
minimal knowledge of numerical methods or computer programming. The
methods are readily adapted to spreadsheets or simple programming.
Although the lack of sophistication associated with this approach is viewed as
an advantage for this course, it is conceded that the methods do not always
converge readily and may require accurate initial estimates of the actual solution to achieve convergence efficiently (or at all).
The general approach involves the following steps:
1. Make initial estimates for all unknown variables (yold)
2. Algebraically manipulate each equation to obtain equations which are
implicit solutions for each variable (meaning each solution equation is a
function of the variable itself)
3. By sequential calculation, determine temporary values for each unknown
(ytemp)
4. Calculate a new estimate of each variable (ynew) as a weighted average
using yold and ytemp
5. Replace all yold values with ynew values
6. Repeat steps 3 through 5 until satisfactory convergence is achieved
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A Process Engineering Approach to Solving Problems
117
The ynew value is calculated from a weighted average as follows:
ynew = ytemp (x) + yold (1 – x)
(3.31)
where:
ynew = latest value of the unknown variable (result of the current iteration)
yold
= value of the unknown variable from the previous iteration
ytemp = value of the unknown variable obtained in step 3 above
x
= weighting factor (varies between 0 and 1)
Values of the weighting factor close to 1 tend to achieve convergence more rapidly for “well-behaved” functions; however, use of small values for x may be
necessary to achieve convergence for those equations that are not well-behaved.
Application of these calculation methods is illustrated in the next example.
EXAMPLE 3.15
The gas-phase reaction
2 HCl + ½ O2 ↔
Cl2 + H2O
has an equilibrium constant KP with a value of 0.0842 at a temperature of
1,800 °F. Assume an initial mixture of gases exists with partial pressures as
follows: HCl = 0.145 atm, O2 = 0.105 atm, H2O = 0.0550 atm, Cl2 = 0.0 atm, and
N2 making up the balance. Calculate the final partial pressure of Cl2 after the
mixture reaches equilibrium. Assume the total pressure is 1.0 atm and does
not change during this equilibrium shift. By definition:
KP =
PCl 2 PH 2O
(PHCl )2 (PO2 )
0.5
SOLUTION
Write the equation, showing the initial concentrations, and the reaction proceeding to equilibrium:
IC
Rxn
Eq
2 HCl
0.145
–2x
0.145 – 2x
+
0.5 O2
0.105
–0.5x
0.105 – 0.5x
→
Cl2
0.0
+x
x
+
H2O
0.055
+x
0.055+x
Thus at equilibrium,
( x ) (0.055 + x )
(0.145 - 2x )2 (0.105 - 0.5x )0.5
= K P = 0.0842
This is a nonlinear function of the variable x, which is best solved iteratively.
Cooper 03.fm Page 118 Thursday, March 3, 2016 9:47 AM
118
Chapter Three
First, rewrite this expression as an implicit solution for x:
2
x=
0. 5
0.0842 (0.145 - 2x ) (0.105 - 0.5x )
(0.055 + x )
Next, make an initial guess for x, substitute it into the right-hand side (RHS)
of the equation, and solve. If the solution equals our guess, we are done. If
not, make a second guess of x and repeat. The second guess for x simply can
be the solved value of the RHS, or can be a weighted average of the initial
guess and the solved RHS. (Note that the first choice is also a weighted average, but with a weighting factor of 1.0.) Again, the calculations are best made
in a spreadsheet.
With a weighting factor of 1.0, the spreadsheet might look like this:
iteration
x
RHS
% error
1
2
3
4
5
6
7
8
0.01
0.006401
0.007647
0.007193
0.007355
0.007296
0.007318
0.00731
0.006401
0.007647
0.007193
0.007355
0.007296
0.007318
0.00731
0.007313
–35.99
19.47
–5.94
2.26
–0.80
0.29
–0.10
0.04
But with a weighting factor of 0.7, the calculations converge much more quickly:
iteration
x
RHS
% error
1
2
3
4
5
6
7
8
0.01
0.00748
0.00732
0.007312
0.007312
0.007312
0.007312
0.007312
0.006401
0.007252
0.007309
0.007312
0.007312
0.007312
0.007312
0.007312
–35.99
–3.06
–0.15
–0.01
0.00
0.00
0.00
0.00
The preceding examples demonstrate the application of numerical solution
techniques to a single linear or nonlinear equation with one unknown. The
numerical approach requires virtually identical programming and computational effort regardless of whether the equation involves more than one term
and/or a nonlinearity. As promised after Example 3.8, we close this chapter
with an example that involves a flowing system with a nonlinear reaction term.
With the numerical methods that have been presented in this chapter, the solution is straightforward.
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A Process Engineering Approach to Solving Problems
119
EXAMPLE 3.16
Given the reaction A → B, rA = –2.5 CA 1.2 / (CA+ 70), where rA is in mg/L-min.
You are studying the reaction in an unsteady-state CSTR reactor; the volume
of the reactor is 300 L. Prior to the start of the experiment, pure water is flowing into and out of the reactor at 15 L/min. The initial and outlet concentrations of A are both 0.0 prior to the start, but at time zero you start injecting
some pure A into the inlet stream, and the inlet concentration jumps up to 120
mg/L and stays there (a step increase).
(a) Starting from Eq. (3.5), derive the finite difference equation, and use
numerical techniques (using a hand calculator) to solve for the outlet concentration after 3 minutes have elapsed; use time increments of 1 min.
(b) Finish the simulation using a spreadsheet to find the outlet concentration
after 90 minutes.
SOLUTION
General equation:
AR = IR - OR + GR
Put in variables,
V dCout
= Q Cin - Q Cout + rAV
dt
Simplify,
dCout Ê Q ˆ
ÊQ ˆ
= Á ˜ Cin - Á ˜ Cout + rA
ËV¯
ËV¯
dt
vert to a finite difference,
Conv
2.5 Cout1.2
DCout Ê Q ˆ
Ê Qˆ
= Á ˜ Cin - Á ˜ Cout ËV¯
ËV¯
Dt
Cout + 70
Drop the subscript "out" for convenience,
2.5 C1.2
DC Ê Q ˆ
Ê Qˆ
= Á ˜ Cin - Á ˜ C ËV¯
Dt Ë V ¯
(C + 70)
Expand the finite difference,
ÏÔÊ Q ˆ
2.5 Cold1.2 ¸Ô
Ê Qˆ
Cnew - Cold = Dt ÌÁ ˜ Cin - Á ˜ Cold ˝
ËV¯
Cold + 70 Ô˛
ÔÓË V ¯
Insert numbers,
-2.5 Cold1.2
Ê 15 ˆ
Ê 15 ˆ
Cnew - Cold = Á
Cold +
120 - Á
˜
˜
Ë 300 ¯
Ë 300 ¯
Cold + 70
Solve for Cnew
Cnew = Cold + Dt ¥ f (Cold , t )
where:
2.5Cold1.2 Ô¸
ÔÏ
f Cold, t = Ì6 - 0.05Cold ˝
Cold + 70 ˛Ô
ÓÔ
(
)
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120
Chapter Three
Set up the table for hand calculations as follows:
Start time
min
Cold
mg/L
f(Cold , t)
mg/L-min
∆t
min
Cnew
mg/L
End time
min
0
1
2
0
6.0
11.4
6.0
5.4
4.9
1
1
1
6.0
11.4
16.3
1
2
3
The concentration after 3 minutes have elapsed is 16.3 mg/L. If this table is
extended to 90 minutes using a spreadsheet, the resulting outlet concentration is 64.7 mg/L (which is essentially the final steady-state concentration for
this problem).
PROBLEMS
3.1
Five million kilograms per day of coal are burned in an electric power
plant. The coal has an ash content of 12% by mass. Forty percent of the
ash falls out the bottom of the furnace. The rest of the ash (the fly ash) is
carried out of the furnace with the hot gases into an electrostatic precipitator (ESP). The ESP is 99.5% efficient in removing the ash that comes into
it. Draw a diagram representing this process and calculate the mass emissions rate of fly ash into the atmosphere from this plant.
3.2
Given the following diagram and data, calculate the quantity of steam
required to regenerate the carbon in step 2.
polluted air
Q = 200 m3/min
C = 500 mg/m3
carbon bed
clean air
C = 10 mg/m3
Step 1
steam
Step 2
In step 1, polluted air flows
through a bed of activated carbon. Pollutant molecules adsorb
onto the carbon and are thus
removed from the air. This step
lasts for 6 hours, and then the
flow is switched to a new bed.
The old bed is steamed for 1 hour
and rested for 5 hours before
being put back into service.
steam
After one bed gets loaded with
+ pollutant pollutant, it is cleaned with
steam. The steam removes all of
the pollutant from the carbon. It
has been found that a ratio of
steam-to-pollutant of 7 kg
steam/1 kg pollutant is necessary
for complete cleaning. Calculate
the quantity of stream required to
do this job, kg/day. Assume the
process runs 24 hr/day.
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A Process Engineering Approach to Solving Problems
3.3
121
The following is a clarifier-thickener system to separate solids and liquids. Ignore the effect of solids content on density of the streams (i.e.,
assume density of solids = density water).
B
A’
A
D
C
E
Complete the following material balance table.
3.4
Stream
Flow rate, L/s
Solids, mg/L
A
B
C
D
E
A′
100
95
3,000
15
6,000
50
A 600 MW coal-fired power plant is burning Illinois bituminous coal with
8% ash content. The plant is 39% efficient, 30% of the ash drops out in the
furnace, and the electrostatic precipitator is 99.0% efficient. A simplified
sketch appears below:
exit gases +
some fly ash
gases +
fly ash
coal
furnace
ESP
bottom ash
collected fly ash
stack
a. Draw an energy balance diagram for the plant and calculate the rate of
heat emitted to the environment, in J/s.
b. Calculate the rate of coal input to the furnace, in kg/day.
c. Calculate the rate of fly ash emissions to the atmosphere, in kg/s.
3.5
Draw a labeled diagram showing raw sludge being fed to an anaerobic
digester that is producing three products: a gas, a liquid supernatant, and
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122
Chapter Three
digested sludge. Also, show the gas product going to a furnace along with
a separate stream of air, and show combustion gases leaving the furnace.
Given the following data, calculate the total molar flow rate of CO2 gas
coming out of the furnace, in gmol/day.
• 1,000 kg/day of raw sludge is fed into the digester
• 40% by mass of the raw sludge is converted to digested sludge
• 58% by mass of the raw sludge is converted to supernatant
• 70% by volume of the digester gas product is CH4, the other 30% is CO2
• The density of the gas product at 1 atm pressure and 25 °C is 1 kg / m3
• Enough air is used in the furnace to burn all of the methane (CH4) to
CO2 and H2O
3.6
Given: A wastewater treatment plant as shown below. Influent TSS = 250
mg/L. Alum is added to help settling at a rate of 30 lb/million gallons.
Effluent requirements call for 90% reduction in TSS.
Find: If the primary settler removes 48% of initial TSS, find the percent
removal required in the secondary settler to achieve the 90% overall
removal requirement. Assume all alum settles out.
primary
settler
influent
alum
secondary
settler
aeration
3.7
The stack gas from a process contains 4.0 g/m3 of PM. The gas flow rate is
5 m3/s. If an electrostatic precipitator removes 1,000 kg of PM per day,
what is the emission rate of PM? Give answer in kg/day.
3.8
A 1,000 MW coal-fired power plant operates at 40% efficiency. It consumes
10,000 tons of coal/day. The plant uses coal that is 4% sulfur and 10% ash.
The ash leaves the furnace as bottom ash or fly ash. Virtually all the sulfur
is converted to sulfur dioxide, and the plant is equipped with a 90% efficient SO2 scrubber. Assume that one-third of the ash in the coal forms bottom ash, and the remainder forms fly ash. Pollution control devices
capture 98% of the fly ash. Calculate the rates (each in tons/day) of:
a. sulfur dioxide emissions
b. bottom ash removed from the furnace
c. fly ash collected
d. emissions of fly ash to the atmosphere
3.9
A mining process discharges a silty water into a clean brook at the rate of
5 cubic meters per minute. The silt content is 200 mg/L. What is the con-
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A Process Engineering Approach to Solving Problems
123
centration in the brook during the rainy season (the brook flows at 50 m3/
min), and during a severe dry spell (the brook flows at 5 m3/minute)?
3.10 Calculate the minimum rate at which cooling water must be pumped
through the condensers of a 800 MW nuclear power plant if the maximum cooling water temperature rise is 15 °F. The efficiency of the plant is
32%. Assume no heat is lost to the atmosphere directly from the system.
Give your answer in pounds/hr, and ft3/sec. Assuming the plant is
designed with five identical pipes in parallel to carry this cooling water,
calculate the diameter of each pipe if the water velocity is 6.0 ft/sec.
3.11 A new state regulation requires that the wastewater (WW) from a tannery
be routed to a municipal wastewater treatment plant (WWTP) for final
treatment. However, the tannery WW must be pretreated to remove certain organic pollutants before it goes to the WWTP. The pretreatment process is a biological anaerobic process that has a limit on total dissolved
solids (TDS) of no more than 20,000 mg/L. But, the tannery WW contains
95,000 mg/L of TDS, so it must be diluted somehow. A proposed treatment process is shown in the accompanying figure. Calculate the flow
rate of the recycle that is required in order to keep the TDS in the mixed
input stream going into the pretreatment process at 20,000 mg/L.
tannery wastewater
Q = 100,000 L/day
TDS = 95,000 mg/L
biological anaerobic
pretreatment process
municipal wastewater
Q = 1,000 m3/day
TDS = 500 mg/L
combined final
treatment process
(WWTP)
recycle
treated
effluent
3.12 A gas containing equal parts (on a molar basis) of H2, N2, and H2O is
passed through a column of silica gel that absorbs 96% of the water and
none of the other gases. The column packing was initially dry, and had a
mass of 2 kg. Following five hours of continuous operation, the pellets are
reweighed and are found to have a mass of 2.18 kg. Calculate the molar
flow rate (in gmols/hr) of the feed gas and the mole fraction of water
vapor in the product gas.
3.13 In order to meet a certain octane number specification, it is necessary to
produce a gasoline containing 83% (by weight) isooctane and 17% n-heptane. How many gallons of a high-octane gasoline containing 92% isooctane and 8% n-heptane must be blended with a straight-run gasoline
containing 63% isooctane and 37% n-heptane to obtain 10,000 gal of the
desired gasoline? The density of each of the liquids is 6.7 lb/gal.
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124
Chapter Three
3.14 The analysis of the exhaust gas from a burner fueled with a natural gas
(essentially pure CH4) is as follows:
N2
72.88 mole percent
O2
3.87 mole percent
CO2
7.75 mole percent
H 2O
15.5 mole percent
What is the ratio of moles of air (assume air is 79% N2 and 21% O2) per
mole of natural gas fed to the burner?
3.15 A stream containing 0.10 gmol/L of reactant A flows into a CSTR at 100
L/min. The reaction proceeds according to:
3
2A Æ B with - RA = k [ A ]
a. Calculate the volume of the reactor if k has a value of:
3.0
L2
(gmol)2 -min
and the concentration of A in the outlet stream is
0.05 gmol/L.
b. Calculate the production rate of B in gmol/day.
3.16 Consider a second-order reaction (2A → B) occurring in a CSTR with a
volume of 250 L. Given an inlet flow rate of 50 L/s, an inlet concentration
of A of 0.20 gmol/L, and an outlet concentration of A of 0.08 gmol/L, calculate the rate constant (with appropriate units).
3.17 For the same reaction and inlet conditions as Problem 3.16 above, but
given a PFR with a volume of 250 L, calculate the expected outlet concentration of A.
3.18 An industrial chemical reaction proceeds according to:
A
→
2B
with a rate expression: RA = –0.60CA2 and RA has units of gmol/(L-min).
Just prior to entering a CSTR, a stream of water flowing at 50 L/min and
containing 0.20 gmol/L of A is mixed with a stream of water flowing at 40
L/min and containing 0.30 gmol/L of A. The combined stream then immediately flows into the CSTR.
a. Draw the labeled diagram and calculate the inlet concentration of A
going into the reactor.
b. The outlet concentration of A is 0.02 gmol/L. How big is the reactor (L)?
c. The product B has a molecular weight of 48 g/mol. What is the production rate of B in kg/day?
3.19 A coal slurry is a mixture of crushed coal and water that can be pumped
via pipeline across the country. A power plant needs 20 million kg/day of
coal (dry basis). A slurry containing 50% coal and 50% water is pumped
in the pipeline to the power plant. When the slurry is received at the
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A Process Engineering Approach to Solving Problems
125
power plant, the coal is separated from the water. However, the separation process is inefficient. The separated water stream contains 2% (by
weight) coal, and the separated coal contains 20% (by weight) water. Fill
in the following material balance table for the flow rates received at the
power plant. (Hint: First draw a diagram of the separation process.)
Mass Flow Rate (millions of kg/day)
Stream
Pipeline Slurry
Separated Water
Separated Coal
Coal
Water
Total
20
3.20 A CSTR is treating a wastewater to remove a toxic compound. The reaction is first order with a rate constant of 0.4/minute. The flow rate is 100
L/min, and Cin = 100 mg/L. Determine the volume of a single CSTR
which is required to achieve 95% removal (an effluent concentration of 5
mg/L).
3.21 For the wastewater described in Problem 3.20 (Flow = 100 L/min, Cin =
100 mg/L, and k = 0.4 per minute) determine the volume of a PFR to
achieve an effluent concentration of 5 mg/L.
3.22 Assume the wastewater described in Problem 3.20 (Flow = 100 L/min, Cin
= 100 mg/L, and k = 0.4 per minute) is to be treated by three equal-volume CSTRs in series. Calculate the total volume of those CSTRs (keep in
mind that each of the three is identical) to achieve the desired effluent
concentration of 5 mg/L.
3.23 Develop a numerical simulation using a spreadsheet for a second-order
reaction destroying compound A in a batch reactor. The initial concentration of A is 100 mg/L, and the rate constant is 0.000220 L/mg-min. What
is the concentration remaining after 2 hours?
3.24 Develop a numerical simulation on a spreadsheet for the reaction
described in Chapter 2, Problem 2.21. What is the concentration remaining after 30 hours?
3.25 Develop a numerical simulation on a spreadsheet for the nonsteady-state
situation described in Example 3.8.
3.26 A 440 MW coal-fired power plant is 40% efficient at producing electricity.
The coal has a heating value of 20,000 kJ/kg, an ash content of 6.0%, and a
sulfur content of 4.0%.
a. What is the daily input rate of coal (kg/day)?
b. If the overall efficiency of ash capture is 98.6%, how much ash is emitted into the air, kg/day?
c. How much heat is emitted to the environment (kJ/day)?
d. If 80% of the heat emitted is to be removed by cooling water, calculate
the flow rate of water (in gal/day) if the cooling water is initially at 65
°F and rises to 95 °F.
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126
Chapter Three
3.27 A reaction will be studied in a CSTR that has a volume of 500 L. At time
zero, the CSTR is filled with pure water. At that instant, the input stream
starts flowing in at 100 L/min, with inlet concentration of A of 60.0 mg/L.
0.75
The reaction is:
A → 2 P with a rate expression: rA =
- 50 ( C A )
(200
+ CA )
and rA has units of mg / (L-min).
By numerical means, calculate the expected outlet concentration of A as a
function of time. Start by deriving, from material balance, an equation to
solve implicitly for Cout. Set up a table showing your calculations and work,
and do three iterations by hand using a ∆t = 1 minute. Using a spreadsheet,
determine the concentration after 10 minutes have passed. Carry your
spreadsheet calculations out to 90 minutes. By observation of your spreadsheet answers, what is your estimate of the steady-state concentration?
3.28 A rinse tank at a chrome-plating shop has a volume of 1,000 liters, and is
used to rinse hubcaps that have been chrome plated. The hubcaps each
drag in 0.10 L of plating solution which contains 90 mg/L of chromium.
The hubcaps get rinsed at a rate of 12 hubcaps per minute. Pure fresh
water is used for the rinse, and flows into the rinse tank continuously at a
rate of 200 L/min. The rinse tank discharges 200 L/min of used rinse
water, and the hubcaps drag out 0.10 L of used rinse water per hubcap.
Calculate the steady-state concentration of chromium in the discharged
rinse water, in mg/L. Be sure to draw your diagram before you start.
3.29 A small river flows at 8 m3/s in the dry season and 30 m3/s in the rainy season. A waste stream flows into the river at 2 m3/s all year. The waste stream
contains solids at 125 mg/L.
a. In the dry season, upstream from where the waste stream joins the river,
the solids concentration in the river is 12 mg/L. Calculate the dry-season concentration of solids in the river after the waste stream has joined
it. Give your answer in mg/L.
b. In the rainy season, the temperature of the small river upstream of the
mixing point is 12 °C. If the temperature of the combined stream is 14
°C, what is the temperature of the waste stream (°C)?
3.30 An industrial chemical reaction proceeds according to:
A +
B → 2D
with a rate expression:
RA = –0.5 (CA)0.75 (CB)1.4
Note: RA has units of gmol/(L-min)
A stream of water flowing at 20 L/min and containing 0.20 gmol/L of A
(and no B) flows into a CSTR. Another stream of water flowing at 30 L/
min and containing 0.30 gmol/L of B (and no A) also flows into the same
CSTR. They mix instantaneously and start to react.
a. Draw the labeled diagram and calculate the molar flow rates of A and
B entering the reactor, each in gmol/min.
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A Process Engineering Approach to Solving Problems
127
b. If the steady-state exit concentration of A is 0.040 gmol/L, and the exit
concentration of B is 0.14 gmol/L, how big is the reactor (L)?
c. What is the concentration of D in the exit stream (gmol/L)?
3.31 A 700 MW coal-fired power plant has a thermal efficiency of 35%. The
coal has a heating value of 22,000 kJ/kg, an ash content of 5.5%, and a sulfur content of 3.8%.
a. What is the daily input rate of coal (kg/day)?
b. If the efficiency of SO2 capture in an APC device is 95%, how much SO2
is emitted into the air, kg/day?
c. If 30% of the ash that comes in with the coal drops out in the boiler as
bottom ash, and if the APC device that treats the exhaust gases is 98.5%
efficient at capturing the fly ash, what are the total fly ash emissions to
the atmosphere, kg/day?
d. Cooling water flows through the heat exchangers at a rate of 225,000 gal/
minute. The water comes into the heat exchanger initially at 75 °F and
gets warmed to 105 °F. How much heat is being removed by the cooling water (Btu/min)? Note that the Cp for water is 1.0 Btu/(lb-°F), and
the density of water is 8.33 lb/gal.
e. The heat calculated in part d is transferred to the air by evaporation of
some of the cooling water. The latent heat of vaporization for water at
this temperature is 1,034 Btu/lb. How much water is evaporated (and
must be replaced) each day, lb/day?
Cooper.book Page 128 Monday, June 23, 2014 9:58 AM
CHAPTER
4
Water Resources
4.1 Water Quantity
As we learned in middle and high school science classes, the hydrologic
cycle is the movement of water through the biosphere. The movement is cyclical (see Figure 4.1) so we can start anywhere in the cycle to describe it. Water in
the ocean is evaporated (and purified) by the energy of the sun, water vapor is
transported from one place to another by the wind, it is condensed and falls to
earth as rain (and snow, etc.), some of the water is evaporated from surfaces on
the land, some gets taken up by plants and transpired through their leaves,
some runs off the land and travels downhill to the sea in rivers or is stored in
lakes, and some infiltrates into the ground and travels down gradient or is
stored underneath the ground. Of course, rain also falls directly back into the
ocean, and that precipitation does not take part in the hydrologic cycle. We
focus only on that which falls on the land and takes part in this grand cycle.
These movements of water are highly nonuniform, both in time and place, and
can lead, in the extremes, to droughts and floods. The nonuniformities are
what make the management of water resources so important and worthy of
engineering study.
Over any period of time, and for any size region of land, the hydrologic
cycle can be analyzed via a mass balance. Drawing a boundary around some
arbitrary region of the land (e.g., a small area or a large watershed), and writing the unsteady-state balance, yields:
S = P – R – ET – I
where:
S = the storage (accumulation) of water within the boundaries
P = precipitation
R = runoff
ET = evapotranspiration
I
= infiltration
128
(4.1)
Cooper.book Page 129 Monday, June 23, 2014 9:58 AM
Water Resources
Figure 4.1
129
Hydrologic cycle.
Evaporation
Evaporation
Evapotranspiration
Vadose
Zone
Precipitation
Ru
Infiltration
no
ff
Lake
Water Table
Ocean
Confined Aquifer
Impermeable Rock
Equation (4.1) is usually written for a defined area of land. Storage refers to
storage in puddles, ponds, or lakes on the surface of the land; precipitation
includes rain, snow, hail, and other forms of moisture deposition onto the area;
runoff refers to water that flows off the area in surface streams, creeks, or other
forms of surface flow; evapotranspiration combines evaporation and transpiration (the uptake of liquid water by the roots of plants and the release of water
vapor through their leaves) as both processes transport water out of the balance volume to the atmosphere, and infiltration refers to the amount of water
that moves through the surface and into ground. Some of that water can be
evaporated, transpired, or stored in the soil near the surface, and so net infiltration refers to water that percolates deeper into the soil or travels out of the area
laterally underneath the ground.
Note that the units of each term in Eq. (4.1) must be consistent. Of course,
we already know that the units must start with mass/time, but we can develop
the equation in other interesting units. For example, it is customary to report
rainfall as so many inches (or cm) of rain. We can then convert the other terms
to a similar linear measure. How can inches be equated to mass? Simply
assume that the inches of rain constitute an average depth of water over a given
surface area. Multiplying that depth by that area gives a volume of rainwater,
which when multiplied by the density of water, gives a mass. Once that is
Cooper.book Page 130 Monday, June 23, 2014 9:58 AM
130
Chapter Four
understood, then all terms in Eq. (4.1) can be expressed in units of inches (or
cm). Example 4.1 illustrates this point.
EXAMPLE 4.1
(a) Orange County, Florida, with an area of 1,004 square miles, gets an average of 50 inches of rain per year. How much rain is that in cubic feet?
(b) A rainstorm drops 1.5 inches of rain on a 20-acre residential area during a
2-hour period. If 55% of the water flows off the area (the runoff) into a
holding pond, and the rest infiltrates into the ground, calculate the volume of water that was added to the pond, in gallons. Ignore evapotranspiration during this short time period. (1 acre = 43,560 ft2 and 1 cubic foot =
7.48 gallons).
(c) How many inches of infiltration occurred in the 20-acre site?
SOLUTION
(a)
Volume = 50 in. ¥ 1, 004 mi 2 ¥
1 ft
5, 280 2 ft 2
¥
12 in.
1 mi 2
11
= 1.166 (10 ) ft 3
(b)
Volume = 1.5 in. ¥ 0.55 ¥ 20 Ac ¥
1 ft
43, 560 ft 2 7.48 gal
¥
¥
12 in.
Ac
ft 3
5
= 4.48 (10 ) gallons
(c) Ignoring evapotranspiration, and assuming no accumulation on the surface of the site,
I = P-R
= P - 0.55P
= 0.45P
I = 0.45 (1.5 in.)
= 0.675 in.
Water is stored in various forms around the globe. This storage is dominated by the salt water in the oceans, which comprises 97% of the earth’s water.
Much of the remaining 3% that is fresh water is incorporated into ice caps and
glaciers, and is therefore not available for societal development. The water
resources which typically are used for domestic, industrial, and agricultural
needs include surface waters (lakes, rivers, and reservoirs) and groundwaters
(shallow groundwater and deep aquifers). Other sources have been used,
including desalination of ocean water, but at much higher costs. Rivers, lakes,
Cooper.book Page 131 Monday, June 23, 2014 9:58 AM
Water Resources
131
and accessible groundwaters represent less than 1% of the total global water
supply; therefore proper management of water resources is vital to assuring
that people have access to reliable water supplies of adequate quantity and
quality for all their needs.
Issues related to excessive water quantity must also be addressed. The
occurrence of extreme precipitation events may result in flooding episodes
with loss of life and property damage. Engineering studies are routinely completed to assess the risks of flooding by studying the history of precipitation
and flood events. This information is used to regulate future development in
flood-prone locations, or to build flood-protection devices in existing situations. In all cases where land development occurs, there is a potential that conversion from one land use (for example, agricultural) to a developed land use
(for example, a shopping center) will result in a greater runoff from the developed site than would have occurred prior to development. Mitigation facilities
(for example, stormwater retention basins) to address this increase in flooding
potential must be included in the site development.
Surface water pollution from stormwater runoff can be a serious problem,
causing great impairment to lakes and rivers. Urban runoff has been shown to
contain toxic metals, organic and inorganic chemicals, nutrients like nitrogen
and phosphorus, and other pollutants. Agricultural runoff may increase nutrient, sediment, and pesticide loadings on surface water resources. Proper stormwater management must consider measures for water quality protection as
well as flood mitigation.
Precipitation Quantities
Rain can fall hard for a short period of time or slowly for a long period of
time. It can rain every day for a week or more, or we can go months without
rain. Knowledge of typical and extreme patterns of rainfall is needed to produce good long-term water management and storage designs that will prevent
shortages of water during droughts and protect against flooding during very
wet periods. We need accurate historical records of rainfall data over many
years to create a reasonable statistical distribution of those patterns.
Some maximum observed rainfall intensities and durations have been
compiled by Dominguez (1996) and are presented in Table 4.1 on the following
page. These “world record” rainfall amounts would not be representative of
expectations for any specific location, so these values have no practical use for
facilities design. The values are instructive, however, to illustrate the relation
between intensity and duration and to demonstrate how extreme some rainfall
events can be.
To properly design for short, very heavy rainfalls (and the subsequent runoff), we must select a design storm intensity (inches per hour) and duration.
The design storm is one that has some fixed probability of occurring in any
future year, based on historical records of how often such size storms have
occurred in the past. We must be able to select a rainfall intensity and duration
that are somewhat extreme (but not outlandish) so we can design cost-effective
facilities that will protect us most of the time. The intensity of the storm is
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132
Chapter Four
Table 4.1
Duration
8 min
15 min
20 min
45 min
2 hour
3 hour
5 hour
18 hour
1 day
5 day
30 day
90 day
Maximum Recorded Precipitation Events
Intensity (in./hour)
Amount (in.)
30
28
24
13
8
7
6
2
1.7
1.3
0.5
0.3
4
7
8
9.8
16
21
30
36
40.8
156
360
648
Location
Fussen, Bavaria
Plumb Point, Jamaica
Curtea de Arges, Romania
Holt, MO
Rockport, WV
D’Hanis, TX
Smithport, PA
Thrall, TX
Baguio, Philippines
Silver Hill Plantation, Jamaica
Cherrapunji, India
Cherrapunji, India
Source: Adapted from Dominguez (1996).
important because it sets the rate at which the rain is falling. The duration of
the storm is important because it determines the total volume of water that
falls. Surface runoff rates do not reach their maximum values instantaneously
when the storm begins, and they continue after the storm abates. The frequency or return period is important because it defines the probability of the
occurrence of an event in a one-year time frame. Thus, engineers need historical data that yields probability distributions of storm intensities, durations, and
frequencies. Based on the historical probabilities for a given area, and the risk
we are willing to assume (the costs we are willing to pay for protection), we can
design facilities to handle the design storm. Sometimes, we are willing to pay
for protecting against the 1-in-20-year (the 20-year, for short) event, and sometimes we are willing to pay more to protect against the 100-year event.
A 20-year event (an event that recurs once every twenty years) would have
an annual probability of occurrence of 0.05. The corresponding annual probabilities of the 25-year and 100-year events are 0.04 and 0.01, respectively. Selection of a design storm for engineering projects will reflect the acceptable risk of
system failure (flooding). For residential development, it is common to require
that all structures be located above the 100-year floodplain. This criterion is
established based on a desire to minimize the risk of residential flooding. In
contrast, the design of storm sewers may be based on a 10-year precipitation
event. Failure of the storm sewers—with consequent flooding of streets—
would be acceptable with a greater frequency than residential flooding. This
kind of decision is based on saving the extra cost of designing storm sewers
large enough to handle the 100-year event.
The interactions between rainfall intensity, duration, and frequency are
commonly presented graphically, as shown in Figure 4.2. These curves are
defined empirically from rainfall records for any specific geographic region. It
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133
7
Rainfall intensity, in inches per hour
6
5
4
3
Frequency, in years
50
2
15
5 10
2
1
Figure 4.2 Example of
rainfall intensity, duration,
and frequency relationships.
(Adapted from ASCE/WPCF 1969.)
0
25
0
20
40
60
80 100 120 140
Time, in minutes
160
180
200
should be noted that the patterns of rainfall (intensity, duration, and frequency)
may change significantly over the years due to climate change. That is, even
though we must rely on historical records, there is no guarantee that the averages of these climate variables will remain fixed in time.
EXAMPLE 4.2
From the intensity/duration/frequency curve in Figure 4.2, compare the effect
of frequency (10-year, 25-year, and 50-year event) on intensity by identification of the design storm intensity (in./hour) for a 20-minute duration event.
SOLUTION
The intensities are obtained from the graph as follows:
Frequency (yr–1)
10
25
50
Intensity (in./hr)
3.6
4.1
4.5
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Chapter Four
As an example of the interpretation of these numbers, we can state that it can
be expected that once every 50 years, a storm will occur that will drop rain at
the rate of 4.5 inches per hour for 20 continuous minutes.
Runoff and Stormwater Management
The determination of runoff for small watersheds can be established by a
material balance approach. As an initial simplified example, assume that rain is
falling on a large asphalt parking lot. In this case, infiltration is zero and if we
ignore evaporation and storage in puddles, 100% of the precipitation volume
becomes runoff (output = input). An equation can be written to calculate the
runoff in volume units as follows:
R
=
PA
(4.2)
where:
R = instantaneous runoff, cfs
P = precipitation intensity, in./hr
A = watershed area, acres
To make Eq. (4.2) exactly correct with the units shown, a units conversion factor is needed:
R (cfs)
=
P (in./hr) A (acres) 1.008 (cfs-hr/acre-inch)
(4.3)
The units conversion factor (1.008) is routinely dropped from calculations so
that the runoff (in units of cfs) is easily calculated as the product of rainfall
intensity (in./hr) and the watershed area (acres).
For a land area that includes soils and plants, not all the rainfall will
become surface runoff. Some precipitation will infiltrate into the ground (and
some of that may later be evaporated or may be taken up and transpired by
plants or may seep into a nearby river or lake), some will evaporate directly
from surfaces, and some will be retained in depressions or ponds. The fraction
of precipitation that is produced as runoff is sometimes empirically shown as
the runoff coefficient, C, which is a function of the rainfall intensity and duration, and the watershed properties such as soil type, soil moisture content, land
slope, vegetation, land use, and others. Equation (4.2) can be modified to reflect
the impact of this runoff coefficient, producing the Rational Equation for estimation of runoff values. This equation is widely used for urban or small rural
watersheds to predict the peak runoff flow rate, or discharge rate:
QP
where:
QP =
C =
i =
A =
=
CiA
peak discharge (cfs)
rational method runoff coefficient (dimensionless)
rainfall intensity (inches/hour)
watershed area (acres)
(4.4)
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Water Resources
A range of values for the
runoff coefficient are reported in
Table 4.2 (Wanielista and Yousef
1993) for some small urban and
suburban areas. Use of the Rational Equation is considered valid
for small watersheds that meet
the following assumptions:
1. The rainfall intensity is
constant for the length of
time required to drain the
watershed. This length of
time is referred to as the
time of concentration.
Table 4.2
135
Runoff Coefficients
Land Use
C
Downtown Business
Neighborhood Business
Residential
single family
multi-family, detached
multi-family, attached
apartment
Light Industrial
Heavy Industrial
Parks and Cemeteries
Unimproved
0.70 to 0.95
0.50 to 0.70
0.30 to 0.50
0.40 to 0.60
0.60 to 0.75
0.50 to 0.70
0.50 to 0.80
0.60 to 0.90
0.10 to 0.25
0.10 to 0.30
Source: Adapted from Wanielista and Yousef (1993).
2. The runoff coefficient remains constant throughout the time of concentration.
3. The watershed area does not change.
Wanielista and Yousef (1993) reported that these assumptions are reasonable
for watersheds with a time of concentration less than 20 minutes. More sophisticated methods are required for analysis of flows for larger watersheds. Reasons include: land slopes and characteristics may differ significantly from one
part of the watershed to another, runoff coefficients may change greatly, there
may be a pond or lake within the watershed, and others.
EXAMPLE 4.3
A 10-acre piece of vacant land in a city is being considered for development
into an apartment complex. Using the Rational Equation, predict the peak discharge from the land prior to development and after development. At present, the land has a time of concentration of 40 minutes, but after development
that time will be reduced to 20 minutes due to site drainage improvements.
Use data from Figure 4.2, and assume a 5-year design event for this analysis.
SOLUTION
Pre-development: From Table 4.2, the runoff coefficient for an unimproved site
is approximately 0.20. From Figure 4.2, the rainfall intensity for a 5-year storm
with a duration of 40 minutes is 2.0 in./hr. The resulting peak discharge is:
Q
=
CiA
=
(0.20) (2.0 in./hr) (10 acres)
=
4 cfs
Post-development: The 5-year rainfall event with a 20-minute duration is 3.0
inches per hour, and the runoff coefficient for the developed site is estimated
to be 0.60. The new peak discharge is:
Q
=
CiA
=
(0.60) (3.0 in./hr) (10 acres)
=
18 cfs
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136
Chapter Four
Comparison of the pre- and post-development peak discharge rates determined in Example 4.3 suggests that the development of the site into an apartment complex may aggravate downstream flooding due to this substantial
increase in peak discharge relative to the pre-development conditions. For this
reason, many new proposed projects must install on-site stormwater detention
ponds to reduce peak discharge rates before the developers can get approval
for development.
Rain storms do not usually start suddenly, maintain a constant intensity,
and then stop suddenly. The “shape” of a particular rainfall event—that is, a
graph of rainfall intensity over time—is called a hyetograph. A hydrograph is
a graph showing the time history of the flow rate of water past a specific point
in space (such as a spot along a river). For any particular place, the runoff produced from a rainfall event may increase rapidly at first and then decrease
slowly over time. The steep rising limb of a runoff hydrograph can be mitigated by detention ponds. Detention ponds work well to reduce peak discharge flows because they can change the shape of the runoff hydrograph. That
is, they accumulate water during the peak intensity of the storm and release
water slowly afterwards. For any rainfall event/pond design, the pond inputs
and outputs can be calculated and an input and output hydrograph can be
drawn. If this is done via a spreadsheet, the engineer can easily change the size
of the pond until satisfactory performance is achieved. The next example utilizes our previous knowledge of numerical methods to aid in the design of a
detention pond.
EXAMPLE 4.4
A rectangular stormwater detention pond is designed to mitigate the runoff
from a development. It has dimensions of 70 feet (width) by 100 feet (length).
For ease of calculations, assume that it has vertical walls, with a depth of 20
feet. The pond receives surface runoff (Qin). For a specific event following a
prolonged dry period (the pond starts empty), the input runoff is defined as
the following function of time:
Qin = 0
if t < 0
Qin = 3 t
if 0 < t < 20
Qin = 80 – t if 20 < t < 80
Qin = 0
if t > 80
where:
Qin = surface runoff into the pond (cfs)
t
= time (minutes)
The discharge from the pond is controlled by a 90-degree V-notch weir that
allows more water to flow out as the level of water in the pond gets higher.
The discharge equation is:
Qout = K H n
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Water Resources
137
where:
Qout = discharge from the pond (cfs)
K
= weir coefficient = 5 ft2.1/sec
n
= 0.9
H
= height of water above the bottom of the weir (ft)
The elevation of the weir notch is set at 2.0 feet above the bottom of the pond
so that water will not start flowing out of the pond until the water level
reaches 2.0 feet. Using a numerical simulation method, determine the flow
rates entering and leaving the pond in response to this storm event, and create an input and output hydrograph. Also, plot the height of water in the
pond. Conduct your simulation for a period of 3 hours (more than twice the
rainfall period).
SOLUTION
The non-steady state mass balance for the pond yields the following equation:
Accumulation = Inputs - Outputs + Generation
d ( rV )
dt
=
rQin
-
rQout +
0
For this pond, the volume (V) is:
V = L×W×H
where H is the height of water above the bottom of the pond, and is a function
of time.
Thus, for constant density and vertical walls (L and W are constant with
height), the material balance equation is simplified as follows:
0.9
dH Qin - K ( H - 2.0 )
=
dt
LW
and for H < 2.0, the output term is ignored.
The finite difference equation becomes:
H new = H old + Dt
Qin - 5 ( H old - 2.0)
0.9
7 , 000
A numerical spreadsheet simulation was done using a Δt of 1 minute. The formula for calculating Qin changed at 20 minutes and again at 80 minutes; also,
a conversion from seconds to minutes was included in the calculations. The
results are shown in Figure 4.3 on the following page.
The previous discussion focused on short-term stormwater management.
Over long periods of time, being able to supply water during the dry season
might be the most important goal. An example would be the sizing of a large
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Chapter Four
Figure 4.3
Input and output hydrograph from Example 4.4.
70
60
Qin , cfs
50
Flow rate, cfs, or
Height, ft
138
40
30
Qout , cfs
20
10
Height, ft
0
0
50
100
150
200
Time, minutes
reservoir to provide drinking water to a city throughout the year. We need historical monthly rainfall data over many years to design (with reasonable certainty) a reservoir that would be big enough to store enough water during the
snow melt/rainy season (building up the water level during that time) to supply the city with adequate water during dry times (by drawing down the reservoir level).
Groundwater Hydrology
So far we have considered only what happens to water on the surface—the
runoff. We have yet to consider the infiltration. As water seeps into the ground
it migrates downward and laterally in response to gravity and geological formations. Much as water on the surface flows downhill, underground water
flows “down gradient.” Moisture and air spaces exist in unsaturated soil (the
vadose zone), which is the usual case near the surface. As depth increases, we
may come to a place where the water fills all available empty spaces. This depth
is called the water table, and the zone below it is called the saturated zone or an
unconfined (or surficial) aquifer. The water table may exist a few feet or hundreds of feet below the surface of the ground. People in rural areas often drill
shallow wells into this groundwater to supply their own drinking water.
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139
Water that is trapped between two impervious layers is called a confined
aquifer. Water in a confined aquifer originates by recharge (net infiltration from
the surface and percolation to groundwater), and may travel laterally hundreds
of miles underground. At some locations, water in a confined aquifer may be
pressurized due to the head of water that is “up gradient” from the location in
question. The head of water can be thought of as basically the height of a water
column above a reference or datum level. (Note, if external pressure is supplied
by an outside force such as a pump, water can be forced to go from a lower elevation to a higher one.) The hydraulic or piezometric head is the total of elevation
and external pressure. Water flows from a higher head to a lower head. With
enough head, water can flow on its own upward through a well that has been
drilled into a confined aquifer; these are called artesian wells. If the piezometric
surface is higher than the ground surface elevation, then an artesian well can flow
water up to ground level under its own pressure. In most cases, though, pumps
are needed to retrieve the groundwater and bring it to the surface (see Figure 4.4).
When considering the flow of water underground, the hydraulic gradient
is the difference in head at two locations divided by the difference in lateral
Figure 4.4
Schematic diagram of groundwater resources.
Recharge
Area
Infiltration
Water Table Well
Vadose
Zone
Infiltration
Water T
able
Impermeable
Layers
Unconfined
Aquifer
Confined
Aquifer
Artesian
Well
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140
Chapter Four
distance (∆h/∆L). Many places in the country obtain their water supplies from
aquifers. In central Florida, most of the drinking water supply comes from the
Floridan Aquifer, with wells reaching down to 800 feet or so. Water in such
aquifers is usually very clean because the water has traveled for many miles
through native sand and rock. It may take water hundreds of years to move a
few miles underground, depending on the hydraulic conductivity of the sand
and rock. The flow rate is proportional to the hydraulic gradient, and the
hydraulic conductivity is the proportionality constant. In future courses, you
will learn much more about these topics.
When water from an unconfined surficial aquifer is withdrawn through a
well, a cone of depression is created around the well (see Figure 4.5). This cone
provides a locally steeper hydraulic gradient to allow the water to flow faster
to the well and meet the demand of the pumped withdrawal from the well. If
too much water is pumped, the level of the water table may drop, and the well
no longer can provide water. Therefore, wells are usually drilled deep beyond
the water table, and have an extended length of screened slotted pipe where
water can flow into the well.
Figure 4.5
Cone of depression near a water well.
Water Production
Well
Ground Surface
Original Water Table
Drawdown
Radius of Influence
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141
The quality of a groundwater resource is directly linked to its hydrology.
Protection of groundwater resources from depletion and from contamination is
critical for preservation of high-quality water resources. Throughout the
United States, there have been numerous cases of contamination of shallow
groundwaters from poor practices of land disposal of toxic substances. Many
of these sites now require extremely expensive remediation. At the same time,
a potential water supply source has been lost. Contaminants move or degrade
extremely slowly in groundwater, so once a source is contaminated, it may
remain so for decades. Source protection is critical to the efficient management
of groundwater resources. We will discuss groundwater remediation in more
detail in Chapter 8.
Water quality concerns may also result simply from excessive groundwater
withdrawals. Removal of groundwater at a rate that exceeds recharge will
result in lowered water table elevations. High population densities in coastal
regions, and their demands for water, often have produced large cones of
depression or even widespread reduction of the water table. Due to these communities’ proximity to the sea, the reduction in the fresh water table has often
been accompanied by salt water intrusion into the aquifers. This phenomenon
has ruined many groundwater supplies in coastal areas, and has required the
development of surface water or inland groundwater resources at great
expense. Management of groundwater resources, including possible forced
recharge of stormwater or reclaimed water (treated wastewater effluent), may
be necessary to prevent depletion.
4.2 Surface Water Quality
Everyone has an idea of what makes a natural body of water “high quality” or “low quality.” A high-quality stream or river is clear and clean; the
water has no odor or color and has a pleasant taste. A “good” river flows fast
and freely over a clean river bed, contains high levels of dissolved oxygen, and
supports abundant fish life. It has a low bacterial count. On the other hand, a
low-quality water body is often characterized by being stagnant, looking
murky, having a foul odor and taste, and containing numerous chemicals and
bacteria. In this section, we will begin to quantify some of these ideas.
Dissolved Oxygen
One of the best measures of the quality of water in the natural environment
is its level of dissolved oxygen. Dissolved oxygen (DO) is crucial to aquatic life.
When water is in contact with air, oxygen from the air dissolves into the water.
Water can only hold a small concentration of DO, but this small concentration
is sufficient to support all fish and other aquatic life. The solubility (saturation
levels) of DO depends mainly on temperature and salinity, as shown in Table
4.3 on the following page.
The transfer of oxygen from the air to the water is a continuous rate process,
and in fact, is one of four rate processes that affect the DO level in natural water
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142
Chapter Four
Table 4.3 Solubility of Oxygen in
Fresh and Sea Water
DO saturation level, mg/L
T, °C
fresh water
sea water
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
15.0
16.0
17.0
18.0
19.0
20.0
21.0
22.0
23.0
24.0
25.0
12.8
12.4
12.1
11.8
11.6
11.3
11.0
10.8
10.5
10.3
10.1
9.9
9.7
9.5
9.3
9.1
8.9
8.7
8.6
8.4
8.2
9.8
9.6
9.4
9.2
9.0
8.8
8.6
8.4
8.2
8.1
7.9
7.7
7.6
7.5
7.3
7.2
7.1
7.0
6.8
6.7
6.6
bodies. The four processes are (1) aeration
(or reaeration), (2) photosynthesis, (3) respiration, and (4) biodegradation of wastes, as
illustrated in Figure 4.6 on the next page.
The two processes that most affect DO levels are reaeration and waste biodegradation. We will analyze these two processes
quantitatively in the last section of this
chapter, after discussing oxygen consumption due to the biodegradation of wastes.
Water Pollutants
Water pollutants can be classified into
four general categories: physical, chemical,
biological, and radiological. Individual pollutants from each of these four general categories can come from either point sources
(municipal and industrial wastewater discharges) or nonpoint sources (runoff from
urban, residential, and agricultural lands).
In many cases, nonpoint sources of pollution contribute more than point sources to
certain receiving bodies of water. The four
general categories can be subdivided into
more specific groupings, including suspended solids, dissolved solids, turbidity
Sources: Fresh water from Metcalf & Eddy (1991);
and color, taste and odor, nutrients, pestisea water from Haefner (1996).
cides, toxic metals, oxygen-demanding
matter, pathogenic organisms, pharmaceuticals, endocrine disruptors, and heat. Each
of these groups of pollutants will be discussed briefly, saving oxygen-demanding matter for the next section, where we explore that topic in detail.
One of the major groupings of water pollutants is suspended solids (particles). These may be natural soil or silt particles or waste products of various
kinds. One measure is total suspended solids (TSS), which refers to particles
that are suspended in the water and can be removed by filtering the water
through a fine filter. The particles may or may not settle by gravity, depending
upon their size and density. Large, dense particles that are carried only by the
turbulence of rapidly moving water can be settled and removed simply by
allowing the water to flow into a large tank and slowing it down. For finer particles, a chemical is sometimes added to the water to assist in the settling. There
will be more on these treatment topics in Chapter 5.
Another grouping is total dissolved solids (TDS). This group includes any
solid materials that have dissolved and cannot be removed by filtering. TDS
includes anything from table salt (NaCl) to a fertilizer (e.g., potassium nitrate—
KNO3), to an organic compound like sugar (C11H22O11), and many others. TDS
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Water Resources
Figure 4.6
143
Rate processes affecting concentration of dissolved oxygen in water.
Reaeration
Respiration
Dissolved
Oxygen
Biodegradation
of Wastes
Photosynthesis
are measured by evaporating a sample of the water and weighing the residue.
Measurements of TSS and TDS (and many other parameters) are routinely carried out in a laboratory following specific procedures (standard methods have
been published for many, many tests). A detailed description of those procedures is not necessary for this textbook, but you should realize that quality control is important to getting good data from a laboratory, and following
standard methods ensures that data from the lab can be used with confidence.
Turbidity and color are two physical parameters that affect water quality.
Turbidity measures the water clarity (which is degraded by very small suspended solids). Color is affected both by suspended and by dissolved material.
Natural waters are often colored by dissolved tannins, humic acids, and other
organic compounds that come from decaying leaves and wood, or by dissolved
minerals such as iron and manganese. Taste and odor are two other physical
properties of water, and are typically assessed for potable water (water used
for human consumption). Even after some degree of treatment, water may
have an unpleasant taste or odor. Obviously, these tests are subjective, and
require the use of human panels.
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Chapter Four
Nutrients primarily refer to nitrogen and phosphorus. These are key elements in fertilizers, and if they wash into a lake or river, they encourage the
growth of aquatic plants. In extreme cases, they can lead to severe eutrophication, and result in the waterway becoming choked with aquatic weeds. Nutrients are crucial for all life and move in cycles through the biosphere. Any
modern society depends heavily on fertilizers to produce the food required by
its population. However, problems may arise when agricultural or residential
runoff gets into surface waters.
Large-scale chicken farms located near the Chesapeake Bay produce tens
of millions of chickens and hundreds of thousands of tons of chicken manure
every year. The manure often piles up near enormous chicken coops on the
farms. Much of it is applied to agricultural lands around the Bay, but when it
rains, the runoff from the large piles of waste at these chicken farms and from
other farm fields carries enormous amounts of excess nutrients and sediments
into the Bay, polluting it heavily. In fact, it is estimated that in 2010, manure
accounted for 19% of the nitrogen and 26% of the phosphorus entering the Bay
(Chesapeake Bay Program 2013). In most cities, homeowners also contribute to
nutrient runoff, although on a much smaller scale. In many urban areas, there
are restrictions for homeowners who live on the shores of small lakes as to
what kind of fertilizers they use on their grass (usually, they must use formulations with zero phosphorus) to prevent such over-fertilization of lake weeds.
Pesticides (insecticides and herbicides) have been widely used for decades
in the United States and other countries to improve agricultural yields or simply to make homeowners’ yards more beautiful. However, the majority of any
pesticide that is broadcast widely (e.g., on crop fields) eventually finds its way
to nontarget organisms. For example, early and widespread use of DDT (a
chlorinated organic molecule that is lethal to insects) led to DDT being carried
from those fields by runoff, and resulted in an accumulation of DDT in estuaries and other water bodies. The DDT concentrated in the food chain, with the
result that the reproductive cycle of fishing birds of prey (like osprey and
eagles) were greatly disrupted, and these birds were threatened with extinction. Since pesticides and herbicides are designed to kill organisms, bacteria
cannot easily biodegrade these substances, so they persist in the environment
for years, causing extensive long-term effects.
Toxic metals that are regulated by the U.S. EPA include arsenic, barium,
cadmium, copper, lead, mercury, nickel, and others. In very small concentrations, some of those metals are essential for life, but in slightly larger concentrations, many of these metals are poisons to humans and to a number of other
organisms. Metals can enter environmental water through continuous industrial discharges to waterways over long periods of time, through large one-time
spills or accidents, through atmospheric deposition of metals that were
released into the air from combustion sources, or even through activities we
would not suspect. For example, mercury is routinely discharged in small
amounts from battery-manufacturing plants, it is exhausted into the air with
the stack gas from a coal-fired power plant, and some even comes from dentists’ offices—as old mercury amalgam fillings are replaced and flushed down
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into the city sewer system. This discharged mercury eventually finds its way
into rivers, lakes, and the ocean. Many people in the United States are exposed
to copper and lead from the leaching of these metals from water pipes and soldered joints in older potable water supply systems and older homes.
Pharmaceuticals and endocrine disruptors are examples of “modern” pollutants. By that we mean that these chemicals did not exist 100 years ago, but
were invented only recently. As such, natural bacteria in the environment have
not evolved to biodegrade these compounds, and so they can be present for
long periods of time in water supplies. Endocrine disruptors are human-made
substances that may mimic or interfere with the function of hormones in the
body. Endocrine disruptors may turn on or off, or otherwise modify, signals
that hormones carry and thus affect the normal functions of tissues and organs.
These compounds have been blamed for a variety of effects on humans, including an increase in breast and prostate cancers between 1969 and 1986, and a
40% drop in sperm count among males from 1940 to 1990 (Birnbaum 2010).
Pharmaceuticals are synthetic or natural chemicals that are found in prescription medicines and over-the-counter therapeutic drugs. Pharmaceuticals can be
introduced into water sources through sewage, which carries the excreta of
individuals who have used these chemicals, from uncontrolled drug disposal
(e.g., discarding drugs into toilets), and from agricultural runoff carrying livestock manure. They have become chemicals of emerging concern to the public
because of their potential to reach drinking water, and not being readily biodegraded by natural bacteria. The effects on people of even very small doses in
water are not fully understood.
Excess heat can be considered a pollutant in some situations. As was seen
in Table 4.3, warmer water holds less oxygen than cooler water. Thus excess
heat may warm the water so much that it reduces the oxygen content enough
to disturb the ecology of the waterway. Mostly, heat is added by industry
(especially electricity generation) when they use local water supplies to dissipate the heat from their operations. In some cases, though, heat may be beneficial; in Florida, manatees often gather at the water discharges from certain
power plants to stay warm during cold snaps.
Often, the warming occurs when a “warm” stream of water is added to a
“cool” river. When making a heat balance on streams of water, two assumptions are made that simplify the calculation. Typically, we assume constant
density and constant Cp for all streams. For no accumulation, and no chemical
reactions, the water flow balance is:
Qww + Qr = Qc
(4.5)
where:
Qww = flow rate of warm water discharge, L/s or cfs
Qr = flow rate of river, L/s or cfs
Qc = flow rate of combined stream, L/s or cfs
The heat balance equation is:
Qww ρ Cp (Tww – Tref) + Qr ρ Cp (Tr – Tref) = Qc ρ Cp (Tc – Tref)
(4.6)
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where:
ρ =
Cp =
T =
Tref =
density of water, kg/L or lb/ft3
specific heat of water, J/kg-°C or Btu/lb-°F
temperature, °C or °F
reference temperature, °C or °F
With our two assumptions, both ρ and Cp divide out of Eq. (4.6). Then when we
substitute Eq. (4.5) into Eq. (4.6) and collect terms, the Tref terms cancel out.
Thus we are left with the very simple equation:
Qww Tww + Qr Tr = Qc Tc
(4.7)
We can even leave the original flow units and temperature units as they are,
even though the units of each term might look odd (e.g., cfs-°F), and solve this
equation directly for the unknown temperature. Example 4.5 demonstrates the
kind of balances that are done to predict the potential effect of warm water discharges into a river.
EXAMPLE 4.5
An industry is discharging a stream of hot water into a river. Upstream from
the point of discharge, the river temperature is 12 °C and is flowing at 80 cfs.
The hot water has a temperature of 75 °C and is flowing at 15 cfs. What is the
temperature of the combined stream? By how much did the DO saturation
level decrease?
SOLUTION
First, make a material balance around the point where the two streams mix.
Even though the densities are slightly different due to the temperatures, we
will assume constant density and make a flow balance.
15 cfs + 80 cfs = 95 cfs
Next make the heat balance around that point and solve for the combined
stream temperature using Eq. (4.7).
15 cfs × 75 °C + 80 cfs × 12 °C = 95 cfs × Tc
Tc = 21.9 °C
Referring to Table 4.3, the DO saturation level dropped from 10.8 to 8.7 mg/L
(about 20%) due to the warming of the river.
4.3 Microbiological Decomposition of Organic Materials in Water
Natural Systems
“Oxygen-demanding wastes” is a catch-all phrase meant to identify all
those organic materials that—when they get into water—would be biode-
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147
graded by aquatic microorganisms. Over millions of years, a very robust ecology has evolved with thousands of different kinds of bacteria (and other
creatures) that utilize complex organic compounds as a food source, and at the
same time biodegrade those compounds back to CO2, H2O, and other simple
compounds. This important role of microorganisms as “decomposers” prevents massive accumulation of organic wastes in any part of our ecosystem.
During this biological process, some or all of the dissolved oxygen (DO) in the
water is consumed, and can result in little or no DO left for fish and other
aquatic life. There are literally millions of organic compounds that can be biodegraded, so rather than dealing with them individually, we lump them all
together and quantify them with one measure—how much oxygen consumption they are responsible for. The term used to describe and quantify such
wastes is biochemical oxygen demand, or BOD. To be clear, BOD is a direct
measure of oxygen usage, and only an indirect measure of the strength (concentration) of organic wastes in water. Despite BOD being an indirect measure
of the strength of organic wastes, it is universally used. Thus, the concentrations of organics in water are reported as BOD in units of mg/L.
Engineered Systems
When humans congregate in cities, one of the chief tasks of caring for their
basic needs is proper waste disposal. For the last 150 years or so, newer cities
installed (and older cities renovated) sanitary sewer systems to carry human
excreta away from living areas to areas where it could be treated and disposed.
One of the most widely used methods in developed countries is the activated
sludge treatment process (which is the topic of a later chapter). Engineers must
also design sanitary landfills to handle the city’s solid wastes, and are sometimes involved with bioremediation projects to “clean up” contaminated soils
or underground water supplies. In all these processes, one of the main goals is
to safely decompose the BOD in the wastes.
With regard to municipal wastewater, the wastewater is transported via
underground sewer pipes to a centralized wastewater treatment plant. The
plant uses microorganisms to biodegrade the wastes inside specially designed
tanks. Needless to say the tanks must be continuously supplied with air! Later,
we will learn how to make simple design calculations to size these tanks and
the other key equipment found at these plants. One measure of effectiveness of
these plants is the destruction of 90% or more of the BOD in the wastewater.
Having a full understanding of biochemical oxygen demand is very important in environmental engineering. The ultimate BOD (BODu) is the amount of
oxygen used to completely oxidize the compound. For many (but not all) compounds, this is identical to the chemical oxygen demand (COD)—the oxygen
needed to chemically oxidize the compound in water. One main advantage of
COD is that COD tests can be completed in a few hours in the lab, whereas the
determination of BODu can take weeks. A major disadvantage is that if the
wastewater contains compounds that are toxic to bacteria, or recalcitrant
(meaning they are not readily biodegradable), the COD test will not show that.
Nevertheless, for any specific organic compound, the theoretical BODu can be
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calculated by writing a chemical equation for the oxidation of that compound,
as shown in Example 4.6. One other test that is often used is the total organic
carbon (TOC), which measures the carbon content of all the organic compounds in the sample.
EXAMPLE 4.6
Calculate the COD and BODu of a solution of 200 mg/L of glucose (C6H12O6 )
in water. Assume glucose is completely biodegradable. Also calculate the TOC.
SOLUTION
Write the oxidation reaction as a chemical reaction and balance it
C6H12O6 + 6 O2
→
6 CO2 + 6 H2O
Insert the chemical masses below the reactants
C6H12O6 + 6 O2
180
192
→
6 CO2 + 6 H2O
Finally, multiply the given compound concentration by the ratio of the mass
of oxygen to the mass of compound (from the balanced chemical equation) to
get the BODu. Since glucose is completely biodegradable, the COD = BODu.
COD = BOD u = 200
mg
192 mg O 2
¥
L
180 mg glucose
= 213 mg/L
The TOC is the fraction of carbon in glucose times its concentration.
TOC =
6 (12 ) 200 mg
¥
= 80 mg/L
180
L
If the compound has nitrogen in it (as many organic wastes do), the nitrogen atom may also be oxidized. However, since this is a biological process, different types of bacteria oxidize nitrogen, so the process starts later and takes
longer. It is customary to talk about the complete oxidation process as occurring in two stages, with the first-stage oxygen usage commonly called the carbonaceous BOD (CBOD) and the second-stage referred to as the nitrogenous
BOD (NBOD). Together they make up the total BOD for the compound. During
the first stage, the carbon is oxidized to CO2 and the water to H2O, but the
organic nitrogen only gets oxidized to ammonia (NH3). In the second stage, the
ammonia is oxidized to nitric acid and water. See Example 4.7.
EXAMPLE 4.7
Calculate the ultimate CBOD, NBOD, and total BOD of a solution of 200 mg/
L of the compound C4H11O2N in water.
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149
SOLUTION
Write the bio-oxidation reaction for the first stage and balance it
C4H11O2N + 5 O2
→
4 CO2 + 4 H2O + NH3
Insert the chemical masses below the reactants and the product ammonia
C4H11O2N + 5 O2
105
160
→
4 CO2 + 4 H2O + NH3
17
Multiply the compound concentration in the sample of water by the ratio of
the mass of oxygen to the mass of compound (from the balanced chemical
equation) to get the CBODu
CBOD u =
200 mg
160 mg O 2
¥
L
105 mg compound
= 305 mg/L
Next write the equation for ammonia oxidation.
NH3 + 2 O2
17
64
→
HNO3 + H2O
Calculate the concentration of ammonia produced from the sample during
the first reaction and then the concentration of oxygen used in biodegrading
that ammonia.
NBOD u =
200 mg
17 mg NH 3
64 mg O 2
¥
¥
L
105 mg compound 17 mg NH 3
NBOD u = 121.9 mg/L
Total BOD u = 305 + 121.9 = 427 mg/L
4.4 BOD Kinetics
Because organics (often called substrate) in water are oxidized in a complex set of microbiological reactions, BOD consumption does not happen
instantaneously. It can take days or weeks to completely biodegrade some substrates. The ultimate BOD is but one number, and it is important to understand
BOD progression (kinetics) and how the BOD test is done.
First, understand that the BODu is usually thought of as the amount of oxygen that was consumed or exerted at the end of the long biodegradation process
of a particular substrate. But, keep in mind that BOD is an indirect measure of the
concentration of that substrate in the water. Thus, it can also be thought of as the
potential amount of oxygen remaining to be used before the biodegradation of the
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Chapter Four
Concentration
Figure 4.7
BOD progression.
yt
BODu
Lt
Time
substrate even starts. Once the process starts, these two concepts are still valid,
and we can apply them at any particular instant of time. At time t we give the
symbol yt to the BOD that has been used and the symbol Lt to the BOD remaining.
At any time, the BOD that has been used plus the BOD remaining to be used must
sum to the BODu as shown in Equation (4.8). As time passes, the BOD is converted
from potential (L) to exerted (y); Figure 4.7 displays this time progression of BOD.
yt + Lt = BODu
(4.8)
The BOD progression curve is a result of a rate process, which is usually
modeled with first-order kinetics as represented by Eq. (4.9).
dL
= - kd L
dt
(4.9)
where:
L = substrate concentration measured as BOD (mg/L)
t
= time (days)
kd = deoxygenation rate constant (days–1)
The differential equation (4.9) can be solved by separation of variables and
integration. Definite integrals are used, corresponding with an initial condition
(t = 0 and L = L0) and a general condition (t = t and L = Lt):
Lt
t
dL
= Ú - kd dt
L 0
(4.10)
Lt
= - kdt
L0
(4.11)
Lt = L0 e - kdt
(4.12)
Ú
L0
ln
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151
The oxygen consumed at any time during the BOD process is equivalent to the
amount of substrate (measured as BOD) that has reacted:
yt = L0 – Lt
(4.13)
where:
yt = BOD exerted at time t (mg/L)
L0 = initial BOD remaining = BODu (mg/L)
Combination of Eq. (4.12) and Eq. (4.13) yields the following general equation to define the BOD progression as a function of time:
(
yt = L0 1 - e - kdt
)
(4.14)
It is noted that the numerical value of kd reflects the ease with which the organics are degraded. For municipal wastewaters, kd can range from about 0.1 to
about 0.23 days–1 (Sawyer et al. 1994). From Eq. (4.14), it is observed that the
BOD exerted at time infinity is equal to the BOD potential at time zero and is
equal to the BODu as shown in Eq. (4.15).
y• = L0 = BOD u
(4.15)
Of all the times in the BOD progression, the five-day BOD exerted (y5 , or
more commonly, BOD5) is the one most often used in testing. There are several
reasons (and some urban legends) explaining how and why the five-day period
of time was picked, but the point is that BOD5 is widely recognized throughout
the world, and five days is the standard time for doing BOD tests. The BOD5 is
also used to set regulatory limits for wastewater treatment plant discharges.
The BOD test is a batch bioassay analysis that measures the oxygen content
of the water at the start of the test and again after five days to determine the
amount of oxygen used. The test is conducted in a closed bottle stored in a
darkened area to prevent oxygen transfer from the atmosphere and exposure to
light (to exclude photosynthesis). The temperature is maintained at 20 °C to
ensure standardization of the results. The sample of the wastewater to be tested
is diluted with seeded dilution water, which contains a viable bacterial seed
and all necessary inorganic nutrients to support the desired biological reaction.
The dilution water does not contain any organic materials, so that the only
source of substrate in the diluted sample is from the water to be tested. The
dilution water must be aerated prior to use to ensure an adequate supply of
oxygen to maintain aerobic conditions throughout the BOD test. That is, for the
test to be valid, the bottle must have at least 1.0 mg/L of dissolved oxygen left
in it after five days.
At the beginning of the batch test, a measured volume of the sample is
placed in a BOD bottle which is then filled with seeded dilution water and
sealed to prevent oxygen transfer from the atmosphere. The dissolved oxygen
is measured carefully at the beginning and end of the test (five days for BOD5).
The difference in the dissolved oxygen concentrations represents the oxygen
consumption. The biological reactions may require in excess of 30 days to reach
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Chapter Four
completion, in which case the five-day oxygen consumption (BOD5) would
represent a fraction of the BODu.
The BOD of the wastewater can be calculated by mass balance on BOD in
the bottle.
Vww BODww + Vdil BODdil = Vbot BODbot
(4.16)
If the dilution water has no substrate in it, then its BOD is zero, and Eq. (4.16)
becomes
BOD ww =
Vbot
BOD bot
Vww
(4.17)
The ratio of the sample volume (Vww) to the bottle volume (Vbot) is called the
dilution factor (P), so Eq. (4.16) is often written as
BOD ww =
BOD bot
P
(4.18)
The BOD of the bottle is calculated by the difference in oxygen levels after five
days.
BODbot = DO0 – DO5
(4.19)
Example 4.8 demonstrates how the BOD test is used, and Example 4.9 demonstrates how knowledge of the BODu and BOD5 can be used to calculate the rate
constant.
EXAMPLE 4.8
A 300-mL BOD bottle was loaded with 5 mL of municipal wastewater and
filled to the top with 295 mL of dilution water. The initial and 5-day dissolved
oxygen concentrations were 9.02 mg/L and 4.13 mg/L, respectively. Determine the BOD5.
SOLUTION
The dilution factor is calculated as the ratio of the sample volume (5 mL) to
the bottle volume (300 mL). The BOD5 of the sample is calculated as the oxygen consumption in the BOD test bottle divided by the dilution factor:
BOD 5 =
9.02 mg/L - 4.13 mg/L
= 293 mg/L
5 mL
300 mL
EXAMPLE 4.9
The glucose solution from Example 4.6 was tested to determine the 5-day
BOD. The BOD5 was 175 mg/L. Determine the value, with appropriate units,
for the rate constant kd.
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153
SOLUTION
Equation (4.14) can be rearranged to solve for the rate constant:
Ê
Ê
175 mg/L ˆ
y ˆ
- ln Á 1 - t ˜ - ln Á 1 213 mg/L ˜¯
L
Ë
Ë
0¯
kd =
=
= 0.345 day -1
t
5 days
Glucose is easily biodegraded; other compounds may be much more difficult, meaning that the rate constant is smaller and the reactions take longer.
Comparing just the 5-day BOD for two wastewaters may not be sufficient to
characterize the potential ultimate impacts of those wastes. Example 4.10 illustrates this point.
EXAMPLE 4.10
Two samples of wastewater are being compared; one is from a domestic
source and the other from an industrial source. The values of kd were determined previously to be 0.35 for the domestic wastewater and 0.15 for the
industrial wastewater. Both samples test out with equal values of BOD5 (150
mg/L). Calculate the BODu for each sample.
SOLUTION
For the domestic wastewater, solving Eq. (4.14), we get
L0 =
150
1- e
-(0.35)( 5)
=
150
= 181.5 mg/L
0.826
=
150
= 284.3 mg/L
0.528
For the industrial wastewater,
L0 =
150
1- e
-(0.15)( 5)
The industrial wastewater would require considerably more oxygen to be
completely biodegraded in the long run.
The BOD test is widely used to characterize the organic content in wastewaters and to establish criteria for effluent discharge into the environment.
Typical municipal wastewaters contain 200 to 300 mg/L of BOD5 (Metcalf &
Eddy 1991). Typical effluent standards for a treated municipal wastewater
effluent range from 5 to 30 mg/L of BOD5. The BOD test has proved to be quite
useful as a measure of substrate in the design and operation of wastewater
treatment facilities, especially for domestic and readily degradable industrial
wastewaters (for example, food-processing wastewaters). However, use of
BOD testing has certain disadvantages, including:
• Simultaneous oxidation of inorganic compounds, which also exerts an
oxygen demand, resulting in falsely high BOD values.
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• Typically, BOD reactions do not go to completion within the normal fiveday test period. Knowledge of the rate constant is required to relate fiveday and ultimate BOD values.
• The test period is long (five days); consequently, the results have almost
no value for real-time process control.
• The presence of toxic materials will interfere with biological activity during the test, resulting in falsely low BOD values. Many industrial wastewaters contain compounds which may be inhibitory or toxic; therefore
BOD data may not be reliable for these wastewaters.
• Many industrial wastewaters require acclimated seed organisms to give
valid BOD results.
The previously mentioned chemical oxygen demand (COD) test may be
used as an alternative to the BOD test for quantification of substrate. There is
one fundamental similarity between the two analytical tests; both tests quantify the oxidation of the material present in an environmental sample. In the
BOD test, the quantity of oxygen required during aerobic biological oxidation
is measured. Under COD test conditions, potassium dichromate (K2Cr2O7)
serves as the oxidizing agent (electron acceptor). The quantity of organic material can be expressed in terms of the equivalent amount oxygen as electron
acceptor. Because very strong oxidizing conditions are provided during the
COD test (elevated temperature, 70% sulfuric acid, etc.), complete oxidation of
most organic compounds to carbon dioxide is achieved within several hours.
The COD test offers some advantages over the BOD test for quantification
of substrate, particularly for industrial wastewaters that may contain toxic
compounds. The COD test can also be completed in a matter of hours, and a
stoichiometric endpoint is reached so that knowledge of a rate constant is
unnecessary. Unfortunately, the COD test is not able to distinguish between
organic materials that are biodegradable and those compounds which are
either toxic or recalcitrant (recall that recalcitrant compounds are not toxic but
are not biodegradable). In addition, NH3 is not oxidized during a COD test,
and some inorganic compounds may be oxidized during the test. These limitations of the COD test are significant when trying to describe biological treatment processes, since a successful measurement of substrate must not be
influenced by the presence of compounds such as toxic organics, recalcitrant
organics, or inorganic species that are not suitable as an energy source for the
organisms. Nevertheless, COD is an oft-used test.
Dissolved Oxygen in Rivers and Lakes
When a wastewater is discharged into a lake or river, it can have serious
implications for oxygen depletion in that water body. Microbiological processes are often “first in line” to consume oxygen, leaving little oxygen for fish
and other aquatic life. In Figure 4.6, we previously saw schematically how four
rate processes occur simultaneously and affect DO levels in water. In this section, we will quantify the effects of a wastewater discharge into a river.
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155
Considering only aeration and biodecomposition of organic wastes as the
two primary processes that affect DO, a model of the DO concentration in the
river at all points downstream from a wastewater discharge can be developed.
It is assumed that the wastewater discharge is introduced into the river and
complete mixing occurs within a very small region (in fact, the mixing is
assumed to occur instantaneously at a point) of the river. It is further assumed
that all data about the wastewater discharge and the river prior to the point of
discharge are known. The concentrations in the mixed stream (river plus
wastewater) of BODu (L, mg/L) and of dissolved oxygen (DO, mg/L), temperature (T, °C), and flow rate (Q, cfs) are determined by steady-state material and
energy balances around the mixing zone. In the following balance equations,
the subscripts R, W, and 0 refer to the river just upstream from the point of discharge, the wastewater discharge (effluent), and the mixed river immediately
downstream from the point of discharge, respectively. The subscript 0 can be
thought of as denoting the point in time or distance that the combined stream
(river plus wastewater) starts. Thus the 0 subscript indicates that moment after
complete mixing but before any reaction or flow away from the mixing point.
For a flowing river, time and distance are related by the velocity of the water as
shown in Eq. (4.20); note that any consistent set of units may be used. A schematic diagram of the system is presented in Figure 4.8.
x = ut
(4.20)
where:
x
= distance downstream from the mixing point, km
u = mixed river water velocity, km/day
t
= time after mixing, days
The flow balance becomes:
Q0 = QR + QW
(4.21)
The BODu balance becomes:
L0 =
Figure 4.8
Mixing zone in a river.
QR LR + QW LW
QR + QW
(4.22)
River
QR DOR
SR TR
Mixture
Q0 DO0
S0T0
Effluent
QW DOW
SW TW
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Chapter Four
The dissolved oxygen balance becomes:
DO0 =
QR DOR + QW DOW
QR + QW
(4.23)
The heat balance becomes:
T0 =
QRTR + QW TW
QR + QW
(4.24)
Just considering the oxidation of organics, the first-order rate is:
dL
= - kd L
dt
(4.25)
Because the concentration of organics is expressed as BODu, the rates of oxygen
depletion and oxidation of organic materials are equivalent:
d (DO ) dL
=
= - kd L
dt
dt
(4.26)
To facilitate solution of the subsequent differential equations, it is useful to
introduce the concept of a dissolved oxygen deficit. The deficit is defined as
the difference between the concentration of dissolved oxygen at saturation
(that is, the concentration in water in equilibrium with the atmosphere) and the
actual concentration in the water. Saturation values are sensitive to temperature and salinity and were provided in Table 4.3.
D(t) = DOsat – DO(t)
(4.27)
where:
D(t) = dissolved oxygen deficit at any time after mixing (mg/L)
DOsat = saturation dissolved oxygen concentration (mg/L)
Taking the derivative of Eq. (4.27), and recognizing that DOsat is a constant, we
see that
dD
dDO
=dt
dt
(4.28)
Thus, Eq. (4.26) may be rewritten:
dD
= kd L
dt
(4.29)
The transfer of oxygen from the atmosphere into the surface water is
referred to as aeration or reaeration. This rate of oxygen transfer is assumed to
follow a first-order mass transfer rate relationship in terms of the dissolved
oxygen deficit. Just considering only the reaeration effects:
dD
= - kaD
dt
(4.30)
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157
where:
ka = reaeration rate constant (days–1)
The overall effect on the deficit of both processes happening simultaneously is given by combining Eqs. (4.29) and (4.30):
dD
= kd L - k a D
dt
(4.31)
Solution of the differential equation in Eq. (4.31) requires specification of the
initial condition for deficit (D0) and ultimate BOD (L0). Again, in these equations, time = zero corresponds with the conditions leaving the mixing zone.
Positive values of time correspond with the travel time downstream from the
point of discharge. The initial deficit is found using Eqs. (4.23) and (4.27):
D0 = DOsat – DO0
(4.32)
The differential equation in Eq. (4.31) may be solved analytically to provide
a general equation for prediction of the deficit at downstream locations (times)
from the point of discharge. This equation was first solved about 100 years ago,
and is known as the Streeter-Phelps Equation:
D (t ) =
kd L0 È - kdt - kat ˘
e
-e
+ D0 e - kat
˚
k a - kd Î
(4.33)
Once D is known at any time (distance) we can calculate DO at that same distance. Recall that for a river flowing at a constant velocity (u), the travel time (t)
and distance downstream (x) are related as noted previously in Eq. (4.20).
The resulting dissolved oxygen profile downstream from the point of discharge is known as an oxygen sag curve; typical sag curves are presented in
Figure 4.9. In general, when we inject a large amount of BOD into a river, we
expect the DO in the river to drop sharply at first, and then to decline for some
distance downstream. If the waste load is not too high, after decomposition of a
portion of the organic material, the rate of deoxygenation slows while the
reaeration rate increases (in response to larger deficits). At some point, the two
rates become equal and the dissolved oxygen reaches a minimum value. This is
called the critical point. At locations downstream from the critical point, the
rate of reaeration exceeds the rate of deoxygenation, and the dissolved oxygen
levels recover, ultimately approaching saturation values. If the minimum DO
concentration drops below some critical value (in many states the critical value
is set at 5 mg/L), survival of desired sport fish species may be impossible. If the
concentration of organic material is excessive, dissolved oxygen levels may
approach zero, resulting in anaerobic (without oxygen) conditions and foul
odors for a portion of the river, as indicated in curve (b) in Figure 4.9 on the following page.
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Chapter Four
Figure 4.9 Dissolved oxygen sag curves: (a) response to moderate waste load; (b)
response to heavy waste load.
DOsat
(a)
(b)
Deficit (D)
DO
DOmin
Critical Point
anaerobic
tc
Time or Distance
The time to reach the critical point is determined by differentiating Eq.
(4.33), setting the derivative equal to zero, and solving for the critical time.
tc =
Èk Ê
D (k - k ) ˆ ˘
1
ln Í a Á 1 - 0 a d ˜ ˙
kd L0
(ka - kd ) ÍÎ kd Ë
¯ ˙˚
(4.34)
where tc = critical time (time to reach minimum DO concentration) (days).
The DO concentration at the critical time can be obtained by substituting tc into
Eq. (4.33), solving for the deficit at that time, and then subtracting the deficit
from the saturation value to get the DO at that point.
The rate constants for decomposition of organics and for reaeration are
functions of temperature (Metcalf & Eddy 1991):
kdT = kd20 (1.135 )(
T - 20)
kdT = kd20 (1.056 )(
T - 20)
where T £ 20 ∞C
(4.35)
where 20 ∞C < T £ 30 ∞C
(4.36)
k aT = k a20 (1.024 )(
T - 20)
(4.37)
Equation (4.35) is valid for temperatures from 4 to 20 °C, and Eq. (4.36) is valid
for temperatures from 20 to 30 °C. In these equations, a subscript of T indicates
the rate constant at T °C, so a subscript of 20 indicates the rate constant at the
reference temperature of 20 °C. Therefore, in solving this type of problem, engineers must first do a heat balance to determine the temperature of the mixed
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159
stream. In general, determination of the dissolved oxygen in a river downstream from the point of discharge of a wastewater requires completion of the
following calculations (Dietz 1996):
1. The conditions at mixing point (where the discharge meets the stream)
are characterized. A mass balance is completed to determine the initial
BODu and the initial DO. A heat balance is completed to determine the
initial temperature. All BOD5 values are converted to BODu values.
Flow, temperature, dissolved oxygen, and ultimate BOD must be known
for the effluent and for the river upstream from the point of discharge.
2. Based on the temperature in the river after discharge, the DOsat concentration is determined and the initial deficit is calculated. Also, values are
determined for the deoxygenation coefficient, and the reaeration coefficient at the temperature of the combined stream.
3. For a known value of time, the deficit is calculated (this can be done for
multiple times using a spreadsheet to determine the oxygen sag curve).
4. To determine the minimum DO concentration, the maximum deficit is
determined at the critical time, and then subtracted from DOsat.
The value for the deoxygenation coefficient may be specific to the wastewater. The reaeration coefficient is highly dependent on local river conditions
(depth and velocity), with a range of values reported from 0.10 per day for
small ponds to 1.15 per day for swift streams (Metcalf & Eddy 1991). Site-specific values for these coefficients are needed for an accurate evaluation of
receiving water dissolved oxygen.
In many practical situations, DO sag calculations are completed by state
regulatory agencies to determine an allowable concentration of BOD5 in an
effluent discharge that will preserve water quality criteria (typically 5 mg/L of
dissolved oxygen). For this wasteload allocation exercise, the concentration of
BOD5 in the effluent is unknown. An iterative calculation procedure is necessary in which successive values of the effluent BOD5 are assumed until the calculated minimum dissolved oxygen concentration is satisfactory. A discharge
permit is then granted for operation of a wastewater treatment facility. Example 4.11 illustrates the numerous but straightforward calculations that are
required for DO sag curve problems.
EXAMPLE 4.11
Determine the minimum dissolved oxygen concentration in a river downstream from a municipal treatment facility discharging at a flow rate of 5.00
MGD. The plant achieves secondary treatment standards (BOD5 = 30 mg/L).
The wastewater temperature is 25.0 °C and the wastewater dissolved oxygen
is 4.0 mg/L. The characteristics of the river upstream from the point of discharge are Q = 20.0 cfs, T = 20.0 °C, BODu = 3.0 mg/L, and DO = 7.0 mg/L.
The deoxygenation coefficient for the wastewater is 0.17 days–1 at 20 °C; the
reaeration coefficient for the river is 0.40 days–1 at 20 °C.
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SOLUTION
The initial conditions (ultimate BOD, DO, and temperature) in the river
immediately after the mixing point must be determined using mass and
energy balances. Any BOD5 values must first be converted to ultimate BOD
using Equation (4.14).
Qmix = 5.00 MGD + 20.0 cfs × 7.48 gal/ft3 × 86,400 s/day
= 5.00 + 12.93
= 17.93 MGD
Tmix =
5 MGD ( 25 ∞C) + 12.93 MGD ( 20 ∞C )
17.93 MGD
= 21.4 ∞C
To get the BODu, we use the kd given at the stated temperature.
BOD u of wastewater BOD u =
=
BOD 5
1- e
-0.17( 5)
30
= 52.4 mg/L
0.573
Note that BODu does not vary with temperature.
BOD u-mix =
5 MGD ( 52.4 mg/L ) + 12.93 MGD ( 3.0 mg/L )
17.93 MGD
= 16.8
8 mg/L
DO mix =
5 MGD ( 4.0 mg/L ) + 12.93 MGD (7.0 mg/L )
17.93 MGD
= 6.16 mg/L
The rate coefficients must be corrected to the final temperature of the river
after mixing:
kd = (0.17 days–1) (1.056)(21.4 – 20) = 0.18 days–1
ka = (0.40 days–1) (1.024)(21.4– 20) = 0.41 days–1
The dissolved oxygen saturation concentration at 21.4 °C is 8.82 mg/L (by
interpolation within Table 4.3).
The initial deficit is calculated:
D0 = 8.82 – 6.16 = 2.66 mg/L
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161
The critical time is calculated with Eq. (4.34):
tc =
È 0.41 Ê
2.66 (0.41 - 0.18 ) ˆ ˘
1
ln Í
1˙ = 2.60 days
Á
(0.41 - 0.18) ÍÎ 0.18 Ë
(0.18)(16.8) ˜¯ ˙˚
The maximum deficit is calculated by substituting the critical time into Eq. (4.33):
D=
(0.18)(16.8) È e-(0.18)(2.60) - e -(0.41)(2.60) ˘ + 2.66e -(0.41)(2.60)
0.41 - 0.18 ÎÍ
˚˙
= 4.62 mg/L
The minimum dissolved oxygen is:
DO = DOsat – D = 8.82 – 4.62 = 4.2 mg/L
SUMMARY
This chapter has presented information on water resources and on water
pollutants. Water resources include both surface and groundwater, and both
are important as sources of drinking water for society. Many types of water
pollutants were discussed, but we focused much attention on BOD. This focus
was necessary owing to BOD’s widespread use as a measure of the strength of
organic water pollutants, and its use by engineers throughout the world in the
characterization, regulation, and treatment of wastewater.
PROBLEMS
4.1
A precipitation event produced 2.0 inches of rainfall in two hours. Using
the rainfall intensity/duration/frequency chart in Figure 4.2, determine
the annual probability of occurrence of this event.
4.2
A precipitation event produced 3.0 inches of rainfall in three hours. Using
the rainfall intensity/duration/frequency chart in Figure 4.2, determine
the annual probability of occurrence of this event.
4.3
A 20-acre site is undeveloped. Using the rainfall data from Figure 4.2,
estimate the peak discharge for a 25-year storm if the time of concentration for the watershed is 30 minutes. State all assumptions.
4.4
The rainfall intensities in Problems 4.1 and 4.2 are identical (1.0 in. per
hour). Would the recurrence interval be the same for each event? Why or
why not?
4.5
Assume that the 20-acre site in Problem 4.3 has been developed as single
family residential. Estimate the peak discharge for a 25-year storm.
Assume that the drainage improvements which occur during site devel-
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opment have decreased the time of concentration for the watershed from
30 to 20 minutes.
4.6
Repeat Example 4.4 using a smaller stormwater pond, but which has a
different exit weir equation. Assume that the dimensions of the pond are
50 feet (width) by 75 feet (length), and that the outflow equation is Qout =
K H n with K = 8 and n = 1.5. All other system parameters remain the
same. Using a spreadsheet, create a graph similar to Figure 4.3. Does this
pond do much good? Why or why not?
4.7
It rains at the rate of 0.5 in./hour for 2.5 hours onto a paved 2-acre parking lot. The parking lot stores water in small puddles on its surface; the
total storage is equivalent to the whole parking lot surface area one-half
inch deep. Calculate the runoff volume from this parking lot for this rain
event. Give your answer in cubic feet.
4.8
Repeat Example 4.4 assuming the output discharge is controlled by a rectangular weir with a length of 10 feet. The outflow equation is:
Qout = C L (H – 2) n
where:
Qout = discharge (cfs)
C
= weir coefficient = 1.20 ft1.2/sec
L
= weir length (ft)
H
= height above bottom of the pond (ft)
n
= 0.8
Display your results in a figure similar to Figure 4.3.
4.9
A lake with surface area of 10,000 m2 sits in the middle of a watershed of
100,000 m2 (including the lake area). A rainfall event occurs that drops 12
cm of rain onto the entire area. Of the rain that falls on the ground, 20%
infiltrates. Assume that evaporation is about zero over the short time of
the precipitation and runoff, and that surface temporary storage is about
zero. Calculate the increase in lake volume.
4.10 An industrial wastewater was analyzed to determine its BOD5. Four different dilutions were completed, with initial and final dissolved oxygen
concentrations as noted below. Determine the BOD5 concentration in the
wastewater.
Dilution Factor
1.0
0.1
0.01
0.001
Initial DO (mg/L)
9.20
9.20
9.20
9.20
Final DO (mg/L)
0.00
0.00
2.05
8.47
4.11 Determine the CBODu , NBODu , and total BODu of a 100 mg/L solution
of alanine (C3H7NO2).
4.12 Determine the CBODu , NBODu , and total BODu of a 100 mg/L solution
of ammonia (NH3).
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4.13 A biodegradable industrial (pulp and paper) wastewater has a COD of
600 mg/L. If the BOD progression follows first-order kinetics with a rate
constant = 0.10 day–1, determine the BOD5.
4.14 A biodegradable industrial (petrochemical) wastewater has a COD of 600
mg/L. If the BOD progression follows first-order kinetics with a rate constant = 0.20 day–1, determine the BOD5.
4.15 A biodegradable industrial (pharmaceutical) wastewater has a COD of
600 mg/L. If the BOD progression follows first-order kinetics with a rate
constant = 0.30 day–1, determine the BOD5.
4.16 A biodegradable industrial (pulp and paper) wastewater has a BOD5 of
600 mg/L. If the BOD progression follows first-order kinetics with a rate
constant = 0.10 day–1, determine the BODu.
4.17 A biodegradable industrial (petrochemical) wastewater has a BOD5 of 600
mg/L. If the BOD progression follows first-order kinetics with a rate constant = 0.20 day–1, determine the BODu.
4.18 A biodegradable industrial (pharmaceutical) wastewater has a BOD5 of
600 mg/L. If the BOD progression follows first-order kinetics with a rate
constant = 0.30 day–1, determine the BODu.
4.19 An industrial wastewater has a 5-day BOD of 370 mg/L and a 10-day
BOD of 500 mg/L. If the BOD progression follows first-order kinetics,
determine the ultimate BOD.
4.20 Repeat Example 4.11 except change the wastewater characteristics as follows: Q = 5.00 MGD, BOD5 = 28 mg/L, DO = 2.0 mg/L, T = 28 °C.
4.21 Consider Example 4.11. Use a spreadsheet to solve the Streeter-Phelps equation for numerous times, and develop the oxygen sag curve for this situation.
4.22 Consider Problem 4.20. Use a spreadsheet to solve the Streeter-Phelps equation for numerous times, and develop the oxygen sag curve for this situation.
4.23 Excess irrigation water runs off a farmer’s field and carries fertilizer and
other pollutants. The field runoff has the following characteristics: flow =
20 m3/min and total suspended solids (TSS) content = 400 mg/L. It combines with a hog pen wash-out stream, with a flow of 5 m3/min and TSS
of 800 mg/L. What is the TSS content of the combined stream (mg/L)?
4.24 The answer to Problem 4.23 is 480 mg/L. A settling basin is built to separate out most of the TSS from the combined stream before it flows into a
nearby river. When the settling basin is operating at steady state, it discharges two streams—a “clean” one to the river and a “dirty” one to a
disposal lagoon. The concentration of TSS in the clean stream going into
the river is 80 mg/L and the concentration of TSS in the dirty stream from
the settling basin to the lagoon is 4,000 mg/L. Calculate the flow rates of
the two discharge streams (m3/min) from the settling basin.
4.25 An agricultural runoff diversion system discharges into a small river. The
runoff has the following characteristics: flow = 100 m3/min, dissolved
oxygen = 3 mg/L, and BOD = 150 mg/L. The river flows at 1,000 m3/min
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Chapter Four
and has zero BOD and 9 mg/L of dissolved oxygen. Given that saturation
DO is 9.5 mg/L, the kd is 0.4 day–1, and the ka is 0.7 day–1, calculate the DO
and the BOD remaining in the stream 1.5 days after the discharge event.
4.26 A river flows steadily at 50 m3/minute into a bay that has a total volume of
85,000 m3. The bay then empties into the ocean through a narrow channel.
The river water has a normal concentration of substance x of 10 mg/L. A
food-processing plant discharges a small stream of polluted water into the
river, but only during the processing season (the months of Nov., Dec., Jan.,
Feb., and March). The plant discharge flows steadily at 5 m3/minute during
those months, and has a concentration of x of 4,000 mg/L. Downstream of
where the processing plant discharges into the river, but before the river
reaches the bay, some river water is withdrawn to irrigate a golf course. The
withdrawal rate averages 3 m3/min, all year. Draw a simplified flow diagram to represent this word problem. Assuming no chemical reactions of x
in the river, calculate the concentration of x in the water that enters the bay
during the processing season. Also calculate the mass loading of x into the
bay during the season, kg/day.
4.27 A 600 MW coal-fired power plant is 40% efficient at producing electricity.
Assume that 80% of the waste heat is removed by cooling water that flows
into the heat exchanger at 25 °C and out at 40 °C. Calculate the flow rate of
water in kg/day and in m3/min. The Cp for water is 4.19 kJ/kg-°C.
4.28 The answer to the above problem is 9.9 (10)8 kg/day. This once-through
cooling water flows into a river that is flowing at 500 m3/s and is at a temperature of 20 °C. By how much does the river water temperature rise?
4.29 A compound C5H12N2O3 can be completely biodegraded by aerobic bacteria. A sample of polluted water contains 75 mg/L of this compound.
First, write the balanced chemical equation depicting the first stage of this
process, and calculate the CBODu for this sample of water. Next, write the
equation for the subsequent oxidation of ammonia, and calculate the
NBODu. Finally, what is the total BODu? Give your answers in mg/L.
4.30 A wastewater is discharged into a river. The wastewater flow rate is 500
L/s; it contains 2.0 mg/L of dissolved oxygen (DO) and 50 mg/L of BOD,
and is at 90 °F. The river (upstream of the discharge point) flows at 5,000
L/s; it contains 7.5 mg/L of DO and 6.0 mg/L of BOD, and is at 70 °F.
a. Calculate the BOD in the river just downstream from the mixing point,
in mg/L.
b. If the DO concentration in the river water at saturation is 8.5 mg/L, calculate the DO deficit just after the wastewater flows into and mixes with
the river. Give your answer in units of mg/L.
4.31 Fifteen (15.0) mL of a sample of wastewater was put into a 300 mL BOD
bottle, and then the bottle was filled to its full volume of 300 mL. The initial dissolved oxygen (DO) was 9.3 mg/L and five days later the DO was
5.4 mg/L.
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165
a. What is the BOD5 of this wastewater, in mg/L?
b. If the BOD decay rate constant is 0.25 day–1, what is the ultimate BOD,
in mg/L?
4.32 Solve this problem via spreadsheet. A new reservoir (lake) is being built;
it will have a surface area of 2.0 square miles, and a maximum average
depth of 18 feet. If the lake starts receiving inflow on Jan. 1, in what month
will it fill? Evaporation from the lake is constant at 0.1 inches per day, and
the outflow seepage rate in cubic feet/day is given by 2,500 H0.5 (where H
is height of water above the bottom, in ft). The monthly average rainfall
data, assumed to fall evenly over the whole watershed of 20 square miles
(including the lake) is as given below. The runoff from the land area of the
watershed into the lake is 45% of the precipitation. If necessary, assume
the rainfall pattern repeats every year.
Jan
Feb
Mar
Apr
2.0 inches/month
1.5 inches/month
2.0 inches/month
3.0 inches/month
May
June
July
Aug
4.5 inches/month
6.5 inches/month
9.0 inches/month
8.0 inches/month
Sept
Oct
Nov
Dec
7.0 inches/month
5.0 inches/month
3.5 inches/month
1.5 inches/month
REFERENCES
ASCE/WPCF. 1969. Design and Construction of Sanitary and Storm Sewers. New York and
Washington, DC: American Society of Civil Engineers and Water Pollution Control
Federation.
Birnbaum, L. 2010. “Testimony on Endocrine Disrupting Chemicals in Drinking Water:
Risks to Human Health and the Environment.” National Institutes of Health.
Accessed June 2013. http://www.hhs.gov/asl/testify/2010/02/t20100225a.html.
Chesapeake Bay Program. 2013. “Agriculture.” Accessed July 2013.
http://www.chesapeakebay.net/issues/issue/agriculture.
Dietz, John D. 1996. “Wastewater Treatment.” In Environmental Engineering: P.E. Examination Guide & Handbook, W. Christopher King (Ed.). Annapolis, MD: American
Academy of Environmental Engineers.
Dominguez, R. 1996. “Hydraulics and Hydrology.” In Environmental Engineering: P.E.
Examination Guide & Handbook, W. Christopher King (Ed.). Annapolis, MD: American Academy of Environmental Engineers.
Haefner, P. A. 1996. Exploring Marine Biology—Laboratory and Field Exercises. New York:
Oxford University Press.
Metcalf & Eddy, Inc. 1991. Wastewater Engineering. 3rd ed. New York: McGraw-Hill.
Sawyer, C. N., P. L. McCarty, and G. F. Parkin. 1994. Chemistry for Environmental Engineers. 4th ed. New York: McGraw-Hill.
Wanielista, M. P., and Y. A. Yousef. 1993. Stormwater Management. New York: John Wiley
& Sons.
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CHAPTER
5
Potable Water Treatment
5.1 Source Water Characteristics
In the previous chapter, both surface water and groundwater resources
were discussed. Either or both types of water supplies can be the source of
drinking water for an individual homeowner or for a large municipality. For
example, the drinking water source for Chicago, Illinois, is Lake Michigan, and
that for Orlando, Florida, is the Floridan aquifer (a porous limestone formation
800 feet below ground). To serve as a supply for municipal potable (drinkable)
water, the source must be plentiful, secure, and economical. By plentiful we
mean that the supply is adequate to meet the city’s water demands, including
the daily, weekly, or seasonal fluctuations, without danger of shortages. By
secure we mean the source is not readily subject to willful or accidental contamination, is not subject to arbitrary curtailment for political or economic reasons,
and such conditions are likely to remain in effect for years to come. By economical we mean that the supply can be treated to acceptable quality and distributed at reasonable cost to all the customers of the water utility or company.
The flow of water from the environment (from any source—e.g., a lake, a
river, an aquifer) to users in a city, and the flow of wastewater from the users
back into the environment is depicted in Figure 5.1. Two types of treatment
plants are shown in this diagram, and although some of the unit operations
they use are similar, the two plants have very different goals. The first type of
plant (potable water treatment) may use any of the different processes that take
environmental quality water and bring it up to potable (drinkable) water standards. We call those facilities simply water treatment plants (or drinking water
treatment plants). The second type of plant (wastewater treatment) may use
any of the different processes that treat municipal wastewater to bring it up to
acceptable quality for safe discharge back into the environment (e.g., a lake, a
river, or the ocean). We call those facilities wastewater (or sewage) treatment
plants. This chapter focuses on potable water treatment, and the next chapter
deals with wastewater treatment.
The specific designs of potable water treatment plants throughout the
country (and indeed the world) depend heavily on (1) knowing where you
166
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Potable Water Treatment
Figure 5.1 Flow
of water from the
environment to
users and back.
167
Potable Water
Treatment
Water
Users
in the
Environment
Wastewater
Treatment
are—that is, knowing the source water characteristics, and (2) knowing where
you want to go—that is, the potable water standards that you are trying to
meet. Standards for potable water treatment are much the same throughout the
country, although a few states may have stricter standards for specific contaminants. For example, Florida maintains a lower standard for sodium to help prevent high blood pressure among the state’s elderly (a large percentage of the
total population). The characteristics of source waters can vary widely; surface
waters are much different from groundwaters, and there can be wide variations within each category.
In general, in comparison with groundwater supplies, surface water supplies tend to be more variable (both in quantity and quality). They tend to have
more suspended solids, more organics, more dissolved oxygen, and more bacteria. They are more easily contaminated by an accidental spill or by polluted
runoff; but on the other hand, are more easily cleaned. An underground water
source, once contaminated, may take years or decades to remediate. Groundwaters tend to have more dissolved solids, especially calcium and magnesium
if they are located within a limestone aquifer. Recall that calcium and magnesium are the primary ions contributing to hardness. Groundwaters also often
contain dissolved gases like carbon dioxide and hydrogen sulfide, and have
nearly zero dissolved oxygen. More details comparing these two types of water
sources are offered in the next two paragraphs.
Surface water sources (lakes, rivers, reservoirs) often have high color associated with naturally occurring organic material (such as leaves and other debris),
as well as high turbidity (organic and inorganic suspended solids). Some surface waters have low concentrations of total dissolved solids (TDS), such as in
mountain streams or lakes, while brackish or salty waters may have very high
concentrations of TDS. River water characteristics may exhibit pronounced seasonal variations, such as wet season and dry season flows or increased sediment
load following precipitation and runoff events, and often carry more nutrients
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Chapter Five
and pesticides during farming season. Lakes and reservoirs may exhibit algal
blooms in the summer and stratification in the winter. These sources commonly
have a high potential to form disinfection by-products (more on this in the next
section) due to the elevated organic concentration. Treatment objectives focus
on removal of turbidity and disinfection. Removal of organic compounds prior
to chlorination, or the use of alternative disinfectants (other than Cl2), may be
necessary to minimize formation of disinfection by-products.
Groundwater sources often exhibit great spatial variability in composition
from one place in the country to another, although the water quality from an
individual aquifer is highly consistent. Even at one geographic location,
though, the groundwater characteristics may vary considerably with depth,
owing to the possibility of different (and separated) aquifer formations. Some
groundwater sources require minimal treatment, perhaps only disinfection, in
order to comply with drinking water standards. Other sources commonly
require removal of iron, manganese, hydrogen sulfide, hardness (calcium and
magnesium), natural organics that are precursors to disinfection by-products,
and/or total dissolved solids. Removal of synthetic organic compounds (for
example, chlorinated solvents) may be necessary in those cases where groundwater pollution from industrial sources has occurred. Agricultural activity may
contaminate shallow aquifers with pesticides and/or nitrate.
Selection of a water source for a public water supply must consider both
the quality of the source water and the reliable quantity that can be utilized.
Surface water sources often exhibit significant seasonal flow variations, necessitating construction of a reservoir to guarantee adequate supplies during
extended dry periods. The yield of groundwater supplies depends on the
hydraulic characteristics of the specific aquifer formation. Recharge of groundwater sources must be adequate to balance withdrawals; if not, mining of the
water may cause a decline in water tables. It must be recognized that the water
quantities that are available for development are finite for either surface or
groundwater supplies. All of these characteristics influence engineering
designs for water treatment plants, as we shall see in Section 5.3. But first let us
discuss drinking water standards.
5.2 Drinking Water Standards
The evolution of drinking water treatment objectives throughout the developed world has followed a similar pattern. Efforts in developed countries in
the early 20th century were directed at preventing acute health effects, in particular transmission of waterborne diseases, including various bacterial, viral,
and protozoan infections. The widespread adoption of chlorination of public
water supplies has been very effective in the control of cholera, typhoid fever,
and other diseases associated with tainted water. In the mid-20th century, the
aesthetic qualities of the finished water were addressed, with a goal to produce
water that had excellent taste, odor, and appearance. In the last 30–40 years,
treatment objectives have focused on chronic health effects, principally those
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169
related to an increased incidence of cancer, associated with ingestion of low
concentrations of suspected carcinogenic compounds (including disinfection
by-products) present in potable water.
The initial focus of water treatment on the prevention of waterborne disease was justified based on the prevalence of illnesses caused by contaminated
water supplies in the mid-to-late 19th century (Okun 1996). It is now commonly accepted that cholera was the predominant epidemic disease of the 19th
century (Sherman 2006). The pioneering work of Dr. John Snow to identify the
cause of a cholera outbreak in London in 1849 is widely cited as the first discovery of disease transmission via public water supplies. Dr. Snow conducted
an epidemiological study (before bacteria were known to exist), and concluded
that the cases centered around a contaminated well. He had the pump handle
removed and the cholera outbreak ended.
Although public water supplies in well-developed countries are now
widely accepted as being free of infectious agents, much of the population of
the undeveloped world still relies on water supplies that are not adequately
treated to prevent waterborne disease. It is noted, however, that good progress
has been made in the last 20 years. In the 1990s, it was reported that worldwide, about 3 million children younger than five died from complications of
diarrhea each year (Otterstetter and Craun 1997). In 2013, worldwide deaths
from diarrhea for children under five were about 760,000 per year according to
the World Health Organization (WHO 2013). After the big earthquake in Haiti
on January 12, 2010, more than 1.5 million people were without adequate shelter, food, or water. Sanitation was abysmal and many people had no recourse
but to drink untreated water from surface water supplies. As of November 20,
2010, there had been 60,240 cases of cholera reported, with 1,415 deaths. By
June 12, 2011, the number of reported cases had risen to 344,623, and the death
toll had climbed to 5,397 (WHO 2011).
Even though the United States is a highly developed country with modern
water treatment plants, our water supplies are not perfect. From 1970 through
1995, a total of 740 outbreaks of infectious disease associated with contaminated water supplies were reported. It is reasonable to assume that many more
episodes went undocumented (Otterstetter and Craun 1997). In 1993, in Milwaukee, Wisconsin, about 400,000 individuals became seriously ill, and more
than 100 died due to exposure to a protozoan pathogen (Cryptosporidium) transmitted in the public water supply (Okun 1996). The public health implications
of these episodes in chlorinated public water supplies have spurred a renewed
focus on the control of pathogens in water supplies.
Concurrent with the development of standards to address chronic health
effects, it was discovered that the use of chlorine for disinfection of water often
results in the formation of chlorinated organic compounds that are suspected
carcinogens. These compounds, referred to collectively as disinfection byproducts, are formed by reaction of benign, naturally occurring organic matter
with chlorine. Studies of trihalomethane (THM) formation in potable water in
the 1970s led to the development of a 1988 EPA regulation and standard for
total trihalomethanes (TTHM) of 80 ppb (US EPA 2012).
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Federal potable water standards have been established by the US EPA pursuant to the Safe Drinking Water Act of 1974. Two sets of limits were listed by
EPA: MCLGs and MCLs. MCLG stands for Maximum Contaminant Level
Goal, and MCL stands for Maximum Contaminant Level. Both are expressed
as mg/L. An MCLG is “the level of a contaminant in drinking water below
which there is no known or expected risk to health” (US EPA 2013). An MCL is
“the highest level of a contaminant that is allowed in drinking water.” MCLs
are set as close to MCLGs as is feasible considering costs and the best available
treatment technology. MCLs are enforceable standards. The MCLG for a
known or probable carcinogen is set at zero. A partial listing of Primary Drinking Water Standards is provided in Table 5.1 (US EPA 2013).
Action levels for public notification have been established for lead (0.015
mg/L) and copper (1.3 mg/L) concentrations measured at the consumer’s tap.
The presence of these metals is normally the result of corrosion of old plumbing
(lead solder) and piping materials (copper pipe) within the customer’s property. Many utilities are engaged in programs to adjust product water quality to
minimize corrosion of lead and copper. These efforts include control of pH and
alkalinity, introduction of corrosion inhibitors (for example, zinc orthophosphates), and control of oxidant levels (dissolved oxygen and chlorine residuals).
The MCLs in Table 5.1 are called primary standards, but public water utilities also strive to meet secondary standards (non-enforceable) to address
potential issues of taste, odor, and appearance of their water. National secondary standards for potable water are summarized in Table 5.2.
Initially, the standards in Tables
5.1 and 5.2 were based on the use of
uncontaminated natural waters as a
Table 5.2 Partial Listing of Secondary
drinking water source. The standards
Drinking Water Standards
were not inclusive of all compounds
that can cause a potential health conLimit
cern. Addition of MCLs for many
Contaminant
mg/L
synthetic organic compounds and
aluminum
0.05 to 0.2
metals that resulted from industrial
chloride
250
activity was necessary to account for
color (color units, cpu)
15
the impact of waste disposal from an
copper
1.0
fluoride
2.0
industrial society. The reuse of treated
foaming agents
0.5
wastewater (reclaimed water) as a
iron
0.3
potential water supply may dictate
manganese
0.05
that other compounds be added to the
odor (threshold odor number)
3
list, especially in light of recent dispH
6.5 to 8.5
coveries of such things as endocrine
silver
0.10
disruptors and pharmaceuticals in
sulfate
250
some of our water supplies. These
total dissolved solids
500
factors will surely result in a continzinc
5
ual evolution of federal drinking
Source: US EPA (2013).
water standards.
MCLG
mg/L
zero
0.003
zero
zero
zero
0.1
zero
0.07
zero
0.007
zero
zero
zero
0.002
0.7
zero
zero
zero
0.0002
0.04
zero
zero
0.1
1
zero
0.05
zero
zero
10
Contaminant
Organic Chemicals
alachlor
atrazine
benzene
carbon tetrachloride
chlordane
chlorobenzene
dibromochloropropane
2,4-D
1,2-dichloroethane
1,1-dichloroethylene
dichloromethane
1,2-dichloropropane
dioxin (2,3,7,8-TCDD)
endrin
ethylbenzene
ethylene dibromide
heptachlor
heptachlor epoxide
lindane
methoxychlor
pentachlorophenol
PCBs
styrene
toluene
toxaphene
2,4,5-TP (silvex)
trichloroethylene
vinyl chloride
xylenes (total)
0.002
0.003
0.005
0.005
0.002
0.1
0.0002
0.07
0.005
0.007
0.005
0.005
0.00000003
0.002
0.7
0.00005
0.0004
0.0002
0.0002
0.04
0.001
0.0005
0.1
1
0.003
0.05
0.005
0.002
10
MCL
mg/L
Table 5.1
MCLG
mg/L
0.006
0.010
7
2
0.004
0.005
0.1
TT
0.2
4
TT
0.002
10
1
0.05
0.002
4.0
4.0
0.8
0.010
1.0
0.060
0.080
MCL
mg/L
Radionuclides
alpha particles
(pCi/L)
beta particles
(mrem/yr)
radium 226 + 228
(pCi/L)
uranium
Physical parameters
turbidity (NTU)
Microorganisms
Cryptosporidium
Giardia lamblia
Legionella
Total coliform
Viruses (enteric)
Contaminant
5
0.03
zero
zero
zero
TT
zero
TT
zero
TT
zero 5% samples positive
zero
TT
TT
4
zero
N/A
15
MCL
mg/L
zero
MCLG
mg/L
Source: US EPA (2013).
Note: TT means treatment technique; *MRDLG and MRDL stand for maximum residual disinfectant level goal
and maximum residual disinfectant level, respectively.
Disinfection by-products
Bromate
zero
Chlorite
0.8
Haloacetic acids
n/a
Total Trihalomethanes
n/a
Disinfectants (MRDLG and MRDL)*
Chloramines (as Cl2)
4
Chlorine (as Cl2)
4
Chlorine dioxide (as ClO2)
0.8
Inorganic Chemicals
antimony
0.006
arsenic
zero
asbestos (million fibers/L)
7
barium
2
beryllium
0.004
cadmium
0.005
chromium (total)
0.1
copper
1.3
cyanide
0.2
fluoride
4
lead
zero
mercury
0.002
nitrate nitrogen
10
nitrite nitrogen
1
selenium
0.05
thallium
0.0005
Contaminant
Partial Listing of Primary Drinking Water Standards
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171
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Chapter Five
5.3 Treatment Processes to Produce Potable Water
A water treatment plant is a combination of several unit operations put
together to achieve specific objectives. Each unit operation can be discussed separately, but it is important to remember that the whole plant is operated as one
process, so all components must work together. Each design is site-specific and
depends on the raw water characteristics. In general, though, certain unit operations are usually found when treating surface waters and others are normally
associated with groundwaters. Several unit operations are common to the treatment of water from both kinds of sources. Figure 5.2 presents two typical treatment trains for a surface water, and Figure 5.3 presents two typical treatment
trains for a groundwater. Each will be discussed briefly here, and then the individual unit operations will be discussed in greater detail in subsequent sections.
Treatment of even relatively clean surface water to produce potable water for
a city or town must, as a minimum, remove turbidity and achieve disinfection.
The most common treatment plant configuration is a series of unit operations for
turbidity removal: coagulation, flocculation, sedimentation, and filtration, followed by a disinfection step. This is the process depicted in Figure 5.2a.
Figure 5.2
Typical process flow diagrams for surface water treatment.
SURFACE WATER
SURFACE WATER
Alum
BAR RACK
Trash
BAR RACK
Trash
FINE SCREENS
Solids
FINE SCREENS
Solids
Alum
RAPID MIX
FLOCCULATION
SEDIMENTATION
FILTRATION
Chlorine
DISINFECTION
POTABLE WATER
(a) conventional
RAPID MIX
FLOCCULATION
Sludge
UF Membrane Processes
Backwash
Chlorine
DISINFECTION
POTABLE WATER
(b) using membranes
Concentrate
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173
Water flows through a bar rack to remove large objects (e.g., sticks or
branches), and then through a finer screen to remove large particles (e.g.,
chunks of soil or pebbles). The tiny particles remaining in the water include
inorganic (clay and silt particles) and organic (algal biomass) components.
These colloidal particles are very small in diameter and only slightly denser
than water. They often carry a negative charge on their surface. The settling
velocity of such particles is extremely low. Settling is enhanced for larger diameter or denser particles, so the unit operations of coagulation and flocculation
are employed.
The presence of negative charges on almost all of the particles results in
electrostatic repulsive forces between the small particles and prevents agglomeration (formation of larger particles from smaller ones). Without chemical
treatment to alter the particle characteristics, gravity sedimentation would not
work very well. A coagulant chemical, commonly alum [ Al2(SO4)3 • 14H2O ],
is added during a rapid-mix process to destabilize the charge on the particle
surface and promote particle interactions. After coagulation, agglomeration of
the destabilized particles is further promoted by gentle mixing (flocculation) in
a second tank. The collisions of particles result in particle “growth,” and the
large particles are now able to settle faster. Subsequent sedimentation is now
more effective in removing most suspended solids. Filtration (often through a
sand filter) is used as a final polishing step to remove all remaining turbidity.
The filtered water flows through a chlorine disinfection process to disinfect the
water from any microorganisms that made it through the filter.
After filtration, some surface waters may still contain natural organic matter that is dissolved and thus not removable by the above-mentioned operations. Dissolved organics often are precursors to THMs. After the disinfection
process, any residual chlorine reacts with organics to form THMs, often in
excess of the MCLs. That fact is one reason why monochloramine often is used to
maintain the chlorine residual needed in water distribution systems (more on
this later in the disinfection section).
Membrane processes used in drinking water treatment are microfiltration
(MF), ultrafiltration (UF), nanofiltration (NF), and reverse osmosis (RO).
Although more details will be given later, at this point let us say simply that
membranes “filter” out tiny particles and even large dissolved organic molecules and remove them from the water. MF and UF are often used to remove
TSS, bacteria, and pathogens. NF and RO are used to remove even smaller particles as well as certain ions and dissolved organic matter prior to disinfection.
Thus, membranes can help prevent the formation of THMs in the finished water.
RO membranes have been used to desalt brackish groundwater for more
than 40 years; NF membranes have been used to remove hardness for more than
20 years. UF membrane processes have become much more popular in the last
two decades, and are now widely used throughout the world. Figure 5.2b represents the schematic process flow diagram for a membrane-based surface water
treatment plant. In cases where the raw water is relatively clean (as indicated by
low turbidity), UF membranes make it possible to avoid building a large sedimentation tank. In some cases, for very low turbidity surface waters, the treat-
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174
Chapter Five
ment process depicted in Figure 5.2b might not even need the flocculation tank,
and would only include a bar rack, a coagulant feed, a rapid-mix tank, the UF
membranes, and disinfection, thus eliminating all sedimentation operations.
Groundwater often has a different set of characteristics from surface water,
and thus requires different unit operations to produce potable water. Many
groundwaters, particularly in the midwestern and western United States, contain substantial concentrations of calcium and magnesium. These ions cause
hardness, which creates particular problems for consumers. It is often desirable
for the water utility to “soften” the water prior to distributing it. This is done
chemically, using lime and soda ash, as will be discussed in detail later. It is
also common practice to aerate groundwater to remove any objectionable
gases. The conventional lime-soda softening process is depicted in Figure 5.3a.
In the past 20 years or so, alternative processes have been developed that also
can remove hardness from water. Such processes include ion exchange and
membrane processes such as nanofiltration or reverse osmosis (Figure 5.3b).
However, there are still many lime-soda softening treatment plants in operation, especially for cities that have a very large water demand.
Figure 5.3
Typical process flow diagrams for groundwater treatment.
GROUNDWATER
GROUNDWATER
Air
AERATION
Lime
Soda Ash
RAPID MIX
H2S, CO2
Chemical
Scale Inhibitor
CARTRIDGE FILTERS
FLOCCULATION
SEDIMENTATION
Sludge
RECARBONATION
Sludge
FILTRATION
Chlorine
DISINFECTION
POTABLE WATER
(a) conventional
NF or RO Membrane Processes
Air
Chlorine
AERATION
DISINFECTION
POTABLE WATER
(b) using membranes
Concentrate
H2S, CO2
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175
5.4 Unit Operations of Water Treatment
As depicted in Figures 5.2 and 5.3, there are a number of individual unit
operations that make up a water treatment plant. It turns out that several of
these unit operations are identical or very similar to those used in treating
wastewater (the subject of the next chapter). In the following discussions of
individual unit operations, the focus is producing potable water, but many of
the principles that are presented will be directly applicable to wastewater treatment as well.
Rapid Mix
The rapid-mixing operation, as the name implies, strives to rapidly and
completely mix a chemical (e.g., alum) with the water. This operation uses
mechanical mixers (motor-driven paddles or propellers) that impart energy
into the water. The water flows continuously through a tank as the chemical is
continuously injected and mixed. The tanks often have baffles to prevent vortices from forming. The detention time in the tank is usually less than one minute, and is defined in Eq. (5.1). The detention time is also known as the
hydraulic residence time (HRT).
θ = HRT = V/Q
(5.1)
where:
θ
= detention time, min
HRT = hydraulic residence time, min
V
= volume of tank, ft3 or m3
Q
= flow rate, ft3/min or m3/min
Another parameter used in the design and operation of rapid-mix tanks is
the velocity gradient. The velocity gradient is an indicator of the degree of mixing and is related to the power (rate of energy) being imparted into the tank,
the volume of the tank, and the viscosity of the water, as shown in Eq. (5.2).
G = 1, 000 P mV
(5.2)
where:
G = velocity gradient, sec–1
P = power imparted to the water, kw
V = volume of tank, m3
µ = absolute viscosity of water, cp
Rapid-mix tanks are open top tanks and can be square or circular; the mixers
are usually top mounted with long shafts that extend into the water. The detention time and velocity gradient are inversely related. That is, as θ goes up, G
goes down. For example, typical values of θ range from 20 seconds to one minute, while corresponding values of G range from 1,000 to 600 sec–1. The absolute viscosity of water varies inversely with temperature, going from 1.55
Cooper.book Page 176 Monday, June 23, 2014 9:58 AM
176
Chapter Five
centipoise (cp) at 40 °F to 0.98 cp at 70 °F to 0.76 cp at 90 °F. The absolute viscosity is also called the dynamic viscosity (but it is not the kinematic viscosity). Eq.
(5.2) must be used with the units given, but it is interesting to note that 1.0 cp =
0.001 Pa-s = 0.001 N-s/m2.
EXAMPLE 5.1
You are designing a rapid-mix tank at a water treatment plant that is treating
5.0 million gallons per day (MGD). You want a square tank with a water
depth = 1.5 times the width of the tank. The detention time is to be 40 seconds
with a velocity gradient of 800 sec–1. The water temperature is 60 °F, at which
the viscosity is 1.13 cp. Calculate the tank dimensions (in ft) and the power
required (in HP).
SOLUTION
From Eq. (5.1), the tank volume is
V = qQ
= 40 s ¥
1 day
1 min 5, 000, 000 gal
1 ft 3
¥
¥
¥
= 310 ft 3
60 s
day
7.48 gal 1, 440 min
Given the stated geometry, the volume equals 1.5 W3, so
W = 3 V 1.5 = 5.9 ft (probably, we will specify this tank to be 6 ft square on the
bottom with 10-ft tall walls). This larger tank increases the detention time slightly.
To use Eq. (5.2), first calculate the new volume of water in this slightly
larger tank, and then convert the volume to m3. Assume the average water
depth is 9 feet.
V = 6 × 6 × 9 = 324 ft3 (note that θ now is actually 41.9 seconds)
V = 324 ft3 × 1 m3/ 35.3 ft3 = 9.18 m3
Finally, rearrange Eq. (5.2) to calculate the power needed in kw, and then convert to HP.
P = (G/1,000)2 × µ V
= (0.80)2 × 1.13 × 9.18 = 6.64 kw or 8.9 HP
(We will probably buy a 10-HP motor for the mixer, or perhaps have two
mixers, each with a 5-HP motor.)
Coagulation/Flocculation
The purpose of coagulation is to allow a chemical to react with water and
change the characteristics of tiny particles that otherwise would not settle in a
reasonable time. For small gravel or large sand particles (which are dense compared with water), gravitational settling velocities are fast. But as particles get
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177
smaller, even if they are
Table 5.3 Settling Times for Various Sized
dense, their settling veParticles* in Water
locities decrease such
that the time required
Particle
Diameter
Time to settle 1 foot
to separate them from
Gravel
1 cm
0.1 sec
water by gravitational
Coarse sand
1 mm
1 sec
settling increases beFine sand
0.1 mm
10 sec
yond reason. Table 5.3
Silt
10 µm
10 min
shows the settling times
Clay
0.1 µm
10 weeks
of various-sized partiColloids
10 nm
20 years
cles in water.
*All particles are assumed to have a specific gravity of 2.5.
Both
aluminum
sulfate (alum) and iron
chloride have been used
as coagulants. Both are readily soluble, and when these compounds dissolve, the
metal cations react with water to form the insoluble hydroxide, as shown below:
Al2(SO4)3 + H2O
FeCl3 + H2O
→
→
Al(OH)3 + other ions
Fe(OH)3 + other ions
(5.3)
(5.4)
For both iron and aluminum hydroxides, the Ksp values are so low that they
essentially pull the OH– ions from water molecules and form “sticky” precipitates. In addition, the metal ions also undergo hydrolysis reactions where a
number of positively charged hydrated ions are formed. Two examples are
shown below:
Fe3+ + 2 H2O
2 Fe3+ + 4 H2O
→
→
Fe(OH)2+ + H3O+
Fe2(OH)24+ + 2 H3O+
(5.5)
(5.6)
Some of these hydrolysis products form long chains or “macro-ions” that are
attracted to the negative surface charges on colloids. They neutralize the charge
on the particle surface, and thus allow the particles to interact and agglomerate. For example, the macro-ion Al13(OH)345+ has been identified. Of course, the
ions that exist are highly dependent on pH, and a detailed review of the chemistry of coagulation is beyond the scope of this text. Suffice it to say that these
coagulants are very effective at agglomerating these tiny particles into small
“floc” particles (relatively large agglomerated particles made up of solids,
coagulants, and water trapped in open spaces). These floc particles generally
have a specific gravity close to that of water and so still settle slowly, unless we
can make them even bigger—this is the purpose of the next step.
Flocculation is the next step toward separating colloidal particles from
water. After the small floc is formed in the rapid-mix tank, the water flows into
a flocculation basin where it is slowly stirred and detained for the time needed
to form larger floc particles that will settle. The slow stirring is important
because if the tips of the paddles move through the water faster than about 1
m/s, the large floc that is forming will be broken up. The mixers are installed
so that the paddle motion is generally in-line with the water flow to help avoid
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178
Chapter Five
breaking up the floc. The hydraulic residence time (HRT) in these tanks varies
between 10 minutes to an hour. The HRT is calculated identically to the detention time defined in Eq. (5.1). The difference is that the longer HRT requires
that a flocculation tank be much larger than a rapid-mix tank.
HRT = V/Q
(5.7)
Often, these large flocculation basins are subdivided with baffles to ensure that
the water does not “short-circuit” the system, and flows through the entire volume. Figure 5.4 shows a rapid-mix tank and a flocculation basin as part of one
unit. Examples 5.2 and 5.3 illustrate some calculations related to the flocculation tank design.
Figure 5.4
Schematic of a combined rapid-mix tank and flocculation basin.
Chemicals
Water
Baffles
To
sedimentation
Tanks
Slow-moving
paddle wheels
Mixer
EXAMPLE 5.2
The HRT design target for a flocculation basin is 45 minutes. If the water flow is
the same as in Example 5.1, what is the required volume of the flocculation basin?
SOLUTION
HRT = V/Q
V=
so
V = Q × HRT
5, 000, 000 gal
1 day
1 ft 3
¥
¥ 45 min ¥
= 20, 890 ft 3
day
7.48 gal
1, 440 min
EXAMPLE 5.3
Specify the dimensions of the flocculation tank of Example 5.2. Assume a rectangular tank with a length-to-width ratio of 4:1, and a length-to-depth ratio of
12:1. The depth has been set at 10 ft.
SOLUTION
The volume is 20,890 ft3, and the dimensions are L, W, and H, so
LWH = 20,890 ft3
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179
But also, L = 4 W and L = 12 H. Substituting, we have
Ê Lˆ Ê L ˆ
L Á ˜ Á ˜ = 20, 890 ft 3
Ë 4 ¯ Ë 12 ¯
6
or L3 = 1.003 (10 ) ft 3
Solving, we get L = 100 ft. In turn, W = 25 ft, and H = 8.34 ft. (Build the side
walls 10 ft and set the outflow weir depth at 8.34 ft.)
Sedimentation
After flocculation, the water must spend some time in a very quiescent
flow regime for the floc to settle; this process is called sedimentation. The fundamentals are best understood with the help of a simple drawing (Figure 5.5).
Visualize water flowing into the front end of a rectangular tank and over the
back wall at the far end of the tank. The water contains various size particles,
but let us assume a “design-size” particle, with a design settling velocity, Vs .
The water is flowing through the tank at an average linear velocity, Vx . If the
particle settles to the bottom of the tank before the water reaches the outlet end
of the tank, then the particle will be collected and removed from the water. If
not, then the particle is not collected, and flows out with the water.
Figure 5.5 Idealized
sedimentation tank.
Q
Vx
Q
H
Vs
L
In discrete settling (rigid, noninteracting spheres), the settling velocity is a
function of the particle size and density, the water viscosity, and flow regime.
For a single spherical particle settling in laminar flow, the velocity is given by:
Vs = g
(rp - rw ) dp2
where:
Vs = settling velocity, m/s
g
= gravitational constant, m/s2
ρp = particle density, kg/m3
ρw = water density, kg/m3
18µ
(5.8)
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180
Chapter Five
dp = particle diameter, m
µ = viscosity of water, kg/m-s
The equation for settling velocity is much different in flocculent settling, but
for now we will use Eq. (5.8) in our explanation. The particle is moving downwards at Vs , a number that we can calculate from the properties of the particle
and the water using Eq. (5.8).
The water velocity in the horizontal direction, Vx , is simply
Vx = Q/WH
(5.9)
where:
Vx = horizontal velocity, m/s
Q = water volumetric flow rate, m3/s
W = tank width, m
H = tank depth, m
The time it takes for a particle that enters the tank near the top of the water surface to fall to the bottom is
tfall = H/Vs
(5.10)
and the time it takes for a particle to flow through the tank along with the
water is
tflow = L/Vx
(5.11)
It is apparent that if tfall is longer than tflow, then the particle will flow out of the
tank before it falls to the bottom and is collected.
A term that is used in the design of settling tanks is the overflow velocity,
Vo, often called the overflow rate (OFR). We can think of the design overflow
rate as the upward velocity of water that is exactly equal to the settling velocity
of the design particle (Vo = Vs.) If water is flowing upward faster than the particle is settling, the particle will be carried out with the water; if the water flows
upward slower than the particle is settling downward, the particle ultimately
will reach the bottom. The OFR is mathematically defined by Eq. (5.12). It can
be derived by setting tfall equal to tflow , substituting Vo for Vs , and using Eq.
(5.9) to substitute for Vx. Both symbols (Vo and OFR) are used for the overflow
rate, which also is sometimes called the surface loading rate.
Vo = OFR = Q/A
(5.12)
where:
Vo
= overflow rate, m3/day-m2
OFR = overflow rate, m3/day-m2
A
= top surface area of the water in the settling basin, m2
For rectangular tanks, A is the product of the length times the width. For circular settling tanks, A is the area of the circle of water at the top of the tank. The
units on Vo can be linear velocity units (e.g., ft/hr or m/day), but more often
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181
are given as volumetric flow rate units divided by area units, like (m3/day)/m2
or gpd/ft2.
Engineers often use one additional parameter to design settling tanks or to
check their operations—the weir loading rate (WLR). A weir is like a small
thin dam located at the outlet end of a rectangular tank or very near the inside
diameter of a circular tank. Clarified water near the top surface flows over the
weir and out of the tank leaving behind water that still has suspended solids in
it. The WLR is given by Eq. (5.13):
WLR = Q/(weir length)
(5.13)
where: WLR = weir loading rate, m3/day-m
The units of weir loading rate are commonly given as volumetric flow rate
over length; for example, (m3/day)/m or gpd/ft. For a rectangular tank, the
length of the exit weir may simply be the tank width, or the weir may extend
some distance along both sides near the far end of the tank. For a circular tank,
the weir is a circular wall located a few inches from the inside diameter of the
tank, and the weir length is about equal to the circumference of the tank. Table
5.4 provides typical values for designing settling tanks for three kinds of flocs
(even though we have not yet discussed lime-soda flocs; we will soon). Since the
bottoms of settling tanks are usually slightly sloped, the depths mentioned in
Table 5.4 are all measured as the water depth at the side walls near the exit weirs.
Finally, as we know from material balance principles, these tanks must also
have an outlet for the concentrated floc that settles out of the water. This stream
exits the tank from the bottom as a stream of water carrying most of the floc
particles. The bottom stream is called the sludge, and it may contain 90% or
more of the solids that came in with the raw water. The stream has a very small
flow—only a few percent of the inlet water flow rate. Even so, this stream is not
very concentrated—it may only contain 1–2% solids (the rest is water). Note
that 1% solids in this kind of stream corresponds to a concentration of 10,000
mg/L. Sludge is often simply sent to an open land area and the water either
Table 5.4
Parameter
Hydraulic residence time, hrs
Overflow rate, m3/day/m2
Overflow rate, gpd/ft2
Weir loading rate, m3/day/m
Weir loading rate, gpd/ft
Rectangular tanks (L:W)
Rectangular tanks (L:D)
Rectangular tanks (depth, ft)
Circular tanks (diameter, ft)
Circular tanks (depth, ft)
Design Criteria for Settling Basins
Alum flocs
Iron flocs
4–6
4–6
20–33
29–41
500–800
700–1,000
150–220
200–270
12,000–18,000
16,000–22,000
2:1–4:1 (all flocs)
10:1–20:1 (all flocs)
10–13 (all flocs)
15–250 ft (all flocs) (in increments of 5 ft)
6–16 ft (all flocs)
Sources: Reynolds and Richards (1996); Mines and Lackey (2009).
Lime-soda flocs
4–8
29–61
700–1,000
270–320
22,000–26,000
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Chapter Five
evaporates or seeps into the ground. Because sludge streams often have quite
different solids concentrations, they are often converted to a mass flow of solids on a dry basis. This simply means that we pay attention only to the weight
of solids contained in the sludge and ignore the water weight. The next example illustrates how changing the concentration of solids in the sludge by only a
few percent can make a significant difference in the total weight (and volume)
of the sludge to be handled.
EXAMPLE 5.4
The sludge stream from a sedimentation tank is flowing at 100,000 gal/day,
and contains 1.5% solids. Assume the solids have a specific gravity close to 1.0.
(a) Calculate the mass flow rate of this sludge stream as is.
(b) Calculate the mass flow rate of sludge on a dry basis.
(c) If the sludge is thickened to 12% solids on a belt filter press, what is the
mass flow rate of the thickened stream?
(d) How much water was removed from the sludge to thicken the sludge
from 1.5% solids to 12% solids?
SOLUTION
Since the sludge solids have a specific gravity close to 1.0, then the density of
the sludge stream is the same as the density of water.
(a) Mass flow rate of sludge, as is:
100,000 gal/day × 8.34 lb/gal = 834,000 lb/day
(b) Mass flow rate, dry basis:
834,000 lb/day × 0.015 = 12,510 lb/day of dry solids
(c) Mass flow rate at 12% solids:
(d) Water removed:
12, 510 lb/day solids
= 104, 250 lb/day
0.12 lb solids/lb sludge
834,000 – 104,250 = 729,750 lb/day
The process design of water treatment facilities must ensure that routine or
emergency outages do not shut down the whole plant, thus shutting off the supply of drinking water to a city. Therefore, multiple units, in parallel, are specified
for reliability and redundancy. Large circular basins (and their mechanisms) are
typically designed and built based on “standard” diameters that are available in
increments of 5 feet. Example 5.5 demonstrates these principles.
EXAMPLE 5.5
Sedimentation tanks for a water treatment plant using iron as a coagulant are
being designed to satisfy the following criteria:
Overflow rate (maximum) = 1,000 gpd/ft2
Hydraulic residence time (minimum) = 4 hours
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183
Flow rate = 4.0 MGD or 4,000,000 gallons per day
Minimum of two sedimentation basins in parallel
Recommend the dimensions for a pair of circular sedimentation basins. Calculate the OFR and HRT for the recommended system.
SOLUTION
First divide the flow by 2 since we will have two units in parallel, each treating half the water. The surface area of one sedimentation tank is determined
using the overflow rate constraint:
A = Q/Vo = 2,000,000 gpd/ 1,000 gpd/ft2 = 2,000 ft2
4A
p D2
, so D =
= 50.5 ft
4
p
Clarifier mechanisms are available in standard size increments of 5 ft. Since
50.5 ft is not a standard size, and if we round down to 50 ft, the OFR will be
too high, we must round up to 55 ft. Therefore, select two units, each with a
diameter of 55 ft. The surface area of each is 2,376 ft2 and the resulting OFR is
842 gpd/ft2, which is within the stated constraint.
The depth is determined by application of the hydraulic residence time
constraint:
The area of a circle is A =
D = HRT ¥
4 hrs ¥ 2, 000, 000 gpd
Q
=
= 18.75 ft
A 2, 376 ft 2 ¥ 24 hr/day ¥ 7.48 gal/ft 3
This is a bit too deep according to Table 5.4. Go back and assume a diameter
of 60 feet. With D = 60 ft, A = 2,827 ft2 and depth = 15.8 ft, which is acceptable.
We must now recalculate the OFR.
Vo = Q/A = 2,000,000/2,827 = 707 gpd/ft2
This is within the range given in Table 5.4.
Our final recommendation is for two circular clarifiers, each with a 60-ft
diameter. Specify a sidewater depth of 16 feet, with an additional freeboard
allowance of two feet for a total tank depth of 18 feet. The actual liquid volume
in each tank is 45,239 ft3, which yields a hydraulic residence time of 4.06 hours.
Rapid Sand Filtration
Sedimentation tanks are not 100% efficient. There are always some particles remaining in the overflow water from such tanks, and filtration is used to
remove this remaining turbidity. Rather than filter only on a surface, the unit
operation of rapid sand filtration is used. Water is allowed to flow by gravity
through a 2- to 3-ft layer of sand, which removes the remaining floc and any
other unsettled particles. The sand is supported on a 1- to 2-ft bed of gravel (to
keep the sand from falling out the bottom of the filter), and the gravel is supported on a steel grate and coarse screen. As the water flows downward
through the layers of the filter, particles are caught throughout the depth of the
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184
Chapter Five
filter. This filtering in depth ensures good removal of particles and allows the
filter to run for a longer period of time before getting clogged than if we simply
used a filter cloth. The water emerging from the bottom of a rapid sand filter is
quite clean.
Sometimes, a 1- to 2-ft layer of charcoal lies atop the sand, which leads to
the name multimedia filter. The charcoal is coarser but lighter than sand, and
both of these properties are important. The purpose of the coarseness of charcoal is to filter out larger particles and prevent them from too quickly clogging
the sand. The purpose of the lightness is so the charcoal will stay on top of the
sand during the backwash procedures. Figure 5.6 is a schematic diaFigure 5.6 A multimedia filter (sand plus
gram of a multimedia filter.
charcoal).
The filter eventually gets plugged
up and must be cleaned. Cleaning of
Water with turbidity
the multimedia bed is done by using
some of the product water to pump
backward and upward through the
filter at a high enough velocity to
expand the beds. The abrasive rubbing of sand particles helps knock off
Water
the small floc particles that had been
caught in the filter, and carries that
material out the top with the overflow
Charcoal
water during backwash, thus cleaning
the bed. This backwash flow is discarded. To allow the plant to keep
Sand
running during backwash events, and
for redundancy in the plant, several
filters are operated in parallel.
Gravel
The sizes of these filters can range
from surface areas of 250–1,000 ft2 and
depths of 10–20 feet. The hydraulic
loading rate (HLR), or flow rate of
Clear water
Water
water divided by surface area of filter,
for a rapid sand filter ranges from 2 to
10 gpm/ft2 depending on the media,
its coarseness, and of course, the particle loading. Backwashes occur as needed
when the filtering rates slow down enough (the head loss through the filter
increases enough), but typically they happen about once per day. Backwash is
achieved by flowing water upward at about twice the HLR for about 10 minutes.
Disinfection
The purpose of disinfection is to kill pathogenic microorganisms; it is not
sterilization (which kills all microorganisms). In water treatment we talk about
primary and secondary disinfection. Primary disinfection is the initial contact
of chlorine or some other strong disinfectant with the water; secondary disin-
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185
fection refers to maintaining a residual concentration of disinfectant in the distribution system. In some cities, treated water may take several days to travel
in the distribution pipes from the treatment plant to the furthest residential
consumers. Although ozone and other chemicals are sometimes used, primary
disinfection is achieved most commonly by chlorination (adding chlorine to
water and allowing time for it to work). Secondary disinfection often is accomplished using monochloramine, which helps prevent the formation of disinfection by-products.
During primary disinfection, the chlorine is mixed into the treated water,
which then flows through a chlorine contact chamber (essentially a channel
that has flow characteristics similar to a plug flow reactor) to ensure there is
adequate time for the chlorine reactions to kill the pathogens. The reaction rate
depends on both the concentration of pathogens and the concentration of chlorine, and is described by the following second-order rate expression:
–r = k CPa CCl
(5.14)
where:
–r
= disinfection rate (organisms per minute per L)
k
= rate constant (L/mg-min)
CPa = pathogen concentration (organisms per L)
CCl = chlorine concentration (mg/L)
Chlorine may be supplied as chlorine gas, which reacts with water to form
hypochlorous acid and hydrochloric acid:
Cl2 + H2O
→
HOCl + HCl
(5.15)
Hypochlorous acid is a weak acid, which dissociates to form a hydrogen ion
and a hypochlorite ion (pKa = 7.5):
HOCl
↔
H+ + OCl–
(5.16)
The total chlorine concentration that works to kill microorganisms is the sum of
the concentrations of hypochlorous acid (HOCl) and hypochlorite ion (OCl–),
and is called the free chlorine residual.
Addition of chlorine to water that contains ammonia will result in the initial formation of a combined chlorine residual. After all of the ammonia has
been oxidized to nitrogen gas, additional chlorine will produce a free chlorine
residual. Provision of sufficient chlorine dose to form a free residual is referred
to as breakpoint chlorination, in which the breakpoint dose is defined as the
chlorine dose that corresponds with the destruction of ammonia. In those cases
where a combined residual is desired, a smaller chlorine dose will suffice.
Breakpoint chlorination may be used to achieve ammonia removal from wastewaters, although other options are generally regarded as more cost effective.
When ammonia is present in the water, chlorine will react with it to form
several chloramines:
NH4+ + HOCl
→
NH2Cl + H+ + H2O
(5.17)
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186
Chapter Five
NH2Cl + HOCl
NHCl2 + HOCl
→
→
NHCl2 + H2O
(5.18)
NCl3 + H2O
(5.19)
The final product of this last reaction is unstable and readily decomposes to
form nitrogen gas, which escapes from solution. The overall reaction for oxidation of ammonia to nitrogen gas is represented below:
2 NH3 + 3 Cl2
→
N2 + 6 H+ + 6 Cl–
(5.20)
The sum of monochloramine (NH2Cl) and dichloroamine (NHCl2) is the
combined chlorine residual. The combined residual is not as strong an oxidant
as the free chlorine residual, and thus is not as effective for primary disinfection. Consequently, longer reaction times are required to achieve disinfection
using a combined residual than with a free chlorine residual. This disadvantage is partially offset by the relative stability of the combined residual in the
distribution system (the pipes that carry the finished water throughout the
city). Combined residuals also have the big advantage of reducing the formation of disinfection by-products during the time (hours to days) that water is in
the distribution system before reaching an end user. Therefore, ammonia may
be added to water after disinfection with chlorine to produce a combined chlorine residual.
The concentration of ammonia in the water is customarily measured and
reported as ammonia-nitrogen (given the symbol NH3-N), similar to the way
we report hardness as calcium carbonate. Expressing all forms of nitrogen
(ammonia, nitrate, nitrite, organic nitrogen, etc.) as nitrogen simplifies calculations because the mass of nitrogen is conserved, but the masses of the various
nitrogen species are not constant due to the formation and destruction of nitrogen compounds with differing molecular weights.
EXAMPLE 5.6
Determine the quantity of chlorine gas (mg/L) required to react with 10 mg/
L of NH3-N.
SOLUTION
The basis for this stoichiometric calculation is the 10 mg/L of ammonia-nitrogen.
Cl 2 needed =
76 mg Cl 2
10 mg / L NH 3 -N
3 mmol Cl 2
¥ 71 mg Cl 2 /mmol =
¥
L
14 mg N/mmol
2 mmol NH 3 -N
EXAMPLE 5.7
Determine the required hydraulic residence time (HRT) in an ideal plug flow
reactor to achieve 99.99% pathogen destruction efficiency. Assume Eq. (5.14)
applies, with k = 2.0 L/mg-min. Also, assume that the concentration of free
chlorine residual stays about constant at 0.2 mg/L during the entire time in
the reactor.
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187
SOLUTION
As we learned in Chapter 3, a steady-state mass balance around a differential
volume of the reactor will yield the design equation. In this case, the reaction
rate is second order, so we cannot simply use the final equation that was
derived in Chapter 3 for a first-order reaction.
Accumulation = Input – Output + Generation
0 = Q CPa – Q (CPa + dCPa) + r dV
For convenience, we drop the subscript Pa from the concentration of pathogens:
0 = –Q dC – k C CCl dV
Recognizing that CCl is constant for the problem conditions, the equation can
be solved by separation of variables and integration:
V
Ú
0
Ce
dV
- dC
-1
= Ú
=
Q C0 kCCCl kCCl
HRT =
Ce
Ú
C0
dC
C
ln (C0 0.0001 C0 )
V ln (C0 Ce )
=
= 23 min
=
Q
kCCl
(2 L mg-min )(0.2 mg/L )
Aeration for Removal of Hydrogen Sulfide from Groundwater
Some compounds that are present in groundwater are volatile (either true
gases or high vapor pressure liquids). These compounds may be transferred
from the water to the atmosphere by aeration (or air stripping). This strategy is
employed for removal of certain industrial contaminants (for example, trichloroethylene, carbon tetrachloride, or gasoline-type hydrocarbons). For such airstripping applications, a packed tower is often necessary to achieve good
removal. A packed tower operates by pumping the water to the top of a cylindrical vessel that is filled with plastic pieces (packing). Air is blown into the
bottom of the tower. As the air and water pass each other, the volatile compounds are transferred out of the water and into the air and are discharged
from the top of the tower with the air. The cleaned water flows out the bottom.
If needed, air pollution control devices are installed on the air exhaust stream
to assure that this solution of a groundwater pollution problem is not achieved
at the expense of contamination of the air.
A far more common application of air stripping in potable water treatment
is for removal of naturally occurring dissolved gases from groundwater, principally hydrogen sulfide (H2S). Hydrogen sulfide is a common contaminant in
many groundwaters. It produces an unpleasant taste and odor with a characteristic smell of rotten eggs. Disinfection, typically with chlorine, is practiced
after aeration because H2S will consume chlorine. For those groundwater
sources that do not require more extensive treatment, a simple process of aera-
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188
Chapter Five
tion followed by chlorination represents a very economical method of treating
water. Aeration for H2S removal is often practiced simply by letting the water
cascade down a series of trays placed near the top of a ground water storage
tank. As H2S is transferred from the water into the air, the breeze blows it away
from the tank. It usually is in such low concentrations that the odor is not
noticeable at the plant boundaries. Aeration also can be used to remove soluble
iron from some groundwaters. Iron in solution that is associated with bicarbonate can be oxidized and removed as iron hydroxide.
Lime-Soda Softening of Groundwater
As mentioned in Chapter 2 and earlier in this chapter, polyvalent metal
ions (mainly calcium and magnesium) contribute to hardness in the water.
These ions are not regulated by either the primary or secondary drinking water
standards; however, removal of these compounds is commonly practiced in
order to minimize customer complaints (mainly scaling problems in the water
distribution system, water heaters, bathtubs, and shower stalls). Scale is the
deposition of an insoluble salt of calcium or magnesium. Many industrial processes (for example, boiler operation) cannot tolerate any hardness due to
problems associated with scale formation. Hardness also contributes to excessive consumption of detergents
during laundry operations.
Neither calcium nor magneTable 5.5 Characterization of Water Hardness*
sium ions in water represent
health concerns for the general
Description
Hardness (mg/L as CaCO3)
population. Target concentrations
Soft
< 60
for hardness are therefore not deModerately hard
60 to 120
fined by federal MCLs. Typical
Hard
120 to 180
local goals for hardness of finVery hard
> 180
ished water range from 60 to 120
*USGS classification system.
mg/L, expressed as calcium carbonate (mg/L as CaCO3). Classification of waters based on
hardness is presented in Table 5.5. Example 5.8 illustrates the conversion of ion
concentrations to concentrations as CaCO3.
EXAMPLE 5.8
Lab analysis of a groundwater provided these results:
Ca2+ = 88 mg/L
Mg2+ = 50 mg/L
Determine the water hardness in units of mg/L as CaCO3.
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189
SOLUTION
Divide each ion concentration by its equivalent weight:
Ca =
88 mg /L Ca 2+ ( 50 mg CaCO 3 )
¥
= 220 mg/L as CaCO 3
meq
20 mg /meq
Mg =
50 mg /L Mg 2+ ( 50 mg CaCO 3 )
= 206 mg/L as CaCO 3
¥
meq
12.15 mg /meq
The total hardness is the sum: 220 + 206 = 426 mg/L as CaCO3. This water
would be classified as very hard per Table 5.5.
Softening is the process of removing from water the ions that cause hardness—primarily calcium and magnesium. In small systems (even as small as
one individual home), water may be softened by ion exchange or membrane
processes, but in large municipal systems it is usually more economical to use a
process known as lime-soda softening. In this process calcium ions and magnesium ions are removed by precipitation as CaCO3 and Mg(OH)2, respectively.
The chemicals that are required for lime-soda softening include lime
(Ca(OH)2), soda ash (Na2CO3), and carbon dioxide (CO2). Lime may be purchased as slaked or hydrated lime (Ca(OH)2), or as quicklime (CaO); both lime
and soda ash are purchased in bulk as dry materials. CO2 is often generated onsite by burning oil or gas. If a very large volumetric flow of water must be softened, the chemical usage may cost millions of dollars per year, and the process
may produce tens of thousands of tons of sludge (the precipitated compounds)
that must be disposed. Therefore it is important to analyze the water and calculate the chemical needs and waste disposal needs prior to embarking on building a multimillion-dollar water-softening plant.
In discussing the lime-soda softening reactions, it is convenient to first construct a bar chart to represent the chemical composition of the water (Viessman
and Hammer 2005). The bar chart follows a certain format with the dissolved
CO2 on the left, the positive ions on top (with calcium first and magnesium second), and the negative ions on the bottom (with bicarbonate first). The reason
for this format is that as we add lime to the water (and we raise its pH), the CO2
is removed first, followed by the calcium ions, and finally the magnesium ions.
The bicarbonate ions play a very important role in these precipitation reactions.
To make the chart, we must have a chemical analysis of the water showing the
concentrations of CO2, Ca2+, Mg2+, HCO3–, and other positive and negative
ions. The concentrations on the chart must be in units of meq/L, but in the
water industry meq/L are often expressed in units of mg/L as CaCO3. One
(1.0) meq/L of any chemical is equal to 50 mg/L as CaCO3. Water is electrically
neutral, so the length of the bar of the positive ions always equals that of the
negative ions when displayed in the above units. This discussion is best illustrated by the following example.
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190
Chapter Five
EXAMPLE 5.9
Given the following analysis of a sample of water, construct a bar chart to represent its chemical composition (note each substance concentration is given as
the substance): CO2 = 17.6 mg/L, HCO3– = 341.6 mg/L, Ca2+ = 80 mg/L, Mg2+
=29.2 mg/L, Na+ = 36.8 mg/L, Cl– = 67.5 mg/L and SO42– = 24 mg/L. Present
your bar chart two ways—one with units of meq/L and the other with units
of mg/L as CaCO3.
SOLUTION
First make a table and calculate the concentrations in meq/L by dividing each
concentration in mg/L by the equivalent weight (mg/meq).
Component
Conc.,
mg/L
Eq.
Wt.
Conc.,
meq/L
Conc., mg/L
as CaCO3
CO2
Ca2+
Mg2+
Na+
HCO3–
Cl–
SO42–
17.6
80
29.2
36.8
341.6
67.5
24
22
20
12.15
23
61
35.5
48
0.80
4.00
2.40
1.60
5.60
1.90
0.50
40
200
120
80
280
95
25
The final charts are as shown below.
meq/L
0.8
0
4.0
Ca2+
6.4
Mg2+
8.0
Na+
CO2
HCO3–
0.8
Cl–
0
5.6
SO42–
7.5
8.0
mg/L as CaCO3
40
0
200
Ca2+
320
Mg2+
400
Na+
CO2
HCO3–
40
0
Cl–
280
SO42–
375
400
In the softening process we mix the lime and soda ash into the water in a
small mixing tank, and then the water flows into a large sedimentation tank to
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191
allow time for the solid precipitates to settle. The various forms of calcium and
magnesium hardness may be classified as carbonate hardness (CH) or noncarbonate hardness (NCH), depending on the anion (either bicarbonate or
others such as sulfate or chloride) which is associated with the calcium or
magnesium. Thus, it is possible for water to have various combinations of calcium carbonate hardness (CaCH), calcium non-carbonate hardness (CaNCH),
magnesium carbonate hardness (MgCH), and magnesium non-carbonate
hardness (MgNCH).
As we add calcium hydroxide, the pH of the water goes up and various
chemical precipitation reactions occur, including the removal of the dissolved
CO2. (Even though CO2 is removed from the water initially during softening, it
is necessary to put some back into the water at the end of the process, as will be
explained later.) Although we add the lime and the soda ash simultaneously,
and all the precipitation reactions occur together in one tank, it is convenient to
think of the reactions as occurring in a certain order as the pH goes up. The first
reactions that occur are all the calcium carbonate precipitation reactions,
because CaCO3 precipitates at a lower pH than Mg(OH)2. We can think of the
calcium carbonate reactions happening in this order: removal of CO2, removal
of CaCH, and removal of CaNCH.
Groundwater that is in contact with carbonate minerals in the subsurface
strata may contain much higher concentrations of dissolved CO2 than surface
waters (they are in contact with the atmosphere). Aeration can be used to
remove much of that excess CO2 before adding lime and soda ash. But tray aerators are not very efficient, and substantial dissolved CO2 may still remain in
the water. Because CO2 is the first substance to react with the lime that we add,
some of our lime unavoidably will be used up, as shown in Eq. (5.21).
CO2 + Ca(OH)2
→
CaCO3↓ + H2O
(5.21)
As long as there are bicarbonate ions present, they will provide the carbonate to help remove calcium ions. We add the lime to raise the pH and shift the
bicarbonate equilibrium to form carbonate ions (as was presented in Chapter 2)
that will then combine with the calcium ions. CaCH precipitates as CaCO3 in
the pH range 9 to 10.3. Hence we say that “we add calcium to remove calcium.” This reaction is shown in Eq. (5.22).
Ca2+ + 2 HCO3– + Ca(OH)2
→
2 CaCO3↓ + 2 H2O
(5.22)
If we don’t have enough bicarbonate present in the water to react with all
the calcium, then we have CaNCH in the water. In that case, we add soda ash
to provide the carbonate ions that will precipitate the CaNCH. This reaction is
shown in Eq. (5.23).
Ca2+ + SO42– + Na2CO3
→
CaCO3↓ + 2 Na+ + SO42–
(5.23)
Note that in Eq. (5.23), the sulfate ions could just as easily have been replaced
by chloride ions or by other anions. Also note that sometimes we do not try to
remove 100% of the calcium; in many cases we want to leave a small amount of
calcium in the water.
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Chapter Five
If there is more than enough bicarbonate to precipitate all the calcium,
there will be some MgCH in the water. We must raise the pH to a range of 10.5
to 11 to precipitate Mg(OH)2. This can be done by using sodium hydroxide, but
NaOH is expensive and adds a lot of sodium to the water, so often we use lime
instead. It is noted that using sodium hydroxide produces less sludge (sodium
carbonate is soluble), so lime is not always used. As we add Ca(OH)2 to water
that still has bicarbonate in it, more CaCO3 will precipitate before the pH gets
high enough to precipitate Mg(OH)2. In this case, all the calcium gets removed,
and we must add excess lime to reach these high pH values. The reaction that
depicts the removal of MgCH is shown in Eq. (5.24). Note that both Mg(OH)2
and CaCO3 precipitate in this step.
Mg2+ + 2 HCO3– + 2 Ca(OH)2
→
2 CaCO3↓ + Mg(OH)2↓ + 2 H2O (5.24)
Once all the carbonate hardness is gone, and if we still have magnesium in
the water, we add more lime to provide the hydroxide ions needed to precipitate the remaining magnesium as Mg(OH)2 as shown in Eq. (5.25). The lime
requirements are based on (1) how much magnesium must be removed, and (2)
an excess quantity to ensure high pH. The excess lime is often approximated as
50 mg/L as CaCO3, although site-specific values should be determined based
on the specific water and target pH.
Mg2+ + SO42– + Ca(OH)2
→
Mg(OH)2↓ + Ca2+ + SO42–
(5.25)
At this point the water has a substantial concentration of calcium ions and
is at very high pH. We must now remove most of the remaining calcium and
lower the pH before distributing the water. Removal of the calcium can occur
by adding carbonate alkalinity, namely either CO2 or soda ash. The decision
whether to use CO2 or soda ash is typically made based on an engineering cost
analysis considering both operating and capital costs.
If we choose to use soda ash to remove the calcium that remains after the
Mg(OH)2 precipitates, the reaction is the same as Eq. (5.23), shown previously, and
we can make use of the same tank in which all the other precipitation reactions
have occurred. Thus, removing MgNCH can be thought of as a two-step process.
Step 1:
Mg2+ + SO42– + Ca(OH)2
Step 2:
Ca2+ + SO42– + Na2CO3
→
→
Mg(OH)2↓ + Ca2+ + SO42–
CaCO3↓ + 2 Na+ + SO42–
(5.25)
(5.23)
Again, chloride, nitrate, or other anions could be shown in these equations in
place of sulfate.
If we want to use CO2 to remove the calcium from the high pH water, we
must first separate the CaCO3 and Mg(OH)2 sludge in a clarifier and let the
clarified water flow into another tank. In this second tank, CO2 is diffused into
the water from the bottom and solid CaCO3 particles form instantly. The water
then flows into another sedimentation tank where this additional CaCO3
sludge is removed. These additional tanks add to the capital cost of the system,
and may make the use of CO2 uneconomical. Using CO2 to precipitate the calcium is called primary recarbonation, and is shown in Eq. (5.26).
Ca2+ + CO2 + 2 OH–
→
CaCO3↓ + H2O
(5.26)
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193
Note the consumption of OH– ions in Eq. (5.26). When CO2 is used to precipitate Ca+, the pH begins to drop. Thus, such precipitation is limited by equilibrium as explained below.
Regardless of whether soda ash or primary recarbonation is used to
remove calcium at the end, the clarified water remains at high pH, and such
water is not acceptable to consumers. The clarified water flows into the next
tank where secondary recarbonation (adding CO2) is practiced to reduce the
pH and to stabilize the water. CO2 is an acid gas; it consumes OH– ions when
dissolved into water and lowers the pH back to an acceptable level. At the
same time, bicarbonate ions are formed. Stabilized water is in a “good” pH
range and contains small concentrations of calcium and bicarbonate. The calcium and bicarbonate ions in equilibrium in the water help prevent scale
(CaCO3) formation in water distribution systems or consumers’ home fixtures,
and also helps protect against corrosion in those systems. This secondary recarbonation reaction is shown as Eq. (5.27).
CO2 + OH–
→
HCO3–
(5.27)
Because there still may be carbonate ions in the high-pH water, another reaction that can occur during secondary recarbonation is the conversion of the
CO32– ions into bicarbonate ions (which further stabilizes the water), as shown
in Eq. (5.28).
CO2 + CO32– + H2O
→
2 HCO3–
(5.28)
In summary, during the process design of a lime-soda softening plant, we
calculate the required chemical doses and sludge quantities, make our mass
balances, and then make the necessary cost estimates. The softening reactions
are summarized in Table 5.6 on the following page, and the design calculations
are illustrated in the several examples that follow. In some of these examples
we will show complete removal of the calcium, and in some cases we will not.
In practice, engineers often do not try to remove all the calcium, but rather
leave some dissolved in the water, thus reducing the need for injecting excess
calcium back into the water.
EXAMPLE 5.10
A groundwater was analyzed as follows:
CO2 = 40 mg/L as CaCO3
Ca2+ = 250 mg/L as CaCO3
Mg2+ = 20 mg/L as CaCO3
Na+ = 50 mg/L as CaCO3
Alkalinity (HCO3–) = 180 mg/L as CaCO3
Cl– = 140 mg/L as CaCO3
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Chapter Five
Table 5.6
Summary of Chemical Reactions that Occur in Lime-Soda Softening
Unit Operation
Action
Chemical Reaction
Remove CO2
Remove CaCH
Remove CaNCH
Remove MgCH
add lime
add lime
add soda ash
add lime
Remove MgNCH
Remove excess Ca2+
using soda ash
using CO2 (primary)
Recarbonate (secondary)
and stabilize
add lime
CO2 + Ca(OH)2 → CaCO3↓ + H2O
Ca2+ + 2 HCO3– + Ca(OH)2 → 2 CaCO3↓ + 2 H2O
Ca2+ + SO42– + Na2CO3 → CaCO3↓ + 2 Na+ + SO42–
Mg2+ + 2 HCO3– + 2 Ca(OH)2 →
2 CaCO3↓ + Mg(OH)2↓ +2 H2O
Mg2+ + SO42– + Ca(OH)2 → Mg(OH)2↓ + Ca2+ + SO42–
Equation #
add soda ash Ca2+ + SO42– + Na2CO3 → CaCO3↓ + 2 Na+ + SO42–
add CO2
Ca2+ + CO2 + 2 OH– → CaCO3↓ + H2O
(5.24)
(5.25)
(5.23)
(5.26)
CO2 + OH– → HCO3–
CO2 +CO32–+ H2O → 2 HCO3–
add CO2
(5.21)
(5.22)
(5.23)
(5.27)
(5.28)
Determine the reactant concentrations (the “basis”) and sludge concentration produced for softening this water just to the point of removing all the
CaCH and 45 mg/L of CaNCH, but leaving 25 mg/L of CaNCH and all the
magnesium in the water. Do not add excess lime. Use 50 mg/L of secondary
recarbonation to stabilize this water after softening. Also show the concentrations of CaCO3 sludge produced by each reaction.
SOLUTION
A bar chart is constructed first.
mg/L as CaCO3
40
0
250 270
Ca2+
Mg2+
320
Na+
CO2
HCO3–
40
0
Cl–
180
320
The results are summarized as follows:
Process Step
Remove CO2
Remove CaCH
Remove CaNCH
Sec. recarbonation
CaCO3 sludge
Equation
Chemical
Added
Basis, mg/L
as CaCO3
Sludge Produced,
mg/L as CaCO3
5.21
5.22
5.23
5.27
Lime
Lime
Soda ash
CO2
40
180
45
50
40
360
45
0
445
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195
Often hard water contains both calcium and magnesium in large concentrations. A case with removal of both cations is reviewed in the next example.
EXAMPLE 5.11
Consider the groundwater of Example 5.9. Determine the basis for stoichiometric calculations using Eqs. (5.21) to (5.27). Calculate the basis assuming
removal of all the calcium and magnesium. Use soda ash to remove all the
CaNCH (which was added when removing the magnesium). Add 50 mg/L
excess lime and use 50 mg/L of secondary recarbonation. Show the basis for
all chemical additions, and show sludge production.
SOLUTION
A bar chart was constructed for this water in Example 5.9; it is repeated here
for use in solving this problem.
mg/L as CaCO3
40
0
200
Ca2+
320
Mg2+
400
Na+
CO2
HCO3–
40
0
Cl–
280
SO42–
375
400
The results are summarized as follows:
Process Step
Remove CO2
Remove CaCH
Remove MgCH
Remove MgNCH
Remove CaNCH
Add excess lime
Sec. recarbonation
Equation
Chemical
Added
Basis, mg/L
as CaCO3
5.21
5.22
5.24
5.25
5.23
5.27
Lime
Lime
Lime
Lime
Soda ash
Lime
CO2
40
200
2(80) = 160
40
40
50
50
Sludge Produced,
mg/L as CaCO3
CaCO3 Mg(OH)2
40
400
160
80
40
40
One final example is provided to illustrate how the basis numbers are
translated into actual amounts and costs of each chemical that must be purchased or the amount of sludge that must be disposed, daily or annually. When
determining the basis for sludge production, it is customary to show CaCO3
and Mg(OH)2 separately even though they precipitate together as one sludge.
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Chapter Five
EXAMPLE 5.12
A groundwater was analyzed as follows:
CO2 = 30 mg/L as CaCO3
Ca2+ = 190 mg/L as CaCO3
Mg2+ = 120 mg/L as CaCO3
Alkalinity = 240 mg/L as CaCO3
SO42– = 70 mg/L as CaCO3
Determine the chemical requirements for softening, using soda ash instead
of primary recarbonation for final calcium removal. Include calculations for
using 50 mg/L of excess lime and 50 mg/L secondary recarbonation.
Express all answers as concentrations of the chemical, not as CaCO3 equivalents. Given that the water production rate is 8,000 m3/day, calculate the
daily quantity of sludge produced in metric tons (tonnes) per day. Also, calculate the annual cost of lime that must be purchased, assuming lime costs
$140/tonne.
SOLUTION
The bar chart looks like this:
mg/L as CaCO3
30
0
190
Ca2+
310
Mg2+
CO2
HCO3–
30
SO42–
0
240
310
The basis is determined as in previous examples, and chemical requirements
and sludge production (all in mg/L as CaCO3) are determined using Equations (5.21) to (5.27) as follows:
Chemicals Required
(mg/L as CaCO3)
Equation
5.21
5.22
5.24
5.25
5.23
Excess lime
5.27
Totals
Sludge Production
(mg/L as CaCO3)
Basis
Lime
Soda
CO2
CaCO3
Mg(OH)2
30
190
50
70
70
50
50
30
190
100
70
0
50
0
440
0
0
0
0
70
0
0
70
0
0
0
0
0
0
50
50
30
380
100
0
70
0
0
580
0
0
50
70
0
0
0
120
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197
Conversion of the chemical doses and sludge quantities from mg/L as CaCO3
to mg/L as the chemical is done simply by multiplying each number by the
ratio of equivalent weight of each chemical to the equivalent weight of CaCO3
(which is 50); those calculations are summarized below:
mg/L as
CaCO3
Equivalent
Weight
mg/L as
Chemical
Ca(OH)2
Na2CO3
CO2
440
70
50
37
53
22
326
74
22
CaCO3 sludge
Mg(OH)2 sludge
580
120
50
29
580
69.6 (~70)
Total Sludge (dry) 650
Chemical
The daily mass rate of sludge production (dry basis) is the volumetric water
flow rate multiplied by the total sludge concentration.
8, 000 m 3 1, 000 L 650 mg 1 tonne 5.2 tonnes
¥
¥
¥ 9
=
day
L
day
m3
10 mg
The annual cost of lime is calculated similarly.
8, 000 m 3 1, 000 L 326 mg 1 tonne 365 day $140 $133, 000
=
¥
¥
¥ 9
¥
¥
yr
day
L
yr
tonne
m3
10 mg
Membrane Processes
Membranes used in potable water treatment are polymeric or cellulosic
materials that allow small molecules (like water) to pass through them but
retain larger substances. The substances that are retained vary depending on
the type of membrane, but can range from suspended fine particles to macromolecules (MW = 1,000,000) to dissolved ions. As noted earlier, membranes are
usually grouped into four categories based on the sizes of substances that are
rejected: microfiltration (MF), ultrafiltration (UF), nanofiltration (NF), and
reverse osmosis (RO). MF and UF can remove particles as small as 0.05 to 0.1
micrometers (microns), including bacterial cells. In the case of RO membranes,
even salt ions can be rejected, and only water molecules pass through.
In general, membranes work in the following manner. The raw water is
pressurized by a pump and forced into multiple pressure vessels that hold the
membranes. The pressure vessels are made of fiberglass-reinforced resin rather
than metal to avoid corrosion. When using RO and NF membranes, each vessel
holds between 6 to 8 membrane elements (an element is a continuous thin-film
membrane manufactured into a spiral-wound configuration). Under high pressure, water molecules pass through the membrane, but dissolved salts and
larger dissolved organic molecules are retained on the membrane surface.
Some of the water does not pass through the membrane but rather flows out
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Chapter Five
from the process train, carrying away the salts and other materials that were
retained. With MF and UF processes, the pressure vessels hold hollow-fiber
membranes that are configured into bundles. Essentially all of the water passes
through the membrane fibers, leaving behind a coating of particles on the
membrane surface.
Figure 5.7 is a photograph of the RO pressure vessels and feed pumps at a
4.5 MGD water-treatment plant for the city of Sarasota, Florida. The plant has
three parallel RO process trains, rated at 1.5 MGD each. The plant treats brackish groundwater that contains about 2,250 mg/L of TDS.
The cleaned water stream that is passed through NF and RO membranes is
called the permeate, while the reject water stream is called the concentrate.
Membrane recovery is defined as the percentage of the raw water that is produced as permeate. NF and RO membrane plants often can recover between 70
and 90% of the raw water as permeate. Because various anions become more
concentrated in the concentrate stream, salts may precipitate on the surface of
the membranes, limiting their recovery. The concentrate must be disposed of
and may require further treatment.
RO and NF processes may have difficulty disposing of their concentrate.
Sometimes the concentrate is simply discharged to a drainage/drying field or
back into the source surface water (but downstream of the intake of the plant).
But more often, this stream must be treated before discharge. The concentration
of rejected ions in the concentrate streams from RO and NF plants can be calculated by material balance.
Figure 5.7 A 4.5 MGD reverse osmosis water-treatment plant for the city of
Sarasota, Florida. (Courtesy of Dr. Steven J. Duranceau, design engineer for this plant.)
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Since the 1980s, membrane-based water treatment plants have become
much more common. Even with the high costs of pressurized pumping, membrane processes are cost competitive with traditional methods. Further, they
can be used to treat source water that may be unsuitable for conventional treatment—such as brackish water or even ocean water. While more widely used for
surface water treatment, they also have been used in softening groundwater. A
large-scale membrane plant has hundreds of individual pressure vessels operating in parallel and series to process enough water to meet the design flow.
Because RO and NF membrane plants retain dissolved ions in the concentrate stream, most of those ions are carried out of the system. But, during operation, even with the concentrate stream carrying away much of the retained
material, some of the slightly soluble solids may precipitate on the surface of
the membrane, forming a scale and fouling the outer surface of the membrane.
Periodically, the membranes must be chemically cleaned by adding acid (to
remove insoluble salts) followed by a base (to remove organic colloids), which
removes most of the foulant from the surface of the membrane. Typically, for
an MF or UF system, membranes are routinely cleaned by backwashing (flowing clean water backward through the membranes, with or without chemical
enhancement). Such cleanings consume some of the permeate water, lowering
the effective recovery of the membranes. Occasionally, a more thorough cleaning is needed (often called “clean-in-place”), which is similar to the methods
used for RO and NF membranes.
The design of a membrane plant begins with determining an appropriate
water flux rate—the flow rate of permeate water divided by the membrane surface area. This rate is dependent on the source water chemistry and on its temperature (which influences its viscosity), and is usually determined in pilot-scale
studies. The units are usually given as gpd/ft2 or L/day-m2. A high flux rate
means that fewer elements will have to be purchased, saving capital costs, but will
require higher pressures and more frequent cleanings, leading to higher operating
costs. A low flux rate means that more elements will be required, resulting in
higher capital costs but lower operating costs. Typical flux rates range between 35
to 55 gpd/ft2 for UF, 15 to 25 gpd/ft2 for NF, and 8 to 20 gpd/ft2 for RO processes.
EXAMPLE 5.13
A reverse osmosis membrane water-treatment plant is being designed to treat
brackish water and to produce 2.0 MGD of permeate. The raw water has a salt
concentration of 5,000 mg/L. The production recovery is expected to be 82%,
and the water flux rate is 12.5 gpd/ft2. The finished water must contain no
more than 500 mg/L of TDS.
(a) How much water should be pumped to the membranes?
(b) What is the flow rate and TDS concentration of the concentrate stream?
(c) If each membrane element contains 400 ft2 of membrane surface and if
each pressure vessel holds 8 elements, how many pressure vessels will be
in operation?
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Chapter Five
SOLUTION
(a) Water flow to membranes = 2.0 MGD/0.82 = 2.44 MGD
(b) Flow rate of concentrate = 2.44 – 2.0 = 0.44 MGD
Concentration of TDS =
2.44 ¥ 5, 000 - 2.0 ¥ 500
= 25, 500 mg/L
0.44
6
(c) Total area needed =
Elements needed =
2.0 (10 ) gal/day
12.5 gpd/ft
2
160 , 000 ft 2
400 ft 2 /element
Pressure vessels needed =
= 160, 000 ft 2
= 400 elements
400
= 50 pressure vessels
8
Treatment and Disposal of Residuals
Water treatment plants produce residuals streams that must be handled
and disposed of properly. Sludges are usually thickened in a gravity thickener,
and then pumped to a lagoon or drying beds. The water content either evaporates or percolates into the soil, leaving the solids behind. Calcium and magnesium sludges may be land applied and may have beneficial effects on crops
and ornamental plants, whereas there are concerns about the long-term effects
of aluminum sludges on food crops. Alum sludge is gelatinous and difficult to
dewater. Millions of tons (dry basis) of lime, iron, and alum sludges are generated in the United States each year. Some of this sludge is land-applied, and
some is disposed of in landfills, especially in urban areas.
PROBLEMS
5.1
It has been demonstrated that HOCl molecules are more effective for disinfection than OCl– ions. Assume you put a certain amount of HOCl into
water and adjust the pH independently with NaOH. Which pH would
produce better disinfection: pH 7 or pH 8? Prove your answer with calculations. The KA for HOCl is 3.2 × 10–8.
5.2
Using a spreadsheet for calculations and plotting, determine the molar
concentrations of HOCl and OCl– for pH values between 6 and 8 (increment by 0.1 pH units). Assume the total free chlorine residual is 10–4
gmol/L. Plot the results on a semi-log scale (log concentration vs. pH).
The KA for HOCl is 3.2 × 10–8.
5.3
A potential groundwater supply for a small community was analyzed as
follows:
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201
CO2 = 10 mg/L
Iron = 1 mg/L
H2S = 0.5 mg/L
Calcium = 40 mg/L
Magnesium = 10 mg/L
Alkalinity = 150 mg/L as CaCO3
pH = 7.1
By comparing this raw water with primary and secondary water standards, determine treatment objectives for development of this source as a
potable supply. List the objectives and provide a very brief rationale for
your list.
5.4
A groundwater was analyzed as follows:
CO2 = 30 mg/L as CaCO3
Ca2+ = 180 mg/L as CaCO3
Mg2+ = 120 mg/L as CaCO3
Na+ = 80 mg/L as CaCO3
Alkalinity = 360 mg/L as CaCO3
SO42– = 10 mg/L as CaCO3
Cl– = 10 mg/L as CaCO3
pH = 7.2
Would softening be recommended for this water? Determine the following quantities (as mg/L of CaCO3): total hardness, calcium hardness
(both carbonate hardness [CH] and non-carbonate hardness [NCH]), and
magnesium hardness (both CH and NCH).
5.5
Describe the purpose of adding excess lime during lime-soda softening.
5.6
Describe the purpose of recarbonation after lime-soda softening.
5.7
A groundwater was analyzed as follows (each concentration is given as
the substance itself except for alkalinity):
CO2 = 35 mg/L (as CO2)
Ca2+ = 200 mg/L (as Ca2+)
Mg2+ = 60 mg/L (as Mg2+)
Na+ = 35 mg/L (as Na+)
Alkalinity = 400 mg/L (as CaCO3)
pH = 7.0
Convert the stated mass concentrations into mg/L as CaCO3. Note that at
pH 7, alkalinity is about equal to the HCO3– ion concentration. Determine
the total hardness of this water. Would softening be recommended for
this water?
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Chapter Five
5.8
Consider the groundwater of Problem 5.7. For a water production rate of
3.0 MGD, determine chemical requirements (lime, soda-ash, and CO2, all
in lbs/day) and sludge production rates (dry basis, lb/day) for lime-soda
softening of this water. Use soda ash instead of primary recarbonation.
Use 50 mg/L of excess lime, and 50 mg/L of secondary recarbonation.
5.9
A groundwater was analyzed as follows (each concentration is given as
the substance itself except for alkalinity):
CO2 = 35 mg/L (as CO2)
Ca2+ = 100 mg/L (as Ca2+)
Mg2+ = 60 mg/L (as Mg2+)
Na+ = 35 mg/L (as Na+)
Alkalinity = 300 mg/L (as CaCO3)
pH = 7.0
For a water production rate of 1.0 MGD, determine chemical requirements (lime, soda, CO2, all in lbs per day) and sludge production rates
(dry basis, lb/day) for lime-soda softening of this water. Use soda ash
instead of primary recarbonation. Use 50 mg/L of excess lime, and 50
mg/L of secondary recarbonation.
5.10 Rapid-mixing facilities in a water treatment plant must be designed to
achieve a minimum HRT of 30 seconds. For a water production rate of
500,000 gallons per day, determine the dimensions of rapid-mixing tanks.
There should be a minimum of two tanks in parallel for reliability. Use a
square tank configuration with the depth equal to 1.25 times the width.
Calculate the actual HRT for your recommended tanks.
5.11 Flocculation basins in a water treatment plant must be designed to
achieve a minimum HRT of 45 minutes. For a water production rate of
475,000 gallons per day, determine the dimensions of flocculation basins.
There should be a minimum of two basins in parallel for reliability. Calculate the actual HRT for your recommended sizes. Use a rectangular tank
configuration with a width equal to one-third of the tank length. The
width and the depth should be equal.
5.12 Sedimentation tanks in a water treatment plant must be designed to
achieve a minimum HRT of 3 hours and a maximum OFR of 750 gal/dayft2. For a water production rate of 500,000 gallons per day, determine the
dimensions of sedimentation tanks. There should be a minimum of two
tanks in parallel for reliability. Use a rectangular tank configuration with
a width equal to one-tenth of the length. For a final water depth of 12 feet,
calculate the actual HRT and OFR for your recommended tanks.
5.13 Sand filters in a water treatment plant must be designed with a maximum
hydraulic loading rate of 5 gal/min-ft2. For a water production rate of
500,000 gallons per day, determine the dimensions of filtration facilities.
There should be a minimum of four filters in parallel for reliability. Calculate the actual HLR for the recommended facilities. Use a square tank con-
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Potable Water Treatment
203
figuration with a filter media (sand) depth of 24 inches. Allow a head
space of 5 feet above the sand media.
5.14 For the specifications of Example 5.7, determine the required HRT for
one CSTR.
5.15 For the specifications of Example 5.7, determine the required total HRT
for a dispersed plug flow reactor that can be modeled as three equal-volume CSTRs in series.
5.16 For the specifications of Example 5.7, determine the required total HRT
for a dispersed plug flow reactor that can be modeled as five equal-volume CSTRs in series.
5.17 Size circular clarifiers to settle the stream from the flocculators in a water
treatment plant. The water flow rate is 5.0 MGD, the design overflow rate
is 725 gpd/ft2, and the hydraulic residence time is 4 hours. Should you
use two clarifiers in parallel or more?
5.18 Using Eqs. (5.9) to (5.11), derive Eq. (5.12).
5.19 A rectangular settling tank is being designed to settle sand particles that
have a settling velocity of 2.0 cm/s. The water flow is 10,000 m3/day. The
tank length should be four times the width and the width is equal to the
depth. Calculate the dimensions (m).
5.20 Given a finished water flow rate of 2.5 MGD, a chlorine contact chamber
is designed as a rectangular cross-section concrete channel that is 5 ft
deep, 10 feet wide, and 75 ft long. Calculate the design HRT (minutes).
5.21 Assuming a constant free chlorine residual of 0.25 mg/L, and a rate constant of 2.0 L/mg-min, how long (in feet) would the chlorine contact
chamber of Problem 5.20 need to be to achieve a 99.99% pathogen
destruction efficiency?
5.22 You are designing a rapid-mix basin at a water treatment plant that is
treating 8.0 MGD. You want a square tank with a water depth = 1.3 times
the width of the tank. The detention time is to be 30 seconds with a velocity gradient of 900 sec–1. The water temperature is 60 °F, at which the viscosity is 1.13 cp. Calculate the tank dimensions (in ft) and the power
required (in HP).
5.23 An ultrafiltration membrane plant is being designed to treat a very low
turbidity surface water. The design flux rate is 50 gpd/ft2, and the recovery is 95%. The plant must provide a potable water production rate of 2.5
MGD. Calculate the membrane area required (ft2). Also calculate the raw
water feed rate (MGD).
5.24 A reverse osmosis plant is being designed to treat a hard groundwater.
The design flux rate is 15 gpd/ft2, and the recovery is 78%. The plant must
provide a potable water production rate of 3.5 MGD. Calculate the membrane area required (ft2). Also calculate the raw water feed rate (MGD).
5.25 Sedimentation tanks are needed for a water treatment plant using alum
as a coagulant. Use the following criteria: OFR (maximum) = 900 gpd/ft2;
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Chapter Five
HRT (minimum) = 4 hours. The flow rate is 5,000,000 gallons per day; a
minimum of two sedimentation basins in parallel is desired. Recommend
the dimensions for a pair of identical circular sedimentation basins. Calculate the final OFR and HRT for the recommended system. What are the
side-water depth and the actual tank wall depth of your clarifiers?
REFERENCES
Mines, Jr., R. O., and L. W. Lackey. 2009. Introduction to Environmental Engineering. New
York: Prentice-Hall.
Okun, D. A. 1996. “From Cholera to Cancer to Cryptosporidiosis.” Journal of Environmental Engineering Div. ASCE, 122:453.
Otterstetter, H., and G. Craun. 1997. “Disinfection in the Americas: A Necessity.” Journal
of American Water Works Assoc., 89(9):8.
Reynolds, T. D., and P. A. Richards. 1996. Unit Operations and Processes in Environmental
Engineering. 2nd ed. Boston: PWS Publishing.
Sherman, I. W. 2006. The Power of Plagues. Washington, DC: ASM Press.
US EPA (Environmental Protection Agency). 2012. “Current EPA Microbial and Disinfection Byproduct Regulations.” Accessed June 2013. http://www.epa.gov/
envirofw/html/icr/regulations.html
US EPA. 2013. “Drinking Water Contaminants.” Accessed June 2013. http://water.epa.gov/
drink/contaminants/index.cfm
Viessman, Jr., W., and M. J. Hammer. 2005. Water Supply and Pollution Control. Upper
Saddle River, NJ: Pearson Prentice-Hall.
WHO (World Health Organization). 2011. “Fighting the Rise in Cholera Cases in Haiti.”
Accessed June 2013. http://www.who.int/hac/crises/hti/highlights/june2011/en/
WHO. 2013 (April). “Diarrhoeal Disease.” Fact Sheet No. 330. Accessed January 2013.
http://www.who.int/mediacentre/actsheets/fs330/en/index.html
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CHAPTER
6
Wastewater Treatment
6.1 Introduction
Figure 6.1 repeats the first figure from Chapter 5. But now our attention
turns to wastewater treatment processes. Wastewater is the term we use to
describe water that is discarded after being used (such as used dishwasher
water from a restaurant or bath water from your home), or that has been used
to carry away various wastes (such as human excreta from homes or businesses). As with potable water treatment plants, the design of a wastewater
treatment plant begins with knowing two crucial pieces of information: (1)
where you are (the flow rate and characteristics of the wastewater that is
received at the treatment plant), and (2) where you want to go (the regulatory
and other standards that must be achieved in the effluent). Once the starting
point and the treatment goals are known, the engineer can design a process to
get from where you are to where you want to go.
As indicated in Figure 6.1, wastewater is collected from numerous places
within a city or county and routed to one of perhaps several large integrated
Figure 6.1 Flow
of water from the
environment to
users and back.
Potable Water
Treatment
Water
Users
in the
Environment
Wastewater
Treatment
205
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206
Chapter Six
treatment facilities in the region. Wastewater treatment plants (WWTPs) are
also called publicly owned treatment works (POTWs); the two terms will be
used interchangeably in this chapter. Large POTWs typically handle flows
from ten million gallons per day (MGD) to hundreds of MGD, but there are
numerous small and medium-sized plants in the United States. According to
the EPA, there are about 533 large plants serving about 62% of the population
of this country, while there are more than 13,000 POTWs of less than 1.0 MGD
capacity serving about 11% of the population (US EPA 2013a). Water usage in a
municipality is mostly nonconsumptive, meaning that most of the water that is
purchased from the city is used and then returned to the city via the wastewater collection system (sewer system). Of course, when the water is returned, it
has been polluted to such an extent that it would be harmful to simply discharge it back into the environment as is. It is worth noting, however, that
municipal wastewater is still about 99% water and only 1% pollutants!
Wastewater treatment plants are designed to remove or transform the various
pollutants that are commonly found in municipal (and in many industrial) wastewater streams. The major feature that makes wastewater treatment so interesting
is that we use natural microorganisms to do most of the work for us. Using these
microorganisms has several advantages—they work very cheaply (only needing
food, air, and nutrients), they don’t complain, and they never go on strike! More
details of how we design these WWTPs will be presented later in this chapter.
Figure 6.2 shows an aerial view of a large municipal WWTP in Orange
County, Florida, whose capacity is 43 MGD. It is the largest of three regional
plants in Orange County (the other two have a combined capacity of 31 MGD).
Figure 6.2 The South Water Reclamation Facility (SWRF) in Orange County, Florida, which has 43 MGD capacity. (Courtesy of Orange County Utilities.)
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These three plants are called water reclamation facilities because their treated
effluent is reclaimed water—that is, nearly 100% of it is used productively. Florida and a few other states (e.g., California, Texas, Arizona) have active programs of reclaiming water for beneficial uses (initially, reclaimed water was
hard to give away, but now it is considered a valuable resource and is sold).
Altogether, throughout the state of Florida in 2007, about 663 MGD of reclaimed
water were produced and utilized, although in some counties much of the
treated effluent was discharged to surface waters. In 2007, reclaimed water was
used in Florida for irrigating golf courses and public areas (59%), providing
cooling water to power plants and industries (14%), providing agricultural irrigation (12%), percolating through sandy soil to recharge groundwater supplies
(11%), or for other uses such as nourishing wetlands or irrigating lawns at private residences (4%) (Toor and Rainey 2009). Signs are often posted to alert people about the reclaimed water (so they won’t drink it)—see Figure 6.3.
In most cases, the major pollutants of concern in municipal wastewater are
total suspended solids (TSS), BOD, and nutrients, although all of the pollutants
mentioned in Chapter 4 have been found in various wastewaters at various
times. Municipal wastewater is collected from residential, commercial (stores,
restaurants, hotels, etc.), and public facilities within the city, whereas industrial
wastewater comes from various smaller, light industries that might be located
within the city (larger, heavy industries usually have their own on-site treatment plants). Small industries may have some on-site pretreatment or may
simply connect directly to the municipal sewer system. In some suburban areas
and in most rural areas, residential wastewater may not be collected at all;
rather it is treated on each homeowner’s property via septic tanks and drainfields located in the back or front yard, usually just a few feet underneath the
Figure 6.3
Sign at a golf course in Central Florida.
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Chapter Six
surface of the ground (although some drainfields may be located above ground
due to a very high water table).
6.2 Characterization of Wastewaters
Raw municipal (domestic) wastewaters exhibit a general consistency in the
types of constituents, but the numeric values of concentrations may vary somewhat depending on geographic location, the affluence of the residential areas,
and the presence or absence of various commercial establishments. Certain
generalizations regarding quantity and quality of municipal wastewater are
often reasonably accurate. Municipal wastewater constituents include BOD,
suspended solids, nutrients, and many other pollutants, and the flow typically
varies during the day (with peak flows usually occurring in the morning).
Industrial and agricultural wastewaters can be extremely variable in composition and flow. Industries often have very high concentrations of BOD and
suspended solids, but also may have one or more specific pollutants of concern
(e.g., chromium from a chrome-plating plant). Flow rates may vary widely due
to plant operating schedules. Agricultural wastewaters often have very high
concentrations of suspended solids (soil, etc.) and fertilizer nutrients (N and P),
but also may carry significant concentrations of BOD and chemical pesticides.
Flow rates may vary extremely due to dry and wet seasons or extreme rainfall
events. If agricultural wastewaters are treated, it is typically done on-site in
lagoons or stabilization ponds.
An engineering handbook or textbook on industrial wastewater treatment,
for example, the Water Environment Federation’s manual (2008), can be very
useful for preliminary characterization of different types of industrial wastewaters, but site-specific sampling is necessary to gather the detailed data
required for design. For an agricultural wastewater, site-specific sampling (in
both dry and wet seasons) is necessary for characterization. Flow rates and
loadings of all three types of wastewater (municipal, industrial, and agricultural) can vary widely due to rainfall events. Loading is a term used to indicate
the mass flow rate of a pollutant—it is simply the product of the wastewater
flow rate and the pollutant concentration.
Municipal Wastewaters
Most of the focus in this chapter is on municipal or domestic wastewater
(WW). Typical concentrations for “weak,” “typical,” and “strong” domestic
wastewaters are reported in Table 6.1. Note that even if you add up all the concentrations of pollutants, WW is still about 99% water and 1% pollution. So, one
of the main concerns in designing facilities to treat wastewater is the cost of moving the water. Engineers often try to take advantage of natural elevation changes
and use gravity to help move water through the plant to keep the total treatment
cost reasonably low. Owing to the variability in flow rates that results from differing water usage in different cities, and to infiltration and inflow and their diluting
effects on the concentrations, the values listed in Table 6.1 can vary significantly.
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Table 6.1
209
Composition of Untreated Domestic Wastewater
Parameter
Alkalinity (as CaCO3)
Ammonia (as N)
BOD5
COD
Fixed Suspended Solids
Grease
Inorganic Phosphorus (as P)
Nitrate (as N)
Nitrite (as N)
Organic Nitrogen (as N)
Organic Phosphorus (as P)
Total Dissolved Solids
Total Nitrogen (as N)
Total Phosphorus (as P)
Total Suspended Solids
TOC
Volatile Suspended Solids
Weak
Typical
Strong
50
12
110
250
20
50
3
0
0
8
1
250
20
4
100
80
80
100
25
220
500
55
100
5
0
0
15
3
500
40
8
220
160
165
200
50
400
1,000
75
150
10
0
0
35
5
850
85
15
350
290
275
Note: All units in mg/L.
Source: Adapted from Metcalf & Eddy (2003).
Infiltration is the seepage of groundwater into the underground sanitary
sewer transport pipes that carry the WW to the WWTP. Infiltration enters
through cracks in the pipes or open joints between sections of pipe that have
settled over time. Inflow refers to stormwater getting into the collection system
through improper direct connection to the sanitary sewer system, sometimes
greatly increasing the flow of WW to be treated. If the engineer does not provide an allowance for infiltration/inflow, the WWTP would be significantly
undersized. Modification of this default value is often appropriate to account
for local infiltration/inflow conditions and water consumption.
Average water usage in municipalities is highly variable, and has been
reported by the USGS as varying from 43 to 177 gallons per capita per day
(gpcd) in 21 selected US cities (Kenny and Juracek 2012). A rule-of-thumb average is about 100 gpcd; nominal residential (both indoor and outdoor) use is
about 85–90 gpcd (US EPA 2013c). The 100 gpcd number includes the average
daily water used in restaurants, hotels, and other commercial facilities within
the city, as well as that used at people’s homes, divided by the population of the
city. A significant portion of the water supplied to all customers may be used
for irrigation or other consumptive uses, so the amount that finally is discharged as municipal WW is lower than the amount supplied. In that regard,
homeowners in many municipalities may install a separate water meter and
lines for lawn irrigation purposes to avoid paying sewer charges on the irrigation water supplied. The total flow rate of WW received at WWTPs (including
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Chapter Six
an allowance of 15 gpcd for normal infiltration/inflow) is commonly estimated
using a factor of 100 gpcd (Ten States Standards 2004). Some simple calculations
dealing with flow rates and loadings are presented in Examples 6.1 and 6.2.
EXAMPLE 6.1
Estimate the average daily flow of WW for a city of 80,000 people, in millions
of gallons per day (MGD); also estimate the daily BOD5 loading going to the
WWTP, in kg/day and in lb/day.
SOLUTION
Lacking any site-specific information, use 100 gpcd and 220 mg/L for the
flow per capita and BOD5 concentrations, respectively.
Daily flow:
100 gpcd × 80,000 people = 8 MGD
BOD5 loading:
6
8.0 (10 ) gal 220 mg 3.785 L
1 kg
¥
¥
¥ 6
= 6, 660 kg/day
day
L
gal
10 mg
orr
6, 600 kg 2.205 lb
¥
= 14, 700 lb/day
day
kg
EXAMPLE 6.2
Wastewater from a food-processing operation is collected from various places
in the facility and then flows into a municipal sewer system. During the dry
season the average flow is 12 m3/min and the TSS is 125 mg/L. During the
rainy season, there is significant inflow and infiltration due to many leaks in
the facility’s old WW collection system, and the flow is 150 m3/min. There is
increased erosion in the rainy season, so despite increased dilution due to
more water, the concentration of TSS is 265 mg/L. Estimate the average daily
loading of TSS being carried into the city’s WWTP in both seasons, in kg/day.
SOLUTION
Dry:
1 kg
1, 440 min
12 m 3 1, 000 L 125 mg
¥
¥
¥ 6
¥
= 2, 160 kg/day
3
min
day
L
10 mg
m
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211
Rainy:
1 kg
1, 440 min
150 m 3 1, 000 L 265 mg
¥
¥
¥ 6
¥
= 57 , 200 kg/day
3
min
day
L
10 mg
m
Industrial Wastewaters
Based on the Pollution Prevention Act of 1990, regulatory emphasis has
been placed on modification of manufacturing processes to alter the characteristics of industrial wastewaters, thereby reducing the quantity (either volume
or mass) and toxicity of materials that require treatment. It is noted that such
emphasis also applies to industrial agricultural sites (such as cattle feedlots,
large poultry-processing plants, etc.). These agribusiness industrial plants are
similar to any other large industry in terms of characterizing their wastewater
problems and designing treatment plants. Waste minimization and pollution
prevention have been widely practiced by industry for many years to minimize costs associated with end-of-pipe wastewater treatment. These pollution
prevention or waste minimization efforts are outlined in the EPA hierarchy
(US EPA 1988) presented earlier in Chapter 1 (Figure 1.5) and repeated in
words below:
1. Source Reduction. Reduce the amount of waste at the source through
changes in industrial processes.
2. Recycling. Reuse and recycle wastes for the original or some other purpose, such as materials recovery or energy production.
3. Incineration/Treatment. Destroy, detoxify, and neutralize wastes into less
harmful substances. (It is noted that incineration is a highly controversial issue in most communities, and it can be very difficult to get an
incinerator permitted.)
4. Secure Land Disposal. Deposit wastes on land using volume reduction,
encapsulation, leachate containment, monitoring, and controlled air and
surface/subsurface waste releases.
6.3 Standards for Wastewater Treatment
After treatment, the treated effluent may be reclaimed/reused or may be
discharged into a body of surface water (river, lake, ocean). Surface waters in
the natural environment have some capacity for self-cleaning. That is, small
concentrations of BOD may be removed by native bacteria in the receiving
water body; TSS may settle out in ponds, marshes, or lakes; nutrients may be
removed via plant growth. However, natural systems can be overloaded, and if
they are, their quality degrades. Thus, federal goals for wastewater treatment
were developed to protect surface waters, and were first formally stated in the
Water Pollution Control Act Amendments of 1972 (PL 92-500) as follows:
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Chapter Six
1. To eliminate the discharge of pollutants.
2. Wherever possible, to have water quality suitable for sustaining fish,
shellfish, and wildlife, and for recreational purposes.
3. To prohibit the discharge of toxic pollutants.
Water quality criteria were established to protect beneficial uses of surface
waters and may be classified as follows (Dietz 1996):
1. Maintenance of adequate dissolved oxygen to support desirable aquatic
life forms. A minimum dissolved oxygen concentration of 5 mg/L is
commonly accepted for support of sport fish species.
2. Reduction of plant and algal nutrient levels to avoid eutrophication
problems. (Note that eutrophication is more likely to be a concern for discharge to lakes and reservoirs than for discharge to free-flowing rivers.)
3. Maintenance of concentrations of toxic substances at values that do not
pose a threat to aquatic species and/or a potable water supply (for
example, the Mississippi River). Toxic substances are more commonly
associated with industrial sources and are typically regulated for each
industrial category. Whole effluent toxicity bioassay testing may be
required to verify the absence of toxic agents in effluents.
4. Elimination of pathogens to control transmission of waterborne diseases. Disinfection of wastewaters which pose a risk of disease transmission is prescribed prior to discharge.
5. Maintenance of suitable aesthetic qualities to foster recreational use of
the surface water resources.
National Pollutant Discharge Elimination System (NPDES)
Section 402 of PL 92-500 established procedures for issuance of discharge
permits for municipal and industrial wastewaters (Nemerow and Dasgupta
1991; US EPA 2010). The NPDES permitting program may be delegated to individual states, but all dischargers are required to have a permit. NPDES permits
allow both municipal and industrial treatment plants to be designed and operated to specific numeric criteria. These permits identify maximum allowable
concentrations of pollutants that may be present in a facility’s discharge. Monitoring and reporting requirements are also prescribed in the permits. The permits may also specify maximum daily loadings that can be added to the
receiving water body, regardless of the concentrations.
Municipal Discharge Standards
In general, when we look at the history of treating municipal wastewater,
we can classify such treatment into three broad categories: primary, secondary,
and tertiary treatment. Prior to the late 1800s–early 1900s, cities simply collected wastewater (often in open ditches) and routed it to the nearest lake or
river. As cities got larger, such direct discharge began to overload and pollute
the receiving water body. Primary treatment plants were built with the main
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213
goal of removing suspended solids and other solid objectionable material, but
the disinfection of the effluent prior to discharge was also a significant benefit
in controlling waterborne diseases.
By the 1940s, people realized that dissolved BOD was having severe
impacts on many rivers and lakes, and secondary treatment was developed.
Secondary treatment removed more of the TSS and utilized microorganisms
and air to oxidize (in the treatment plant) most of the organic compounds that
otherwise would consume oxygen in the receiving waters. In the 1970s and
1980s, awareness had grown and technology had developed so that we could
begin practicing biological nutrient removal (sometimes called tertiary treatment) to remove the nutrients (nitrogen and phosphorus) that were stimulating plant and algae growth and negatively impacting the ecology of the
receiving waters. In some modern plants, nutrient removal is practiced as part
of secondary treatment, but in many others, nutrient removal is achieved using
separate unit operations/processes that follow conventional secondary treatment—hence the name tertiary treatment.
The level of treatment required of POTWs will vary as needed to maintain
receiving water quality standards. If the DO concentration in the receiving lake
or river can be maintained above the regulatory criterion (typically 5 mg/L),
secondary treatment often is adequate. Secondary treatment standards as
shown in Table 6.2 include effluent concentrations of 30 mg/L of BOD5 and 30
mg/L of TSS. These standards
are typically achieved by
standard biological treatment
Table 6.2 Secondary Treatment Standards
processes, as will be discussed
Maximum Concentrations, mg/L
in detail in a subsequent section. Reclaimed water may
Parameter
Avg. Monthly
Avg. Weekly
not need to meet DO or nutriBOD5
30
45
ent standards because it is not
TSS
30
45
discharged directly to rivers
pH
6 to 9
—
or lakes. Disinfection would
—
Removal
85% of BOD5 and TSS
typically also be required,
Source: US EPA (2010).
which is accomplished by
chlorination in the United
States or by ozone treatment in Europe. Dechlorination may be practiced prior
to discharge to a receiving water to mitigate any toxic effects associated with the
disinfectant residual.
In those cases where discharge of a secondary effluent would not satisfy
receiving water quality criteria, a greater removal of BOD would be required.
The limits in Table 6.2 are maximum concentrations that cannot be exceeded.
Actual numerical limits are site specific, depending on receiving water characteristics, and may be set lower than those in Table 6.2 by local and state regulators. Removal of additional BOD5 and reduced nitrogen compounds (TKN)
may be specified to maintain adequate dissolved oxygen levels in the receiving
water. Also, the discharge permit may contain total nitrogen and phosphorus
limits (see next paragraph).
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Chapter Six
For wastewater discharge into lakes or impoundments, removal of nutrients (nitrogen and phosphorus) may be prescribed to minimize algal growth
potential. This is often called advanced wastewater treatment, and may be part
of the secondary treatment system or may
have separate (tertiary) unit operations/
processes. Specific numerical effluent conTable 6.3 Effluent Standards for
centration limits are site specific. One posAdvanced Wastewater Treatment
sible set of effluent standards which is
(Nutrient Removal)
common in Florida for advanced wastewater treatment is noted in Table 6.3. Tertiary
Maximum Concentration
Parameter
mg/L
treatment can be thought of as a combined
system composed of secondary treatment
BOD5
5
(which may achieve a significant fraction of
TSS
5
Total nitrogen
3
nutrient removal) followed by biological
Total phosphorus
1
processes (nitrification/denitrification, biological phosphorus uptake) and chemical
processes (phosphorus precipitation) to
achieve these limits. Nitrification is the microbial oxidation of ammonia to
nitrate, and denitrification is the microbial reduction of nitrate to nitrogen gas.
Terminal filtration, disinfection, and dechlorination would typically be provided also if the treated effluent is discharged to a receiving water body.
Industrial Discharge Standards
Industrial facilities with direct surface water discharge are subject to specific discharge requirements established in an NPDES permit issued to the
facility. In addition to regulation of conventional pollutants (pH, BOD5, TSS, oil
and grease), standards have been established for each industry defining Best
Available Technology (BAT) for reduction of toxic compounds that may be specific to each industry. Pretreatment standards for industries that discharge to a
POTW are also established for the toxic substances. Again, the federal standards represent minimum pretreatment requirements; individual municipalities may establish pretreatment standards that are more restrictive than the
federal mandates. The specific effluent standards faced by industrial dischargers are quite variable in light of different wastewater characteristics and local
sewer use ordinances.
6.4 Unit Operations/Processes of Domestic Wastewater Treatment
Overview
As discussed in the previous section, wastewater treatment objectives may
vary significantly among POTWs and industrial WWTPs due both to the effluent standards and to the raw wastewater characteristics. From here forward, the
focus is on domestic wastewater and the design and operation of POTWs. We
will focus on large-scale systems, but acknowledge that small “package” plants
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215
and even septic tanks treat a considerable percentage of domestic wastewater.
At the end of this chapter, we will provide a few details on septic tank systems.
A generalized flow diagram of domestic wastewater treatment is shown in
Figure 6.4. The typical POTW combines and utilizes unit operations and processes from primary, secondary, and tertiary treatment as needed, some of
which are very similar to those utilized in potable water treatment. Referring to
Figure 6.4, raw wastewater enters the treatment plant from the collection system and is routed through bar screens whose function is to remove large
objects and floating materials (such as plastics, wood, cigarette butts, etc.). The
water flows into a grit chamber where large dense inorganic particles are
removed. There may be an equalization basin (these are common at industrial
plants) whose purpose is to “smooth out” hourly variations in flow and composition of the wastewater to maintain a steadier flow to the rest of the plant.
All of the treatment through equalization is called pretreatment.
The wastewater next may flow through a primary clarifier where a significant portion of the TSS (and BOD associated with the organic solids) is
removed. This is the traditional primary treatment. It is noted that many modern
plants have eliminated the primary clarifier, allowing the whole wastewater to
enter secondary treatment directly. In classic secondary treatment the wastewater
flows into the “heart of the process”—the activated sludge treatment process.
This is the biological-based treatment part of the plant where most of the BOD
is converted by microorganisms, and then most of the microorganisms (biomass) and remaining TSS are separated from the water. The secondary clarifier
separates the sludge (biomass and other suspended solids) from the effluent
water. The effluent flows to either tertiary treatment or directly to chlorination
and discharge. Some of the sludge is removed from this part of the plant and
sent for further treatment, but most is recycled back into the aeration tank. The
sludge that is discharged from the system goes to further treatment which can
include thickening, anaerobic or aerobic digestion, dewatering, incineration, or
Figure 6.4
Schematic process flow diagram of wastewater treatment.
PRETREATMENT
Raw
WW
Grit
Chamber
Bar screen
Debris
to
Disposal
Grit
to
Disposal
Equalization
Tank
PRIMARY
TREATMENT
Primary
Clarifier
(optional)
SECONDARY
TREATMENT
Aeration
Tank
Recycled
Sludge
Primary Sludge
Secondary
Clarifier
Waste
Sludge
Treatment
Disposal
Effluent
Tertiary
Treatment or
Disinfection
Discharge
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gasification, with final disposal of the inorganic solid residues. All of these
operations will be discussed in the following sections of this chapter.
Bar Screen
The bar screen (or bar rack) is simply a sturdy steel grate that is designed to
remove large objects that would damage or clog downstream equipment (such
as pumps or valves). These include rags, branches, plastic pails, and other
items. The rack often is manually cleaned and the debris is landfilled. Often, a
finer screen is located after the first, coarser screen to remove smaller floatable
objects. The finer screen is cleaned automatically.
Grit Chamber
The grit chamber is a relatively small detention tank where heavy dense
particles settle out as the flow slows down. Items that are removed here
include sand, pebbles, broken glass, small metal objects like nails or screws,
and others. These chambers usually have a device such as a grit pump or clamshell system that continuously removes the grit from the bottom of the tank.
The grit is deposited into a dumpster that is periodically emptied; the debris is
hauled away to a landfill. Typical detention times (θ) or hydraulic residence
times (HRT) in grit chambers are about one minute or so, based on the same
principle that was introduced in Chapter 5, and repeated below in Eq. (6.1).
HRT = θ = V/Q
(6.1)
where:
HRT = hydraulic residence time, min
θ
= detention time, min
V
= tank volume, gal or m3
Q
= flow rate, gal/min or m3/min
Equalization Tanks
Most unit operations and processes run better and do a more efficient job if
they are run at or near steady-state conditions. For operators, if the equipment
and biological processes can be kept at a nearly constant flow rate with nearly
constant concentrations, the treatment plant is much easier to run and produces smoother and better treatment than if things are rapidly changing. However, such operations rarely occur in practice by themselves. For domestic
wastewaters, it is well known that both the flow and strength follow the daily
activities of the population. Both tend to be higher than average during the day,
and much lower late at night. For industrial wastewaters, the hourly variability
can be very significant, and it is very worthwhile to install an equalization
tank after the grit chamber. For domestic wastewaters it can still be economically attractive to include equalization in the design.
Wastewater (WW) flows into the equalization tank directly from the grit
chamber at the rate at which it is received, but is pumped out of the tank to the
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rest of the plant at a constant rate that matches the historical daily average flow
rate. The level in the tank will rise when the WW flows in faster than it is
pumped out, and the level will drop when the inflow rate is less than the
pump-out rate. If the tank is well-mixed, the concentrations of all the pollutants in the WW will be smoothed out as well.
Another reason to include equalization is that a POTW must be designed to
handle the peak flows when they occur. Without equalization, the plant would
have to be built with larger tanks and equipment, incurring a larger capital cost.
During rain events, inflow/infiltration might be quite large, and a large spike in
flow would upset the plant operations. An equalization tank can make good economic sense for industrial WWTPs, but most domestic WWTPs do not have one
unless they have very significant inflow/infiltration issues. The larger the tank, the
more smoothing can be done, but a huge tank is neither practical nor economical.
The analysis of an equalization tank begins with two mass balances—one
on the total flow and one on the pollutant of interest (say, BOD). Since WW is
99% water, we typically assume the density of all streams is constant, so the
flow mass balance becomes a volume balance:
dV/dt = Qin – Qout
(6.2)
where:
V
= volume of liquid in the tank, m3
Qin
= flow rate into the tank, m3/hr
Qout = flow rate out from the tank, m3/hr
The BOD balance is:
dM/dt = Qin Cin – Qout Cout
(6.3)
where:
M
= mass of BOD in the tank, g
Cin
= concentration of BOD in the inflow to the tank, mg/L
Cout = concentration of BOD in stream coming out from the tank, mg/L
Note that the mass of BOD in the tank is simply the product of the volume of
liquid in the tank and the concentration of BOD in the tank at any given time,
and since the tank is well-mixed, the concentration in the tank is the same as
the exiting concentration. Thus:
M = V Cout
(6.4)
The first step in sizing the equalization tank is to gather data on the hourly
pattern of flow and concentration in the incoming WW. Next, solve for the
daily average flow and weighted average strength of the WW. After that, it is
straightforward to set up a spreadsheet to solve Eqs. (6.2) to (6.4) for each hour.
It should be pointed out, however, that the volume of liquid in the tank cannot
go to zero. Pumps need a minimum liquid head above their centerline elevation to
work, so there must always be some amount of liquid in the tank. Further, the pipe
connections from the tank are usually set a short distance above the very bottom
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Chapter Six
level to avoid drawing in a high concentration of solids that may accumulate near
the bottom. (Even though the tanks are well-mixed and are not meant to be used
for settling, some settling does occur.) In addition, when designing and operating
tanks, it is prudent to keep enough liquid in the tank to keep the plant running for
an hour or two if something upstream of the tank goes wrong. (Likewise, a tank
should never be allowed to get 100% full; in case something downstream of the
tank should go wrong, the tank will overflow). Finally, it is convenient to start our
calculations at the point when the tank is at the minimum level; that is, after the
inflow rate has been lower than average for a long time and just starts exceeding
the average. The sizing process is best illustrated in the following example.
EXAMPLE 6.3
The following measurements and “grab samples” of wastewater flow rates
and BOD5 concentrations were taken at the top of each hour over a 24-hour
period. Based on these samples, and assuming they would repeat day after
day, size an equalization tank and demonstrate its effects on “smoothing out”
the flow rate and concentrations going into the downstream equipment.
Assume that the liquid remaining in the tank when it is at its lowest level is
equivalent to about 1.5 hours of pumping at the average rate. Size the tank for
20% more than the calculated maximum volume to allow for contingencies.
Use a spreadsheet to solve this problem numerically.
Hour
Flow m3/hr
BOD5 mg/L
1
2
3
4
5
6
7
8
9
10
11
12 (noon)
13
14
15
16
17
18
19
20
21
22
23
24 (midnight)
250
200
160
150
140
165
220
340
385
410
425
435
455
440
450
475
495
518
480
450
425
405
360
300
104
78
52
42
55
67
90
115
135
140
142
156
168
140
131
118
129
146
188
215
237
183
165
117
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219
SOLUTION
Average hourly flow rate:
Qavg =
 Qi
= 355.54 m 3 /hr
24
Weighted average BOD5 concentration:
BOD 5 -avg =
 Qi Ci
= 143.0 mg/L
 Qi
Select the time when the liquid level in the tank reaches its minimum:
Inspection of the data shows that the flow rate is below average from midnight (hour 24) through 8:00 AM. So 8:00 AM is when the liquid level in the
tank would be at its minimum; the level begins to rise in hour 9, so we start
our spreadsheet at the beginning of hour 9 (9:00 AM).
Volume of liquid at minimum: 1.5 hr × 355.54 m3/hr = 533.3 m3
Assume that the BOD5 concentration in the tank at 9:00 AM starts at the
weighted average or 143 mg/L. (If, after solving the equations, you find that
BOD5 at the end of the 24 hours is not 143, simply replace the starting concentration with the one achieved after 24 lines of calculations and the spreadsheet will recalculate immediately.)
The volume balance on the tank is
dV/dt
=
Qin – Qout
or
∆V/∆t
= Qin – Qout
Keep in mind that Qout is constant at the 24-hour average hourly flow rate.
Solving numerically for the volume at the next hour, we get
Vnew
=
Vold + ∆t × (Qin – Qout)
Similarly, the BOD5 balance is
d(VCout)/dt
=
Qin Cin – Qout Cout
which can be solved for Cout as follows
Vnew Cout-new
=
Vold Cout-old + ∆t × (Qin Cin – Qout Cout-old)
or
Cout-new
=
Vold Cout-old + Dt ¥ (Qin Cin - Qout Cout-old )
Vnew
Set up the spreadsheet to start at 9:00 AM, with an “old” volume and concentration already in the tank. Using a ∆t of 1 hour, solve the equations sequentially around the clock to 9:00 AM the next morning (the end of the hour that
begins with 8:00 AM). The reason to solve to 9:00 AM is because it is a good
check on your calculations. The 9:00 AM volume should be identical to your
Cooper 06.fm Page 220 Thursday, March 3, 2016 10:10 AM
220
Chapter Six
assumed starting value, and the concentration should be close to your
assumed value. The solution appears below:
Hour
Qin
m3/hr
Cin
mg/L
Old Vol.
m3
Old C
mg/L
New V
m3
New C
mg/L
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1
2
3
4
5
6
7
8
385
410
425
435
455
440
450
475
495
518
480
450
425
405
360
300
250
200
160
150
140
165
220
340
135
140
142
156
168
140
131
118
129
146
188
215
237
183
165
117
104
78
52
42
55
67
90
115
533.3
562.8
617.2
686.7
766.1
865.6
950.0
1044.5
1164.0
1303.4
1465.9
1590.3
1684.8
1754.3
1803.7
1808.2
1752.6
1647.1
1491.6
1296.0
1090.5
874.9
684.4
548.8
143.0
137.5
139.2
140.9
149.5
159.2
150.3
142.0
132.2
131.0
136.3
151.9
168.8
185.3
184.8
180.8
169.9
159.9
148.9
137.0
123.9
112.9
101.8
97.1
562.8
617.2
686.7
766.1
865.6
950.0
1044.5
1164.0
1303.4
1465.9
1590.3
1684.8
1754.3
1803.7
1808.2
1752.6
1647.1
1491.6
1296.0
1090.5
874.9
684.4
548.8
533.3
137.5
139.2
140.9
149.5
159.2
150.3
142.0
132.2
131.0
136.3
151.9
168.8
185.3
184.8
180.8
169.9
159.9
148.9
137.0
123.9
112.9
101.8
97.1
108.5
Note that in this first iteration the final concentration is 108.5 (and not 143.0)
mg/L. The reason for this disparity is because we assumed a non-zero starting volume in the tank. But if we now insert 108.5 into the cell at the top to
replace the old concentration at 9:00 AM, the spreadsheet immediately recalculates and the two concentrations will match perfectly.
From the spreadsheet, we can see that the tank must be big enough to hold
1,808.2 m3 at its fullest, which would occur at the end of hour 23 (11:59 PM).
We must allow some further volume for flow variations, downstream equipment outages, etc. Allowing for 20% more volume, we get 2,170 m3. Note that
while the liquid volume in the tank swings dramatically during the day, the
flow rate out of the tank is constant at 355.5 m3/hr. Also, while the BOD5 concentrations coming into the tank range from 42 to 237 mg/L, the range of
BOD5 concentrations coming out of the tank is only 108 to 185 mg/L. We have
achieved our goals.
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Wastewater Treatment
221
Primary Treatment
As was seen in Figure 6.4, primary treatment is an optional next step in the
process. In a great many POTWs, primary treatment is no longer practiced. But
in larger plants, it still can serve an important function. A primary clarifier is a
settling tank, which may be rectangular or circular, and operates much like the
clarifiers discussed in the previous chapter. At a WWTP the solids to be settled
are the ones that are light enough to make it through the grit chamber but are
still heavy enough to settle in a much larger tank, with a longer detention time.
Primary clarifiers do not have a significant effect on the final effluent quality but are used because they reduce oxygen requirements and save operating
costs. The solids that are removed have substantial organic components, so a
sizeable amount of BOD is removed. Thus, they reduce sludge production in
the downstream activated sludge process. The sludge that is removed from the
bottom of the primary clarifier is sent for further treatment because it still contains objectionable organic waste material. A properly designed and operated
primary clarifier can remove roughly one-third of the BOD5 and two-thirds of
the TSS in the WW (Metcalf & Eddy 2003).
There are three main considerations in the design of a primary clarifier:
overflow rate, detention time, and weir loading rate. These terms were all
introduced in the previous chapter, and all have essentially the same definitions here. The overflow rate (OFR) is the volumetric flow rate of WW divided
by the top surface area of the clarifier:
OFR
=
Q/A
(6.5)
where:
OFR = overflow rate, gpd/ft2 (or m/day)
Q
= WW flow rate, gpd (or m3/day)
A
= surface area of clarifier, ft2 (or m2)
Typical values for the OFR in a primary clarifier are 600–1,200 gpd/ft2 (Spellman 2009) or 25–50 m/day (m3/m2-day). During design, the WW flow rate is
known, and an OFR value is selected. The total surface area of the clarifier is
calculated by rearranging Eq. (6.5).
The detention time or hydraulic residence time (HRT) is calculated by
dividing the volume of the tank by the flow rate:
HRT = V/Q
(6.6)
where:
HRT = hydraulic residence time, hours
V
= volume of clarifier, ft3 (or m3)
Q
= flow rate, ft3/hour (or m3/hr)
Typical values of HRT in a primary clarifier are from 1.5–2.5 hours. During
design, the HRT is selected and—with the known flow rate—the total volume
is calculated from Eq. (6.6). The surface area of the clarifier can be obtained
Cooper.book Page 222 Monday, June 23, 2014 9:58 AM
222
Chapter Six
from Eq. (6.5), and the depth of the clarifier is simply the volume divided by
the surface area. For a circular clarifier, the diameter is obtained from the formula for the area of a circle. It is noted that sludge rake mechanisms are available in standard 5-foot increments, so clarifiers are always designed to the
nearest 5-foot diameter.
The weir loading rate is the volumetric flow rate divided by the length of
the exit weir:
WLR = Q/L
(6.7)
where:
WLR = weir loading rate, gpd/ft (or m3/day-m)
L
= length of weir, ft (or m)
Q
= flow rate, gpd (or m3/day)
The length of the weir in a circular clarifier is approximately equal to the circumference of the inside wall. For rectangular clarifiers, the weir may extend
across the width and part of the length near the end opposite from where the
wastewater enters. Typical values of WLR are 10,000–20,000 gpd/ft (Spellman
2009) or 40–80 m3/day-m. During design, after the clarifier is sized using Eqs.
(6.5) and (6.6), the WLR is checked to ensure it falls within the correct range. If
the WLR is not acceptable, the clarifier design can be changed by varying the
depth or the diameter (which also changes the HRT or the OFR).
Often, more than one primary clarifier is built at a plant (especially large
plants, where a lengthy tank outage could prove catastrophic). The calculations
above are for the total area and volume of the clarifier(s), so designers must
take that into account if more than one tank is being specified to operate in parallel. Example 6.4 illustrates this point.
EXAMPLE 6.4
Design two equal-sized circular primary clarifiers to operate in parallel to
handle a nominal flow of 4.5 MGD, keeping the OFR, HRT, and WLR within
typical ranges.
SOLUTION
Because we want two identical tanks, we simply design one tank for half the
flow, keeping all the criteria the same.
A=
2, 250 , 000 gpd
Q
=
= 2, 250 ft 2
OFR 1, 000 gpd/ft 2
V = HRT ¥ Q
= 2 hrs ¥ 2, 250 , 000 gpd ¥
= 25, 067 ft 3
1 day
1 ft 3
¥
7.48 gal 24 hrs
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Wastewater Treatment
Depth =
Diameter =
The final WLR =
223
25, 067
= 11.14 ft
2, 250
4 ¥ 2, 250
= 53.5 ft (round up to 55 ft)
p
2, 250 , 000
= 13, 022 gpd/ft
p ¥ 55
2
The final area =
The final OFR =
p ( 55)
4
= 2, 376 ft 2
Q 2, 250 , 000 gpd
= 947 gpd/ft 2
=
A
2, 376 ft 2
Keeping the HRT at 2 hours, the final water depth is
25, 067 ft 3
2, 376 ft 2
= 10.6 ft
Activated Sludge Process
Now that we have transported the wastewater to the WWTP, screened out
large and floatable objects, settled out the sand in the grit chamber, and
removed the BOD associated with the influent organic suspended solids in the
primary clarifier, we still have a very large volume of wastewater with dissolved organics (BOD) and small suspended solids (both organic and inert)
that must be treated. How do we approach this enormous task, and do it cost
effectively? The answer is that we let nature work for us. We harness microorganisms to decompose the BOD, and we allow gravity to settle the resulting
sludge in large tanks. Biological processes have been used around the world
for municipal and industrial wastewater treatment to not only remove BOD
and suspended solids, but also nitrogen, phosphorus, and many synthetic
organic compounds.
The heart of a wastewater treatment plant is the activated sludge process.
The key parts of the activated sludge process are the biological reactor or aeration tank, and the secondary clarifier. Equipment to supply oxygen to the biological reactor, and facilities to pump the recycle and waste sludge are also
required. A process flow schematic with the labeling we will use in this discussion is shown in Figure 6.5 on the following page. A photograph showing parts
of the aeration tanks and one of the three secondary clarifiers at a WWTP in
Sanford, Florida, is shown in Figure 6.6, also on the next page.
Biological processes depend on a diverse population of microorganisms,
principally bacteria, but also protozoa, parasites, viruses, and others, which consume organic material as part of their life processes. Aerobic bacteria need oxygen and will metabolize organics much faster than anaerobic bacteria, which
die in the presence of oxygen. However, anaerobic bacteria are utilized in treat-
Cooper.book Page 224 Monday, June 23, 2014 9:58 AM
224
Chapter Six
ing the waste sludge from the aerobic activated sludge system, mainly because
they greatly reduce the volume of solids (more on this later), and because they
produce methane that can be harnessed as an energy source. Microorganisms
require food (BOD), nutrients, air (or the absence of air), and controlled temperature and pH of their environment to survive. If these conditions are met, these
microorganisms (biomass) in the activated sludge aeration tank grow, and reliably convert particulate and dissolved BOD into CO2, H2O, and more biomass.
Influent
Qi Si Xi
Aeration
Tank
V
Mixed Liquor
Qa Sa Xa
Secondary
Clarifier
Effluent
Qe Se Xe
Figure 6.5 The
activated sludge
process flow
diagram.
Air
Recycled Sludge
Q r S r Xr
Waste Sludge
Qw Sw Xw
Figure 6.6 Aeration tanks (foreground) and a secondary clarifier (background)
near Lake Monroe (far background) in Sanford, Florida. The clarifier diameter is 85
feet, and there are three clarifiers at this plant. (Courtesy of CPH Inc., Sanford, FL.)
Cooper.book Page 225 Monday, June 23, 2014 9:58 AM
Wastewater Treatment
225
Biomass (the mass of all the cells of microorganisms) is given the symbol X
in this text. Biomass grows in the aeration tank as the microorganisms metabolize the BOD. In practice, biomass is measured as volatile suspended solids
(VSS). A very common term denoting the biomass concentration in the aeration tank is the mixed liquor volatile suspended solids (MLVSS). The term
volatile refers to the fact that the biomass is mostly organic material that can be
oxidized, and distinguishes it from inert material (such as fine sand). The total
suspended solids (TSS) is termed mixed liquor suspended solids (MLSS) and
includes both inert and volatile matter. Typically, MLVSS is about 70–80% of
the MLSS.
Biomass is slightly denser than water, so it can be separated from the effluent by gravity settling. The BOD is called the substrate and given the symbol S.
Most of the BOD in the WW enters as dissolved or colloidal matter. After most
of the BOD has been decomposed in the reactor (the aeration tank in Figure
6.5), the wastewater is routed to the secondary clarifiers to settle and remove
the biomass. Such removal is not 100% efficient, and since the majority of the
solids is biomass, it can contribute to the BOD loading in the effluent stream.
Thus, we must not only remove soluble BOD in the reactor, but also must
remove as much biomass as possible in the clarifiers.
The design of biological municipal WWTPs may be accomplished with the
help of empirical guidelines developed from experience, or by using material
balances and kinetic information. Guidelines are well established for municipal
POTWs due to the similarity in domestic wastewater characteristics (Ten States
Standards 2004). The design of industrial WW treatment facilities often
requires a more fundamental approach based on data on the reaction kinetics
for the specific wastewater. The kinetic results are integrated into the design
using mass balances and reactor engineering principles. Even if we make use of
empirical guidelines for municipal WW, the fundamental approach presented
in Chapter 3 is always valid.
Some parameters that are commonly used in the design and operation of
activated sludge systems include: hydraulic residence time (HRT), solids residence time (SRT), organic loading or food-to-microorganism ratio (F:M), recycle ratio (R), clarifier hydraulic loading rate (HLR), and clarifier solids loading
rate (SLR). Each of these is defined in the following discussion.
The hydraulic residence time (HRT) is the average time that water spends
in the aeration tank; it has the usual mathematical definition (note that it is
based on only the raw wastewater influent and does not include the recycle). It
has units of time and is typically in the range of 4–10 hours:
HRT = V/Qi
(6.8)
The solids residence time (SRT), sometimes called the mean cell residence
time (MCRT), is a key parameter. It is the average amount of time that biomass
stays in the system. By recycling biomass, we decouple the residence time of
the microorganisms from the residence time of the wastewater. This allows us
to make use of the biomass over and over again. Like the HRT, the SRT has
units of time, but unlike the HRT, the SRT typically has values in the range of
Cooper.book Page 226 Monday, June 23, 2014 9:58 AM
226
Chapter Six
5–15 days. The SRT is calculated by dividing the total amount of biomass in the
aeration tank (XaV) by the total biomass discharge rate (ΔX/Δt); see p. 229.
SRT =
X aV
D
( X Dt )
(6.9)
The F:M ratio (also called the organic loading) is the ratio of substrate
entering the system to the amount of biomass in the aeration tank. The use of
organic loading is attractive for attached growth systems (trickling filters and
rotating biological contactors) due to difficulties in measurement of biomass
concentration in these systems. It appears to have units of day–1, but the units
really are mg of BOD5/mg of biomass-day. Both the SRT and F:M ratio are controlled by operators by adjusting the waste sludge rate. A high F:M corresponds to a low SRT and indicates an excess of food for the bacteria, which
results in a low treatment efficiency. A low F:M ratio corresponds to a high SRT
and indicates that the bacteria are starving, which results in a greater conversion of substrate. Values of F:M typically range between 0.1 to 1.0. However,
there are many other factors that complicate operations (such as nitrification
[the oxidation of ammonia to nitrates], sludge “settleability,” power consumption, etc.) and make it impossible to simply use one parameter to judge operations or make designs.
F:M =
Qi Si
X aV
(6.10)
The recycle ratio (R) is simply the ratio of the recycle sludge stream flow
rate (Qr) to the influent rate (Qi). Many plants try to operate with a recycle ratio
in the range of 0.5 to 1.0.
There are two key parameters that deal with the secondary clarifier:
hydraulic loading rate (HLR) and solids loading rate (SLR). The HLR is the
flow rate of influent divided by the surface area of the secondary clarifier, and
is calculated the same as the overflow rate for a primary clarifier. In calculating
the HLR, we exclude the recycle flow rate; it is simply Qi/A. The SLR is the
total flow rate of solids to the clarifier divided by the surface area, and must
include the recycle flow rate, as shown in Eq. (6.11). Further discussion of these
two parameters will be presented later in this chapter.
SLR =
Qa ¥ MLSS a
A
(6.11)
where:
SLR
= solids loading rate, lb/ft2-day or kg/m2-day
Qa
= mixed liquor stream flow rate leaving the aeration tank,
gal/day or m3/day
A
= top surface area of clarifier, ft2 or m2
MLSSa = total concentration of suspended solids (both biomass and inerts)
in the aeration tank, lb/gal or kg/m3
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Wastewater Treatment
227
Some empirical guidelines are shown in Table 6.4 for municipal wastewater applications. A minimum solids retention time (SRT) of five days is normally selected to avoid problems with the sludge settling in the clarifier
(Bisogni and Lawrence 1971), even if kinetic information indicates that a lower
SRT would suffice. Significantly higher values of SRT (> 15 days) may be
required to achieve compliance with specific effluent standards. Secondary
clarifiers are designed based on the HLR and SLR using the average annual
daily flow (AADF), but also must be evaluated for the peak hour flow (PHF).
Table 6.4
Typical Design Guidelines for Municipal Activated Sludge Systems
Parameter
Value
SRT (days)
5 to 15
Ê lb BOD5 ˆ
Organic Loading Á
Ë lb VSS-day ˜¯
Ê lb BOD ˆ
5
Organic Loading Á
˜
Ë 103 ft3 -day ¯
MLVSS (mg/L)
HRT (hours)
R (Qr/Qi) (fraction)
Clarifier HLR (gpd/ft2)
Clarifier SLR (lb/ft2-day)
0.2 to 0.6
10 to 120
2,500 to 4,000
3 to 8*
0.25 to 1.25
400 to 800*
18 to 30**
*Based on fresh influent flow rate only (excludes recycle flow rate).
**Based on total flow rate to clarifier (includes recycle flow rate), and total solids concentration (MLSS).
Source: Metcalf & Eddy (2003) and others.
6.5 Material Balance Approach
Design of a complete-mix activated sludge facility begins with fundamental material balances. Equations can be written for conservation of the mass of
substrate around the entire system, for conservation of the mass of biomass
around the secondary clarifier, or the reactor, or the entire system. Knowledge
of appropriate kinetic relationships for substrate removal and biomass production is also necessary. Development of key design equations is presented here
for the activated sludge process represented previously in Figure 6.5. We start
with a mass balance diagram for the overall system (Figure 6.7 on the following page). Note that the stream flowing from the reactor to the clarifier is called
the mixed liquor.
Cooper.book Page 228 Monday, June 23, 2014 9:58 AM
228
Chapter Six
Influent
Qi Si Xi
Aeration
Tank
V
Mixed Liquor
Qa Sa Xa
Effluent
Qe Se Xe
Secondary
Clarifier
Figure 6.7
Material balance
diagram on the
whole system.
Air
Recycled Sludge
Q r S r Xr
Waste Sludge
Qw Sw Xw
Overall System Balances
A steady-state flow balance around the whole system yields Eq. (6.12):
Qi = Qe + Qw
(6.12)
where:
Qi = influent wastewater flow rate, L/d
Qe = effluent flow rate, L/d
Qw = waste sludge flow rate, L/day (sometimes called the sludge wasting rate)
A steady-state mass balance for substrate around the whole system yields
Eq. (6.13):
0
=
Qi Si – Qe Se – Qw Sw + rSV
(6.13)
where:
Si = substrate concentration in the influent, mg/L
Se = effluent substrate concentration, mg/L
Sw = waste sludge substrate concentration, mg/L
V = aeration basin volume, L
rS = substrate generation rate (which has a negative value), mg/L-day
Note, by recognizing that Sw = Se and that Qi = Qe + Qw, Eq. (6.13) can be simplified to
–rSV = Qi (Si – Se)
(6.14)
The term –rSV is the negative of the substrate generation rate and is simply the
substrate destruction rate (which is a positive number). It is the rate at which
BOD is destroyed in the reactor (usually reported in kg/day or lb/day). It is
exactly equal to the product of the flow rate into the plant and the difference in
the influent and effluent substrate concentrations as shown in Eq. (6.14). The
substrate destruction rate is often given the symbol ∆S/∆t.
A steady-state biomass mass balance around the whole facility yields Eq. (6.15):
0
=
Q Xi – Qe Xe – Qw Xw + rXV
(6.15)
Cooper.book Page 229 Monday, June 23, 2014 9:58 AM
Wastewater Treatment
229
where:
Xi = biomass concentration in the wastewater influent, mg/L
Xe = effluent biomass concentration, mg/L
Xw = waste sludge biomass concentration, mg/L
rX = biomass generation rate (which is positive), mg/L-day
Even though there are billions of bacterial cells in each liter of influent, their
concentration is negligible compared with that of the cells in the aeration tank
due to growth of biomass in the tank and the recycling of biomass in the recycle
sludge stream. Assuming that Xi is essentially zero, and rearranging Eq. (6.15)
yields Eq. (6.16):
Qe Xe + Qw Xw = rXV
(6.16)
The sum Qe Xe + Qw Xw is known as the total biomass discharge rate, also
given the symbol ∆X/∆t. The effluent and the waste sludge are the only two
streams where biomass can exit the system, and ∆X/∆t defines the discharge
rate. Eq. (6.16) is very important because it tells us that, for the system to
remain at steady state, the total biomass discharge rate must equal the total
growth rate of biomass in the aeration tank. In other words, we must discharge
biomass as fast as it grows.
An important operating parameter is the observed yield coefficient. The
observed yield coefficient (Yobs) is defined as the net amount of biomass grown
per unit amount of substrate destroyed, and is calculated from Eq. (6.17).
Yobs =
DX Dt rX
=
DS Dt - rS
(6.17))
where:
Yobs = observed yield, mass of biomass grown per mass of substrate used
The observed yield is always less than 1.0, and for aerobic processes Yobs
has values in the range of 0.4 to 0.8. Yobs is often presented as a dimensionless
number, but it should be understood that it has units of biomass grown per unit
of BOD removed. Sometimes it is presented in units of (∆ mg/L VSS)/(∆ mg/L
of BOD), where VSS stands for volatile suspended solids and indicates the biomass, and BOD of course indicates the substrate.
Reactor Balances
Let us next consider the aeration tank or biological reactor. The material
balance diagram for the aeration tank is shown in Figure 6.8. The flow balance
around the aeration tank is:
Qi + Qr
where:
Qa = mixed liquor flow rate, L/day
=
Qa
(6.18)
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230
Chapter Six
A steady-state mass balance for substrate around the aeration tank yields
Eq. (6.19).
0
=
Qi Si + Qr Sr – Qa Sa + rSV
(6.19)
where:
Sa = mixed liquor substrate concentration, mg/L
A steady-state mass balance for biomass around the aeration tank yields
Eq. (6.20).
0
=
Qr Xr – Qa Xa + rXV
(6.20)
where:
Xa = mixed liquor biomass concentration, mg/L
It is noted that the mixed liquor biomass concentration is often referred to in
other texts as the mixed liquor volatile suspended solids (MLVSS) or the mixed
liquor suspended solids (MLSS). However, there is an important difference.
The MLVSS is a measure of biomass whereas the MLSS is a measure of the total
suspended solids and includes inert solids as well as biomass.
Influent
Qi Si Xi
Figure 6.8 Material balance diagram for
the aeration tank (reactor).
Aeration
Tank
V
Mixed Liquor
Qa Sa Xa
Air
Recycled Sludge
Q r S r Xr
Aeration tanks can be quite large, and even though we do not spend much
time in this text discussing the aeration equipment, it is important that such
equipment be sized properly to distribute enough air to all parts of the tank to
ensure that the microbes always have an adequate supply of oxygen. Aerobic
microorganisms need at least 0.5 to 1.0 mg/L of dissolved oxygen to remain
metabolically active, and a rule of thumb is to provide air contact rates of about
30–50 m3 of air per kg of BOD removed (480–800 ft3 of air/lb of BOD removed).
We do not want a lack of oxygen to limit the growth rate of the microorganisms
(and slow down the destruction rate of substrate). Different types of aeration
systems are used—surface aerators, which mechanically entrain air into the
mixed liquor in the tank, and subsurface (diffused air) aeration systems that
utilize compressors and blow air in through perforated pipes mounted at the
bottom of the aeration tank. Both types of aeration systems require a significant
amount of electrical energy. Figure 6.9 is a photograph of a surface aerator in
Cooper.book Page 231 Monday, June 23, 2014 9:58 AM
Wastewater Treatment
231
Figure 6.9 Surface aerator in operation at a wastewater treatment plant. (Courtesy
of Purestream Inc., Walton, KY.)
operation, and in Figure 6.6 (presented earlier) the bubbles from a diffused aeration system can be seen on the wastewater surface in an aeration tank.
Clarifier Balances
The material balance diagram around the secondary clarifier is shown in
Figure 6.10. The steady-state flow balance becomes:
Qa
=
Qe + Qw + Qr
(6.21)
where:
Qa = flow rate of mixed liquor, L/day
Figure 6.10 Material balance diagram
for the secondary clarifier.
Mixed Liquor
Qa Sa Xa
Recycled Sludge
Qr Sr Xr
Secondary
Clarifier
Effluent
Qe Se Xe
Waste Sludge
Qw Sw Xw
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232
Chapter Six
Once the mixed liquor leaves the reactor, and the remaining dissolved oxygen is consumed, it is assumed that all reaction stops. Since there is no reaction
in the clarifier, the substrate concentration does not change further, and Sa = Se
= Sw = Sr. However, the solids settle in the clarifier, so the solids concentrations
do change. A steady-state biomass balance is:
Q a Xa
=
Qe Xe + Qw Xw + Qr Xr
(6.22)
where:
Xa = concentration of biomass (MLVSS) in the mixed liquor, mg/L
Because we are only splitting the underflow from the clarifier into two streams,
their properties remain identical and Xw = Xr. But obviously the concentration
Xe will be much lower than Xa, while Xw will be greater than Xa.
Secondary clarifiers at a WWTP can be quite large. They are usually circular, and may be 100 feet or more in diameter. The bottom is sloped slightly
towards the center, and there is a mechanical rake at the bottom of the tank that
slowly moves the settled solids towards the outlet pipe (located in the floor
near the center of the tank). The weir over which the clarified effluent flows is
often a trough that is attached to the tank wall and which collects the clarified
effluent and directs it outside the tank. It is usually 6 to 12 inches in width.
There can be multiple clarifiers at a large WWTP.
EXAMPLE 6.5
Consider an activated sludge WWTP. Influent comes in at 20 million L/day,
and effluent flows out at 19 million L/day. The recycle flow rate is 22 million
L/day. The aeration basin has a volume of 8.0 million liters. The influent and
effluent BOD5 are 400 mg/L and 20 mg/L, respectively. The biomass (MLVSS)
concentrations in the waste sludge and clarifier effluent are 4,600 mg/L and
15 mg/L, respectively. Calculate:
(a) the total biomass discharge rate, kg/day
(b) the observed yield
(c) the biomass concentration in the aeration tank, mg/L
(d) the recycle ratio
(e) the SRT
SOLUTION
(a) From the overall flow balance, Qw = Qi – Qe
Qw = 20 - 19 = 1.0 million L/day
mg ˆ
mg
1 kg
DX Ê
6 L
6 L
¥ 4, 600
¥ 6
= Á 19 (10 )
¥ 15
+ 1.0 (10 )
˜
day
L ¯ 10 mg
Dt Ë
day
L
= 4, 885 kg/day (of MLVSS)
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233
Notice that when multiplying a flow in millions of L/day times a concentration in mg/L, and then converting the product to kg/day, the factors of 106
divide out, so we can simply multiply the numbers straight away to get the
answer in kg/day. That is,
DX = (19 × 15 + 1.0 × 4,600) = 4,885 kg/day (of MLVSS)
Dt
(b)
mg
1 kg
DS
6 L
= 20 (10 )
¥ ( 400 - 20 )
¥
= 7 , 600 kg / day (of BOD 5 )
Dt
day
L
106 mg
or
DS
= 20 ¥ ( 400 - 20 ) = 7 , 600 kg/day (of BOD 5 )
Dt
Yobs =
kg of biomass
DX Dt 4 , 885
= 0.64
=
kg of BOD 5
DS Dt 7 , 600
(c) First, do a flow balance around the aeration tank.
Qa = Qi + Qr
= 20 + 22 = 42 million L/day
Next, start with an MLVSS balance around the clarifier
Qa Xa = Qe Xe + Qw Xw + Qr Xr
and solve for Xa
Xa =
19 (15 ) + 1 ( 4, 600 ) + 22 ( 4, 600 )
= 2, 526 mg/L
42
(d)
R = Qr/Qi = 22/20 = 1.10
6
(e)
SRT =
2, 526 mg/L ¥ 8.0 (10 ) L
X aV
= 4.14 days
=
DX Dt 4 , 885 kg/day ¥ 106 mg/kg
(it is noted that this SRT is a bit low)
EXAMPLE 6.6
Consider Example 6.5. You want to increase the SRT, but you are already recycling a lot of sludge, so you decrease the sludge wasting rate (waste sludge
flow rate) to 400,000 L/day. After a new steady state is established, the biomass concentration in the effluent has not changed, but the concentrations of
biomass in the waste sludge and in the aeration tank are now 5,700 mg/L and
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234
Chapter Six
3,200 mg/L respectively. Calculate the new SRT. Calculate the F:M ratio at
these new conditions and compare it with the old conditions.
SOLUTION
From a flow balance around the whole system, the new Qe = Qi – Qw
Qe = 20 – 0.4 = 19.6 million L/day
The new
DX
= (19.6 ¥ 15 + 0.4 ¥ 5, 700 ) = 2, 574 kg/day
Dt
The new SRT is
3, 200 ¥ 8.0
= 9.95 days
2, 574
6
F:M new =
Qi Si 20 (10 ) L/day ¥ 400 mg/L
=
6
X aV
3, 200 mg/L ¥ 8.0 (10 ) L
= 0.31 mg BOD 5 /mg biomass-day
6
F:M old =
Qi Si 20 (10 ) L/day ¥ 400 mg/L
=
6
X aV
2, 526 mg/L ¥ 8.0 (10 ) L
= 0.40 mg BOD 5 /mg biomass-day
6.6 WWTP Biological Kinetics
In the previous development, we did not give any details of the kinetic
reaction rates of substrate (rS) or biomass (rX). After a plant is built and operating, material balances (such as those presented earlier) can be used to analyze
and optimize operations. Furthermore, with enough experience, empirical
guidelines can be used with simple material balances to allow a reasonable
approach to the design. However, early study of the reaction kinetics in the lab
can be very helpful in optimizing the design. Biological processes for wastewater treatment achieve substrate removal both by oxidizing some of the substrate to CO2 and H2O, and by converting some of it to biomass. The kinetics of
removal has direct relevance to the sizing of the reactor. Also, this assimilative
removal of substrate simultaneously produces biomass. The rate and quantity
of biomass produced directly affects the design of many unit operations and
processes, including sludge-handling facilities. Knowledge of both rates therefore is important.
Various empirical models have found widespread use for analysis of biological kinetics. The rate of substrate conversion is dependent on the biomass
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235
concentration. The most common kinetic model for substrate utilization rate is
the Monod equation because it is flexible enough to describe both zero-order
(high concentrations of S) and first-order (low concentrations of S) kinetics:
- rS =
kSX
KS + S
(6.23)
where:
k = maximum specific substrate utilization rate,
(mg/L substrate)/(mg/L biomass-time)
KS = half saturation constant, mg/L
S = substrate concentration, mg/L
X = biomass concentration, mg/L
It is common to define a specific substrate utilization rate (q) as the rate of
substrate utilization normalized by the biomass concentration (–rS /X). Four
models for the specific substrate utilization rate are presented in Eqs. (6.24) to
(6.27) representing zero-order, first-order, variable-order (Monod), and inhibitory kinetics, respectively:
q=
- rS
= k0
X
(6.24)
q=
- rS
= k1S
X
(6.25)
q=
q=
- rS
kS
=
X
KS + S
- rS
=
X
kS
S2
KS + S +
KI
(6.26)
(6.27)
The inhibitory model, Eq. (6.27), includes an inhibition constant KI, and is appropriate for specific organic compounds in industrial waste treatment situations
where the rate of biological activity is reduced at very high substrate concentrations. The inhibitory model degenerates to give the Monod model in the limit as
the inhibition constant approaches infinity. Various synthetic organic compounds
have been reported to exhibit inhibition at high concentrations (Grady 1990).
The kinetic expressions can be tied in with the material balance equations
to link laboratory data with design equations. For example, the specific substrate utilization rate is linked to the substrate mass balance by solving Eq.
(6.14) for –rS and then dividing both sides by X, yielding:
q=
- rS Qi (Si - Se )
=
X
VX
(6.28)
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Chapter Six
The specific biomass growth rate (µ) is the normalized biomass growth rate
and is defined as:
m=
rX
Ê -r ˆ
= Yobs Á S ˜
Ë X ¯
X
(6.29)
The specific growth rate, µ, has units of inverse time, and can be linked to the
SRT as follows. We know from Eq. (6.16) that the biomass growth rate is exactly
equal to the total biomass discharge rate (rXV = ∆X/∆t). From Eq. (6.9), we
know that the SRT = XaV/(∆X/∆t). Inverting Eq. (6.9), and substituting rXV for
∆X/∆t yields the relationship between the design variable, SRT, and the kinetic
variable, µ:
r V
r
1
= X = X =m
SRT X aV X a
(6.30)
By substituting Eq. (6.26) for –rS /X into Eq. (6.29), we see that
m = Yobs
kS
KS + S
(6.31)
Often the Monod equation is presented as
m=
mmS
KS + S
(6.32)
where: µm = maximum specific growth rate, days–1
Thus it would seem that µm is the product of Yobs and k. But because
microorganisms not only grow but also die, we must include an endogenous
decay (death) rate (ke) which can be taken as proportional to the concentration
of biomass.
Thus, the specific biomass growth rate is fully represented as
m = Ymax
kS
- ke
KS + S
(6.33)
where:
ke
= endogenous decay rate, time–1
Ymax = maximum yield, (mg/L biomass grown)/(mg/L BOD5 used)
The observed yield varies with system operation and is not constant. The
maximum yield is independent of system operation and may be obtained from
laboratory data. Both yields are commonly reported as dimensionless, since the
units appear to cancel. But formally, the units are mg/L of biomass per mg/L
of substrate, or mass of biomass per mass of substrate. The maximum yield is
related to the observed yield as follows:
Ê
k ˆ
Ymax = Yobs Á 1 + e ˜
m¯
Ë
(6.34)
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For municipal WW, typical values of the maximum yield and other kinetic
constants are: Ymax = 0.60 mg/L biomass per mg/L BOD5 , ke = 0.06 days–1 , k =
(3.0 mg/L BOD5)/(mg/L biomass-day), and KS = 60 mg/L BOD5 (Metcalf &
Eddy 2003), but each parameter can vary plus or minus 50% from those values.
From kinetic parameters then, the desired design values may be obtained,
including aeration basin volume, biomass concentration (Xa), and SRT.
EXAMPLE 6.7
Determine the required solids residence time for an activated sludge system
to achieve compliance with an effluent BOD5 limit of 12 mg/L. Assume that
Ymax = 0.60 mg biomass/mg BOD5, ke = 0.06 days–1, k = 3.0 mg BOD5/mg biomass-day, and KS = 60 mg/L BOD5.
SOLUTION
Using Eq. (6.33) to solve for µ
m=
Ê
ˆ
mg biomass ˆ Ê
mg BOD 5
ÁË 0.60 mg BOD ˜¯ ÁË 3.0 mg biomass-day ˜¯ (12 mg/L BOD 5 )
5
(60 + 12) mg/L BOD 5
- 0.06 days-1
= 0.24 days-1
From Eq. (6.30), µ = 1/SRT,
SRT =
1
= 4.2 days
m
The preceding kinetic equations (substrate removal and biomass production) can be coupled with mass balance equations and solved simultaneously
to establish a relationship between the substrate concentration and the SRT. For
variable-order kinetics:
S=
KS (1 + ke SRT )
SRT Ymax k - (1 + ke SRT )
(6.35)
or
SRT =
KS + S
S (Ymax k - ke ) - ke KS
(6.36)
It is evident from Eqs. (6.35) and (6.36) that the effluent quality (represented by
the effluent substrate concentration, S) is directly related to the SRT (inverse of
the specific growth rate). As the specific growth rate (µ) is decreased (as SRT is
increased), the removal of substrate is improved as was recognized many years
ago (Lawrence and McCarty 1970).
For complete-mix suspended growth systems without solids recycle (such
as aerated lagoons, aerobic digesters, or anaerobic digesters) at steady state,
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Chapter Six
there is no separation of HRT and SRT. The growth rate is related to the reactor
hydraulic residence time. If the aerated lagoon is modeled as a CSTR without
recycle (and of course with no clarifier), the steady-state biomass balance is:
0 = 0 – QX + rXV
(6.37)
where:
Q = flow rate into and out of the lagoon, L/day
X = biomass in the lagoon (and outlet stream), mg/L
Rearranging Eq. (6.37),
Q rX
=
V X
or
1
=m
HRT
(6.38)
EXAMPLE 6.8
Determine the required volume of an aerated lagoon treatment of a 5.0 MGD
municipal wastewater to achieve compliance with an effluent BOD5 standard
of 12 mg/L. Lab studies have shown that Ymax = 0.65 mg biomass/mg BOD5,
ke = 0.06 days–1, k = 1.5 mg BOD5/mg biomass-day, and KS = 80 mg/L BOD5.
Assume that an aerated lagoon may be described as a completely mixed reactor without solids separation and recycle.
SOLUTION
The specific substrate utilization is
q=
mg BOD 5
kS
1.5 ¥ 12
=
= 0.196
mg biomass-day
KS + S (80 + 12)
The required growth rate for the specified kinetics is:
µ = Ymax (q) – ke = 0.65 × 0.196 – 0.06
µ = 0.0674 days–1
But HRT = 1/µ, so the required hydraulic residence time is 14.8 days.
Finally, the volume of the lagoon is HRT × Q:
V = 14.8 days × 5.0 MGD = 74 million gallons
The next example deals with a specialized waste, one that contains only
one easily biodegraded compound. It illustrates a point that we have not discussed yet, and that is the need for nutrients. Bacteria are living creatures and
need not only food (the BOD) but also nutrients like N and P. In this next example, no nitrogen compounds are present. A source of nitrogen must be available
to support biomass cell synthesis, so we must add nitrogen (in a form that can
be used by bacteria) to achieve successful biological treatment. Municipal
wastewaters almost always have enough nutrients; however, certain industrial
wastewaters, particularly high-carbohydrate food-processing wastewaters,
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may be deficient in nitrogen and/or phosphorus. If we know an approximate
empirical composition of bacterial cells, we can calculate the stoichiometric
nutrient requirements just like any other component in a chemical reaction.
EXAMPLE 6.9
A food-processing wastewater contains 1,000 mg/L of glucose (C6H12O6). The
flow rate is 2.0 million gallons per day (MGD). Ammonium bicarbonate is
added to provide a source of N for the bacteria. Biological treatment with the
activated sludge process achieves removal of 90% of the glucose according to
the following reaction:
25 C6H12O6 + 3 NH4HCO3 + 135 O2 →
3 C5H7O2N + 138 CO2 + 147 H2O
Determine
(a) the quantity of biomass (C5H7O2N) which is produced (lb/day)
(b) the quantity of oxygen that must be supplied by the aeration equipment
(lb/day)
(c) the observed yield for this biological process (lb VSS/lb glucose)
(d) the consumption of ammonium bicarbonate (NH4HCO3) (lb/day)
SOLUTION
The basis for this stoichiometric calculation is the mass of glucose removed:
Ê 8.34 lbs/MGAL ˆ
Glucose = (1, 000 mg/L ) ( 2 MGD ) Á
˜ (0.90 removal )
mg/L
Ë
¯
= 15, 012 lbs glucose/day
Stoichiometric quantities of cell mass and oxygen are calculated as follows:
(a) Cell mass =
15, 012 lbs glucose/day
3 lbmol cells
113 lb cells
¥
¥
180 lb/lbmol
25 lbmol glucose lbmol cells
= 1, 131 lbs biomass/day
(b)
Oxygen mass =
15, 012 lbs glucose/day 135 lbmol oxygen 32 lb oxygen
¥
¥
180 lb/lbmol
25 lbmol glucose lbmol oxygen
= 14 , 412 lbs oxygen/day
(c) The observed yield is the ratio of rate of biomass production to rate of substrate removal:
Yobs =
1, 131 lbs VSS/day
= 0.075 lb VSS/lb glucose
15, 012 lbs glucose/day
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Chapter Six
(d)
NH 4 HCO 3 mass =
15, 012 lb glucose 3 lbmol NH 4 HCO 3 77 lb NH 4 HCO 3
¥
= 770.6 lb/day
¥
25 lbmol glucose
lbmol
180 lb/lbmol
Secondary Clarifier
The design of a secondary clarifier is very similar to a primary clarifier, the
main difference being that there is also a design criterion for the solids loading
rate (SLR). The criteria for HLR and WLR are similar to those for primary clarifiers; the HLR should be in the range of 400–800 gpd/ft2, and the WLR should
be between 10,000 and 30,000 gpd/ft. Note that, by convention, these guidelines are based on only the flow rate of the fresh influent to the WWTP, and do
not include the recycle flow rate. At first, this might not seem appropriate since
the clarifier receives a larger total flow rate (Qa = Qi + Qr). But consider that the
rate of water flow out the top of the clarifier, the clarified effluent (Qe), is very
close to equal to the influent flow rate (Qi). The recycle flow (Qr) goes out the
bottom of the clarifier. Thus, the HLR (or the OFR) is essentially the upward
velocity of water near the top of the clarifier. (If the upward water velocity is
slower than the particle-settling velocity, the particle will settle downward.) So
the fresh influent is indeed the appropriate flow rate to consider for the HLR
for a secondary clarifier.
A secondary clarifier must handle a much higher flow rate of solids than a
primary clarifier, so the solids loading rate (SLR) is also important. The SLR is
calculated as the product of the total flow to the clarifier and the solids concentration in the mixed liquor. The SLR was defined earlier in Eq. (6.11) (repeated
below) as the mass flow rate of solids per unit amount of surface area.
SLR =
Qa ¥ MLSS
A
(6.11)
A type of settling called zone settling occurs in the bottom half of the clarifier, and this is where the sludge starts to thicken. The design criterion for SLR
is that it should be in the range of 18–30 lb/ft2-day. Notice that the surface area
of the clarifier appears in both the SLR and the HLR equations. During the
design process, when using empirical guidelines, it is customary to calculate
the surface area using both the HLR and the SLR independently, and then
select the larger of the two areas.
Disinfection
After clarification and before discharge, the effluent must be disinfected. In
the United States, the most common process to achieve disinfection is chlorination. Many states recommend that the design provide for a chlorine contact
time of at least 30 minutes at the average daily flow rate or 15 minutes at peak
hourly flow rate. Chlorine contact basins should be designed to minimize
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241
short-circuiting and longitudinal dispersion, and ideally approach plug-flow
behavior. Actual requirements are often dictated by state regulatory agencies
or conditions of discharge permits, which may specify a range of acceptable
chlorine residuals and hydraulic residence times.
After passing through the chlorine contact chamber, the disinfected water
may need to be dechlorinated (remove any residual chlorine) to prevent later
formation of THMs in natural waters. Dechlorination, if required, is often
achieved by adding sulfur dioxide to the water. The free chlorine in the water is
in the form of HOCl, and the balanced dechlorination reaction is:
SO2 + HOCl + H2O
→
Cl– + SO42– + 3 H+
(6.39)
EXAMPLE 6.10
Given the balanced redox reaction for dechlorination of water using sulfur
dioxide, Eq. (6.39), calculate the amount of SO2 used to dechlorinate 25 MGD
of water that contains 1.2 mg/L of HOCl. Give your answer in kg/day.
SOLUTION
From the balanced equation,
SO2 + HOCl + H2O
64
52.5
→
Cl– + SO42– + 3 H+
we see that 64 mg of SO2 are needed to react with 52.5 mg of HOCl. Therefore,
6
25 (10 ) gal
day
¥
1 kg
138 kg
1.2 mg HOCl
64 mg SO 2
3.785 L
¥ 6
=
¥
¥
day
L
gal
52.5 mg HOCl
10 mg
Alternative disinfection processes exist, including treatment with alternative
chlorine compounds such as bleach (sodium hypochlorite—NaOCl), chlorine
dioxide (ClO2), or calcium hypochlorite [Ca(OCl)2], or non-chlorine processes
such as ozonation or UV radiation.
6.7 Septic Tanks
It has been estimated that almost 20 million people in America use septic
tanks to treat their household wastewater (US EPA 2013b). Septic tank systems
are found where homes are located inconveniently far from a public sewage
collection system, or in older areas within an urban service area that have not
converted to the public system. A septic tank system consists of pipes to carry
the wastewater from the home, a septic tank (and perhaps a separate pumping
tank), a drainfield, and the soil itself (US EPA 2002). Anaerobic microbes within
the septic tank biodegrade most of the organic wastes, and reduce the wastes to
a small amount of solids that accumulates in the bottom of the tank. The
treated effluent (water) that flows out of the tank still has BOD and nutrients in
it, but it is then distributed throughout the drainfield into the soil. Naturally
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Chapter Six
occurring microbes in the soil biodegrade most of the remaining BOD, along
with much of the nitrogen. Phosphorus is mostly adsorbed onto the soil.
The septic tank is a watertight tank made of concrete, fiberglass, or plastic
that is buried on the homeowner’s property. It is large enough to hold wastewater long enough to allow solids to settle, grease and oil to float to the top
water surface, and many of the microbial decomposition processes to proceed
to completion. Common sizes of septic tanks for private homes are between
1,000 and 2,500 gallons, depending on the size of the house. An elbow-shaped
outlet allows water to exit the tank without carrying away grease and oils.
Many septic tanks utilize gravity to allow the treated wastewater to flow from
the tank into the drainfield, but in some systems (especially those close to
lakes), the drainfield must be located some distance from the lake. This usually
means that it is uphill from the septic tank. An uphill drainfield may also occur
in areas with very high water tables. In those cases, the septic tank discharges
to a treated water collection tank from where the water is pumped to the drainfield. The tanks have a removable access cover to allow occasional pump-out
by a commercial company when the solids accumulate to a high level, impairing the tank’s operation. It is recommended that a septic tank be pumped every
three years or so, but some systems may operate for 10–15 years before the tank
needs pumping.
The tank can be overloaded with too much water, especially if there are
leaking fixtures in the house. Careful attention is needed to prevent foreign
objects (dental floss, feminine hygiene items, condoms, cigarette butts, etc.)
from getting flushed down the toilet and impairing the tank’s operation. Also,
limiting the amount of food wastes that go through the kitchen garbage disposal and down the drain helps keep a septic system operating well for many
years. Obviously, flushing items like household cleaning chemicals, gasoline,
pesticides, and paint will kill or severely damage the microbial populations living in the septic tank, and could potentially contaminate groundwater under
the drainfield.
The drainfield is an underground area in the homeowner’s back or front
yard which has been prepared by installation of clean coarse sand, and either
clay tiles or perforated plastic pipes to allow the water to be distributed over a
large area of soil. The drainfield can eventually become plugged if the tank
allows solids or grease to pass through. Also, in rainy weather or if the household is discharging too much water, the drainfield can become flooded which
can cause sewage to flood the ground, or cause the whole system to back up
and stop functioning. The soil is the final step in the treatment process, and a
good, well-drained soil is necessary. The soil provides for final removal of
BOD, harmful bacteria, viruses, and some nutrients. In many places, after years
of use, the soil may become saturated with phosphorus and allow leaching of
this nutrient into local water bodies, contributing to eutrophication and other
problems. Only grass should be planted over the drainfield as roots from trees
or large bushes might clog or damage it. Heavy equipment driving over the
drainfield might crush some of the pipes or compact the soil and thus prevent
good drainage.
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6.8 WWTP Sludge Treatment and Disposal
Up to now, we have not mentioned what happens to the waste sludge
streams—both the one from the secondary clarifier and (if applicable) the one
from the primary clarifier. Of course we know that it just doesn’t go away—
something must be done with it. Let us focus on the waste activated sludge—
that produced in the aeration tank and settled in the secondary clarifier. The
sludge that is produced in the aeration tank is substantial. Even a mediumsized WWTP may produce thousands of pounds (dry basis) of solids each day.
Furthermore, keep in mind that the solids are contained in a stream that is still
mostly water (recall that, at the usual specific gravity of about 1.0, a concentration of 10,000 mg/L is only 1% solids). Three thousand pounds per day (dry
weight) of a sludge that has a concentration of, say, 7,000 mg/L (0.7% solids) is
really a liquid sludge stream with a total mass flow rate of nearly 430,000 lb/
day, 99.3% of which is water! Treatment of this wet sludge stream is required,
followed by ultimate disposal of the residuals.
There are numerous treatment and disposal options for WWTP sludge. The
most common method is anaerobic digestion, but others exist including aerobic digestion, composting, incineration, and gasification. With anaerobic digestion, the raw sludge stream flows into a large tank where microorganisms
begin to decompose the aerobic biomass. Any dissolved oxygen remaining in
the sludge is consumed very quickly, and then anaerobic microorganisms dominate. The biological reactions occur in three main phases: hydrolysis of organics, acid fermentation, and methane formation.
Under anaerobic conditions, most of the organic material is converted to
CH4 and CO2 in a slow biological process, involving many different types of
microorganisms. There is no recycle stream, and the hydraulic residence time is
on the order of 15–40 days for an anaerobic digester, depending on the temperature within the digester. The tank must be kept warm to promote the anaerobic reactions, and typically it is not well-mixed. The lack of mixing allows the
digested solids to settle and any scum or grease to rise to the top of the liquid.
Naturally, the gases rise to the top of the tank. In some operations, two tanks
are used in series, with the first tank being mixed to promote faster reactions,
and the second tank being used to finish the reactions and to separate the various streams.
Generally, three streams are removed from the anaerobic digester: digested
solids, a supernatant stream, and a gas stream. Sometimes, intermittently, a
scum/grease layer is also removed, as shown in Figure 6.11 on the following
page. The ultimate fate of the digested solids is disposal to a landfill, incineration, or land application. To be land-applied as a fertilizer supplement, the
digested sludge must meet Class A or Class B standards (low or no pathogens,
low heavy metals, etc.).
The gas contains about 50–70% methane and can be burned directly as a
low-grade fuel, or it may be processed to purify it into a high-grade fuel gas. In
many locations, it is burned to provide heat to keep the digester warm. The
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244
Chapter Six
Figure 6.11 Schematic diagram
of an anaerobic digester.
Gas
Scum
(intermittent)
Raw
WWTP
Sludge
Supernatant
Anaerobic
Digester
Digested
Sludge
observed yield for this process is quite low, on the order of 0.05 to 0.10, so the
mass of the digested solids is much smaller than for aerobic digestion, and that
stream often is routed to nearby sand drying beds for dewatering, and later the
digested solids are sent to a sanitary landfill. The supernatant is mostly water
and typically is sent back to the front end of the WWTP. In cases where the
plant has a very tight effluent nitrogen standard, this stream may be treated
separately before being recycled. The scum layer is a very small stream and at
many plants is removed only intermittently and, after additional treatment,
may be sent to a landfill.
Incineration of digested sludge (called biosolids) is accomplished in multiple-hearth incinerators in which the solids are dried and then burned in one
vessel. Incineration is expensive due to the fuel use and the cost of the air pollution control (APC) equipment that is necessary to control the various pollutants emitted from the incinerator (more about APC in the next chapter). It
would take an enormous amount of heat and be a huge waste of fuel to evaporate the 99% of the stream that is water, so often the sludge is dewatered prior
to digestion. First the sludge is thickened in a gravity thickener and later the
biosolids are filtered in a filter press to remove much of the water prior to routing the biosolids to an incinerator. The waste sludge goes from less than 1% solids to perhaps 15–18% solids. A stream that is 18% solids is still 82% water, but
a lot of water is removed in the dewatering (thickening and filtering) operations; to see how much, study the next example.
EXAMPLE 6.11
A wastewater sludge is processed to prepare it for incineration. The sludge
flows into the thickening/filtering operation at 450,000 liters per day and has
a solids concentration of 6,500 mg/L. It gets dewatered to 18% solids.
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245
(a) Calculate how much water is removed, kg/day.
(b) Water has a heat of vaporization of about 2,400 kJ/kg. Calculate how
much heat is saved in the incinerator by not evaporating the water that
was removed mechanically in the thickening/filtering operations.
(c) If heat is worth $5.00/million kJ, how much money is saved annually by
dewatering the sludge prior to incineration?
SOLUTION
(a)
Mass of dry solids =
1 kg
450, 000 L 6, 500 mg
¥
¥ 6
= 2, 925 kg/day
day
L
10 mg
Mass of water in original sludge =
450, 000 L 1 kg
¥
- 2, 925 = 447 , 075 kg/day
day
L
Mass of water remaining in filtered sludge =
2, 925 kg/day
- 2, 925 kg/day
0.18
= 13, 325 kg/day
Mass of water removed = 447 , 075 - 13, 325 = 433, 750 kg/day
(b) Heat saved =
433, 750 kg 2, 400 kJ
9
¥
= 1.04 (10 ) kJ/day
day
kg
9
(c) Cost savings =
1.04 (10 ) kJ $5.00 365 days
¥ 6
¥
= $1.9 million/year
day
year
10 kJ
Gasification of waste activated sludge and biosolids is a relatively new process. As with incineration, the sludge/biosolids must be dewatered. But for the
gasification process to work efficiently, the biosolids stream must be very dry
(about 80–90% solids, or 10–20% water). This further drying is accomplished in
specifically designed sludge dryers. The dried biosolids are next routed to the
gasifier in which a small amount of air is admitted. The amount of air is between
20 and 40% of what would be needed for stoichiometric combustion. The air
starts the oxidation of a portion of the biosolids, which provides heat to power the
rest of the gasification reactions. During gasification, the biosolids are converted
to a synthesis gas (syngas), which has substantial amounts of combustible gases
(CO, CH4, H2, and higher hydrocarbons) as well as CO2, N2, and water vapor.
The syngas exits from the gasifier and can then be burned in a closely-coupled thermal oxidizer to recover heat. The syngas is usually burned directly onsite to recover enough heat to dry the biosolids and make the process energy
neutral (it may even produce net energy). Alternatively, the syngas can be
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246
Chapter Six
cleaned up and routed elsewhere (e.g., internal combustion engines) to produce heat, power, and work. Gasification tends to have lower air pollutant
emissions than incineration, but both types of plants must have good APC systems. As of 2013, a company in Sanford, Florida (Maxwest Environmental Systems), had the only operating full-scale gasifier in the country known to this
author. Figure 6.12 presents a photograph of the Maxwest gasifier.
A big advantage of both gasification and incineration is the significant
reduction in mass and volume of residuals for disposal. In addition, any pathogens in the sludge will be destroyed in either process. The ash from both is dry
Figure 6.12 An operating biosolids gasifier. (Courtesy of Maxwest Environmental
Systems, Inc.)
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247
and inert and can be used as a soil amendment in pasture lands, as a soil stabilizer at construction sites, or it can be sent to a landfill.
PROBLEMS
6.1
6.2
6.3
6.4
A city in Florida (population 60,000) wants to start reclaiming water from
its POTW. Town council members set what they think is an ambitious
goal to reclaim 100,000,000 gallons of water per year. What percentage of
their effluent is this? Assume the Ten States Standards average per capita
wastewater flow rate (gpcd).
A POTW was designed to treat 2.0 MGD, using the Ten States Standards
for average per capita wastewater flow rate. Assuming that infiltration/
inflow is about 15 gpcd, estimate how much of this capacity is needed to
treat the groundwater and stormwater that comes into the system inadvertently (in MGD).
A POTW is being designed to treat 2.5 MGD. Size the grit chamber (calculate the volume—do not specify dimensions).
Given the following data, determine the volume (gallons) of an equalization tank to provide flow equalization. Also, calculate and plot the incoming and outgoing BOD5 concentrations over a 24-hour period. Assume
the tank liquid never drops below 125,000 gallons.
Hour
Flow gpm
BOD5 mg/L
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1,000
800
650
600
575
580
760
1,210
1,350
1,640
1,700
1,730
1,810
1,840
1,800
1,900
1,960
2,050
1,920
1,890
1,760
1,610
1,450
1,220
95
78
52
35
48
62
90
115
125
140
149
156
172
140
131
112
134
146
188
215
256
188
165
114
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Chapter Six
6.5
You are thinking about buying a used tank from a shut-down treatment
facility, with the idea of dismantling it and moving it to your plant site,
where you will reassemble it. It has a diameter of 15 feet and a depth of 10
feet. Will this tank be satisfactory as a primary clarifier for your small
WWTP which treats 0.50 MGD? (Hint: check the HRT and OFR.)
6.6
You are part of a team designing a WWTP to treat 3.5 MGD of municipal
wastewater. Using the guidelines given in this chapter, specify the depth
and diameter of a single tank to serve as a primary clarifier.
6.7
Using the characteristics for a typical municipal wastewater from Table
6.1, and assuming a per capita wastewater generation rate of 100 gallons
per day, determine the per capita generation rate of BOD5 and suspended
solids. Express the answers in lbs per capita-day.
6.8
a. Using the variable-order kinetic information from Example 6.8 for a
municipal wastewater, determine the effluent BOD5 concentration for a
complete-mix activated sludge facility operated at a five-day solids
residence time.
b. For a wastewater flow rate of 2 million gallons per day with characteristics of a typical wastewater from Table 6.1, determine the discharge
rate of biomass (lbs/day).
6.9
Determine the required hydraulic residence time for removal of 95% of
the BOD5 from an industrial wastewater in an aerated lagoon (well-mixed
with no recycle). The kinetics were determined as follows:
Variable order (Monod) kinetics
k = 1.2 mg BOD5/mg VSS-day
KS = 90 mg BOD5/L
Ymax = 0.5 mg VSS/mg BOD5
ke = 0.04 day–1
The influent wastewater BOD5 is 800 mg/L.
6.10 A municipal WW activated sludge facility has the following characteristics:
Flow = 3.0 MGD
Influent BOD5 = 250 mg/L
Aeration basin volume = 100,000 ft3
Two circular clarifiers (each with diameter = 60 feet and depth = 12 feet)
Return activated sludge flow rate = 1.0 MGD per clarifier
Mixed liquor suspended solids concentration = 3,500 mg/L
Waste sludge flow rate (from clarifier underflow) = 100,000 gal/day
Waste sludge VSS concentration = 5,100 mg/L
Effluent suspended solids concentration is negligible
Determine the following parameters: organic loading rate, aeration tank
hydraulic residence time (excluding recycle), percent recycle, clarifier
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249
hydraulic loading rate, and solids residence time. Compare these values
with the empirical guidelines presented in Table 6.4 for a municipal activated sludge system.
6.11 A municipal WW activated sludge facility has the following characteristics:
Flow = 7.2 million liters per day
Influent BOD5 = 200 mg/L
Aeration basin volume = 3,000 m3
Two circular clarifiers (each with diameter = 20 m and depth = 4.0 m)
Return activated sludge flow rate = 3.5 million L/day per clarifier
MLVSS concentration = 3,500 mg/L
Effluent VSS is negligible
Sludge VSS concentration = 6,800 mg/L
Determine the daily volume of sludge (liters per day) that must be
removed to maintain a solids residence time of 5 days.
6.12 A municipal wastewater activated sludge facility has the following characteristics:
Flow = 5 MGD
Influent BOD5 = 250 mg/L
Aeration basin volume = 506,400 ft3
Two circular clarifiers (each with diameter = 80 feet and depth = 12 feet)
Return activated sludge flow rate = 2.0 MGD per clarifier
MLVSS concentration = 3,300 mg/L
Effluent VSS concentration is negligible
Sludge VSS concentration = 6,800 mg/L
Determine the daily volume of sludge (gallons per day) that must be
removed to maintain a solids residence time of 5 days.
6.13 An anaerobic sludge digester must be designed to achieve a minimum
hydraulic residence time of 20 days. The input sludge mass is 5,000 lbs
(dry weight) per day and the suspended solids concentration is 10,000
mg/L. Determine the required digester volume (ft3).
6.14 For municipal wastewater that receives treatment with the activated
sludge process, plot the effluent BOD5 as a function of solids residence time
using values of SRT that range from 1 to 20 days. The influent BOD5 concentration is 220 mg/L. Assume the following wastewater characteristics:
Variable order kinetics
k = 5 mg BOD5/mg VSS-day
KS = 60 mg BOD5/L
Ymax = 0.6 mg VSS/mg BOD5
ke = 0.06 day–1
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6.15 For municipal wastewater being treated with the activated sludge process, recommend an operating value for the SRT to achieve compliance
with secondary treatment standards. The influent BOD5 concentration is
220 mg/L. Assume the following wastewater characteristics:
Variable order kinetics
k = 5 mg BOD5/mg VSS-day
KS = 60 mg BOD5/L
Ymax = 0.6 mg VSS/mg BOD5
ke = 0.06 day–1
6.16 For the conditions in Problem 6.13, it has been proposed to thicken the
sludge prior to digestion. It is expected that thickening will achieve an
increase in the sludge solids content to 3% solids. Determine the required
digester volume (ft3) to achieve a 20-day hydraulic residence time with a
thickened sludge input.
6.17 For the conditions in Problem 6.13, it has been proposed to further dewater the sludge prior to digestion. It is expected that gravity belt thickeners
will achieve an increase in the sludge solids content to 15% solids. Determine the required digester volume (ft3) to achieve a 20-day hydraulic residence time with this dewatered sludge input.
6.18 For the biological growth reaction in Example 6.9, consider if ammonia
(NH3) was used to provide the nitrogen requirements. Determine the
ammonia addition rate (pounds of ammonia per day). The balanced reaction for production of cell mass using ammonia is:
25 C6H12O6 + 3 NH3 + 135 O2
→
3 C5H7O2N + 135 CO2 + 144 H2O
6.19 Given a typical activated sludge wastewater treatment plant, with sludge
wasting from the clarifier underflow, and given the following data:
Qi = 11.8 million L/day
Qe = 11.5 million L/day
Si = 400 mg/L
Se = 20 mg/L
Xr = 6,500 mg/L
Xe = 15 mg/L
a. Draw a simple sketch of the process, labeling, as appropriate, the
parameters listed above.
b. Calculate the waste sludge volumetric flow rate, Qw, L/day.
c. Calculate the total biomass discharge rate (ΔX/Δt), kg/day.
d. Given that the biomass concentration in the aeration tank is 2,000 mg/
L, calculate the sludge recycle flow rate, L/day.
e. Given that the aeration tank volume is 4.0 million liters, calculate the
F:M ratio.
f. Calculate the SRT, days.
6.20 Given a typical activated sludge wastewater treatment plant, with sludge
wasting from the clarifier underflow, and given the following data:
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Wastewater Treatment
Qi = 9.8 million L/day
Qe = 9.5 million L/day
Si = 350 mg/L
Se = 16 mg/L
Xr = 5,500 mg/L
Xe = 14 mg/L
251
Qr = 5 million L/day
a. Draw a simple sketch of the process, labeling, as appropriate, the
parameters listed above.
b. Calculate the waste sludge volumetric flow rate, Qw, in L/day.
c. Calculate the total biomass discharge rate (ΔX/Δt), in kg/day.
d. Given that the aeration tank volume is 4.0 million liters, calculate the
F:M ratio.
e. Calculate the solids residence time (SRT) in days.
6.21 Given a typical activated sludge wastewater treatment plant, with sludge
wasting from the clarifier underflow, and given the following data:
Qi = 8.8 MGD
Qe = 8.6 MGD
Si = 350 mg/L
Se = 16 mg/L
Xr = 5,500 mg/L
Xe = 12 mg/L
a. Draw a simple sketch of the process, labeling, as appropriate, the parameters listed above.
b. Calculate the waste sludge volumetric flow rate, Qw, in L/day.
c. Calculate the total biomass discharge rate (ΔX/Δt), in kg/day.
d. Calculate the observed yield (Y) of biomass.
e. The aeration tank has a volume of 12.0 million liters. If the solids residence time (SRT) is to be set at 6 days, calculate the biomass concentration that must be maintained in the aeration tank, mg/L.
6.22 Using at least two identical secondary clarifiers, calculate the diameter of
each clarifier for an activated sludge WWTP that treats 10 MGD. The
MLSS is 3,000 mg/L, and the recycle sludge rate is 4.0 MGD. Assume an
HLR and SLR that are mid-range of the typical values given in Table 6.4.
Calculate the area using each guideline, and pick the appropriate area.
Specify the number of clarifiers and the diameter (ft) of each. Keep in
mind that diameters must be specified in 5-ft increments.
6.23 Using at least two identical secondary clarifiers, calculate the diameter of
each clarifier for an activated sludge WWTP that treats 8 MGD. The MLSS
is 2,600 mg/L, and the recycle sludge rate is 5.0 MGD. Assume an HLR
and SLR that are mid-range of the typical values given in Table 6.4. Calculate the area using each guideline, and pick the appropriate area. Specify
the number of clarifiers and the diameter (ft) of each. Keep in mind that
diameters must be specified in 5-ft increments.
6.24 What is the nominal HRT for a 1,500-gallon septic tank installed at a
home for a family of four people? Assume that the family generates
wastewater at an average rate of 70 gpcd.
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REFERENCES
Bisogni, J. J., and A. W. Lawrence. 1971. “Relationships Between Biological Solids Retention Time and Settling Characteristics of Activated Sludge.” Water Research, 5:753.
Dietz, John D. 1996. “Wastewater Treatment.” In Environmental Engineering P.E. Examination Guide & Handbook, edited by W. Christopher King. Annapolis, MD: American
Academy of Environmental Engineers.
Grady, C. P. L. 1990. “Biodegradation of Toxic Organics: Status and Potential.” Journal of
Environmental Engineering Div. ASCE, 116:805.
Kenny, J. F., and K. E. Juracek. 2012. “Description of 2005-2010 Domestic Water Use for
Selected U.S. Cities and Guidance for Estimating Domestic Water Use.” U.S. Geological Survey Scientific Investigations Report 2012-5163.
Lawrence, A. W., and P. L. McCarty. 1970. “A Unified Basis for Biological Treatment
Design and Operation.” Journal of Sanitation Engineering Div. ASCE, 96:757.
Metcalf & Eddy, Inc. 2003. Wastewater Engineering: Treatment and Reuse. New York:
McGraw-Hill.
Nemerow, N. L., and A. Dasgupta. 1991. Industrial and Hazardous Waste Treatment. New
York: Van Nostrand Reinhold.
Spellman, F. R. 2009. Handbook of Water and Wastewater Treatment Plant Operations. 2nd
ed. Boca Raton, FL: CRC Press, Taylor & Francis Group.
Ten States Standards. 2004. Recommended Standards for Wastewater Facilities. 2004 ed.
Albany, NY: Health Research Inc./Health Education Services.
Toor, G. S., and D. P. Rainey. 2009. “History and Current Status of Reclaimed Water Use
in Florida.” Institute of Food and Agricultural Sciences, University of Florida, Doc.
#SL308. Accessed July 2013. https://edis.ifas.ufl.edu/ss520
US EPA (Environmental Protection Agency). 1988. Waste Minimization Opportunity
Assessment Manual. EPA/625/7-88/003.
US EPA. 2002. “A Homeowner’s Guide to Septic Systems.” EPA-832-B-02-005. Washington, DC: EPA Office of Water.
US EPA. 2010. “Technology-Based Effluent Limits.” Chapter 5 in NPDES Permit Writers’
Manual. EPA-833-K-10-001. Accessed July 2013. http://cfpub.epa.gov/npdes/
writermanual.cfm
US EPA. 2013a. “Basic Information about Water Security.” Office of Groundwater and
Drinking Water, Water Security Division. Accessed September 2013.
http://water.epa.gov/infrastructure/watersecurity/basicinformation.cfm#ww
US EPA. 2013b. “U.S. Census Data on Small Community Housing and Wastewater Disposal Practices.” Accessed July 2013. http://water.epa.gov/infrastructure/
wastewater/septic/census_index.cfm
US EPA. 2013c. “Water on Tap—What You Need to Know.” Accessed December 2013.
http://water.epa.gov/drink/guide/upload/book_waterontap_full.pdf
Water Environment Federation (WEF). 2008. Industrial Wastewater Management, Treatment, and Disposal. 3rd ed. New York: McGraw-Hill.
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CHAPTER
7
Air Resources
7.1 Management of Air Resources
When our early ancestors first discovered fire they also first experienced
air pollution, especially if that first fire was built inside a cave without good
ventilation! Air pollution is tied directly to the burning of fuels, and can be pervasive in large, densely populated urban areas. Air pollution has caused local
problems for many centuries (the use of coal in London was banned for several
years starting in 1307 by King Edward I), but it did not become a global problem until the advent of the industrial revolution. Now, air pollution is a serious
problem in many countries, with the highest concentrations being in the
world’s megacities, including Jakarta (Indonesia), São Paulo (Brazil), Bangkok
(Thailand), Cairo (Egypt), Santiago (Chile), Los Angeles (United States), Mexico City (Mexico), and others.
Air pollution in Beijing, China, is among the world’s worst. It was so bad
during the summer Olympic Games of 2008, the Chinese government closed a
number of factories and severely limited automotive traffic to try to improve
local air quality temporarily. In March 2014, the French government instituted
an odd-even driving ban in Paris to combat poor air quality in that city
(Petroff 2014). An odd-even driving ban is a local ordinance that restricts
motorists to only driving their cars on odd-numbered days if their license
plate ends in an odd number, or on even-numbered days if their license plate
ends in an even number.
Since the early 1800s, human population has increased from about 500 million to more than 7 billion, or more than one order of magnitude. World energy
consumption has increased by more than two orders of magnitude, with much
greater than average increases occurring in the industrialized centers. With
such great increases in energy consumption, air pollution emissions have
increased enormously since the 1800s, notwithstanding the efforts of the last
forty years to control pollution. Based on the principles of material balance,
with increased emissions, and with no change in the natural rates of removal
from the atmosphere, it should not be surprising that accumulations of pollutants in the air have occurred frequently, resulting in harmful levels of air pollu253
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tion at various times and places. In this chapter we briefly address some major
air pollution problems facing the world and review modern approaches to the
management and control of these problems.
Air resource management (ARM) involves several steps, including identifying the problem pollutants, determining appropriate air quality standards,
identifying and controlling major sources of emissions, conducting air quality
modeling, monitoring actual air quality, and assessing program effectiveness.
The purpose of ARM is to provide good air quality to all citizens while balancing other social, cultural, political, and economic needs. Because the atmosphere knows no boundaries, pollutants emitted in one location can easily be
transported to another. This same transport dilutes pollution and makes it difficult to identify polluters, so in the past, many industries simply discharged
their wastes into the air and counted on the atmosphere to carry them away.
However, in very large urban areas, there is no “away,” so it becomes much
more important to reduce emissions and prevent pollution.
There are several well-known episodes where very high air pollution concentrations in cities caused thousands to get sick and resulted in many deaths
(Donora, Pennsylvania, in 1948 and London, England, in 1952, for example),
but air resource management is more than just emergency action planning.
Modern ARM involves making plans and taking actions to reduce everyday
concentrations and to provide healthy air on a long-term basis.
An important prerequisite to ARM is to know what the problem is. That is,
an emissions inventory is needed to identify problem pollutants and polluters.
An emissions inventory (EI) is a comprehensive and quantitative listing or
estimation of major sources of pollution within a geographic area. An EI is very
helpful in deciding where to focus the efforts of a city or state in controlling air
pollution. The EI is developed from measured or estimated emissions data
about all major pollutants and for all sources, including industry, mobile (or
transportation) sources, area sources, and natural sources.
The agencies in charge of air quality must organize and present their data
effectively to encourage political support and public participation. They must
analyze the emissions data in conjunction with the ambient air quality data to
propose the best strategies for reducing certain pollutants. These strategies
might include emissions limits on certain industries, on-site inspections and/
or fines, vehicle inspections, and traffic reduction ordinances and/or programs. Such agencies must then follow up and measure or estimate the results
of their efforts.
In this chapter, we will learn about the major components of air resource
management. This knowledge will enable us to identify and quantify the major
pollutants (and their causes, sources, and effects), to become familiar with the
legislative and regulatory standards that have been developed for effective
management of local and national air quality, to select appropriate engineered
equipment for air pollution control (from both stationary and mobile sources),
and to conduct dispersion modeling of air pollution from proposed or existing
sources to predict pollutant concentrations in ambient air at locations downwind of those sources.
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255
7.2 The Major Air Pollutants
Air pollution can be defined as foreign matter contained in the air in high
enough concentrations to cause harm to people, plants, animals, or things. Primary air pollutants (those emitted directly to the air) and secondary pollutants
(those formed by reactions in the atmosphere) are both serious problems.
Major pollutants are defined in this text to mean those emitted in very large
quantities (millions of tons per year in the United States) or those of national
concern because of their health effects. The six major primary pollutants are
particulate matter (PM), sulfur oxides (SOx), nitrogen oxides (NOx), volatile
organic compounds (VOCs), hazardous air pollutants (HAPs), and carbon
monoxide (CO). The most significant secondary pollutant is ground-level
ozone (O3). All of the above except VOCs and HAPs are also called criteria pollutants, because the US EPA in the 1970s established ambient standards based
on measurable health effects (the criteria for the standards).
Particulate matter (PM) is the term used to describe very small diameter
solids or liquids that remain suspended in the atmosphere. Two sizes are regulated: PM-10 and PM-2.5, referring to particulate matter less than 10 and 2.5 µm
in diameter, respectively. Particles are emitted from a variety of sources, including fossil-fuel combustion, metals and mineral processing, fugitive dusts from
roads, agricultural fields, and many others (US EPA 2013). In the 1970s, the
EPA established health-based air quality standards for total PM, but changed
the standards to PM-10 in the late 1980s, and in 1997 added PM-2.5 in recognition of the more serious health effects of smaller particles. The effects of PM-2.5
include an adverse impact on human health (mainly related to the respiratory
system), reduction in visibility due to haze (small particles scatter light and
make things appear hazy), and soiling of buildings and other materials.
Nitrogen oxides (NOx) are formed whenever any fuel is burned in air at a
high enough temperature. In high-temperature combustion gases, the nitrogen
and oxygen in the air combine to form NO and NO2. Also, nitrogen atoms present in some fuels can form NOx. In the US, the 2012 annual NOx emissions from
mobile sources (both on-road and off-road vehicles) were significantly greater
than those from electric utilities and industrial furnaces (US EPA 2013). Nitrogen
oxides can contribute to a reduction in visibility (NO2 absorbs light), can be injurious to plants and animals, and can have adverse effects on human health.
Nitrogen oxides also contribute to acid deposition. Perhaps most importantly
for urban areas, NOx reacts with VOCs in the presence of sunlight to form
ground-level ozone, a significant secondary pollutant with serious health effects.
Sulfur oxides (SOx) are produced whenever a fuel that contains sulfur is
burned. The main source is fossil-fuel combustion, although nonferrous metal
smelting is also a major source. The main sulfur oxide is SO2; however, some
SO3 is formed during combustion, and some SO3 is formed by oxidization of
SO2 in the atmosphere. SO2 and SO3 form acids when they combine with
water, and these acids can have detrimental effects on aquatic and terrestrial
ecosystems, and on statues and buildings. In addition, gaseous SO2 has been
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associated directly with human health problems and can cause damage to
plants and animals.
Volatile organic compounds (VOCs) include any organics with an appreciable vapor pressure such that they vaporize when exposed to air. A major
source of VOCs are automobiles and other mobile sources, from which small
amounts of unburned fuel are exhausted to the air. Petrochemicals production;
petroleum refining, transport, and storage; and the pumping of gas into consumers’ cars account for substantial VOC emissions, as does evaporation of
solvents (such as those in oil-based paints or printing inks). Some VOCs are
carcinogenic, but the major problem with VOCs is that they react photochemically with NOx in the atmosphere to form ozone.
Hazardous air pollutants (HAPs) are certain compounds (such as benzene,
formaldehyde, vinyl chloride, lead, mercury, and many others) that were specifically identified by the US EPA in the 1990 Clean Air Amendments. HAPs
are emitted from a wide variety of sources, both combustion and noncombustion. Most HAPs are not emitted in very large quantities, but are considered
serious because of their potential to severely damage human health.
HAPs are emitted not only from process industries like refineries and
chemical plants, but also as unwanted by-products of combustion from any
industry that burns fuel in boilers or process heaters (Federal Register 2013). So
the emission rates by source categories are not as well documented as with criteria pollutants. EPA’s regulatory efforts have been paying off, though. For
example, the benzene (a known carcinogen) content of gasoline has been
reduced significantly from 1990 to present. EPA estimates that the nationwide
emissions of benzene from on-road motor vehicles has dropped from 257,000
tons in 1990 to 171,000 tons in 1996, and will drop to 68,000 tons in 2020 (US
EPA 2000).
While the emission of HAPs from certain industrial sources has long been
recognized and regulated by the EPA (e.g., mercury, vinyl chloride, benzene,
asbestos), the regulation of HAPs from combustion sources is relatively new; in
fact, EPA published its final rule on HAPs Emission Standards on January 31,
2013. This final rule regulates, among other things, boiler and process heater
emissions of HCl, mercury, CO, and PM for a variety of combustion equipment
(Federal Register 2013). The CO and PM are surrogates for organic HAPs (like
formaldehyde, benzene, and others) and inorganic HAPs (like lead, arsenic,
and others), respectively. There are different numerical limits for new boilers
and process heaters versus existing equipment.
Carbon monoxide (CO) is a colorless, odorless, tasteless gas that results
from the incomplete combustion of any carbonaceous fuel. Usually, power
plants and other large furnaces do not emit much CO because they are
designed and operated carefully enough to ensure fairly complete combustion.
Thus, the major sources of CO are on-road and non-road mobile sources. Automobiles, trucks, buses, airplanes, trains, boats, construction equipment, lawn
and garden equipment, and farm vehicles exhausted more than 34 million tons
of CO in 2012 (US EPA 2013). CO reacts with the hemoglobin in blood to block
oxygen transfer. Depending on the concentration of CO and length of expo-
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257
sure, effects of polluted air on humans may range from slight impairment of
some psychomotor functions to dizziness and nausea. Of course, in enclosed
spaces with poor ventilation (such as a garage), CO can build up to very high
concentrations and can cause death.
Ozone and other oxidants are not emitted from a source per se, but are
formed by complex photochemical reactions in the atmosphere involving
mainly VOCs, NOx, and sunlight. The classic term “smog” is defined as a mixture of smoke and fog, and originated in the early part of the twentieth century
when the major problems were PM and SOx. The term “smog” as used today
describes the complex mix of air pollutants found in many cities, including
high levels of ozone but also PM, NOx, VOCs, SOx, and many other compounds; it often appears from a distance as a visible layer of material hanging
over the city. Thus, urban smog is caused both by direct emissions of pollutants
and by photochemical reactions in the air, and often is more of a problem during the warm months of the year. The photochemical reactions also form small
particles that contribute to haze.
The ozone and other oxidants in smog attack plants and materials, and
cause serious human health effects including eye, nose, and throat irritation, as
well as premature “aging” of the lungs. Reduction of ozone problems focuses
on controlling emissions of VOCs or NOx, depending on which compound is
present in the air in the “critical” concentration. Detailed emission inventories
and modeling are required to determine the best strategy for each urban area.
The preceding discussion of the causes, sources, and effects of air pollution
is summarized in Table 7.1 on the following page. Recent estimates of US
annual emissions of air pollutants are presented in Table 7.2 on p. 259.
7.3 Global Issues
There are three global air pollution issues discussed in this section: acid
deposition, ozone layer depletion, and global climate change. The reader must
recognize in advance that the following discussions are brief, and that there
have been volumes written about each of these issues. Although we cannot
devote much space to these topics in an introductory text, this in no way
diminishes their level of importance to the world community.
Acid Deposition
Acid deposition refers to the fallout of acid rain, acid snow (or any other
material that is acidic) from the atmosphere onto the earth. The acids are usually sulfuric or nitric acids that are formed in the atmosphere as a result of SOx
and NOx emissions. Acid deposition is a serious issue and has been studied
and debated for many years due to two important characteristics: (1) acid
deposition has seriously damaged lake and forest ecosystems, and (2) the acids
tend to be transported long distances before they are deposited. For example,
acids originating in the upper midwestern region of the United States have
acidified lakes in Canada as well as damaged forests from New England to Vir-
Power plant boilers,
construction,
industrial processes
Boilers, furnaces,
vehicles
Burning coal,
crushing, grinding
Reaction of N2 with
O2 at high
temperature
Burning any fuel
with sulfur; processes using SO2 or
H2SO4
Incomplete burning of fuels; evaporation of solvents
Similar to causes
for VOC emissions,
plus others
Incomplete combustion of carbon
fuels
Chemical reaction
of VOCs, NOx, and
sunlight
Particulate
Matter (PM-10
& PM-2.5)
NOx
SOX
VOCs
HAPs
CO
O3
N.A.
Vehicles and industrial processes
HAPs are emitted
from industrial &
commercial processes, furnaces,
and mobile sources
Motor vehicles,
industrial processes
Boilers and furnaces, industrial
processes, smelting
Source
Cause
Irritating and damaging to lungs,
eyes, nose, and
throat
Poisonous; reacts
with blood
hemoglobin
HAPs are suspected or known
carcinogens or
have other toxic
health effects
Certain VOCs are
toxic or
carcinogenic
Acts with particulates (synergism)
Respiratory irritant
Bronchitis, emphysema, cancer, etc.
Human Health
Severe damage to
plants
None to plants
Certain HAPs have
similar toxic effects
on animals
Minor
Necrosis, chlorosis
Minor
Damage to leaf
structure, toxic
effects
Plants, Animals
Corrosion,
oxidation,
bleaching
None
Some HAPs are
acids
None
Corrosion
Corrosion
Soiling,
corrosion
Materials
Effects On:
Summary of Major Air Pollutants—Causes, Sources, and Effects
Photochemical
smog
Eventually oxidizes to CO2. Contributes to global
warming
Some HAPs bioaccumulate in the
environment
Reacts with NOx to
form photochemical smog
Forms H2SO4 in
atmosphere; adds
to haze and smog
Reacts with VOCs
to form photochemical smog
and forms HNO3 in
atmosphere
Haze, smog
Other
258
Pollutant
Table 7.1
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Table 7.2
259
US Emissions Estimates of Major Pollutants, 2012 (millions of tons/yr)
CO
SOx
NOx
VOC
PM-10
PM-2.5
Fuel Combustion
Electric Utilities
Industrial Furnaces
Other Combustion
Subtotal: Fuel Combustion
0.72
0.89
2.75
4.36
3.31
1.07
0.29
4.67
1.72
1.39
0.58
3.69
0.04
0.11
0.41
0.56
0.40
0.19
0.39
0.98
0.30
0.15
0.38
0.83
Industry & Area Sources
Chemical and Allied Manuf.
Metals Processing
Petroleum & Related Industries
Other Industrial Processes
Solvent Utilization
Storage & Transport
Waste Disposal & Recycling
Subtotal: Industry & Area Sources
0.18
0.84
0.26
0.42
0.01
0.02
1.40
3.13
0.18
0.18
0.15
0.25
0.00
0.01
0.02
0.79
0.06
0.08
0.42
0.42
0.01
0.01
0.10
1.10
0.09
0.04
1.74
0.36
3.30
1.19
0.18
6.90
0.03
0.08
0.03
1.07
0.01
0.05
0.24
1.51
0.02
0.06
0.02
0.34
0.01
0.02
0.20
0.67
22.77
11.40
34.17
12.20
8.91
62.77
0.01
0.01
0.02
0.07
0.14
5.69
3.86
2.26
6.12
0.10
0.16
11.17
1.95
1.57
3.52
2.85
1.85
15.68
0.27
0.19
0.46
1.18
18.48
22.61
0.15
0.17
0.32
1.00
3.01
5.83
Mobile Sources
Highway Vehicles
Off-Highway
Subtotal: Mobile Sources
Wildfires
Other Miscellaneous
Total All Above
Source: US EPA (2013).
ginia; acids from many countries in western Europe have impacted lakes in
Scandinavia, as well as destroyed forests in Germany. Because of this characteristic, there have been acrimonious debates between countries about how to
control acid deposition and who should do what to mitigate its effects. It is
clear that acid deposition can be reduced only if acid gas emissions are
reduced. The United States and other developed countries have made a major
commitment to this goal. Over a 22-year span, US emissions of SOx and NOx
have decreased remarkably: SOx emissions dropped from 23 million tons in
1990 to 5.6 million tons in 2012, and NOx emissions dropped from 25 million
tons in 1990 to 11 million tons in 2012 according to the US EPA (2013). However, they continue to increase in many countries in the world.
Stratospheric Ozone Depletion
The ozone layer, located about 15 to 30 km above the earth’s surface, is a portion of the atmosphere containing relatively high concentrations of ozone. This
stratospheric ozone is formed naturally by photo-dissociation of oxygen molecules, and is chemically identical to the ground-level ozone that pollutes many of
our cities. However, ozone in the stratosphere is extremely important to life on
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260
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earth because ozone absorbs ultraviolet rays, preventing this harmful radiation
from reaching the surface. The small fraction of the solar ultraviolet radiation that
makes it to ground level is responsible for sunburns, skin cancer, and cataracts.
In the mid-1970s, scientists discovered that certain chlorine-containing
compounds that had been emitted into the atmosphere were slowly working
their way up through the lower layers of air into the ozone layer. These compounds are collectively known as chlorofluorocarbons (CFCs). Once in the
stratosphere, with its higher levels of ultraviolet radiation, the CFCs break
down—releasing free chlorine atoms that decompose ozone. The process is catalytic, meaning that one free chlorine atom can break down thousands of ozone
molecules before eventually becoming inactivated.
For years, CFCs were used in aerosol spray cans, in automobile air conditioners, as foaming agents in making plastics, as fire suppressants, as degreasing compounds for certain metal parts before electroplating, and in many other
uses. The high usage level resulted in emissions of CFCs to the atmosphere with
significant long-term damage to the earth’s protective ozone layer. In the late
1970s scientific evidence was gathered to prove the thinning of the ozone layer,
especially in the Antarctic region. In 1987, representatives from 46 countries met
in Montreal and signed an international accord aimed at limiting the production
and use of these compounds. This accord, known as the Montreal Protocol,
resulted in significant decreases in the 1990s in the use and manufacture of these
and other ozone-depleting chemicals. The original signatory countries pledged
to stop the manufacture and use of CFCs and Halons (and later HCFCs) with a
specific timetable. For example, under the agreement, each country pledged to
reduce the manufacture of the refrigerants CFC-11, CFC-12, CFC-113, and others
from 100% of the 1986 level to 50% in 2005, 15% in 2007, and to completely stop
producing these chemicals in 2010 [United Nations Environment Programme
(UNEP) 2013]. More and more countries joined and accelerated efforts to protect
the ozone layer. As of 2013, 197 countries have signed the agreement (UNEP
2013). As a result, CFC manufacture went from over 1 million tons per year in
1986 to zero in less than 30 years! Substitutes such as hydrochlorofluorocarbons
(HCFCs) were developed to replace CFCs in air conditioning (HCFCs are less
damaging than CFCs), but they still have damaging effects. Work is under way
to replace the HCFCs, and HCFC production is starting to decline.
Global Climate Change
Global climate change (GCC) is perhaps the greatest environmental threat
yet faced by humankind. GCC refers to the change in climate caused by emissions of various compounds into the atmosphere and the resulting change in the
earth’s energy balance. It is well known that the sun provides energy to the earth
that warms the surface during the day, drives the winds and hydrologic cycle,
and is the basis for plant photosynthesis. It is also known that the earth cools off
during the night by radiating infrared energy (heat) out to space. Based on a simple energy balance, we know that over any substantial period of time, the
amount of energy given off during the nights must exactly equal the amount of
energy absorbed during the days or else the heat content of the earth will change.
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261
During the past two hundred years, human activity has disrupted this
energy balance. Emissions of fine particles and sulfuric acid mists have contributed to more reflection of sunlight, preventing some energy from reaching the
earth’s surface. On the other hand, gases in the atmosphere (mainly carbon dioxide, methane, CFCs, and nitrous oxide) absorb infrared radiation, thus retaining
heat and preventing the earth from cooling as much as it otherwise would. This
latter effect seems to be outweighing the former, and the earth’s average temperature is increasing. GCC is popularly called the greenhouse effect.
Carbon dioxide (CO2) is responsible for the majority of the greenhouse effect,
but the other gases—methane, CFCs, and nitrous oxide—all contribute substantially. The latter three are each more “powerful” than CO2 in that their molecules
absorb infrared radiation more effectively. However, the production rate of CO2
is much greater, and it continues to be emitted in great quantity by fuel combustion. Because CO2 has no short-term toxic or irritating effects, and because it is
abundant in the atmosphere and is necessary to plant life, it traditionally was not
considered a pollutant. However in 2009, EPA designated CO2 and other greenhouse gases as pollutants, making them subject to regulation (US EPA 2009).
From material balance considerations, we know that excess emissions of
CO2 into the atmosphere will result in an increase in its overall concentration.
Scientists and engineers for years have had proof of the steady increases in the
atmospheric concentration of CO2 (see Figure 7.1). Notice that CO2 concentrations in this graph rise and fall within each year. Since the measurements are
taken at a remote mountain observatory in Hawaii, this is not due to local
Figure 7.1 Concentration of atmospheric CO2 measured at Mauna Loa, Hawaii.
(Drawn from data from Mauna Loa, Hawaii, Observatory, NOAA 2013.)
400
CO2 Concentration, ppmv
390
380
370
360
350
340
330
320
310
300
1955
1965
1975
1985
Year
1995
2005
2015
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influences. Rather, it reflects the summer/winter plant growth cycle in the
northern hemisphere, with CO2 reaching annual minimums in early Fall, and
annual maximums in early Spring. Despite the variations during each year,
growth in CO2 concentrations in the atmosphere has been very steady, and in
May 2014 topped 400 ppm.
EXAMPLE 7.1
Assume that worldwide combustion of fossil fuel accounts for an energy use
of 300 quadrillion kJ/year. Also, assume that all fossil fuels can be represented by the formula C3H5 with energy content of 40,000 kJ/kg. Finally,
assume that air is 79% N2 and 21% O2, and has a “molecular weight” of 29
kg/kgmol.
(a) Estimate the annual release of CO2 to the atmosphere by burning fossil
fuels, in kg/yr.
(b) If all that CO2 entered the atmosphere (and none was removed), estimate
the increase in atmospheric concentration, in ppm by volume. Assume the
atmosphere contains 5.0 (10)18 kg of air.
SOLUTION
(a) The mass of fossil fuels combusted per year is:
15
300 (10 ) kJ/yr
12
= 7.5 (10 ) kg/yr
40, 000 kJ/kg
Since the carbon mass fraction of the fuel is 36/41 = 0.878, then the carbon
combusted is
0.878 × 7.5(10)12 = 6.58 (10)12 kg/yr
A balanced reaction for the combustion of carbon is
C + O2
→
CO2
which shows that CO2 is produced mole for mole. Therefore, the moles of
CO2 produced equal the moles of carbon burned:
12
6.58 (10 )
kg/yr
12 kg C/kgmol
11
= 5.49 (10 )
kgmol/yr
which translates into a mass of CO2 of:
44
kg CO 2
11
13
¥ 5.49 (10 ) = 2.41 (10 ) kg of CO 2 per year
kgmol
(b) The moles of air in the atmosphere are:
18
5.0 (10 ) kg
17
= 1.72 (10 ) kgmol of air
29 kg air/kgmol
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263
Therefore the increase in concentration is:
11
5.49 (10 )
kgmol CO 2
17
1.72 (10 )
kgmol air
¥ 106 = 3.2 ppm
Of course, the carbon balance for the earth is much more complicated than
implied by Example 7.1. There are many sources of CO2 (including human and
animal respiration, burning of trees and brush, escape of CO2 from ocean
water, volcanic eruptions, and biodegradation of organic matter), and many
sinks for CO2 (such as photosynthesis, absorption of CO2 into ocean water,
mineralization into carbonates, and storage in living or dead organic matter).
All the sources and sinks are interconnected, many in nonlinear ways. Modeling the system is difficult and uncertain, but the net result of this biogeochemical cycle is the steady increase of atmospheric CO2 as shown in Figure 7.1. In
fact, evidence from analyzing gases trapped in old ice cores indicates that the
carbon dioxide content of the atmosphere has varied between about 185 and
295 ppm for the last 400,000 years, and has remained relatively stable at 260–
290 ppm for the last 10,000 years. Through analysis of the ice core gases and
carbon isotope ratios, a strong correlation has been demonstrated between low
CO2 levels and the ice ages, and high CO2 levels and warmer interglacial periods. CO2 was about 280 ppm as recently as 1750, had increased to 315 ppm in
1958, and is now about 400 ppm. The change from 1958 to the present is an
amazing 27% in just 55 years, an enormous change over a tiny interval of time.
It appears that we humans are conducting a massive experiment with our
atmosphere, and we don’t know what the outcome will be.
The greenhouse effect means much more than just a gentle warming of the
earth as the name might imply. As the atmospheric heat content increases, it is
inevitable that weather and climate will change, perhaps significantly. Rainfall
patterns will change, hurricanes may become more frequent and more severe,
crop-growing regions may shift northward, plant and animal (especially
insect) habitats may be affected, and sea level will increase. Predictions of an
average warming of 1 to 2 °C within 50 years are not uncommon. While this
might not sound alarming, remember that during the great ice ages, the average temperature was only about 7 to 9 °C less than it is today.
It has been argued for years that increases in atmospheric CO2 could
increase global temperature and change the climate, both regionally and globally. In fact, such an effect was first predicted in a technical publication by the
famous Swedish chemist Svante Arrhenius more than one hundred years ago
(Arrhenius 1896). However, the earth is a massive system; changes are slow
and are not well understood. The effects of change in one parameter are
masked and counteracted by hundreds of other factors. One thing is clear—
increased heat retention will result in warmer temperatures. One common
measure is the average global temperature (AGT). Unfortunately, this is not the
most sensitive measure and the record is very “noisy” (erratic), making conclu-
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264
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sive trends more difficult to spot (see Figure 7.2). Thus, a number of people
have remained unconvinced that global climate change is a real concern, and
arguments and debates over the validity of GCC have continued. However,
most technical experts have come to agree that global climate change is definitely occurring and may cause dramatic and unpleasant changes in the environments of many regions of the world.
Clearly, with more energy now contained in the warmer oceans and atmosphere than in past years, we can expect more extreme weather. Observation of
actual weather events offers strong anecdotal evidence that weather has indeed
become more extreme (Derevianko and Balentine 2013). The 2010–2011
drought in Texas was the worst in that state’s history (Amico et al. 2013). The
devastation and loss of life wrought by the massive tornadoes in Joplin, Missouri, on May 22, 2011, and in Moore, Oklahoma, on June 1, 2013, were among
the worst in those states’ histories. The widespread damage and huge cost of
superstorm Sandy in New York and New Jersey, on October 25, 2012, was
unprecedented. On June 30, 2013, a temperature of 129 °F was recorded in
Death Valley, California—the hottest temperature ever recorded anywhere in
the United States for the month of June. Heat waves can be particularly hard on
people with medical conditions and on the elderly. In Chicago in 1995, several
hundred people died from heat (Whitman et al. 1997), and more than 70,000
died in Europe during the 2003 heat wave (Robine et al. 2008).
In addition to anecdotal evidence, climate modelers also predict more
extremes of weather, especially heat waves (Gao et al. 2012). It has been projected that an increase in the mean atmospheric CO2 concentration from its
present value of about 400 ppm to about 600 ppm will occur sometime this century. The potential consequences of such an increase include other effects in
Figure 7.2
Average global
temperature for
the last century.
(Drawn from data
from http://
data.giss.nasa.gov/
gstemp/graphs/)
Average Global Temperature, °C
16.0
15.8
15.6
15.4
15.2
15.0
Base period:
1930–1980
14.8
14.6
14.4
14.2
14.0
1880
1900
1920
1940
1960
Year
1980
2000
2020
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265
addition to those previously mentioned, such as altered rainfall patterns and
crop-growing regions, increased insect and bacterial infestations, and a rise in
sea level and subsequent flooding in low-lying coastal areas. In fact, many
large cities in the coastal US (e.g., Boston, Miami) have already begun planning
on how to deal with the various effects on their infrastructure (roads, sewer
systems, drinking water supplies).
7.4 Legislative and Regulatory Standards
It has been said that something that is everybody’s responsibility often
ends up being nobody’s responsibility. So it was with our air resources up until
the 1950s, when the federal government began to recognize its role and to
assume its responsibility. Because air knows no political boundaries, air quality
management appropriately begins at the national level. But, in addition to
national management, state and local efforts are essential to the attainment and
maintenance of good air quality.
Air-pollution standards were mandated by Congress and established by
the US EPA to protect the health and promote the well-being of individuals and
communities. These standards were set by government with input from professional organizations as a result of increased awareness of pollutants and their
effects upon living organisms, especially people. Federal legislation and regulations have been developed over a period of more than fifty years with input
from many interested groups. Some of these laws were the Air Pollution Control Act of 1955, the Motor Vehicle Air Pollution Control Act of 1965, the farreaching Clean Air Act Amendments (CAAA) of 1970, the CAAA of 1977, and
the comprehensive CAAA of 1990. Compliance with these laws requires not
only proper environmental engineering design and operation of pollutionabatement equipment, but careful analysis and accurate measurements of specified pollutants and environmental quality parameters.
There are two types of standards: ambient air quality standards (AAQSs)
deal with concentrations of pollutants in the outdoor atmosphere, while source
performance standards (SPSs) limit the emissions of pollutants from specific
sources. Ambient air quality standards are always written in terms of concentration (µg/m3 or ppm), while SPSs are often written in terms of mass emissions per unit of time or unit of production (g/min or kg of pollutant per ton of
product produced).
National ambient air quality standards (NAAQS) were set by the EPA for
several of the major pollutants at levels to protect the public health. The current
standards are presented in Table 7.3 on the following page. It should be noted
that some states have set their own standards, which are stricter than those
listed. Note also that some pollutants have more than one standard (depending
on the averaging time, or time of exposure). Source performance standards (or
emissions standards) are numerous because of the variety of sources. Some
examples are given in Table 7.4 on p. 268. The new source performance standards listed in Table 7.4 were derived either from materials balance consider-
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266
Chapter Seven
Table 7.3
National Ambient Air Quality Standards
Pollutant
Averaging Time
Primary Standard
PM-10
24-hour average
150 µg/m3
PM-2.5
Annual arithmetic mean
24-hour average
12 µg/m3
35 µg/m3
CO
1-hour average
8-hour average
35 ppm
9 ppm
SO2
1-hour average
75 ppb
NO2
Annual arithmetic mean
1-hour average
53 ppb
100 ppb
O3
3-year average of annual
4th highest daily
8-hour maximum
0.075 ppm
Lead
3-month average
0.15 µg/m3
ations or from actual field tests at a number of industrial plants. The principle
use of an SPS, of course, is to set a legal limit on the amount of pollution that a
particular facility is allowed to emit. One other use of these standards is as emissions factors to estimate emissions from plants yet to be constructed. Yet another
use is to estimate emissions from sources that are difficult to measure in the
field. The next few examples illustrate some calculations with these standards.
EXAMPLE 7.2
The 1-hour National Ambient Air Quality Standard for CO is 35 ppm. Calculate
the corresponding concentration in mg/m3, at 25 °C and atmospheric pressure.
SOLUTION
For gases in air, ppm refers to mole fraction or volume fraction. So 35 ppm is
equivalent to 35 ml of CO per million ml of polluted air, or 35 ml of CO per m3
of air. The mass density of pure CO in mg/ml can be derived from the ideal
gas law:
n
P
=
V RT
or
m P ( M.W .)
=
V
RT
1.0 atm ( 28g/gmol )
m
=
V 0.08206 L-atm 298K
gmol-K
or
m
= 1.145 g/L or 1.145 mg/ml
V
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267
Thus, the concentration of CO in air (in mg/m3) that equates to 35 ppm is:
CO =
35 ml CO
3
m air
¥
1.145 mg CO
= 40 mg/m 3
ml CO
Alternatively, we could have used Eq. (2.5) to solve directly:
C=
35 ¥ 1, 000 ¥ 28
= 40, 080 µg/m3 or 40 mg/m 3
24.45
EXAMPLE 7.3
Calculate the daily SO2 emissions that a 200-ton-per-day sulfuric acid plant
would emit if emitting at the maximum allowable rate (per Table 7.4).
SOLUTION
The standard is a maximum of 2 kg SO2 per metric ton of sulfuric acid.
2 kg SO 2
1 MT
T
¥
¥ 200
= 363 kg SO 2 /day
MT acid 1.102 T
day
EXAMPLE 7.4
Calculate the daily emissions of particulates (PM) and SO2 from a 1000-MW coalfired power plant which meets the performance standards listed in Table 7.4.
SOLUTION
First calculate the gross energy output for this plant:
Eout = 1, 000 MW ¥
24 hr
MWh
= 24, 000
day
day
Next, calculate PM and SO2 emissions, separately:
PM = 0.090
SO 2 = 1.0
lb
MWh
1 ton
ton
¥ 24, 000
¥
= 1.08
MWh
day
2, 000 lb
day
lb
MWh
1 ton
ton
¥ 24, 000
¥
= 12.0
MWh
day
2, 000 lb
day
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268
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Table 7.4
Selected Examples of New Source Performance Standards (NSPSs)
1. Steam electric power plants (coal-fired), newly constructed
a. Particulate Matter: 0.090 lb/MWh gross energy output
b. NOx: 0.70 lb/ MWh gross energy output
c. SO2: 1.0 lb/MWh gross energy output or 95% reduction
d. Hg: 0.002 lb/GWh gross energy output
2. New large (>250 tons/day) municipal solid waste (MSW) combustors: There are individual
standards for dioxins/furans, cadmium, lead, mercury, HCl, particulate matter, NOx and SO2.
Four examples are:
a. PM: 20 mg/dscm* corrected to 7% O2
b. HCl: 25 ppm dry volume, corrected to 7% O2
c. Hg: 50 µg/dscm corrected to 7% O2
d. NOx: 180 ppm dry volume, corrected to 7% O2
3. Nitric acid plants: The standard is a maximum 3-hr average NOx emission of 0.5 lb/ton of
100% acid produced. All NOx emissions are to be expressed as 100% NO2. Also, the stack
gases must meet 10% opacity (where 0% opacity represents perfectly clear stack gas, and
100% opacity means completely opaque).
4. Sulfuric acid plants: The standard is a maximum 3-hr average emission of SO2 of 2 kg/metric ton of 100% acid produced. An acid mist standard is a maximum 3-hr emission of 0.075
kg SO2 per metric ton of acid produced. Also, the stack gases must meet 10% opacity.
5. Primary copper smelters: The particulate emission standard is 50 mg/dscm, the SO2 standard is 0.065% by volume, and the opacity is limited to 20%.
6. Wet-process phosphoric acid plants: The total fluorides emission standard is 10.0 g/metric
ton of P2O5 feed.
7. Iron and steel plants: Particulate discharges may not exceed 0.10 lb PM/ton of metal
charged, and the opacity must be 10% or less except for 2 minutes in any hour.
8. Sewage sludge incinerators: The particulate emission standard is 0.65 g/kg sludge input
(dry basis). The opacity standard is 20%.
9. Hospital/medical/infectious waste incinerators: large ( >500 lb/hr of waste feed). There are
individual standards for PM, CO, dioxins/furans, HCl, SO2, NOx, and several metals, all corrected to 7% O2. Examples include:
a. PM: 25 mg/dscm
b. CO: 11 ppmv
c. Dioxins/furans: 9.3 ng/dscm total CDD/CDF or 0.054 ng/dscm TEQ
d. HCl: 6.6 ppmv or 99% reduction
e. Selected metals: Pb—0.036 mg/dscm, Cd—0.0092 mg/dscm, Hg—0.018 mg/dscm
*dscm means dry standard cubic meter.
Source: US EPA, various regulations.
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269
7.5 Air Pollution Control: Stationary Sources
The most effective control often is simply a step or steps to prevent pollution
from being formed. In recent years, industries have increasingly taken such
steps. As discussed previously, since the mid-1980s industries have worked
hard to find replacement cleaning techniques and have significantly reduced
their use of (and emissions of) CFCs. Oil refineries have continually increased
their standards of maintenance over the last forty years and have significantly
reduced leaks of petroleum products (this not only prevents air pollution but
results in the recovery of more products that can be sold). Nevertheless, no process can be made 100% efficient. Therefore, there will always be some air pollution emissions that must be controlled. Engineers have developed several large,
interesting, and important air pollution control (APC) devices for industrial
sources. These devices fall into two broad categories—those that capture and
remove particles and those that capture or destroy gases.
The discussion of APC technology is organized as follows: First we discuss
the various types of devices that control particulate matter (PM)—small solids
or liquid droplets that are suspended in a flowing stream of air or exhaust
gases. After the discussion of PM controls, we present information on control
equipment for gases, including volatile organic compounds (VOCs), carbon
monoxide (CO), sulfur oxides (SOx), and nitrogen oxides (NOx).
Particulate Matter Control Devices
There are several APC devices for removing particulate matter from
exhaust gases before the gases are emitted into the atmosphere. These include
cyclones, electrostatic precipitators, baghouses, and scrubbers. In the following
few pages we give a brief description of each device, then compare their
advantages and disadvantages for removing particulate matter. In comparing
particulate control devices we should consider collection efficiency and cost
(both capital and operating) simultaneously. Collection efficiency is defined as
h=
Mass rate of particles collected
Mass input rate of particcles
(7.1)
where: η = collection efficiency, fraction
A cyclone is designed to remove particles by causing the entire gas stream
to spin in a vortex. The centrifugal force acting on the larger particles flings
them toward the wall of the cyclone where they impinge and then slide to the
bottom of the cyclone. The gas flows out through the top of the cyclone and the
particles can be removed from the bottom. Advantages of cyclones are that
they are simple, rugged, and inexpensive. Also, they collect the PM in a dry
form so that the solids can be re-used. The major disadvantage is that the efficiency is usually not high enough to be able to use the cyclone as a final control
device. Also, moving the gas through a cyclone at high enough velocities to
collect a reasonable fraction of the PM creates a substantial pressure drop
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270
Chapter Seven
(which means an increase in operating costs). Figure 7.3 presents a
schematic diagram of a cyclone.
Cleaned gas out
An electrostatic precipitator
(ESP) removes particulate matter
Vortex-finder tube
from a gas stream by creating a
Tangential
high voltage drop between elecinlet duct
trodes. A gas stream carrying particles flows into the ESP and
between sets of large plate electrodes; gas molecules are ionized,
Gas flow path
Dusty gas in
the resulting ions stick to the particles, and the particles acquire a
charge. The charged particles are
attracted to and collected on the
oppositely charged plates while
the cleaned gas flows through the
device. While the gas flows between the plates at velocities in the
range of meters per second, the
particles move toward the plates at
a velocity (called the drift velocity)
which is in the range of a meter
per minute. During the operation
Dust out
of the device, the plates are rapped
periodically to shake off the layer
of dust that builds up. The dust is collected and disposed of or recycled. A cutaway view of an electrostatic precipitator is presented in Figure 7.4.
ESPs are large and expensive, but collect particles with very high efficiencies. A major advantage is that they present very little resistance to gas flow
and therefore cause only a slight pressure drop even when operating on flows
as large as a million cubic feet per minute. Because of this advantage, their
operating costs are not as great as one might expect. Many coal-fired power
plants use ESPs. The design of ESPs is complicated, but the collection efficiency
is reasonably well-modeled by the simple Deutsch equation:
Figure 7.3 Schematic flow diagram of a cyclone.
η = 1– e–Aw/Q
where:
η = collection efficiency, fraction
A = area of the collection plates, ft2
w = effective drift velocity of particles toward the plates, ft/min
Q = gas volumetric flow through the ESP, ft3/min
(7.2)
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Air Resources
271
Figure 7.4 Cutaway view of an electrostatic precipitator. (Courtesy of the Babcock &
Wilcox Company, Barberton, OH.)
EXAMPLE 7.5
(a) Calculate the collection plate area needed for an ESP that must treat
500,000 cfm of air and achieve 98.8% efficiency. The drift velocity is
expected to be 18 ft/min.
(b) After the ESP is built with the plate area calculated in part (a), it is discovered that the actual efficiency is only 97.5%. Calculate the actual drift
velocity.
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272
Chapter Seven
SOLUTION
(a) Rearranging Eq. (7.2), we get
A=
=
-Q
ln (1 - h)
w
-500, 000
ln (1 - 0.988 )
18
= 122, 857 ft 2
or 122, 900 ft 2
(b) Rearranging Eq. (7.2) again, we get
w=
=
-Q
ln (1 - h)
A
-500 , 000
ln (1 - 0.975)
122, 900
= 15.0 ft/min
A baghouse can be thought of as a giant multiple-bag vacuum cleaner. The
average air velocity through the bags is an important design parameter and is
calculated by dividing the total gas flow by the total cloth area of the bags (V =
Q/A). Typically, V must be kept very low (1.5 to 8 ft/min), therefore a baghouse may require many bags in parallel. For example, to filter a flow of
500,000 cfm at an average velocity of 2.5 ft/min using bags that have a surface
area of 20 ft2 each, 10,000 bags would be required. Gas containing the particulates flows through cloth filter bags. The dust is filtered from the gas stream,
while the cleaned gas passes through the cloth and is exhausted to the atmosphere. The bags are periodically cleaned (usually by shaking the bags or by
blowing clean air backwards through them) to knock the dust down to the bottom hoppers where it can be removed to be either recycled or disposed. A cutaway view of a baghouse is presented in Figure 7.5.
The capital costs of baghouses are high, but their efficiencies are so high
that they have become very popular as final control devices. Many power
plants, and a variety of dry-process industries, use baghouses. Baghouses have
been used to control medical waste incinerator emissions. When powdered
lime and activated carbon are injected into the gases before flowing into the
baghouse, the system will control not only particles, but also HCl gases and
mercury fumes. The biggest operating cost comes from forcing large volumetric
flows of air or combustion gases through the bags, which creates a substantial
pressure drop. An equation relating pressure drop to energy consumption is:
W=
Q DP
h
(7.3)
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Air Resources
273
where:
W = the energy consumed by a fan, kw
Q
= gas flow rate, m3/s
∆P = pressure drop, kPa
η
= fan and motor efficiency (not particle collection efficiency)
Figure 7.5 Cutaway view of a shaker baghouse. (Courtesy of Siemens Energy, Inc.,
Orlando, FL.)
Shaker
mechanism
Clean air
side
Outlet
pipe
Filter
bags
Inlet
pipe
Baffle
plate
Cell
plate
Dusty air
side
Hopper
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274
Chapter Seven
EXAMPLE 7.6
Assume one very large fan is moving the air of Example 7.5. Assume that
electricity costs 9 cents/kwh, and that the fan is 75% efficient. Compare the
annual cost of electricity to move the air through an ESP and ductwork vs. a
baghouse and ductwork. The pressure drop for the ESP and ductwork is 2.5
inches of water whereas the pressure drop for the baghouse and ductwork is
8.5 inches of water (note 1.0 kPa = 4.0 inches of water). Assume the fan runs
8,000 hours/year, and is down for maintenance the rest of the time.
SOLUTION
First convert cfm to m3/s:
Q = 500, 000 cfm ¥
1 m3
35.3 ft
3
¥
1 min
= 236.1 m 3 /s
60 sec
Next use Eq. (7.3) to calculate the power used by the fan in kw:
For the ESP:
236.1 m 3 /s ¥ 2.5 in. H 2 O ¥
W=
1.0 kPa
4.0 in. H 2 O
0.75
= 197 kw
For the baghouse:
236.1 m 3 /s ¥ 8.5 in. H 2 O ¥
W=
1.0 kPa
4.0 in. H 2 O
0.75
= 669 kw
Finally, multiply the power usage by the time and the unit cost:
ESP Cost = 197 kw × 8,000 hrs/yr × $0.09/kwh = $142,000/year
Baghouse Cost = 669 kw × 8,000 hrs/yr × $0.09/kwh = $482,000/year
Wet scrubbers operate on the principle of collision between particles and
water droplets, collecting the particles in the larger, heavier water drops. The
water falls through the upward-flowing gases, collides with and removes particles, and accumulates in the bottom of the scrubber. Later, the “dirty” water is
treated to remove the solids.
Advantages of wet scrubbers include being able to handle flammable or
explosive dusts, providing cooling of the gases, and neutralizing acid mists
and vapors. Disadvantages include a high potential for corrosion, a high use of
water, and a liquid or sludge effluent that must be treated and/or disposed.
The capital and operating costs of wet scrubbers vary considerably with type
of scrubber, efficiency desired, and local availability of water and local wastewa-
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Air Resources
275
ter discharge standards. Figure
Figure 7.6 Cutaway view of a spray tower
7.6 presents a cutaway view of
scrubber. (Courtesy of Midwest Air Products Co.,
one of the many types of spray
Owosso, MI.)
tower scrubbers.
In summary, there are several different types of particulate matter control devices,
with varying efficiencies and
costs. Each has its own advantages and disadvantages; sitespecific engineering is needed
to make the best choice. Figure
7.7 compares the collection efficiencies of these devices over
a range of particle sizes.
Two important observations can be made from Figure 7.7. First, notice that in
general, as particle size decreases, efficiency decreases.
Second, notice that in general
the baghouse is the most efficient device over the widest range of particle sizes; the electrostatic precipitator
is second, and the scrubber is third. The efficiencies of these three devices are
all high, while cyclones typically have lower efficiencies. As engineers, we always want to try to accomplish our objectives in the most cost-effective man-
100
D
C
B
Collection efficiency, percent
Figure 7.7 Typical collection efficiencies for various
types of particle collectors:
A-ordinary cyclone; B-spray
tower; C-electrostatic precipitator; D-baghouse.
80
A
60
40
20
0
5
10
Particle diameter, μm
15
20
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276
Chapter Seven
ner. The cyclone is by far the cheapest device but does only a moderate job of
removing particles. The other devices are all quite good at removing particulates but are expensive. However, each application is slightly different from all
others. Proper engineering analysis and design are required to ensure the right
choice for the job.
EXAMPLE 7.7
Calculate the overall efficiency of a particulate control system composed of a
cyclone (75% efficient) followed by an electrostatic precipitator (95% efficient).
SOLUTION
The overall system looks like this:
C
25%
A
100%
.95
.75
B
75%
E
1.25%
D
23.75%
where streams A, B, C, D, and E represent particulate flow rates into and out
of the equipment. The ESP collects 95% of the particles that pass through the
cyclone. The overall collection efficiency is the sum of B and D, or 98.75%.
In Example 7.7, we analyzed each piece of equipment, then added streams
B and D to get the total collection efficiency. Let us define fractional penetration
as one minus the fractional efficiency:
Pt = 1 – η
(7.4)
Then, it should be obvious that penetration is the fraction of particle passthrough and that overall penetration for two devices in series is
Ptoverall = Pt1 × Pt2
(7.5)
Thus, the overall efficiency of collection for two devices in series is
ηoverall = (1 – Ptoverall)
(7.6)
Equations (7.5) and (7.6) allow us to solve Example 7.7 directly.
Gaseous Emissions Control Devices
As with particulate matter, there are several control devices and processes
for removing or destroying gaseous pollutants before they are emitted into the
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Air Resources
277
atmosphere. These include carbon adsorbers, vapor incinerators, scrubbers,
and biofilters. In the next few pages we give a brief description of each device.
Volatile organic compounds usually are controlled by one of two methods:
carbon adsorption or vapor incineration. Both are effective, and while carbon
adsorption is a bit more complicated, it usually allows for recovery and reuse
of the organic compounds. Incineration may be preferred for highly toxic or
very odorous compounds.
In the carbon adsorption process, the contaminated air stream is passed
through a bed of granular activated charcoal. The organic molecules are
adsorbed onto the surfaces of the highly porous carbon pellets, while the
cleaned air flows through the bed. When all the surfaces of the carbon have been
covered, the air stream is switched to an identical bed. While the air is being
treated on the new bed, the old bed is regenerated by passing steam through it.
The hot steam desorbs the organics and carries them out of the bed, thus renewing the carbon for another cycle of adsorption. The steam and organic vapors are
separated by condensing the steam and vapors, and then decanting the two
immiscible liquids. The costs include the capital cost of the system, the fan
power to move air through the bed, the amount of steam used for regeneration,
and the costs for disposal/replacement of carbon. If enough VOC is recovered
and if it has a high value, this type of system can actually generate revenue. Figure 7.8 presents a schematic diagram of a carbon adsorption system.
Figure 7.8
Process flow diagram for a fixed-bed carbon adsorption system.
vent
vent
T-1A
carbon
T-1B
carbon
adsorbers
steam / organics
steam
E-3
condenser
CW
T-2
decanter
T-3
organics
E-1
C-1
CW cooler
main blower
water
layer
P-1
T-4
waste
air/solvent
mixture
P-2
waste pump
pump
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278
Chapter Seven
Vapor incineration (often called thermal oxidation) is a simple process.
The contaminated air stream is heated by burning a fuel gas in air. In the hot air
stream, the VOCs are oxidized by reaction with the oxygen already present in
the air stream. Figure 7.9 presents a cutaway view of a small unit. The desired
end products of the reactions are CO2 and water vapor; however, other products are possible. For example, if the organic waste is a chlorinated compound,
HCl will be formed as a product of combustion. Furthermore, if combustion is
not complete, CO and organic HAPs will be formed. It is only necessary to provide enough temperature, time, and turbulence (the “three Ts” of incinerator
design) to ensure good performance of the thermal oxidizer. Because the operating temperatures are 650 to 1100 °C, good designs almost always provide for
heat recovery. Catalytic oxidizers also are used on relatively clean gas streams.
The catalyst allows the reactions to occur at a lower temperature, saving fuel.
Another major use of wet scrubbers (besides removing particles) is to
absorb a pollutant gas from a mixture of gases. The rate and extent of absorption are commonly assisted by a chemical reaction in the absorbing medium. A
widespread example is the scrubbing of SO2 from power-plant combustion
gases by an alkaline solution. The most widely used method in this country
uses a lime or limestone slurry to scrub the SO2, producing a CaSO3 plus CaSO4
sludge that is usually discarded.
The chemistry of limestone scrubbing (often called flue gas desulfurization or FGD) is complex and only a brief synopsis is presented here. The
absorption of SO2 gas is a two-step process: first, the SO2 gas is absorbed into
the liquid, forming bisulfite ions (HSO3–), and second, the bisulfite ions react
with calcium ions to precipitate solid calcium sulfite (Henzel et al. 1981). The
net reaction is
CaCO3(s) + SO2
→
CaSO3(s) + CO2
(7.7)
The pH in the scrubber must be kept fairly low (less than about 6) to prevent
precipitation in the scrubber itself. The possible plugging of the scrubber is the
reason that a separate vessel, the effluent hold tank (shown in Figure 7.10 on p.
280), is provided. More limestone is added here, and the precipitation reactions
occur in this tank. With excess oxygen in the scrubber there is some oxidation
of sulfite, so some of the sludge produced is CaSO4. In fact, because CaSO4 can
be dewatered much more easily than CaSO3 , and because it is a more stable
sludge in a landfill, a popular process modification, called forced oxidation, has
been developed to purposely oxidize the sludge before disposal. To summarize, the sulfur enters as a gas (SO2) and leaves as a sludge (CaSO3 or CaSO4).
FGD using lime or limestone is a “throwaway” process because of the discarded sludge. This process accounts for more than 90% of all scrubbing systems in the United States because it is the cheapest and easiest to use. However,
there are many other processes, some of which recover the SO2 as either H2SO4
or elemental sulfur, both of which are valuable products.
Other countries use recovery-type processes. In Japan, for example, SO2
scrubbing with sulfur recovery is mandated by the government and results in
significant production of sulfuric acid, which is marketed within that country
Exh
Cooling air
induction system
(adjustable)
Blower
Insulation
Sample port
Temperature sensor
Unitized skid mounting
air
d
e
t
llu
g
el
Fu
as
in
Pressure tap
Straightening
vanes
Po
Gas system control
Sample
port
Control panel
(remote optional)
Flame sensor
Burner
Refractory
s
se y
ga ver
Insulation
t
s o
au t rec
h
Turbulent expansion zone
Ex hea
to
Steel shell
Compression zone
A sectional view of a direct-flame thermal oxidizer. (Adapted from KTI Gas Processors, Inc., Santa Ana, CA.)
Return
au heat
s
to t gas
sta
ck es
Figure 7.9
in
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Air Resources
279
Cooper.book Page 280 Monday, June 23, 2014 9:58 AM
Chapter Seven
280
Reheater
Clean flue gas
to stack
Mist eliminator
washwater
Flue
gas
Figure 7.10
Schematic process flow
diagram for limestonebased SO2 scrubbing
system. (Cooper and
Alley 2011; adapted
from Henzel et al. 1981.)
Scrubber Scrubbing
slurry
Ground
limestone
slurry
Make-up
water
Effluent
hold tank
Thickener
Fly
ash
To
disposal
Mixer
Thickener
overflow
tank
Vacuum filter
as well as exported. As native sulfur supplies decline and become more expensive, and as disposal costs climb, these regenerative processes should become
more widely used in the future. No matter what type of process is used, SO2
scrubbing is a complex, large-scale, expensive undertaking. This type of chemical processing is unfamiliar to most power-plant operators, and most utilities
initially were slow to adopt FGD scrubbing. However, federal law requires it on
all new coal-fired power plants, and it certainly does reduce acid gas discharges
to the atmosphere. The Clean Air Act Amendments of 1990 mandated significant reductions in SO2 from many existing power plants. Many more scrubbers
were built in the 1990s and 2000s, and power plant SO2 emissions declined from
15.9 million tons/year in 1990 to 3.3 million tons/year in 2012 (US EPA 2013).
Another example of using water (enhanced with alkaline chemicals) to
remove a pollutant gas is that of HCl scrubbing. HCl is formed whenever a
waste that contains chlorine is incinerated. Two common examples of such
wastes are municipal solid wastes and chlorinated pesticide wastes. All newly
built incinerators of these wastes have wet scrubbers or equivalent air pollution control devices to absorb and neutralize HCl emissions.
Still another example of alkaline scrubbing is the technology being developed for control of carbon dioxide emissions from power plants. As mentioned
earlier in this chapter, global climate change (caused mainly by excessive emissions of CO2) is a very serious problem. One of the emerging technologies
being considered for CO2 capture is scrubbing with various amine solutions.
CO2 is a weakly acidic gas, and amines (organic compounds based on the
Cooper.book Page 281 Monday, June 23, 2014 9:58 AM
Air Resources
281
ammonia structure) have the ability to neutralize the carbonic acid but have the
added advantage that they can be regenerated for re-use in the scrubbing process. The scrubbers being contemplated for such applications at power plants
are huge towers—perhaps 50 feet in diameter and two hundred feet tall.
Gases such as HCl and SO2 can be scrubbed out of exhaust gases with alkaline aqueous solutions, and this is currently being done throughout the world.
But gases like NO2 or NO, which are relatively insoluble, are very difficult to
remove once they have been formed. Therefore, the principal control mechanism for NOx in the United States has been to minimize its formation through
proper design and operation of burners and furnaces. However, a process to
chemically reduce the NOx to N2 gas, known as selective catalytic reduction
(SCR), has become commercialized and is in use at many power plants
throughout the world. Even though it is expensive, SCR became more widely
adopted in the US after the Clean Air Act Amendments of 1990, and NOx emissions from US power plants have declined from 6.7 million tons/year in 1990
to 1.7 million tons/year in 2012 (US EPA 2013).
Biofiltration is a process in which air containing a low concentration of
pollutants is passed through a loosely-packed bed of wetted material (such as
bark or wood chips or even plastic supports). Microbial films attach to the
media and grow within the water layer that resides on the solid material. As
the air passes through the bed, the pollutants absorb into the water film and are
metabolized by the microorganisms and converted into innocuous products
like CO2 and H2O. Pollutants that are soluble in water and easily biodegradable by bacteria and other microorganisms are best suited for biofiltration.
Examples include ethanol, ammonia, H2S, and many odorous VOCs. Biofilters
are finding widespread use at municipal wastewater treatment plants to control odor emissions.
7.6 Air Pollution Control: Mobile Sources
Mobile Sources—A Global Problem
As stated earlier, mobile sources emit all the criteria pollutants, but especially large amounts of CO. Mobile sources are major producers of VOCs and
NOx, which participate in the formation of ground-level (tropospheric) ozone,
a significant problem in many of the world’s largest urban areas. In addition,
the transportation sector (including on-road, non-road, air, rail, and water)
accounts for nearly one-fourth of all the anthropogenic CO2 emissions in the
world (IEA 2013). Mobile sources include cars, trucks, buses, airplanes, boats,
locomotives, construction equipment, lawn and garden equipment, and other
items. The only criteria for inclusion as a mobile source is the presence of a fossil-fuel fired engine and the ability to move around. Their mobility and their
individually small sizes make air pollution control from this source category
difficult. However, the extremely large and growing numbers of mobile sources
make such control imperative.
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282
Chapter Seven
Mobile sources traditionally burn gasoline or diesel fuel—both of which
are made from crude oil. The exploration, drilling, production, refining, and
marketing of oil is a huge global business, which consumes energy and creates
emissions before its products are ever used. The subsequent use of these fuels
also creates emissions, and thus mobile sources are a major sector in the world
economy responsible for millions of tons of air pollution each year.
The traditional pollutants (VOCs, NOx, and CO) all respond to technological controls on vehicles, such as fuel injection, catalytic converters, carbon canisters, and computer control of engine operations. Cars today are much cleaner
than they were in the past; in the US, vehicle emissions of VOCs, NOx, and CO
per mile driven were about 90% lower in 2010 than they were in 1970. Other
countries have been striving hard to reduce fuel consumption and emissions
from mobile sources as well.
One problem facing developing countries is that as their economies grow,
their people tend to want more cars—clearly, an automobile is a status symbol
around the world! Moreover, in developing countries, the affordable cars are usually older and not as well equipped for pollution control. Table 7.5 displays the
2010 registrations of road vehicles in various countries and regions of the world.
Carbon dioxide emission from mobile sources is directly related to fuel
consumption (inversely related to fuel economy). Gasoline and diesel are about
80% carbon by weight, so the less fuel a vehicle burns per mile, the lower the
CO2 emissions will be. In recent years, many countries have been pushing hard
to improve their fleet fuel economy standards and thus reduce their CO2 emissions (IEA 2013), as well as to reduce the emissions of the traditional air pollutants. Table 7.6 displays selected vehicle emissions limits in several of the
countries with large numbers of vehicles.
Table 7.5
On-Road Vehicles Around the World, 2010
Millions of vehicles
Country or Region
Cars
Trucks & Buses
Total
USA
Canada + Mexico
Japan
Europe
Latin America*
Africa
China
India
Rest of Asia
Middle East
Pacific Region
World total
118.9
41.1
58.4
293.7
44.3
17.6
34.4
13.3
48.4
22.3
15.3
707.8
120.9
10.4
15.5
46.4
7.9
9.1
43.6
7.5
24.7
9.3
3.9
307.5
239.8
51.4
73.9
340.1
45.6
26.7
78.0
20.8
73.1
31.6
19.2
1,015.3
*Includes Central and South America and Caribbean Islands; excludes Mexico.
Source: Adapted from Sousanis (2011).
Cooper.book Page 283 Monday, June 23, 2014 9:58 AM
Air Resources
283
Table 7.6 Fuel Economy Standards and Air Pollution Emission Limits for
Gasoline-Powered Automobiles and Light Trucks in Selected Countries, 2010
Fuel Econ. Stds
Emissions, g/km
Country
mpg
km/L
NMHC*
NOx
Emissions, g/km
CO2
USA
China
Europe
Japan
25.3
35.0
43.0
44.0
10.7
14.8
18.2
18.6
0.06
0.15
0.07
0.05
0.04
0.08
0.06
0.04
250
175
142
138
*NMHC stands for non-methane hydrocarbons, which is a nearly identical classification of mobile-source
pollutants to VOCs.
Sources: Transport Policy.net (2013); An et al. (2011); and others.
Characteristics of Engines and Fuels
The internal combustion engine operating on gasoline or diesel is the
prime mover of most mobile sources. A well-operated gasoline engine brings
in precisely controlled amounts of fuel and air into the cylinder, ignites the mixture with a spark from the spark plug, and combusts the gasoline almost completely. An idealized equation showing the stoichiometric combustion of octene
(one of the many compounds in gasoline) in air is shown in Equation (7.8).
C8H16 + 12 O2 + 45.1 N2
→
8 CO2 + 8 H2O + 45.1 N2
(7.8)
However, even in an extremely well-designed car, the combustion reactions are
not perfect, and small amounts of CO and NOx are formed and are emitted from
the cylinders along with small amounts of unburned gasoline fragments (VOCs).
The actual air-to-fuel ratio, the mixing, and other factors greatly influence the relative amounts of these pollutants that escape from the engine. Furthermore,
adjustments to reduce one pollutant often result in increasing another. For example, tuning an engine to run “richer” (more fuel and less air) may reduce NOx
but will increase CO and VOCs (see Figure 7.11 on the following page).
Modern cars have a variety of pollution controls: computer control of the
engine operation (air-to-fuel ratio, spark timing, etc.) to minimize the formation of pollutants, catalytic converters (a small catalytic oxidizer) to further
react pollutants while they are still in the vehicle exhaust system, carbon canisters to capture evaporative emissions of VOCs, and others. However, to reap
these benefits the equipment must not be tampered with and must be properly
maintained. In addition to equipment wearing out, there are many older cars
without these pollution-control devices still being driven—everyone has seen
“old clunkers” going down the street emitting visible smoke (as well as invisible CO and NOx). In many US urban areas, it has been found that more than
half the vehicle pollution is caused by less than 15% of the vehicles. Of course,
in many countries pollution-control equipment either was never installed, or
was broken and then removed.
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Chapter Seven
3500
14
3000
12
NO
10
2500
8
2000
6
1500
CO
HC
4
1000
S
500
2
0
10
12
14
16
18
20
Air/fuel ratio, weight per weight
Nitric oxide concentration, parts per million
Hydrocarbon concentration, parts per million C6 × 10–2
Carbon monoxide concentration, percent
284
Figure 7.11 Effects
of air/fuel ratio on
hydrocarbon, carbon
monoxide, and nitric
oxide exhaust emissions. (Adapted from
Agnew 1968.)
0
22
Fuel quality is very important. Based on mass balance, we know that what
goes in must come out. So any sulfur and/or chlorine contained in the fuel
comes out as acid gases, and any metal impurities are emitted as metal vapors
or metal salts. Diesel fuel is lower quality fuel than gasoline and has more sulfur. Also, because of the nature of the fuel and the design of diesel engines, a lot
more smoke is produced during acceleration. In the less-developed countries
(LDCs), drivers often can only afford the cheaper, lower quality fuels (more
sulfur and other impurities). Moreover, some countries still use leaded gasoline, which is a cheap way to improve engine performance. Lead is a toxic
metal that has been linked with reduction of IQ in children. Also, lead quickly
deactivates a catalytic converter and renders it useless in controlling other pollutants. In many LDCs, there is a high usage of motor scooters and small threewheeled vehicles that use two-stroke engines (in which the oil and gasoline are
pre-mixed and burned together), which results in more pollution per vehiclemile. Higher densities of smaller vehicles and more crowded cities also result
in higher local concentrations of pollutants.
Environmental conditions (such as temperature and altitude) and driving
conditions (such as average speed and smoothness of traffic flow) can affect
engine operation and thus the rates of emissions. (In general, emissions from
vehicles are measured in grams per mile driven; these emissions go up rapidly
as the average speed goes down, and as there are more stops and starts.) The
US EPA has developed a large computer program to predict the emissions from
roadway traffic for any set of conditions—speeds, temperatures, and many
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285
other factors. The program, called MOVES, requires user input for speed, geographic location, year, vehicle types and ages, road grade, and many other conditions; it contains much built-in data as well. MOVES predicts an average
emission factor for these input conditions, in g/veh-mile, and can directly estimate total emissions in a county or state, or just along one particular roadway.
This model is widely used throughout the United States; other countries have
similar emission models for their roadway vehicles.
Strategies for Pollution Prevention/Control from Mobile Sources
The vehicle manufacturer has the responsibility to produce vehicles that
meet all applicable federal and state laws with regard to air pollution emissions. As soon as vehicles are sold, however, it becomes the owners’ responsibility to operate them properly and to maintain the vehicles such that they will
continue to drive cleanly. Often this goal is not achieved. Early on, it was recognized that organized efforts by state and local governments are necessary to
ensure that we achieve clean air in our urban areas. Several strategies have
evolved for controlling or preventing pollution from vehicles, and are discussed in the next few paragraphs.
Inspection and maintenance programs have been implemented in metropolitan areas where emissions levels violate the national ambient air quality
standards. Such programs involve annual or biannual vehicle inspections in
which a computer-linked probe is inserted into the exhaust pipe to measure
emissions. Vehicles that pass get a driving permit (sticker). Vehicles that do not
pass are required to be repaired and then must get reinspected.
Transportation control measures (TCMs) and traffic management techniques (TMTs) can be effective. TCMs include providing mass transit options
(more buses, exclusive bus or car-pool lanes, park and ride lots), economic incentives and penalties (subsidies for van pools, higher priced parking in downtown
areas), and regulatory steps (parking bans, restricted driving zones, or even
restricted driving days, e.g., the Paris example cited at the start of this chapter).
Traffic management techniques refer to the control of traffic flow on the streets
and the management of travel demand. A very successful TMT is the proper
timing and sequencing of traffic signals on the main thoroughfares of a downtown area. Other techniques to reduce congestion include creating one-way
street pairs, adding protected left turn lanes, planning truck routes, and others.
Travel demand can be altered by staggered work hours, four-day work weeks,
or allowing for employee work-at-home programs (electronic commuting).
Changes in motor vehicle fuels can have a significant effect on air pollution. One of the great success stories in this regard is the EPA’s mandating the
phase-out of tetraethyl lead from gasoline in the mid-1970s. In 1970, lead emissions from motor vehicles were 200,000 metric tons per year; by 1990 they were
only 2,000—a 99% decrease! Other fuel changes that have reduced air pollution
are (1) reducing the vapor pressure of gasoline (cuts down on evaporative and
engine emissions of VOCs) and (2) using some oxygenated fuels, like ethanol
or ether (reduces CO emissions). In recent years, hybrid-electric and all-electric
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286
Chapter Seven
vehicles have become popular; in the near future we may see more compressed
natural gas and hydrogen fueled vehicles. Over the long term, solar-powered
vehicles might provide a substantial answer to this vexing problem of mobile
source emissions.
7.7 Meteorology and Atmospheric Dispersion of Pollutants
The release of pollutants into the atmosphere is a time-honored technique
for “disposing” of them. One of the reasons that the air did not become completely unusable long ago is the atmosphere’s ability to quickly disperse high
concentrations of pollutants to lower, relatively harmless concentrations. Of
course, the atmosphere’s ability to disperse pollutants is not infinite and varies
from quite good to quite poor, depending on meteorological and geographical
conditions. Thus, with the onset of the industrial revolution and the subsequent exponential growth of human population and energy consumption, we
have had to begin installing pollution-control equipment to prevent indiscriminate abuse of the atmosphere. However, we still rely heavily on atmospheric
dispersion for final disposal of many pollutants.
Meteorology and Atmospheric Stability
Harmful effects occur when pollutants build up to high concentrations in a
local area. The accumulation of pollutants in any localized region is a function
of emission (input) rates, transport and dispersion (output) rates, and generation or destruction rates (by chemical reaction). The dispersion of pollutants is
almost entirely due to natural conditions like geography and local meteorological conditions such as wind, rainfall, and atmospheric stability. Therefore,
meteorology is extremely important. Meteorology includes both horizontal and
vertical movements of the atmosphere, on a local, regional, and global scale.
Atmospheric stability is a term used to describe the mixing tendencies of
the air. Air is termed unstable when there is good vertical mixing. This occurs
when there is strong solar insolation and consequent heating of the layers of air
near the ground. The warm air rises and is replaced by cooler air, which in turn
is heated and rises. This process is good for diluting pollutants and carrying
them away from the ground.
Stable air results when the surface of the earth is cooler than the air above
it (such as on a clear, cool night). The layers of air next to the earth are thus
cooled and no vertical mixing can occur. In stable air, pollutants tend to stay
near the ground, and can build up in the air to high, sometimes unhealthful,
concentrations.
If there is a strong wind or good vertical mixing in the atmosphere, pollutants will be dispersed quickly into a large volume of air, resulting in low concentrations. If the wind is weak and there is an inversion layer (defined as a
layer of warm air above a layer of cooler air, which prevents the vertical mixing
of air in the region), then pollutant concentrations can increase. The modeling
of air pollution dispersion is an important part of managing our air resources.
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287
The Box Model
The simplest model that can be developed to predict pollutant concentrations under an inversion is called a box model. In this model we picture an area
being covered by a rectangular box (see Figure 7.12). The height of the box is
determined by the inversion layer and the box is aligned with the wind direction. Pollutants can be carried into the box from upwind or be emitted from
ground-based sources within the box. Pollutants are carried out of the box by
the wind. If the air in the box is assumed to be well mixed (like a CSTR), then
the analysis is straightforward. One input of pollutant is the material carried
into the box with the incoming wind. The volumetric flow into the box is determined by the wind velocity times the front face area of the box, so the mass
input of pollution is uWHC0. Another input comes from source emissions
within the box, which can either be identifiable point sources or area sources
that can be viewed as a flux of pollutants coming from the “floor” of the box.
Reaction is certainly possible, but first consider the steady-state material balance on a nonreactive pollutant:
0 = uWHC0 + ΣQi + qLW – uWHCe
(7.9)
where terms are as depicted in Figure 7.12. Equation (7.9) can be solved for the
outlet concentration as follows:
Ce = C0 +
qL
 Qi
+
uH uWH
(7.10)
It should be recognized that the box model is fundamentally a material balance model that can be used for estimating the concentration of a pollutant in
the air for any situation where the air is “trapped” (e.g., in a town covered by an
inversion layer, in a valley with steep walls, or inside a structure). The prediction of concentrations inside rooms or buildings is a common tool when assessing indoor air quality issues (which is discussed in more detail in Chapter 9).
Example 7.8 illustrates a typical use of the box model in an outdoor situation.
Figure 7.12
The box
model for
predicting
pollutant concentrations.
L(m)
W(m)
Wind
u(m/s)
C0(μg/m3)
H(m)
Point sources
Qi (μg/s)
Area source emissions
q(μg/m2–s)
u(m/s)
Ce(μg/m3)
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288
Chapter Seven
EXAMPLE 7.8
A town is covered by an inversion layer 500 m above the ground. The town’s
area is approximately rectangular with dimensions 9 km by 12 km. The
incoming air is clean and the wind is blowing at 3 m/s parallel with the 12km side of the town. The emissions of SO2 in this town come from a number
of small sources that average 0.0008 mg/m2-s and from one industrial plant
that averages 2,150 kg/day. Assuming that the air above the town is wellmixed horizontally and vertically up to the inversion layer, calculate the
steady-state concentration of SO2 in the town’s air.
SOLUTION
First, convert both emission rates into appropriate units for use in Eq. (7.10):
0.0008 mg/m2-s × 103 µg/mg = 0.80 µg/m2-s
2,150 kg/day × 109 µg/kg × 1 day/86,400 s = 2.49 (10)7 µg/s
Now substitute into Eq. (7.10):
7
Ce = 0 +
2.49 (10 )
0.8 ¥ 12, 000
= 8.24 mg/m 3
+
3 ¥ 500
3 ¥ 9, 000 ¥ 500
In most cases with the box model, we assume that the entire box is wellmixed. However, in some cases the physical situation demands that we take a
different approach. Consider, for example, a traffic tunnel where cars are emitting pollution all along its length, and we want to know how the resulting concentration may vary with length. In this case, we can model the tunnel as a
series of very thin, well-mixed boxes in which the air flows from one to
another, and the pollution concentration continuously increases along the
length of the tunnel. We make our balance on a differential element and then
integrate as shown in Example 7.9.
EXAMPLE 7.9
A 2,000-foot, one-way, two-lane rectangular tunnel filled with cars is blocked
by an accident at the exit. The longitudinal ventilation system involves large
exhaust fans drawing ambient air into one end of the tunnel and blowing it
out the other end. Assume that the tunnel fills with cars, all with engines
idling. One car occupies 20 linear feet of space in one lane in the tunnel and
emits 2.8 grams of CO per minute with the engine idling. The tunnel is 30 feet
wide and 12 feet tall. The fans blow 300,000 ft3/min of ambient air (which contains 2.0 mg/m3 of CO) into the tunnel entrance. Derive an equation to predict
the steady-state concentration of CO as a function of position in the tunnel.
Calculate the distance where the CO concentration surpasses 40 mg/m3.
SOLUTION
The problem is diagrammed schematically as follows.
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289
L
W
Q
Q
H
q
Assume that emissions can be represented as a continuous line source along
the length of the tunnel, and calculate q.
q=
2.8 g CO
0.28 g CO
1 car
¥ 2 lanes ¥
=
minute-car ft-minute
lane-20 ft
Choose an arbitrary volume element in the tunnel located x feet from the
entrance.
Δx
Qx
Qx + Δx
Cx
Cx+ Δx
q
A steady-state material balance for CO on this volume element (assuming it is
well-mixed vertically and horizontally) is
0 = QxCx + qΔx – Qx+Δx Cx+Δx
Since Q is essentially constant, we get
–Q (Cx – Cx+Δx) = qΔx
Rearranging and dividing by ∆x, we get
- Cx ˆ
ÊC
Q Á x+Dx
˜¯ = q
Ë
Dx
Taking the limit as ∆x → 0 yields the definition of a derivative:
Ê dC ˆ
QÁ
=q
Ë dx ˜¯
This equation is easily integrated as follows:
CL
Q
L
Ú dC = q Ú dx
C0
0
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290
Chapter Seven
or
Q (CL - C0 ) = qL or CL = C0 +
q
L
Q
Solving for L and its value where CL = 40 mg/m3:
L=
Q (CL - C0 )
q
and substituting, we get:
300, 000
L=
mg
1 m3
ft 3
¥ ( 40 - 2 ) 3 ¥
min
35.3 ft 3
m
0.28 g 1, 000 mg
¥
g
ft-min
= 1, 153 ft
The Gaussian Dispersion Model
Releases of large amounts of pollution (as from a large power plant) are
directed through tall stacks to allow some time and space for the pollutants to
disperse before reaching ground level where we live (and where the NAAQS
apply). Although the equations to describe the dispersion of pollutants in the
open atmosphere can be developed from material-balance considerations, it is
beyond the scope of this text to do so. Rather, we will qualitatively describe this
process and then present the equation that is widely used to calculate concentrations at points downwind from the source.
Pollutants released from a stack into the atmosphere form a “plume” that
bends over and travels with the mean wind. They also spread out in both the
horizontal and vertical directions from the center line of the plume. The rate of
spread in each direction is a complex function of micrometeorological conditions, the characteristics of the pollutants, and local geographical features. Figure 7.13 schematically portrays the time-averaged behavior of a plume being
emitted into the air and bent over by a steady wind. Figure 7.14 is a photograph
of condensing water vapor plumes from two tall stacks and two cooling towers
at a coal-fired power plant.
Earlier, we introduced the concept of atmospheric stability. This concept is
quantified by the use of atmospheric stability classes. Atmospheric stability is
broken into six classes, arbitrarily labeled A through F, with A being the most
unstable. Stability classes depend on the solar insolation and on the wind
speed near the ground; Table 7.7 on p. 292 depicts these relationships. Once the
stability classification is determined, this can be combined with other information to predict the concentration anywhere downwind from the source using
the Gaussian dispersion model. Researchers have correlated values of the dis-
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291
Figure 7.13 The spreading of a bent-over plume, with x, y, z axes. (Adapted from
Turner 1970.)
z
x
(x, –y, z)
x(x, 0, 0)
(x, –y, 0)
y
H
h
Mean
wind direction
Figure 7.14 Photograph showing condensing water vapor plumes from two exhaust
stacks and two cooling towers at a coal-fired power plant. (Courtesy of Anthony Crawford.)
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292
Chapter Seven
Table 7.7
Atmospheric Stability Classifications
Day
Incoming Solar Radiation
Night
Cloudinesse
Surface
Wind Speeda
m/s
Strongb
Moderatec
Slightd
Cloudy
(> 4/8)
Clear
(<3/8)
<2
2–3
3–5
5–6
>6
A
A–B
B
C
C
A–Bf
B
B–C
C–D
D
B
C
C
D
D
E
E
D
D
D
F
F
E
D
D
Notes:
a Surface wind speed is measured at 10 m above the ground.
b
Corresponds to clear summer day with sun higher than 60° above the horizon.
c
Corresponds to a summer day with a few broken clouds, or a clear day with sun 35–60° above the horizon.
d Corresponds to a fall afternoon, or a cloudy summer day, or clear summer day with the sun 15–35°.
e Cloudiness is defined as the fraction of sky covered by clouds.
f
For A–B, B–C, or C–D conditions, average the values obtained for each.
A = Very unstable
D = Neutral
B = Moderately unstable E = Slightly stable
C = Slightly unstable
F = Stable
Regardless of wind speed, Class D should be assumed for overcast conditions, day or night.
Source: Adapted from Turner (1970).
persion coefficients with atmospheric stability and distance away from the
source. These correlations are presented in Figures 7.15 and 7.16.
It has been shown that the spread of pollutants can be approximated by a
Gaussian or normal distribution (Turner 1970). The equation that models the
normal dispersion of a gaseous pollutant from an elevated source is given
below in a form that predicts the steady-state concentration at a point (x, y, z):
C=
Ê 1 y 2 ˆ ÏÔ
Ê 1 ( z - H )2 ˆ
Ê 1 ( z + H )2 ˆ ¸Ô
Q
exp Á exp
exp
+
˜
Á
˜
Á˜˝
Ì
ÁË 2 s z2 ˜¯
ÁË 2 s z2 ˜¯ Ô
ÁË 2 s y2 ˜¯ Ô
2p us ys z
Ó
˛
(7.11)
where:
C
= steady-state concentration at a point (x, y, z), µg/m3
Q
= emissions rate, µg/s
σy , σz = horizontal and vertical dispersion parameters, m
u
= average wind speed at stack height, m/s
y
= horizontal distance from plume center line, m
z
= vertical distance from ground level, m
H
= effective stack height (H = h +∆h, where h = physical stack height,
and ∆h = plume rise), m
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Air Resources
Figure 7.15
Horizontal dispersion
coefficient as a function of downwind distance from the source.
(Adapted from
Turner 1970.)
293
10,000
1000
C
D
E
F
y
, meters
A B
100
10
0.1
1
10
Distance downwind, km
100
Refer to any text on air pollution (Cooper and Alley 2011) for a review of the
development of Equation (7.11) and the assumptions implicit in its use. Keep in
mind some general relationships indicated by Equation (7.11):
1. The downwind concentration at any location is directly proportional to
the source strength, Q.
2. The downwind ground-level concentration is generally inversely proportional to wind speed. (H also depends on wind speed in a complicated fashion that prevents a strict inverse proportionality.)
3. Because σy and σz increase as the downwind distance x increases, the
plume center-line concentration continuously declines with increasing
x. However, ground-level concentrations increase, go through a maximum, and then decrease as one moves away from the stack.
4. The dispersion parameters σy and σz increase with increasing atmospheric turbulence (instability). Thus, unstable conditions decrease
downwind concentrations (on the average).
Cooper.book Page 294 Monday, June 23, 2014 9:58 AM
Chapter Seven
294
Figure 7.16 Vertical
dispersion coefficient
as a function of downwind distance from
the source and atmospheric stability class.
(Adapted from
Turner 1970.)
5000
1000
A
C
D
100
E
z
, meters
B
F
10
1.0
0.1
1
10
Distance downwind, km
100
5. The maximum ground-level concentration calculated from Equation (7.11)
decreases as effective stack height (H) increases. The distance from the
stack at which the maximum concentration occurs increases as H increases.
The Gaussian dispersion equation is extremely important in air pollution
modeling work, and is used in almost all of the computer programs developed
by the US EPA for atmospheric dispersion modeling. The relationships in Figures 7.15 and 7.16 have been curve-fit for use in computer programs (Martin
1976); the curve-fit equations and constants are presented in Table 7.8 on p. 295.
EXAMPLE 7.10
(a) Using a hand calculator, calculate the ground-level downwind center-line
concentration of SO2 10 km from the source. The effective stack height is
100 m, the wind speed at stack height is 6 m/s, and the emissions rate is
0.25 kg/s. The atmosphere has a neutral stability.
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295
(b) Using a spreadsheet, calculate the concentration at enough downwind
distances to be able to draw a curve depicting how the ground-level concentration changes with distance from the source. Find the maximum
ground-level concentration, and where it occurs.
SOLUTION
(a) For D stability at x = 10 km, the values of σy and σz (as read from Figures
7.15 and 7.16) are 550 m and 130 m, respectively. In air modeling parlance,
“ground-level” means z = 0 and “center-line” means y = 0. Substituting
into Equation (7.11) we get:
C=
9
ÏÔ
Ê 1 100 2 ˆ ¸Ô
Ê 1 100 2 ˆ
0.25 (10 ) µg/s
= 92.7 ¥ 1.49
exp
+
1) Ìexp Á (
Á2˜˝
2˜
m
Ë 2 130 ¯ ˛Ô
Ë 2 130 ¯
2p 6 550m 130m
ÓÔ
s
= 138 µg/m3
(b) A portion of the spreadsheet is shown in Figure 7.17 on the following page.
Increments of 100 m were chosen over the range of 100 m to 20 km. This
choice gives 200 points with which to plot the curve. If, after developing the
spreadsheet, the resulting curve is deemed not acceptable, then either
smaller increments or a wider range can be implemented easily. From Table
7.8 and Figure 7.17, note that the calculation formula for the sigma values
switches at x = 1 km. Also, note that at 10 km, the spreadsheet answer is 141
µg/m3 as compared with the 138 µg/m3 obtained in part (a) above. The difference is due to errors incurred in visually reading Figures 7.15 and 7.16.
Figure 7.18 on p. 297 displays the plot that was obtained directly from the
spreadsheet. The plot clearly shows how, for an elevated point source, the
downwind ground-level concentrations start low, increase through a maximum, and then decline with distance. From the curve, the maximum is
estimated as 350 µg/m3 and occurs 2.7 km downwind.
Table 7.8
Curve-Fit Equations and Constants for Dispersion Parameters
σy = axb; σz = cxd + f
(x must be in units of km in both equations)
x<1 km
Stability
a
A
B
C
D
E
F
213
156
104
68
50.5
34
b*
c
d
f
c
d
f
440.8
106.6
61.0
33.2
22.8
14.35
1.941
1.149
0.911
0.725
0.678
0.740
9.27
3.3
0
–1.7
–1.3
–0.35
459.7
108.2
61.0
44.5
55.4
62.6
2.094
1.098
0.911
0.516
0.305
0.180
–9.6
2.0
0
–13.0
–34.0
–48.6
*b = 0.894 for all stability classes and values of x.
Source: Adapted from Martin (1976).
x>1 km
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296
Chapter Seven
Figure 7.17
A portion from the spreadsheet solution for Example 7.10.
Define input values
H=
u=
Q=
Class =
xstart =
x inc =
x end =
100 m
6 m/s
0.25 kg/s
D
100 m
100 m
20 km
For x < 1 km
Stability
a
A
B
C
D
E
F
213
156
104
68
50.5
34
c
440.8
106.6
61
33.2
22.8
14.35
For x > 1 km
d
f
c
d
f
1.941
1.149
0.911
0.725
0.678
0.74
9.27
3.3
0
–1.7
–1.3
–0.35
459.7
108.2
61
44.5
55.4
62.6
2.094
1.098
0.911
0.516
0.305
0.18
–9.6
2
0
–13
–34
–48.6
Note: b = 0.894 for all classes and all distances.
Calculations and results
For a more general solution, pick coefficients for each class and distance using IF statements. For this
example, coefficients were hand-selected and copied into the right cells
Calc No.
x, km
c
d
f
sigma y
sigma z
Conc
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0.1000
0.2000
0.3000
0.4000
0.5000
0.6000
0.7000
0.8000
0.9000
1.0000
1.1000
1.2000
1.3000
1.4000
33.2
33.2
33.2
33.2
33.2
33.2
33.2
33.2
33.2
44.5
44.5
44.5
44.5
44.5
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.5160
0.5160
0.5160
0.5160
0.5160
–1.7
–1.7
–1.7
–1.7
–1.7
–1.7
–1.7
–1.7
–1.7
–13
–13
–13
–13
–13
9
16
23
30
37
43
49
56
62
68
74
80
86
92
5
9
12
15
18
21
24
27
29
32
34
36
38
40
0.0
0.0
0.0
0.0
0.0
0.2
1.8
7.4
19.8
40.1
65.7
95.2
126.3
157.3
Stability class plays a strong role in the dispersion of air pollution and the
concentrations that result. When the atmosphere is more stable, there is less
vertical movement, and pollution tends to stay closer to the ground yielding
higher concentrations. When the air is unstable, the pollutants more easily mix
upward away from the ground yielding lower concentrations. Example 7.11
illustrates this point.
Cooper.book Page 297 Monday, June 23, 2014 9:58 AM
Air Resources
297
Figure 7.18 The spreadsheet plot of concentration versus distance for Example 7.10.
400
Concentration, μg/m3
350
300
250
200
150
100
50
0
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18
Distance, km
EXAMPLE 7.11
Repeat part (a) of Example 7.10, except use stability class C and then E.
SOLUTION
For C stability, from Figures 7.15 and 7.16, σy = 825 m and σz = 500 m. Thus,
C=
9
ÏÔ
Ê 1 100 2 ˆ
Ê 1 100 2 ˆ ¸Ô
0.25 (10 ) mg/s
1) Ìexp Á +
p
ex
= 16.1 ¥ 1.96
(
˜
Á2
2 ˜˝
6m
Ë 2 500 ¯
Ë 2 500 ¯ Ô˛
2p
825 m 500 m
ÓÔ
s
= 31.5 mg/m 3
For E stability, σy = 380 m and σz = 77 m. Thus,
C=
9
ÏÔ
Ê 1 100 2 ˆ ¸Ô
Ê 1 100 2 ˆ
0.25 (10 ) mg/s
exp
1) Ìexp Á +
(
Á˜ ˝ = 227 ¥ 0.861
˜
6m
Ë 2 77 2 ¯ ˛Ô
Ë 2 77 2 ¯
2p
380 m 77 m ÓÔ
s
= 195 mg / m 3
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298
Chapter Seven
Residuals from APC Systems
Air pollution control technology has improved greatly since the passage of
the Clean Air Act Amendments of 1970. Emissions of all the major pollutants
have decreased by 50 to 80% (depending on the pollutant) over that time
period. However, the control of air pollutants from large industrial sources
often results in the production of some solid or liquid wastes that must be disposed of. Ash that has been collected by an ESP or baghouse is dry, and often
can be sold (or given away) to make low-grade concrete or road base. Sometimes, material remains that must be landfilled, either on-site or at a nearby
landfill designed for such material. The sludges from FGD scrubbers can be
more problematic. Large volumes are produced daily and the solids settle very
slowly, so very large holding ponds are necessary. Sometimes, the FGD sludge
is mixed with waste fly ash to help stabilize both materials before being deposited in the holding ponds. As we know by now, no process is 100% efficient, so
there are always low levels of gaseous or particulate pollutants contained in
the exhaust gases from APC devices. These low-concentration gas streams are
simply exhausted to the atmosphere as a means of final dilution and disposal.
PROBLEMS
7.1
Assume that gasoline is combusted with 99% efficiency in a car engine
with 1% remaining in the exhaust gases as VOC. If the engine exhausts 16
kg of gases (molecular weight = 30) for each kilogram of gasoline (molecular weight = 100), calculate the fraction VOC in the exhaust. Give your
answer in ppm.
7.2
What fraction of 2012 US NOx emissions were contributed by highway vehicles? By off-highway vehicles? By electric utilities? By industrial furnaces?
7.3
Wind blows down a trapezoidal valley at 8 m/s. The valley depth is 800
m, the floor width is 1,000 m, and the width at the top is 2,000 m. A
smelter emits SO2 at a steady rate of 10,000 kg/day. The valley is capped
by an inversion. Calculate the steady-state concentration a long way
downwind from the smelter, where the pollutant is uniformly spread
across the width and height of the valley.
7.4
In Problem 7.3 above, consider the same smelter to be on level, open
ground. The stack is 200 m tall and the plume rise is 100 m. The wind is 4
m/s and the stability class is C. Estimate the ground-level SO2 concentration 6,000 m directly downwind. What is the concentration 300 m off the
centerline at this same x value?
7.5
The concentration of SO2 was measured as 1,300 µg/m3 for a 3-hour averaging time. Calculate the equivalent concentration in ppm.
7.6
A power plant burns 10,000 kg/hr of 3.00% sulfur coal. Approximately
90% of the SO2 formed must be removed prior to release of the stack
gases. A limestone system is to be used to remove the SO2. Estimate the
minimum limestone requirements for this plant (kg/hr).
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Air Resources
7.7
7.8
7.9
7.10
7.11
7.12
7.13
7.14
299
A particulate removal system consists of a cyclone followed by an electrostatic precipitator. The cyclone is 65% efficient and the ESP is 95% efficient. Calculate the overall efficiency of the system.
A particulate removal system must achieve 99.4% overall efficiency. Calculate the required efficiency of an ESP if it is preceded by an 80% efficient cyclone.
Assuming compliance with federal NSPS, predict the rate of emissions (in
tons/day) of NOx from a coal-fired power plant producing 880 MW of
electrical power.
Which emits more SOx per unit amount of electricity produced, a coal-fired
plant with a 90% efficient SO2 scrubber or an oil-fired plant with no scrubber? Assume the coal is Illinois bituminous with 3.5% sulfur and 11,500 Btu
per pound heating value. Assume the oil has a heating value of 6.0 million
Btu per barrel, contains 0.9% sulfur, and has a specific gravity of 0.92.
Calculate the annual emissions of CO2 from two cars, each of which travels 10,000 miles in a year. The first car is a small gasoline-fueled sedan,
and gets 35 miles/gallon. The second car is a large diesel-fueled SUV and
gets 18 miles/gallon. The CO2 emission factors are 19.4 and 22.2 lb/gal
for gasoline and diesel, respectively.
Estimate the daily emissions of particulates from a solid waste incinerator
emitting at 20 mg/dscm. The incinerator burns 250 tons/day and
exhausts gases in a ratio of 20 kg dry gases per kilogram of feed. Assume
that STP is 25 °C and 1 atm. Assume that the gases have an average
molecular weight of 30. Give your answer in kg/day.
Estimate the daily emissions of SO2 (in kg/day) from a sulfuric acid plant
emitting at 80% of the maximum standards. The production rate is 300
metric tons/day.
The diagram given below illustrates a coal-fired power plant with an SO2
scrubber in which an overall SO2 removal efficiency of 85% is required.
Rather than treat the entire stream to 85% removal, the company proposes to treat part of the flue gas to 95% removal, and to bypass the
remainder. The reblended stream must still satisfy the overall removal
requirement. Calculate the fraction of the flue gas stream that can be
bypassed around the scrubber.
Stack
Coal
Air
Furnace
boiler
ESP
Bottom ash
Fly ash
SO2
scrubber
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Chapter Seven
7.15 Referring to the diagram in Problem 7.14, the gas temperature required in
the stack for good buoyancy is at least 80 °C. The gases bypassing the
scrubber are at 200 °C and the gases exiting the scrubber are at 50 °C. For
a bypass percentage of 10%, calculate the temperature of the mixed stack
gases, ignoring the water which evaporates in the scrubber. Will the
blended stack gases have good buoyancy?
7.16 A carbon adsorption system has been proposed to treat a stream of contaminated air flowing at 17,090 ft3/min (T = 104 °F, P = 1 atm) that contains 1,000 ppm hexane vapor. Assume that we can recover 95% of the
inlet hexane as liquid hexane, which can be reused. (Some hexane vapor
is lost into the exhaust air when we switch beds, and some hexane liquid
is lost into the wastewater stream). If hexane costs $1.50 per liquid gallon
(which has a density of 5.50 lb/gal), calculate how much money we can
potentially save by this operation. Give your answer in $/year assuming
that the plant runs 24 hours per day, 240 days per year.
7.17 China and India combined are adding an average of one new coal-fired
power plant each week to meet their increasing electricity demand.
Assume that in 2012 they added 52 new 800-MW power plants, each at
32% efficiency, and each burning low-grade coal with a heating value of
9,000 Btu/lb and 45% carbon content. How much CO2 did those 52 plants
add to atmosphere in 2013 (tons)?
7.18 An ESP is being designed to treat 10,000 m3/min of gas flow to remove
PM with a 98.7% efficiency. If the drift velocity is estimated to be 0.15 m/
s, how much plate area is needed (m2)?
7.19 If the ESP of Problem 7.18 incurs a pressure drop of 1.1 in. H2O, and the
fan is 70% efficient, what is the annual cost of fan power due to the ESP
alone? Assume electricity is worth 7.5 cents per kwh at this plant, and the
fan runs 8,300 hours per year.
7.20 A carbon adsorption system is being designed to capture benzene vapors
from a stream of air flowing at 10,000 acfm (actual cubic feet per minute)
(at 90 °F and 1 atm). The benzene concentration is 600 ppm. What is the
maximum expected rate of capture of benzene, lb/min?
7.21 A thermal oxidizer is being considered to destroy the benzene of Problem
7.20. The air must be heated to 1,900 °F to ensure 99.99% benzene destruction. How much heat is required to be added to the air to raise its temperature to 1,900 °F (Btu/min)? The Cp of air is 0.24 Btu/lb-°F, and its
molecular weight is 29.
7.22 Assume that you burn natural gas (essentially, pure methane with a heating value = 21,560 Btu/lb) to get the heat needed in Problem 7.21. You
simply burn the natural gas in the polluted air stream, using the oxygen
in the polluted air stream to produce CO2 and H2O from the methane.
Calculate the mass flow rate of methane needed (lb/min). Ignore the heat
required to raise the mass of methane to 1,900 °F, ignore the heat released
by burning the benzene, and ignore heat losses.
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301
7.23 Pentane (C5H12) is evaporating at the rate of 1.44 kg/min into a stream of
pure air (78% N2, 21% O2, and 1% Ar) that is flowing at 754 kg/min. The
mixture is sent to a thermal oxidizer where 16.0 kg/min of methane is
added. The mixture is then ignited, burning all of the methane and pentane
completely to CO2 and H2O. After combustion, the exhaust gases come out
at 980 °C and 1 atm. Calculate the volumetric flow rate of the exhaust
gases, in m3/min. Also, calculate the mole percent O2 in the exhaust gases.
7.24 If the temperature of the mixture of gases in Problem 7.23 before ignition
was 25 °C, calculate the heat losses from the oxidizer. The heats of combustion are 50,150 kJ/kg for methane and 45,320 kJ/kg for pentane.
Assume that the average Cp for all gases is constant over this temperature
range at 1.06 kJ/kg-°C. Give your answer as kJ/min, and then express it
as a percent of the heat released by burning the methane.
7.25 Sulfur dioxide is emitted at 0.10 kg/s from a stack with a physical height
of 75 meters and a plume rise of 25 meters. The wind speed at 100 m
above the ground is 6 m/s on an overcast day. Calculate the ground level
centerline concentrations at the following distances (all directly downwind from the stack): 2 km, 5 km, and 10 km.
7.26 Rework Problem 7.25 with the following changes. Plume rise is 75 meters,
and it is mid-afternoon on a warm sunny summer day (Class B stability).
Wind speed at 150 m is 6 m/s. Calculate answers only for the 2 km and 5
km downwind distances.
7.27 Assume that 140,000 vehicles per day travel on an interstate highway that
passes through an urban area. The length of the section of highway inside
the urban area boundaries is 25 miles and, at the speeds traveled, the
annual average NOx emission factor is 3 grams per mile for the average
vehicle on the road. Calculate the annual emissions of NOx emitted from
vehicles on this piece of interstate. Give your answer in metric tons.
7.28 A power plant is using a limestone FGD system to remove 100,000
pounds per day of SO2. How much pure CaCO3 must be added (lb/day)?
If the limestone is 94% CaCO3 and 6% sand, how many lb/day of limestone must you buy?
REFERENCES
Agnew, W. G. 1968. Research Publication GMR-743. Warren, MI: General Motors Corporation.
Amico, C., D. DeBelius, T. Henry, and M. Stiles. 2013. “State Impact Texas,” published
by KUT Austin, KUHT Houston, and NPR. Accessed July 2013.
http://stateimpact.npr.org/texas/drought/
An, F., R. Earley, and L. Green-Weiskel. 2011. “Global Overview on Fuel Efficiency and
Motor Vehicle Emission Standards: Policy Options and Perspectives for International Cooperation.” United Nations Dept. of Economic and Social Affairs, Commission on Sustainable Development, CSD 19/2011/BP3 (May).
Arrhenius, Svante. 1896. “On the Influence of Carbonic Acid in the Air upon the Temperature of the Ground.” Philosophical Magazine and Journal of Science, 5 (41).
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Cooper, C. David, and F. C. Alley. 2011. Air Pollution Control: A Design Approach. 4th ed.
Long Grove, IL: Waveland Press.
Derevianko, G., and H. Balentine. 2013. “The New Climate Normal: Observed Changes
in the Intensity and Frequency of Extreme Weather Events.” Paper presented at the
106th Annual Conference & Exhibition of the A&WMA, Chicago, IL, June 25–28.
Federal Register. 2013. “National Emission Standards for Hazardous Air Pollutants for
Major Sources: Industrial, Commercial, and Institutional Boilers and Process Heaters; Final Rule.” 40 CFR Part 63. FR 78 (21) January 31.
Gao, Y., J. S. Fu, J. B. Drake, and J. F. Lamarque. 2012. “Projected Changes of Extreme
Weather Events in the Eastern United States Based on a High Resolution Climate
Modeling System.” Environmental Research Letters, 7.
Henzel, D. S., B. A. Laseke, E. O. Smith, and D. O. Swenson. 1981. Limestone FGD Scrubbers: User’s Handbook. EPA-600/8-81-017 (August).
IEA (International Energy Agency). 2013. “CO2 Emissions from Fuel Combustion—
Highlights.” IEA Statistics—2012 Edition. Paris: International Energy Agency.
Martin, D. O. 1976. “The Change of Concentration Standard Deviation with Distance.”
Journal of the Air Pollution Control Association, 26(2).
NOAA (National Oceanic & Atmospheric Administration). 2013. “Trends in Atmospheric Carbon Dioxide.” Earth System Research Laboratory. Accessed July 2013.
http://www.esrl.noaa.gov/gmd/ccgg/trends/
Petroff, A. 2014. “Paris Pollution Leads to Car Ban.” CNN Money, March 18. http://
money.cnn.com/2014/03/17/news/paris-pollution-traffic/index.html
Robine, J. M., S. L. K. Cheung, S. Le Roy, H. Van Oyen, C. Griffiths, J. P. Michel, and F. R.
Herrmann. 2008. “Death Toll Exceeded 70,000 in Europe during the Summer of
2003.” Comptes Rendus Biologies, 331.
Sousanis, J. 2011. “World Vehicle Population Tops 1 Billion Units.” Accessed March
2014. http://wardsauto.com/ar/world_vehicle_population_110815
TransportPolicy.net. 2013. “Global Comparison: Light-duty Emissions.” Accessed July
2013. http://transportpolicy.net/index.php?title=Global_Comparison:_
Light-duty_Emissions
Turner, D. B. 1970. Workbook of Atmospheric Dispersion Estimates. Washington, DC: Environmental Protection Agency Publication AP-26.
UNEP (United Nations Environment Programme), Ozone Secretariat. 2013. “Status of
Ratification.” Accessed July 2013. http://ozone.unep.org/new_site/en/
treaty_ratification_status.php
US EPA (Environmental Protection Agency). 2000. “Technical Support Document: Control of Emissions of Hazardous Air Pollutants from Motor Vehicles and Motor Vehicle Fuels,” EPA 420-R-00-23 (December).
US EPA. 2009. “EPA: Greenhouse Gases Threaten Public Health and the Environment.”
EPA Newsroom, Dec. 7. Accessed July 2013. http://yosemite.epa.gov/opa/
admpress.nsf/0/08D11A451131BCA585257685005BF252
US EPA. 2013. “National Air Pollution Emissions Inventory, updated June 6, 2013.”
Accessed July 2013. http://www.epa.gov/ttn/chief/trends/index.html
Whitman, S., G. Good, E. R. Donoghue, N. Benbow, W. Shou, and S. Mou. 1997. “Mortality in Chicago Attributed to the July 1995 Heat Wave.” American Journal of Public
Health, 87.
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CHAPTER
8
Solid and Hazardous
Waste Management
Previous chapters in this text have focused on the problems of pollution of
our waterways and atmosphere, and have presented engineering approaches
to controlling that pollution. In this chapter we focus on the third medium—
the land. The term “solid wastes” includes such diverse materials as household
garbage, discarded office paper and plastic, batteries, construction and demolition debris, tree branches, restaurant oil and grease, broken glass, industrial
sludges, worn-out stoves and refrigerators, drums of toxic chemicals, tailings
from coal mines, fields of corn stalks, animal manure, and many others. A logical approach is to subdivide these diverse wastes into classes such as municipal
solid waste, industrial solid waste, agricultural waste, and mining wastes.
Agricultural and mining wastes typically occur as large volumes of “singletype” materials, and have very specific management techniques that usually
result in disposal near their points of origin.
This chapter will focus on municipal and industrial solid wastes. The first
section of this chapter deals with municipal solid wastes (MSW)—their composition, quantities, collection, treatment, and disposal. Industrial solid wastes are
further subdivided into hazardous wastes and nonhazardous wastes, and a
later section of this chapter will delve into the problems of management, treatment, and disposal of hazardous wastes. Interestingly, radioactive wastes are
not defined under US solid waste regulations as hazardous wastes—they have
their own special category. Radioactive wastes will be discussed in Chapter 9.
8.1 Municipal Solid Wastes
Quantities and Composition
In 1978, Americans generated about 140 million tons of MSW, enough to
fill the Superdome in New Orleans from floor to ceiling twice a day, every day
of the year (Council on Environmental Quality 1978). But, despite years of
effort to minimize waste, in 2011 we generated 250 million tons (US EPA
303
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304
2013a)! The United States is a “throwaway” society, and growth in the amount
of waste has been linked to population growth and expansion of the economy.
Many of our consumer products are wrapped in plastic, and then sealed in
boxes, and, when purchased, the boxes are placed in bigger plastic bags so we
can carry them out of the store. For many items, it is cheaper to buy a new
replacement than it is to fix the old item.
Every year, the EPA conducts a study of the composition and quantity of
MSW generated in the United States, and the most recent study reported that
previously mentioned number (250 million tons in 2011). But that number was
actually down from 256 million tons in 2007. In 2011, the US average rate of MSW
generation per person was 4.40 pounds per capita per day (ppcd). The per capita
generation rate steadily increased from 1960 (2.7 ppcd) to 2000 (4.74 ppcd), but
has declined since then due mainly to increased recycling (US EPA 2013b).
Figure 8.1 presents the average US composition of MSW (as generated, before recycling) on a weight basis. The composition of waste varies with location,
season, economic conditions, and demographics (and, if other countries are considered, with culture). As can be seen in Figure 8.1, MSW in the United States is
predominantly paper and plastics, which is typical of highly developed countries
(due to the pervasive use of packaging and containers, as well as our high percentage of office jobs). The largest categories of MSW are organic materials—paper and cardboard, food wastes,
and yard trimmings. Together,
Figure 8.1 Composition (by weight) of MSW as
these three categories account for
generated in the United States in 2011. (Drawn
over 50% of the weight of MSW.
from data from US EPA 2013a.)
Fortunately, these and other materials can be recycled—and to a
250 Million Tons (before recycling)
large extent, they are.
Other
3.3%
Food
waste
14.5%
Yard
trimmings
13.5%
Wood
6.4%
Paper &
paperboard
28.0%
Rubber,
leather &
textiles
8.2%
Plastics
12.7%
Metals
8.8%
Glass
4.6%
Recycling
Recycling is one of EPA’s fundamental strategies for managing
MSW. As presented in Figure 1.5
in Chapter 1, the EPA hierarchy of
solving pollution problems includes recycling as the action that
should precede treatment and disposal. Recycling allows beneficial
re-use of materials, prevents
greenhouse gas emissions (created during the biodegradation of
wastes), and is a good way to save
space in a landfill (see Section 8.2)
and increase its useful life.
In 2011, recycling was effective, and about 87 million tons of
MSW (about 35% of the amount
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305
generated) were recycled (including yard trash and food wastes recovered for
composting) (US EPA 2013b). The per capita recycling amount was 1.54 ppcd.
In 2011, the recycling rates for newspaper and yard trimmings (through composting) were 72% and 57%, respectively (US EPA 2013b). Other materials with
high recycling rates included auto batteries (96%), steel cans (71%), aluminum
cans (55%), tires (45%), and glass containers (34%) (US EPA 2013b). The overall
recycling rate has increased steadily from only 6.4% in 1960 to 35% in 2011 (US
EPA 2013b). Figure 8.2 on the following page presents the average composition
of MSW discards (after recycling has removed some materials). Notice how the
discarded waste composition is different from that of the generated wastes,
because different materials have different rates of recycling. Example 8.1 illustrates the types of calculations that are done to determine overall recycling percentages, and the revised composition of the discarded waste.
EXAMPLE 8.1
Determine the percentage of generated waste that is recycled given the following information. The waste generated is 30% paper, 12% cardboard, 18%
yard waste, 8% glass, 5% plastic, 3% aluminum, 9% ferrous metal, and 15%
miscellaneous. Of that, 50% of the paper, 30% of the cardboard, 40% of glass,
20% of plastic, 50% of aluminum, and 10% of ferrous metal is recycled. Determine the final composition of the waste after recyclables are removed.
SOLUTION
The results of calculations necessary to solve this problem are provided in the
table below.
Waste
component
Paper
Cardboard
Yard waste
Glass
Plastic
Aluminum
Ferrous metal
Miscellaneous
Total
Generated
waste
composition,
lbs/100 lbs
30
12
18
8
5
3
9
15
100
Recycling
efficiency, %
Weight
disposed,
lbs/100 lbs
generated1
Composition
after recycling,
% by weight2
50
30
0
40
20
50
10
0
—
15
8.4
18
4.8
4
1.5
8.1
15
74.8
20.1
11.2
24.1
6.4
5.3
2.0
10.8
20.1
100
1 Column
4, the mass of each component remaining after removing the recycled material, is generated
by multiplying Column 2 by [1 – (Column 3 / 100)].
2
To determine the composition of the waste disposed (Column 5), divide each entry in Column 4 by the
Column 4 total (74.8) and express as percent.
The percentage of waste recycled is determined by subtracting the total of
Column 4 from 100: Percent Recycled = 100 – 74.8
= 25.2%
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Chapter Eight
Recycling can be accomplished by single-stream or dualstream recycling methods. Singlestream recycling refers to collect164 Million Tons (after recycling and composting)
ing all the wastes as one mixed
stream and then sorting the
waste in a materials recovery
Other
facility (MRF). This has the
4.4%
Food
advantages of easier disposal for
waste
residents, and easier collection
21.3%
for solid waste collection workers, but has the disadvantages of
Yard
contaminating various wastes
trimmings
with each other and incurring
8.8%
Paper &
paperboard
higher sorting costs later. A MRF
14.8%
is designed to accept mixed
Wood
MSW, and through a combina8.4%
Glass
tion of manual sorting, mechani5.1%
cal shredding, air classification of
Rubber,
leather &
paper and plastics, optical sorttextiles
Metals
ing of glass, magnetic separation
10.6%
8.8%
of steel, and other processes, can
Plastics
separate MSW into many recy17.8%
clable materials.
Dual-stream recycling refers
to presorting some of the easily
recycled materials (e.g., glass, paper, plastics, and aluminum) at the point of
generation, and collecting them separately. Community residents can greatly
assist in the municipality’s efforts by presorting some wastes and having them
available for separate pick-up. Drop-off sites and buy-back centers (in bottlebill states) also provide a significant quantity of recovered material. Sometimes, revenue can be generated from sale of recovered materials, but that revenue is extremely variable—subject to supply and demand forces, distance from
markets and other factors, and the degree of contamination—and is usually
insufficient to cover the costs of the MRF. Figure 8.3 shows a conceptual flow
diagram for the processes that occur at an MRF.
Figure 8.2 Composition (by weight) of MSW
discards in the United States in 2011. (Drawn from
data from US EPA 2013b.)
Collection
Collection of MSW in a municipality is a key part of the MSW management
system, and almost always is the mostly costly part. Considering the capital cost
for trucks, the fuel costs, and the labor costs, collection can consume 70% or
more of the municipality’s budget for solid waste management. Collection is
either accomplished by the public agency, or more frequently, by private companies that have been given exclusive contracts to collect wastes within the city (or
county). Collection from homes is usually done once or twice per week using
large compactor trucks (“garbage trucks”) that move from house to house, pick-
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Solid and Hazardous Waste Management
Figure 8.3
307
Simplified process flow diagram at a materials recovery facility.
Mixed MSW
Source-separated
bulk recyclables
Receiving area
First-stage
manual presorting
Cardboard
Bulky items
White goods
Other contaminants
Mixed recyclables
in bags or bins
Second-stage
manual presorting
Cardboard
Other large items
Bagged
commingled
recyclable
materials
Bag breaker
First stage
manual sorting
Paper
Cardboard
Plastics
Glass
Aluminum cans
Tin cans
Screening
Oversize material
Magnetic
separation
Second stage
manual sorting
Ferrous
metals
Shredding
Landfill
Incineration
Compost for intermediate
landfill cover
Paper
Plastics
Glass
Aluminum cans
Tin cans
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Chapter Eight
ing up waste left at the curb by the homeowners. Source-separated recyclables
like newspapers, glass bottles, steel cans, and aluminum cans are generally
picked up by a different collection vehicle and placed in separate bins without
compaction. Compaction of recyclables is not desirable as it can increase contamination from broken glass, which can reduce the value of the materials.
A typical garbage truck can hold anywhere from 10 to 40 cubic yards of compacted waste. The waste is usually compacted by a ratio between 2.0 to 3.0, meaning that the volume occupied by the waste after compaction is one-half to onethird of the volume before compaction. Such compaction allows the truck to carry
more weight in the same volume. Thus, the volume of the truck can be multiplied
by the compaction ratio to determine the volume of uncompacted waste that it can
carry. For example, a 30-cubic-yard truck can carry 60 to 90 cubic yards of uncompacted waste. Figure 8.4 shows a compactor truck unloading MSW at a landfill in
Polk County, Florida. The truck is a rear-loader with a capacity of 25 cubic yards.
Residential service trucks may have a crew of 1, 2, or 3 people and can service at most 300–400 homes before they get filled. They must then go directly to
a landfill or incinerator, to an MRF, or to a transfer station. A transfer station is
used when the place of final disposal (e.g., a landfill or an incinerator) is distant—collector trucks have notoriously bad fuel economy (1.5 to 4.0 mpg).
Advantages of transfer stations include cost savings associated with the use of
larger volume long-haul trucks with better gas mileage and a single driver for
the longer trip to the disposal site, shorter turnaround for collection vehicles,
opportunity for inspection of loads for hazardous wastes, and ease of locating
Figure 8.4 A compactor truck unloading collected municipal solid waste at a Florida landfill. (Courtesy of Dr. Tim Townsend, Professor of Environmental Engineering,
University of Florida.)
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309
disposal sites in less populated areas. To save fuel and labor costs, the collection routes must be carefully planned. There are different types of trucks for
commercial facilities like hotels and restaurants. For certain situations (such as
hotels or shopping malls), mechanically loaded compactor trucks can be used
to lift a container (dumpster) up and empty it into the truck from the top (as
opposed to a back-end loaded compactor truck). Alternatively, for construction
sites and the like, hoist trucks can be used, which simply hoist a large filled
dumpster (also known as a roll-off container) onto the truck and take it directly
to the landfill. An empty dumpster is returned to the site.
EXAMPLE 8.2
Compare the time required to collect waste generated at a large shopping center using a 30-yd3 mechanically loaded compactor truck vs. a hoist truck,
which hauls each roll-off container separately to the landfill. The waste at the
shopping center is stored in 12 containers, 10 yd3 (7.6 m3) each. Use the following information:
Compaction ratio:
2.5
Average time to travel to the landfill:
30 min
Average time at the landfill:
10 min
Time to pick up and unload container (compactor): 7 min
Time to load container onto hoist truck:
5 min
Time to drive between containers:
2 min
SOLUTION
Hoist Truck
Time per trip
= time to load container + time to landfill +
time at landfill + time to return from
landfill
= (5 min + 30 min + 10 min + 30 min) per trip
= 75 min/ round trip
Number of trips (hoist truck) = 12
Total time
= 75 × 12 = 900 min
Compactor Truck
Number of trips (compactor) = total volume waste/vehicle capacity
Volume of waste per trip
= volume × compaction ratio
= 30 × 2.5
= 75 yd3/trip
Number of trips
Total Time
=
120 yd 3
75 yd 3 /trip
= 1.6 trips (round this up to 2 trips)
= Number of containers × pick-up/unload
time + drive between containers ×
(number of containers – 1) + number of
whole trips × (haul time + at site time)
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= 12 containers × 7 min/container +
2 min/drive × 11 drives between containers + 2 trips × (30 + 10 + 30) min/trip
= 246 min
In this example, a compactor truck requires approximately one-fourth of the
time that a hoist truck requires. However, capital and operating costs for a
large compactor vehicle are much greater than for a hoist truck; all the facts
must be considered in an economic analysis of collection systems.
8.2 Landfill Disposal of MSW
The two major options for disposal of MSW are landfills and incineration.
According to the US EPA (2013b), 54% of the MSW generated in the United
States in 2011 was sent to landfills for final disposal and 12% was sent to incinerators with energy recovery (the rest was recycled). There are other types of
specialized landfills—construction and demolition (C & D) debris landfills, ash
landfills, and others, but this section focuses on MSW landfills. In the next section we will address MSW incineration.
Description of a Sanitary Landfill
A sanitary landfill is defined as an engineered piece of land where MSW is
deposited and operations are conducted to protect the environment from the
effects of the MSW. A site is excavated, and the bottom is prepared by installing
a clay or geotextile liner to prevent leakage of contaminated water (called leachate) into groundwater or local streams. A network of pipes is installed to collect
the leachate, and convey it from the bottom of the landfill to a treatment process. The solid waste is brought in by trucks and is deposited daily, spread into
layers, and then compacted by heavy equipment. At the end of each day, the
waste is covered with a 4-to-6-inch layer of soil, which is commonly referred to
as daily cover (see Figure 8.5). The soil stops the wind from blowing paper wastes
away from the site, prevents vectors (insects, birds, rodents) from gaining access
to the waste, and helps absorb some of the odorous gases. Often, perforated
pipes are installed as the layers of waste are built up; these pipes are later connected to collect the gas that is generated by the waste decomposition reactions.
The site must be large enough to accumulate waste for many years (sometimes reaching heights of several hundred feet above ground), but eventually it
becomes filled. Then it is capped with a geotextile liner, covered with two feet
of soil (the final cover), and grassed over. The geotextile helps retain methane
and other gases to allow for more efficient gas collection, and the grass cover
helps prevent erosion by rainfall and excess leachate formation. Often, monitoring wells are installed around the landfill to monitor for leachate leakage or
gas migration. Leachate treatment can be complicated by the fact that the
leachate is highly variable over time in both rate of generation and strength
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Figure 8.5 Schematic diagram for operation of a sanitary landfill. (Adapted from
Washington State Dept. of Ecology, 1987.)
FINAL
COVER
DAILY
COVER
INTERMEDIATE
COVER
UNDISTURBED
SOIL
LINER
REFUSE
(the concentrations of various organic and inorganic pollutants in the leachate).
Landfill gas has two major components: methane and carbon dioxide (in about
equal volumes), but also contains trace amounts of ammonia, hydrogen sulfide, and many hazardous volatile organic compounds (e.g., benzene, xylene,
vinyl chloride, dichloroethane, and many others). Therefore, any plan to utilize
the gas for its energy content must include some sort of gas clean-up facility.
Gas generation may continue for many years after the landfill stops accepting
new waste. A schematic of a typical MSW landfill is provided in Figure 8.6 on
the following page.
Over the years, the MSW biodegrades anaerobically. The land may eventually be reclaimed for use again, mainly for recreational purposes such as golf
courses, parks, and so forth. Much of the park and bike trails on the very eastern edge of Chicago next to Lake Michigan are built on top of the debris that
was placed there after the great Chicago fire of 1871, and many parks and several other parts of New York City are built on old landfills.
Figure 8.7 (on the next page) shows a portion of the active operations at a
regional landfill located in Sampson County, North Carolina, a private landfill
operated by Waste Industries, Inc. to serve multiple counties in the state.
Notice the large MSW transport trucks that are unloading here and all the
heavy equipment used at this landfill. Also, notice the plastic liner in the top
right part of the photo. This landfill is open Monday–Saturday, and accepts
MSW, scrap tires, C&D (construction and demolition) wastes, and commercial
yard waste (Sampson County 2014). On average, in 2010, the landfill accepted
about 3,500 tons per day (tpd) of solid wastes (Waste Industries 2010), but in
late 2013 the company proposed a significant increase in the waste disposal
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Figure 8.6
Schematic diagram of a typical MSW landfill after closure.
Gas control
Methane
monitoring Final cover
Groundwater
monitoring
WASTE
Liner
(synthetic
or natural)
Leachate collection system
Figure 8.7 A portion of an MSW landfill showing typical activity. (Courtesy of
Dr. Hamid Amini.)
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rate to accommodate an additional 500-1,500 tpd of MSW from New Hanover
County (Curran 2013).
As rain falls on a landfill, and the water percolates through the waste, biological, chemical, and physical processes occur which promote the degradation
of wastes and result in the production of leachate and gases. The leachate contains high concentrations of a variety of organic and inorganic pollutants
(BOD, COD, acids, metals, etc.), which must be controlled to prevent contamination of nearby ground or surface waters. The liner system at the bottom of a
modern landfill often consists of a layer of clay or a geotextile at the bottom, on
top of which is a porous layer of gravel or sand with leachate collection pipes
embedded in that layer. The leachate percolates downward into the porous
layer at the bottom and is retained there due to the liner. The leachate is then
withdrawn from the bottom of the landfill and (most commonly) is pumped to
a traditional WWTP for treatment. Alternatively, the leachate is pumped back
to the top of the waste to provide more moisture to the wastes and thus hasten
the biodegradation process (this type of landfill is called a bioreactor landfill).
As mentioned previously, modern landfills have gas collection wells. There
may be hundreds of vertical or horizontal perforated pipes in the landfill, and
eventually they are all connected to a gas compressor that collects all the gas.
The collected gas may be burned directly in a flare to destroy the methane
(methane is a greenhouse gas that is 25 times more powerful than carbon dioxide) and odorous gases, or the gas may be cleaned and then burned in internal
combustion engines or in a turbine to generate electricity. In addition to dealing
with leachate and gas, actions are undertaken at modern landfills to preserve
space in the landfill and to protect the environment, such as separating and
recycling large items, tires, e-waste like old TVs and computers, and household
hazardous wastes like cans of bug spray, old paint cans, and so forth.
As an example of the previous discussion, consider the landfill that is
located in a remote area of Seminole County, Florida (Seminole County 2013).
The landfill property includes more than 6,000 acres, but the waste area itself is
only about 230 acres. It is underlain by an emplaced clay liner. Phase I of the
landfill is capped, and is 131 feet high. Phase II is the active landfill, and has a
permitted final height of 270 feet. Gas recovery and leachate removal systems
are installed on the site to ensure environmental compliance. Four internal combustion engines coupled with electricity generators have been installed, each of
which now generates about 1 MW of electrical power. Whole tires and items like
stoves and other large appliances are separated for later recycling. Yard waste is
ground up and used for mulch or biomass fuel (Seminole County 2013).
Preliminary Design of an MSW Landfill
The design of a landfill begins with a calculation of the amount of waste to
be received over a period of time. The design life of the landfill should be many
years (20 or more) if enough land can be found in an acceptable location. The
amount of land required is dependent on the total amount of waste to be
received, and is determined based on the projected population and an average
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MSW generation rate (e.g., 4.4 ppcd in the US). Such projections are developed
based on engineering studies, and must be approved by city or county leadership prior to undertaking any detailed design calculations.
Selecting an appropriate site for an MSW landfill requires extensive evaluation of potential locations with consideration of distance from waste generation, climate, surface and groundwater hydrology, topography and geology,
seismic activity, and neighborhood acceptance. This last item may be the most
difficult to achieve—very few people are willing to accept a landfill in close
proximity to where they live. Assuming that a site in a relatively remote location can be found, the next step is to calculate the overall amount of land
needed to be purchased. The size of the site is a function of the amount of waste
to be disposed over the life of the landfill, and the additional land area required
for support facilities such as maintenance and administration buildings, access
roads, and ancillary waste management operations (composting; handling of
tires, recyclables, and bulky items; household hazardous waste collection; construction and demolition debris disposal; gas-to-energy facilities, etc.).
Once the total waste mass is determined, the volume occupied by that
mass is calculated, and then the maximum acceptable height of the landfill
must be set. The waste that had been compacted by the trucks during pick-up
is further compacted by heavy equipment at the landfill. The average density
of compacted MSW in a landfill is estimated to be between 750 to 1,250 lb/yd3
(US EPA 2013c). Knowing the waste volume and final height of the landfill, the
area of waste disposal (the footprint) can be calculated. The actual land area for
waste disposal is only a portion of the whole landfill site. As stated above,
there must be room for access roads, operation and administrative buildings,
and so forth, and, if desired, a buffer zone around the active landfill (to help
ease potential future impacts of noise and odor on the surrounding areas).
Excluding the buffer zone, the nonwaste disposal portion of the landfill may be
about 40–50% of the total land area. The next example illustrates these calculations for the simple case of a stable population.
EXAMPLE 8.3
Determine the land area that must be purchased to provide ten years of landfill disposal capacity for waste generated by a stable population of 250,000.
Use the following information:
Per capita daily generation rate:
In place density:
Average landfill depth:
Area required for buildings, roads, etc:
4.4 ppcd
950 lb/yd3
40 ft
40% of total
SOLUTION
Total mass of waste = population × per capita generation × landfill life
= 250,000 × 4.4 ppcd × 10 yrs × 365 days/yr
= 4 × 109 lb
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Volume required
315
= mass of waste/ density of waste
=
4 ¥ 109 lb
950 lb/yd 3
= 4, 226,000 yd3
= 4, 226, 000 yd 3 ¥ 27
Waste footprint
yd 3
= 114, 111, 000 ft 3
= volume/ depth
= 114,111,000 ft3/ 40 ft
= 2,853,000 ft2
= 2, 853, 000 ft 2 ¥
Total land area
ft 3
=
1 acre
43, 560 ft 2
= 66 acres
66 acres
= 110 acres
1 - 0.40
8.3 Thermal Destruction of Waste
Incineration of MSW is used to reduce the volume of waste (a solid residue
[ash] remains that typically must be landfilled) and to destroy organics. Incineration with heat recovery is a way to make beneficial use of the energy content
of the paper, plastics, and other combustible materials in the waste. Thermal
destruction of waste involves the high temperature oxidation of combustible
matter, typically with the addition of excess air to ensure complete combustion.
Some revenue may be obtained from energy sales, but MSW incineration
requires expensive equipment and uses auxiliary fuel, so there is still a net cost
to the process. In the United States in 2011, 11.7% of MSW was burned with
energy recovery, and that amounted to about 29 million tons (US EPA 2013b).
Most MSW incinerators are mass-burn devices, which means that the asreceived waste is pushed into a large chamber and ignited. Some facilities burn
shredded wastes, and some separate the paper and plastics and burn just those
high-energy-content materials. All MSW incinerators must meet stringent air
pollution control regulations. Figure 8.8 on the following page provides a conceptual flow diagram of the incineration process.
The waste incinerator is designed to provide a high temperature chamber
with adequate time for combustion and good mixing (turbulence) of air and
waste gases within the incinerator. The incinerator is often a two-chamber
device, with the solid waste burning incompletely in the first chamber to form
a variety of gases (including carbon monoxide and organic compounds), and
then those gases burning completely in the second chamber. Good combustion
is ensured by the introduction of auxiliary fuel and excess air (exceeding stoichiometric oxygen requirements) using forced-air or induced-draft blowers.
Excess air is often used to help control temperature and residence time as well.
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Chapter Eight
Figure 8.8
Conceptual process flow diagram of MSW incineration.
CLEAN GASES
TO ATMOSPHERE
WASTE
Waste
processing
Acid gas control
Auxiliary
fuel
Waste
feeding
Combustion
air
Combustion
Chamber
Hot
gases
Particulate removal
Heat recovery
Solids
and/or
liquids
Ash
removal
Solids
Residuals treatment
TO DISPOSAL
The refractory-lined furnace is sized to provide adequate volume to retain
waste and combustion products for a sufficient amount of time to complete all
of the combustion reactions. Often, regulations prescribe a minimum temperature and a minimum amount of time that is required. In many units, energy
recovery is possible for subsequent steam production or electric power generation. Energy recovery can be accomplished by turning the incinerator into a
boiler by installing closely spaced steel tubes along the primary combustion
unit walls and circulating water through them. Also, more energy can be recovered using boiler tubes suspended within the secondary chamber. Mass and
energy balances are made by the engineers to evaluate an MSW incinerator
before it is built, as illustrated by the following series of examples.
EXAMPLE 8.4
Draw a sketch showing the energy and material inputs and outputs for an
MSW waste-to-energy incinerator. The energy released during combustion is
used to produce steam that drives a steam turbine to generate electricity.
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SOLUTION
Heat
losses
Hot gases to
APC equipment
Energy losses
during power
generation
Waste input
Steam turbine for
MSW
Auxiliary fuel
(as needed)
incinerator/boiler
Steam
electricity generation
Electric
power
Air input
Condenser
Unburned
carbon
in ash
Hot ash
(out) Cooling
(in) water
Boiler feed water
EXAMPLE 8.5
Given the quantitative information in the following table regarding the
energy and mass balance for the WTE system shown in Example 8.4, calculate
the major energy outputs from just the incinerator/boiler.
Input/Output
Value
MSW feed rate
Energy content of waste
(higher heating value, HHV)
Heat losses from incinerator walls
Moisture content of waste
Moisture generated during combustion
Latent heat of water evaporation
Ash content
Carbon content of ash
Heat capacity of ash
Energy content of carbon in ash
Temperature of hot gases going to APC
Grate temperature
Ambient temperature
Power conversion rate1
Exit gas flow rate
Heat capacity of air and exit gases
100 tons/day
5,000 Btu/lb
1
0.5% of input energy
20% of waste input
15% of waste input
1,040 Btu/lb
15% of waste input
5% of ash
0.25 Btu/lb-°F
14,000 Btu/lb
370 °F
850 °F
70 °F
7,000 Btu/kwh
20 lb/lb waste
0.25 Btu/lb-°F
Includes turbine and generator mechanical and electrical losses, and heat lost to cooling water.
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SOLUTION
1. Calculate energy content (HHV) of incoming waste stream.
5, 000 Btu 100 tons 2, 000 lb
¥
¥
= 1 ¥ 109 Btu/day
lb
day
ton
2. Calculate heat losses from furnace walls.
0.005 × 1 × 109 Btu/day = 5×106 Btu/day
3. Calculate heat “loss” to evaporate moisture. Moisture is the liquid water
that was in the waste initially (20% of waste weight), and the water that
was generated as a result of oxidation of organic hydrogen in the waste
(water produced = 15% of waste weight). Both heats must be subtracted
from the HHV of the waste.
(0.2 + 0.15) lb water ¥ 100 tons ¥ 2, 000 lb = 70, 000 lb/day
lb waste
day
ton
70,000 lb 1, 040 Btu
¥
= 7.3 ¥ 107 Btu/day
day
lb
4. Calculate heat associated with ash leaving incinerator.
0.15 lb ash 100 tons 2, 000 lb
¥
¥
= 30 , 000 lb ash/day
lb waste
day
ton
30 , 000 lb ash
0.25 Btu
¥ (850 - 70 ) ∞F ¥
= 5.9 ¥ 106 Btu/day
day
lb - ∞F
5. Calculate energy losses due to the fact that waste is not completely burned
and carbon remains in the ash (this carbon did not combust and did not
release its heating value).
0.05 lb carbon 30, 000 lb 14, 000 Btu
¥
¥
= 2.1 ¥ 107 Btu/day
lb ash
day
lb
6. Calculate heat lost in hot gases
20 lb gases 100 tons 2, 000 lb
0.25 Btu
= 3.0 ¥ 108 Btu/day
¥
¥
¥ ( 370 - 70 ) ∞F ¥
lb waste
day
ton
lb-∞F
7. Calculate heat contained in the steam sent to the turbine/generator
Heat in steam = Input Heat – Summation of outputs (losses)
= 1×109 – (5×106 + 7.3×107 + 5.9×106 + 2.1×107 + 3.0×108)
= 6.0×108 Btu/day
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319
EXAMPLE 8.6
Calculate the energy efficiency of the waste incinerator/boiler described in
Example 8.5.
SOLUTION
Efficiency =
Useful energy
Input energy
Ê 6.0 ¥ 108 ˆ
¥ 100
=Á
9˜
Ë 1.0 ¥ 10 ¯
= 60%
EXAMPLE 8.7
Calculate the electric power generation from the steam turbine described in
Examples 8.4 and 8.5.
SOLUTION
The energy available for power generation is 6.0 × 108 Btu/day. Calculate the
electric power generated.
6.0 ¥ 108 Btu/day 1 day
¥
= 3, 571 kw (3.6 MW)
7 , 000 Btu/kwh
24 hr
EXAMPLE 8.8
Calculate the overall energy efficiency of power generation from the MSW
incinerator system described in the previous several examples.
SOLUTION
% Efficiency =
Electricity out
¥ 100
Energy in
3, 571 kw ¥
= 100 ¥
= 29%
12 Btu 24 hr
3, 41
¥
kwh
day
1 ¥ 109 Btu
day
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Chapter Eight
Stack emissions from MSW incinerators are stringently regulated (US EPA
2006). Standards have been set for specific pollutants and are based on different
sizes of incinerators, and whether they were newly constructed or existing at
the time of implementation of the regulations. Pollutants of concern include
particulate matter (PM-10), acid gases (primarily HCl), nitrogen oxides, mercury and other metals, and products of incomplete combustion (such as carbon
monoxide and certain hazardous organic compounds such as chlorinated dioxins and furans). These pollutants are controlled using good combustion practice and air pollution control devices.
Good combustion practice requires operation of incinerators under optimum time, temperature, and turbulence conditions, with proper monitoring of
operational conditions and emission levels. In addition, elimination of certain
items from the waste stream may also minimize the production of undesirable
emissions (such as the mercury from fluorescent bulbs, thermometers, and certain small batteries). APC devices typically used to control emissions from
waste incinerators include wet or dry scrubbers, and electrostatic precipitators
or fabric filters. In addition, innovative processes such as catalytic and noncatalytic reduction of nitrogen oxides, and powdered lime and carbon injection followed by fabric filtration for dioxin and mercury control are used. Many of
these air pollution control processes were described in Chapter 7.
8.4 Hazardous Wastes
Introduction
One of the most contentious environmental issues of the 1960s and 1970s
was that of hazardous wastes. There are many horror stories of chemical
wastes in drums being buried or even just abandoned in fields, and eventually
rusting and leaking toxic chemicals into the soil, the water, or even into nearby
homes. One such story that particularly resonates is that of Love Canal (US
EPA 1979). A brief overview is presented here, and more details are given later
in the chapter.
In the early 1900s, Love Canal was a partially completed excavation (a
would-be canal originally envisioned to connect the Niagara River to Lake
Ontario) near Niagara Falls, New York. The project was abandoned and eventually, in the 1920s, the excavation was converted into a dumpsite for industrial
chemicals and municipal waste. In the 1950s, the ditch (filled with wastes) was
covered over with dirt and clay, and the land was sold to the city for one dollar.
As the city grew and the area developed, a public school and homes were
built near the old canal site, and by the 1970s, chemical wastes were leaking
everywhere. Many different chemicals (including 11 suspected carcinogens)
were discovered in the school yard and in the basements of the homes. Corroding drums could be seen breaking through the ground in some backyards, and
puddles of noxious chemicals were found in basements or in yards. Children
playing in some areas outdoors received acid burns, and there were cases of
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321
birth defects in the community. This particular horror story, as bad as it was,
had one positive outcome: it galvanized people and Congress into action.
Laws and Regulations
Significant regulation of land-based waste disposal did not occur until the
Resource Conservation and Recovery Act (RCRA) was passed by Congress on
October 21, 1976. RCRA laid out the framework for regulating hazardous wastes
and required EPA to develop a comprehensive set of regulations for the management, treatment, storage, and disposal of hazardous wastes. EPA took this responsibility seriously and, on May 19, 1980, published the first RCRA regulations for
solid and hazardous wastes. Under RCRA, solid waste was defined as “any solid,
liquid, semi-solid, or contained gaseous discarded material, which is no longer
useful for its intended purposed or is an unintended or unusable by-product.”
RCRA regulations go into more detail; they distinguish between nonhazardous and
hazardous solid wastes and treat them in very different ways. We have already discussed municipal solid wastes, and this section focuses on hazardous wastes.
The regulatory system for managing hazardous wastes can be found in
Title 40 of the Code of Federal Regulations, Parts 260–282; the key sections are in
Parts 260–264. The term hazardous waste is defined as “. . . a solid waste, or
combination of solid wastes, which may, because of its quantity; concentration;
or physical, chemical, or infectious characteristics . . . pose a substantial present
or potential hazard . . . when improperly treated, stored, transported, or disposed of, or otherwise mismanaged.” This definition is structured to give the
US EPA broad authority to identify and regulate hazardous wastes. The regulations address all hazardous waste activities regardless of whether historical
evidence exists for poor handling of hazardous wastes or not. Additionally,
RCRA does not consider cost as a basis for failing to institute regulatory controls. In other words, RCRA regulations must be followed regardless of economic impact (Fortuna and Lennett 1987). Failure to do so can result in
criminal charges against high-level facility managers and owners.
Under RCRA regulations, EPA provides a system to track hazardous
wastes from when they are first created to when they are finally disposed. In
other words, the key to hazardous waste management under RCRA is a “cradle-to-grave” approach. The regulations have the following major components.
• identification and listing of hazardous wastes
• a manifest system to track wastes from “cradle to grave”
• standards of performance for hazardous waste generators; transporters;
and treatment, storage, and disposal (TSD) facilities
• a permit system for TSD facilities
Based upon RCRA definitions, a solid waste is designated as a hazardous
waste if it falls within one of the following four categories, and is not specifically excluded (40 CFR 261):
• The waste is listed as hazardous if it has specific hazardous constituents,
or if it comes from a listed process operation.
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• Wastes not specifically listed are hazardous if they exhibit one or more of
the four characteristics of a hazardous waste.
• If the waste is a mixture of a listed hazardous waste or characteristic
waste and any other material, it is hazardous if the mixture exhibits hazardous characteristics.
• If the waste is derived from the treatment, storage, or disposal of a listed
waste, then the waste is hazardous.
A listed hazardous waste is one that has been specifically identified under
RCRA regulations as a hazardous waste. There are thousands of listed wastes,
some of which are provided in Table 8.1, and some of which are cited in this
paragraph. Listed wastes are found in four different lists. The F list identifies
specific wastes that can come from a variety of industrial processes, the K list is
for mixed wastes from specific industries, the P and U lists are for specific
chemicals (single compounds). Some examples from the F list are spent halogen solvents such as carbon tetrachloride or cholorobenzene, and spent solvents such as acetone, methanol, or mixed xylenes. Examples from the K list
are wastewater sludge from the production of chrome pigments, distillation
bottoms from the production of acetaldehyde, or baghouse dust from the production of ethylenebisdithiocarbamic acid. The P list includes the pesticide
aldrin, arsenic compounds, barium chloride, and many others. The U list
includes compounds like acetaldehyde, aniline, benzene, chlorobenzene, and
many others.
A characteristic waste is one that is not a listed waste but rather exhibits
one or more hazardous characteristics defined as corrosivity, reactivity, ignitability, and/or toxicity (as summarized in Table 8.2). Toxicity is defined as the
ability to leach certain heavy metals and/or toxic organic compounds out of
the waste by a standardized laboratory test—the Toxic Characteristic Leaching
Procedure (TCLP) test (40 CFR 261). A characteristic waste can be rendered
nonhazardous by eliminating the characteristic that caused it to be hazardous.
However, a listed waste and any waste derived from its treatment continues to
Table 8.1
Examples of Listed Hazardous Wastes
Hazardous Waste
Basis for Listing
Spent halogenated solvent used in degreasing
Spent nonhalogenated solvents
Wastewater treatment sludge from the
production of chrome yellow and orange pigments
Stripping still tails from the production of methyl ethyl pyridines
2,6-dichlorophenol waste from the production of 2,4-dichlorophenol
Ammonia still lime sludge from coking operations
Cyanogen chlorine
Thiophenol
Emission control dust/sludge from secondary lead smelting
Toxicity
Ignitability
Toxicity
Toxicity
Toxicity
Toxicity
Acute Toxicity
Acute Toxicity
Toxicity
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Table 8.2
Hazardous
Characteristic
Corrosivity
Ignitability
Reactivity
Toxicity
323
Hazardous Waste Characteristics*
Definition
Waste that is highly acidic or alkaline (pH < 2 or > 12.5)
Waste that is easily ignited and poses a fire hazard during routine management
Waste that is capable of potentially harmful, sudden reactions
Waste capable of leaching any of 8 heavy metals (As, Ba, Cd, Cr, Pb, Hg, Se, Ag) and/
or 32 specific pesticides and/or organic compounds into slightly acidic water
*The US EPA has specified laboratory procedures for each characteristic.
be regulated as a hazardous waste even if the treated stream no longer exhibits
hazardous characteristics.
To help implement the “cradle-to-grave” approach in managing hazardous
wastes, EPA identifies generators, transporters, and treatment, storage, and
disposal facilities (TSDFs). This system allows state and federal regulators to
track the waste from the time it is formed until it reaches its final disposal site.
RCRA regulations divide generators into three categories: large quantity generators, small quantity generators, and conditionally exempt small quantity generators. Large quantity generators (LQGs) produce more than 2,200 lbs (1,000
kg) of hazardous waste each month. LQGs are required to test their waste to
determine whether the waste is a hazardous waste. If the waste is hazardous
and will be shipped off-site, the generator is required to acquire an EPA identification number, start a manifest (a set of forms that allows each waste shipment to be tracked) for the waste, containerize and label the waste, and notify
the TSD facility that the waste is being shipped to them. Examples of labels for
a hazardous waste shipment are shown in Figure 8.9 on the following page.
The generator may store hazardous waste on the production site for up to 90
days without acquiring a TSD permit. The 90-day time clock starts from the
time the first bit of waste enters the storage/shipment container.
Generators
In 2011, there were 16,447 LQGs in the United States who collectively produced over 34 million tons of RCRA hazardous waste (US EPA 2013d). These
numbers represent large decreases over just two decades: a decrease of over
6,000 large generators, and a decrease of more than 200 million tons of hazardous wastes since 1993. In 2011, the largest hazardous-waste-generating states in
the US were Texas, Louisiana, Mississippi, Ohio, and Kansas. These five states
accounted for 73% of the national total of hazardous waste generated.
To be classified as a small quantity generator (SQG) under RCRA, a generator must only produce between 220 and 2,200 lbs (100 and 1,000 kg) of hazardous
waste per month. SQGs basically have the same RCRA requirements as LQGs
except for some minor changes in storage time limits. SQGs can store hazardous
waste on-site for up to 180 days as long as the total weight does not exceed
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Chapter Eight
Figure 8.9
Hazardous waste shipping labels—EPA.
PROPER D.O.T.
SHIPPING NAME ___________________________________________________
HAZARDOUS WASTE, LIQUID N.O.S. NA9189
ORM-E
HAZARDOUS WASTE
FEDERAL LAW PROHIBITS IMPROPER DISPOSAL
IF FOUND, CONTACT THE NEAREST POLICE, OR
PUBLIC SAFETY AUTHORITY, OR THE
U.S. ENVIRONMENTAL PROTECTION AGENCY
GENERATOR INFORMATION:
NAME _______________________________________________________________________
ADDRESS ____________________________________________________________________
CITY _______________________________________ STATE ____________ ZIP _________
EPA
ID NO. ________________________________
EPA
WASTE NO. _______________________
ACCUMULATION
START DATE _________________________
MANIFEST
DOCUMENT NO.
_________________
HANDLE WITH CARE!
CONTAINS HAZARDOUS OR TOXIC WASTES
STYLE WM-10
HAZARDOUS
WASTE
FEDERAL LAW PROHIBITS IMPROPER DISPOSAL
IF FOUND, CONTACT THE NEAREST POLICE, OR
PUBLIC SAFETY AUTHORITY, OR THE
U.S. ENVIRONMENTAL PROTECTION AGENCY
PROPER D.O.T.
SHIPPING NAME ________________________________ UN OR NA#________
GENERATOR INFORMATION:
NAME _______________________________________________________________________
ADDRESS ____________________________________________________________________
CITY _______________________________________ STATE ____________ ZIP _________
EPA
ID NO. ________________________________
EPA
WASTE NO. _______________________
ACCUMULATION
START DATE _________________________
MANIFEST
DOCUMENT NO.
_________________
HANDLE WITH CARE!
CONTAINS HAZARDOUS OR TOXIC WASTES
STYLE WM-6
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325
13,200 lbs (6,000 kg). SQGs that ship wastes off-site are still required to start a
manifest and to ensure a signed copy is received from the TSD facility within 35
days. It has been noted that historically, SQGs produce only 1% of US hazardous
waste. In 2011 there were 2,185 SQGs in the United States (US EPA 2013d).
Conditionally exempt small quantity generators (CESQGs) produce less
than 220 lbs (100 kg) of hazardous waste each month. These generators are not
subject to RCRA Subtitle C requirements, but they are required to properly
manage their solid waste under RCRA Subtitle D (nonhazardous waste regulations). Because of the potential liability associated with improper disposal of
hazardous waste, CESQGs often dispose of their hazardous waste at an RCRA
permitted TSD facility. In general, they typically store waste on site for long
periods of time until they accumulate enough volume to warrant a trip to a disposal site.
Treatment, Storage, and Disposal (TSD) of Hazardous Wastes
Much of the hazardous waste generated is treated on site by the producer
of the waste. This practice avoids the possible liability that comes with shipping hazardous wastes to others. In 2011, there were 460 facilities that received
hazardous waste shipments from generators totaling about 6.1 million tons (US
EPA 2013d). In 2011, the hazardous waste management techniques employed
by off-site facilities (and the percentages handled by each type of practice)
included landfill (15%), energy recovery (13%), metals recovery (13%), deepwell injection (10%), fuel blending (9%), incineration (9%), and several others
(US EPA 2013d).
The 1984 amendments to RCRA mandated treatment of hazardous waste
prior to land disposal. Hazardous waste TSD facilities (TSDFs) are regulated
and permitted under RCRA Subtitle C, and have their own operating requirements including record keeping, monitoring and inspection, contingency
plans, personnel training, and financial responsibility. Additionally, the TSDF
owner/operator must monitor groundwater in order to detect and evaluate
any migration of contaminants from the property.
Landfilling of hazardous wastes must take into account the nature of the
wastes (segregate different types of waste into different parts of the landfill),
and must of course have a secure site to ensure that toxic materials do not get
leached out of the landfill. As of 2013, there were 21 commercial EPA-permitted
hazardous waste landfills in the United States that were operating as TSDFs as
shown in Figure 8.10 on the next page (EHSO 2013).
Treatment of hazardous wastes involves physical, biological, and chemical
processes. Some TSDFs have all three processes on site as well as an on-site
chemical landfill for disposal of residuals. Physical processes are used to concentrate contaminants or effect a phase change so that the hazardous constituent can be more conveniently processed or disposed. These processes include
activated carbon adsorption, stripping, sedimentation, dewatering, and distillation. Biological treatment is usually applied for the destruction of organic
hazardous constituents in dilute wastewaters using activated sludge processes,
trickling filters, stabilization ponds, and anaerobic digestion. Chemical treat-
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326
Chapter Eight
Figure 8.10
Commercial TSDFs in the United States. (Adapted from EHSO 2013.)
Commercial
Waste Management
(CWM)
MAX
Peoria
Wayne
Disposal CWM Disposal Environmental
CWM
EnviroSafe
EnviroSafe
Laidlaw
U.S. Ecology
CWM
Laidlaw
Laidlaw
Laidlaw
Laidlaw
Waste Control
Specialists Laidlaw
CWM
Laidlaw
Texas
Ecologists
CWM
ment actually alters the hazardous constituents through chemical reactions that
render the waste nonhazardous or immobile. Chemical processes include oxidation and reduction, neutralization, precipitation, and solidification.
Recall that in Chapter 2 we said chemistry is important to environmental
engineering. There are many types of wastes, so there are many types of chemical treatment processes (as well as physical and biological). We choose to illustrate chemical treatment with two examples from the electroplating industry.
The first deals with chrome plating. Chromic acid (H2CrO4) is used in an acidplating bath to deposit hard chrome onto metal surfaces (for example, hubcaps
for cars). Hexavalent chromium (Cr6+) is very toxic, so waste chromic acid is
first chemically reduced to Cr3+ and then precipitated as Cr(OH)3. The solid
precipitate can then be sent to a secure landfill. One common reducing agent is
sulfur dioxide gas. The first reaction takes place in one tank in a dilute solution
of sulfuric acid (pH 2 to 3). The net balanced reduction reaction is shown below.
2 H2CrO4 + 3 SO2
→
2 Cr3+ + 3 SO42– + 2 H2O
(8.1)
After the first reaction is completed, as measured by an oxidation-reduction
potential (ORP) sensor, the liquid flows to a second tank, where caustic is
added to conduct the precipitation reaction (at about pH 9 to 10). This second
(balanced reaction) is shown below.
2 Cr3+ + 3 SO42– + 6 NaOH
→
2 Cr(OH)3↓ + 6 Na+ + 3 SO42–
(8.2)
The precipitate is removed as a sludge which is then filtered to produce a
solids residual. The solids are sent to a secure landfill for final disposal. Note
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327
that if lime [Ca(OH)2] is used instead of sodium hydroxide, there will be solid
CaSO4 that co-precipitates with the Cr(OH)3. There is no way to separate the
hazardous chromium hydroxide from the nonhazardous calcium sulfate, so the
whole sludge amount must be sent to the hazardous waste landfill, increasing
the disposal cost.
The second example comes from the gold-plating business. The plating of
gold onto metal objects occurs in an aqueous solution containing gold complexed with cyanide (CN–) to make it soluble. Once the plating is done, the
wastewater contains cyanides, a poisonous material. The cyanide can be oxidized and destroyed using alkaline chlorination, which occurs in two stages,
and often in two separate tanks. During the first stage, the pH must be kept
high (10.5 to 11.5) to prevent the formation of hydrogen cyanide gas (HCN),
which could escape the tank and poison workers. For the process using bleach
(sodium hypochlorite—NaOCl) as the oxidant, the cyanide is converted into
cyanate as shown below:
→
NaCN + NaOCl
NaCNO + NaCl
(8.3)
Of course, the ionic compounds shown in reaction (8.3) above (and in the
reaction below) exist as ions in aqueous solutions. After the first reaction is
complete (as measured by an ORP probe), the liquid is transferred to a second
tank. There, the cyanate ion is destroyed by adding more bleach.
→
2 NaCNO + 3 NaOCl + H2O
3 NaCl + N2 + 2 NaHCO3
(8.4)
The second tank is operated at a pH of about 7.5 to 8.0. When the pH is
lowered prior to discharge (say with HCl), the bicarbonate ion may combine
with hydrogen ions and form CO2 gas as shown below.
HCl + NaHCO3
→
NaCl + CO2 + H2O
(8.5)
The overall process thus converts toxic cyanide into water, salt, and the
innocuous gases CO2 and N2. The overall balanced reaction is obtained by multiplying reaction (8.3) by two, then adding it to reaction (8.4), and then adding
the result to reaction (8.5) multiplied by two.
2 NaCN + 5 NaOCl + 2 HCl
→
7 NaCl + N2 + 2 CO2 + H2O
(8.6)
Incineration is used for hazardous organic wastes (it does not do any good
to incinerate heavy metal wastes). Incineration can quickly convert toxic materials made up of large complex organic molecules into CO2 and H2O and other
low molecular weight compounds. Good examples of hazardous organic
wastes well suited for incineration are concentrated solutions of pesticides,
organic solvents, and even army nerve agents. However, proper incineration
entails more than simply igniting the waste inside a small furnace. Integrated
hazardous waste incineration facilities are complex operations that include
waste-receiving facilities, incinerators, air pollution control equipment, and
laboratory facilities. They may include different kinds of incinerators to handle
different forms of waste. Figure 8.11 depicts such a facility.
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Chapter Eight
328
Figure 8.11
Integrated hazardous waste incineration facility. (Adapted from Oppelt 1987.)
Solid
Waste
Weight
P
Instrumentation
Legend
P Pressure
T
F
O2
CO2
CO
T Temperature
Rotary
Kiln
Fuel
F Flow Rate
F
POHC
Metals
TCLP
Air
F
ΔP Differential Pressure
T
Ash
Water
Air
T
F
T O2
Aqueous
Waste
Afterburner
F
F
F
Stack
ΔP
T
Heat
Recovery
Fuel
F
Air
Liquid
Waste
ΔP
F
F
T
F
Absorber
ΔP
T
Liquid
Injection
Gaseous
Waste
F
F O2
F
Venturi
F
Fuel
8.5 Site Remediation (Soil and Groundwater Cleanup)
Introduction—A Case Study
As mentioned earlier, there were many instances of improper disposal of hazardous wastes that came to light in the 1960s and 1970s. These incidents opened
the eyes of the American public to the problems associated with historical methods of waste disposal. One of the most widely publicized problems occurred at
Love Canal. This case study serves as a good introduction to this section.
The abandoned Love Canal was 60 feet wide, 10 feet deep, and 3 blocks
long. It had been used as a local swimming hole for many years until it was
purchased in the 1920s by the Hooker Electrochemical Company for use as a
waste dump. The site was well suited for waste dumping because the clay soil
in which the canal had been dug had a very low hydraulic conductivity.
Hooker Electrochemical placed some 21,800 tons of chemical wastes in the
north and south sectors of the canal over a 10-year period. Municipal solid
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Solid and Hazardous Waste Management
329
wastes were placed in the central sector. The dump was closed in 1953, at
which time Hooker installed a clay cap to seal the wastes.
Shortly after closing the dump, Hooker was pressured by the City of Niagara Falls Board of Education to deed the site to the city. Hooker, under threat of
seizure, sold 16 acres to the city for one dollar, while warning the city that hazardous chemicals had been placed in the dump. In 1954, the Board of Education constructed an elementary school in the center of the site, with associated
sewers and roads. Surrounding land was sold for residential development, and
a large road was constructed. As early as 1958, children were complaining of
chemical burns while playing in road building debris. Vegetation blackened
and died. In 1976 residents began complaining of odors and black oozing material in their basements. Lois Gibbs established a homeowners’ association that
actively pursued governmental intervention, and brought media attention to
the problem.
In 1978, the State of New York Commissioner of Health declared a health
emergency and relocated 236 families. In 1980, an additional 800 families living
in the second ring around the canal were relocated. It soon became apparent
that cleanup of the site was going to be quite expensive. There were no provisions at the time for federal funding (like the Superfund of today), therefore, in
1981, President Jimmy Carter declared Love Canal a disaster area to make federal funds available through the Federal Emergency Management Act. Another
2,500 families were relocated at a cost of $30 million.
Site investigation and cleanup continued over the next few years. Nearly
100 monitoring wells were dug to determine subsurface hydrologic conditions
and to monitor contaminant transport. The site investigation determined,
among other things, that the original clay cap had been disturbed by construction and had allowed rainwater to leach into the canal. The underlying clay
kept the water in contact with the waste, which caused the water to become
highly contaminated. After the water had filled the old canal, it began to flow
out through the topsoil and along municipal sewer trenches to basements of
homes all around the site. A study of health effects in 1976 revealed children
with chemical burns, increased incidents of miscarriage, various kinds of birth
defects, and others. In 1978, more than 400 chemicals were identified in the surrounding soil, surface water and groundwater, and air (in basements), some at
concentrations 5,000 times maximum safe levels.
The immediate response was to relocate families, install a new 22-acre clay
cap, construct a drainage system to divert groundwater, close the school, and
fence off the area. Long-term remediation was designed for hydraulic containment and cleanup following a four-step plan: (1) contain, collect, and treat contaminated groundwater and leachate, (2) dewater the canal over time, (3) cover
the site with a permanent impermeable cap, and (4) monitor the progress until
remediation was completed.
Collection of leachate and drainage of the canal was accomplished using
7,000 feet of French drains, installed 15-20 feet below grade surrounding the
canal. The drains were filled with gravel, perforated clay pipe, and sand. The
drainage was pumped to an on-site treatment facility. The treatment consisted
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Chapter Eight
of pH adjustment, clarification, and activated carbon adsorption of organics.
The treated effluent was then routed to the city sewer and was further treated
at the local WWTP. Ultimately, the site was closed with a 40-acre cap consisting
of two 3-ft layers of clay separated and overlain with geosynthetic membranes.
The total cost of the investigation and cleanup was $325 million. Ironically, the
cost for constructing a secure chemical landfill in 1953 would have been about
$4 million.
CERCLA
As the Love Canal incident illustrates, the cleanup costs of contaminated
sites can be enormous. In the past, companies responsible for contaminations
have gone out of business, leaving government to bear the cost of cleanup. The
Comprehensive Emergency Response, Compensation, and Liability Act (CERCLA) was passed by Congress in 1980 (with major amendments in 1986) to
help address that shortcoming. CERCLA deals with the cleanup of past hazardous waste sites that are contaminating the environment, and defines the liability of parties responsible for the contamination. RCRA had no provisions to
deal with abandoned hazardous waste sites, and EPA had no authority to
respond to spills, leaks, and explosions until CERCLA was passed. CERCLA is
often called Superfund because it also provided an ongoing “special tax” on
chemicals that could, if spilled or improperly disposed, cause contamination of
soil or groundwater.
CERCLA granted EPA the authority to take any necessary short-term and
emergency steps to address hazards to human health and the environment
caused by burning, leakage, or explosion of hazardous substances; imminent
contamination of food chains; or pollution of a drinking-water source (Wentz
1989). Under CERCLA, EPA also can undertake long-term actions (greater than
6 months) at a complex hazardous waste site.
To be eligible for Superfund dollars for site cleanup under CERCLA legislation, a contaminated site must first appear on the National Priority List
(NPL). Sites are placed on the NPL following an evaluation that assesses the
relative risk to the public and the environment from hazardous substances in
groundwater, surface water, air, and soil. There are potentially tens of thousands of Superfund sites, but as of May 2013, there were 1,320 sites on the NPL
(US EPA 2013e); 348 sites have been cleaned up and removed from the list since
the passage of CERCLA.
The approach to remediation is to define the nature and extent of the contamination, identify potential remedial alternatives, and choose the best remedial design alternative. There are four steps outlined in the National
Contingency Plan that must be completed for any Superfund remedial action:
• site evaluation and scoping of response actions
• initial evaluation and screening of technologies
• detailed evaluation of technologies
• selection of remedy
Following the selection of a remedial technology, the remedial action begins.
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CERCLA is also unique in its definition of liable parties. CERCLA imposes
liability on all potentially responsible parties (PRPs) for their involvement in
the improper disposal of hazardous substances. PRPs may be past owners or
operators of the site, generators of the hazardous substances that have polluted
the site, and/or transporters who brought the hazardous wastes to the site.
PRPs are retroactively liable for hazardous waste problems under CERCLA. In
other words, even if waste was disposed in accordance with regulatory
requirements in effect at the time of the disposal, and the contamination shows
up years later, the PRPs still can be held liable for the cleanup of the waste.
CERCLA provides only three limited defenses to liability. These statutory
defenses include an act of God, an act of war, or acts of omissions of contractually unrelated third parties where the defendant(s) exercised due care and took
appropriate actions in response to the threatened or actual release. US courts
have consistently upheld these provisions as the only defenses available to
CERCLA defendants (Freilich 1992). In addition, the nation’s courts have recognized that PRPs are “jointly and severally liable” for cleanup of polluted
sites, which often results in those with the “deepest pockets” paying the most.
Remediation Technologies
Superfund sites are small in number compared with the total number of
contaminated sites. Called brownfields, these contaminated and often abandoned sites do not present the high level of risk that NPL sites do, but still are a
serious problem. Despite years of remediation and clean up, as of 2012 there
were an estimated 450,000 brownfields in America (US EPA 2012).
There are several techniques that have been used to remediate both contaminated soils and groundwater. They are roughly classified as ex situ or in
situ methods. Among the ex situ techniques are excavation and incineration of
contaminated soil with on-site mobile soil-burners, soil vapor extraction and
treatment, groundwater pumping and air stripping, activated carbon adsorption (both for water and for air), solidification/stabilization (the addition of
chemicals to immobilize the hazardous contaminants either in place or in a
land disposal site), biological treatment of excavated sludges and sediments,
soil washing, thermal desorption of contaminants from soils and sediments,
and solvent extraction of contaminants from soils and sediments. In situ techniques include biological treatment of soils and groundwater by injecting special microbial cultures, essential nutrients, and oxygen to the subsurface.
Treatment of soils and groundwater is also possible in situ through the introduction of heat, steam, chemicals, microwaves, or bacteria to the subsurface.
Soil vapor extraction (SVE) and air stripping (AS) of contaminated water,
with or without carbon adsorption, are popular processes that are often used to
remove volatile hydrocarbons from subsurface soil or groundwater, respectively (see Figure 8.12). Both SVE and AS are minimally invasive and are frequently used to clean up underground areas near old gas stations or fuel
terminals that were contaminated with fuel spills or by leaking underground
storage tanks.
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Chapter Eight
SVE utilizes a vacuum pump to create a vacuum in a well pipe that has
been inserted into a contaminated region of soil. Air is drawn through the surrounding soil into the pipe, and from there it flows to the surface. Many components of gasoline are quite volatile (gasoline is actually composed of
hundreds of compounds), and the compounds vaporize and are transported
via the moving air. Some of the gasoline components adsorb strongly onto the
soil particles, so not all of the compounds will be removed. If the region of contamination is fairly shallow, air from the atmosphere can diffuse downward to
replace the air and vapors, and allow the flow to continue. If the region is deep,
other pipes might be inserted, and air can be forced down those pipes and into
the soil to “complete the circuit.”
Air stripping is a method of remediating contaminated groundwater. As
briefly discussed in Chapter 5, the contaminated groundwater is pumped to the
surface and sprayed into the top of a stripping tower. The water flows downward
inside the tower due to gravity. At the same time, atmospheric air is blown into
the bottom of the tower and flows upward due to pressure differential. The tower
is filled with plastic packing (lightweight, oddly shaped pieces that provide a lot
of surface area for air-water contact). As the air and water pass each other in the
tower, the contaminant is stripped from the water and transferred into the air in
response to Henry’s Law. Recall from previous chapters that several common
gases have Henry’s Law constants. But a number of organic liquids have Henry’s
Law constants as well. These are shown in Appendix B. The clean water coming
out the bottom of the tower can be discharged to the surface or pumped back
underground, and the contaminated air from the top of the tower is sent to a
treatment unit (see Figure 8.12). The “treatment unit” seen in Figure 8.12 often is
a carbon adsorption unit (see Chapter 7 for a description of carbon adsorption),
where the VOC is removed from the air and stored on the carbon. Later the spent
carbon is transported to a site where the VOC can be recovered or destroyed.
Sometimes, if the VOC contaminant concentration is fairly low, and the contaminant is not toxic, the VOC-in-air stream might simply be discharged into the
atmosphere to allow atmospheric dispersion to “dispose” of the contaminant.
Other times the contamination is so extensive that a non-aqueous phase
liquid (NAPL) forms. This is a layer of liquid organics that does not fully mix
with the groundwater. If the compounds are denser than water, they form a
dense NAPL (or DNAPL) that may exist near the bottom of the groundwater
layer, under which lies clay or rock. In cases where an NAPL or DNAPL exists,
it must be removed first. Otherwise, as the groundwater is pumped and
treated, more organics dissolve into the water, and very little progress is made.
It is much more efficient to pump out the NAPL first (if it can be located).
A more passive remediation approach is intrinsic attenuation, where natural phenomena such as dilution, biodegradation, sorption, and advection are
permitted to reduce contaminant concentration to levels that do not pose significant risk to human or ecological health. This is of course a much slower process, but is not invasive. However, periodic monitoring should be conducted to
ensure that adequate progress is being made. Contaminated site remediation is
presently an exciting and dynamic area of environmental engineering.
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Solid and Hazardous Waste Management
333
Figure 8.12 Schematic diagram of soil vapor extraction and groundwater air-stripping
methods of remediating contaminated underground sites.
Clean air
Air
bleed Vacuum
line
pump
VOC
Treatment
Unit
Pressure
gauge
Air + VOC
Air
stripper
Air
Air + VOC
Clean water
Vapor
well
Groundwater
well
Vapor
flow
Contaminated
soil
Vapor
flow
Groundwater table
Contaminated
groundwater
Water
flow
Well
pump
Water
flow
EXAMPLE 8.9
Approximately 500 gallons of liquid gasoline have contaminated the soil
underneath an existing gasoline station. The SVE system air flow is designed
at 100 cfm. Calculate the amount of time required to remove 80% of the gasoline if the average concentration of gasoline vapors in the extracted air is 450
mg/m3. The density of liquid gasoline is 6.6 lb/gal.
SOLUTION
1. Calculate the mass of gasoline in the spill.
500 gal × 6.6 lb/gal = 3,300 lb
= 1.50 × 106 grams
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Chapter Eight
2. Calculate the mass rate of removal.
100
mg
g
ft 3
m3
¥ 0.0283 3 ¥ 450 3 ¥ 0.001
= 1.274 g/min
min
mg
ft
m
3. Calculate the time required to remove 80% of the 1.5 × 106 grams.
0.8 ¥ 1.5 ¥ 106 grams
= 942, 000 min (654 days)
1.274 g/min
PROBLEMS
8.1
A city has a stable population of 250,000 and is running out of room at its
MSW landfill. Estimate the land area (in acres) that must be set aside for a
new landfill given the following conditions: MSW generation rate for this
city is 5.0 ppcd, residents maintain a 20% recycling rate, the density of
compacted MSW in the landfill is 800 lb/yd3, the average depth of waste
in the landfill is 50 ft. Assume that 50% of the total land area will be used
for roads, buildings, buffers, etc. Design for a landfill life of 20 years.
8.2
A city with a population of 40,000 is growing at 5% per year. For a net
generation rate of 2.9 ppcd, a compacted density of 900 lb/yd3, an average depth of 80 ft, and allowing for 40% of the area to be used for buildings, roads, etc., calculate the land area (in acres) that must be set aside for
a new landfill for this city. Assume it must last for 20 years, and that population growth continues at the same rate for the full 20 years. Also
assume that the net generation rate remains constant for the full 20 years.
8.3
The existing landfill for a city will be filled in three more years. It receives
MSW at the steady rate of 73,000 tons per year. The present landfill site is
130 acres (80 acres for wastes and 50 acres of land for roads, buildings,
buffer, etc.). The present landfill design calls for stopping at an average
height of 150 feet. How many more years will the landfill last if the final
average height could be raised to 200 feet? Assume that compacted waste
has an average density in the landfill of 1,000 lb per cubic yard.
8.4
A landfill for 90,000 people has been designed to last for 15 years based
on the 2011 US average MSW generation and recycling rates. If the population remains stable, and the generation rate remains constant, but recycling can be increased by 20%, how much longer would the landfill last?
8.5
Estimate the potential annual electricity generation (in units of kwh per
year) from an MSW incinerator with turbine/generator that burns
182,500 tons/year of MSW. Assume that the MSW has a heat content of
2,500 Btu/lb and that overall thermal efficiency of 30% can be achieved. If
an average household uses 1,500 kwh per month, how many households
can be powered by this MSW incinerator?
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8.6
Use the Internet to identify a Superfund site in your state. Summarize
your findings: Is it still on the National Priority List? What were the costs?
What impacts were there on the surrounding area? What technologies
were used? How long did it take to complete the cleanup?
8.7
Find the critical distance from a city to a landfill. The critical distance is
that distance where it becomes economical to buy and use long-haul
trucks to take the MSW to the landfill (in this alternative, the city also must
build and operate a transfer station in the city) rather than using the city’s
MSW collection trucks to drive the extra distance to the remote landfill.
Use only the following data to make your estimate. Collection trucks hold
25 yd3 of MSW and get 4 miles/gallon (mpg) of fuel. Long-haul trucks
hold 100 yd3 of waste and get 10 mpg. Fuel costs $4.25/gal. The city presently operates 30 MSW collection trucks that each drive their route once
per day. To build and operate a transfer station will cost the city $500,000
per year, and the annualized cost of owning and operating long-haul
trucks is $75,000/yr per truck plus fuel costs. The city already owns the
collection trucks, but assume that the extra labor, maintenance, and depreciation costs associated with driving them the extra distance to the remote
landfill will add another $15,000 per year per truck to that alternative.
Assume that MSW operations occur 6 days per week, 52 weeks per year.
8.8
Discuss the pros and cons of waste recycling and reuse considering the
environmental and economic implications.
8.9
Compare the efficiency of a fleet of 20-yd3 vs. 30-yd3 collection vehicles to
pick up a community’s waste by calculating the total number of hours
required to collect the waste per week for each type truck, and the number of trucks of each type required. Use the following data: 25,000 collection locations, 2.4 minutes/location to pick up (includes drive time
between locations), compaction ratio is 2.0 for the 30-yd3 truck and 2.5 for
the 20-yd3 truck, collections are once per week, 0.24 yd3/location, a 20minute drive time (one-way) to or from landfill, 12 minutes spent at the
landfill, and a 40-hour work week. What size trucks are preferred?
8.10 If you were required to start up a hazardous waste management program in
a country that had no regulations, would you follow the approach to define
hazardous wastes used in the United States (i.e., list the waste and characteristics) or would you develop a different approach? Why or why not?
8.11 Determine the recycling efficiency of a community with the waste composition shown in Figure 8.1, and that recycles 10% of its plastics, 50% of its
metals, and 30% of its paper and paperboard products.
8.12 How would you convince your community to accept an MSW landfill,
hazardous waste treatment facility, or other “undesirable” land use?
8.13 For a US city with a stable population of 100,000, generating solid waste
at a “typical average US” rate, calculate the mass (lb) and volume of
waste (cubic feet) generated during one year. The typical density of compacted waste is 30 lb/ft3.
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Chapter Eight
8.14 A community with 100 homes is considering switching from its present
hauled-container system to a weekly pickup at each home by a compacting vehicle. Compare the total labor requirements for the two systems
using the following data: five 9-yd3 containers clustered together are
presently filled to 70% of their capacity on average. The haul truck spends
12 minutes at the landfill, which is 20 minutes from the community. It
takes 4 minutes to load each container. It takes 30 minutes to get to and
from the community from the dispatch site. The compactor vehicle holds
27 yd3, has a compaction ratio of 2.0, and it takes 1.0 min at each house.
The compactor truck spends 6.0 minutes per trip at the landfill. The compactor truck has a crew of 2; the haul truck only has a driver.
8.15 A chrome-plating shop must treat 50 L/min of a wastewater containing
600 mg/L of chromic acid. Calculate the amounts of SO2 and NaOH
required. Calculate the amount of sludge produced (dry basis). If the
sludge (as produced) is 40% solids, how much sludge must be disposed
of? Give all your answers in kg/day.
8.16 Rework Problem 8.15 but use Ca(OH)2 instead of NaOH to do the precipitation.
8.17 Over the years, 1,500 gallons of dry-cleaner fluid (tetrachloroethylene—
C2Cl4) have contaminated the soil underneath an old laundry and drycleaner shop. A SVE system will be used to remove the vapors. The air
flow is designed for 140 cfm. Calculate the amount of time (years)
required to remove 85% of the C2Cl4 if the average concentration of C2Cl4
vapors in the extracted air is 650 mg/m3. The density of liquid C2Cl4 is
1.62 g/cm3.
8.18 Suppose you must landfill an average of 80,000 tons of municipal solid
waste (MSW) each year for 20 years. You can build your landfill 50 feet
high. How many acres of land should you set aside for this landfill to last
20 years? Plan on keeping about 25% of the total land area free of waste
for roads, buildings, etc. Assume that compacted solid waste has a density of 900 lb/yd3.
8.19 A groundwater has been contaminated with trichloroethylene at a level
of 150 ppm (by mass). Assuming equilibrium with the air in the soil right
above the groundwater table, calculate the concentration of trichloroethylene in the air space. Give your answer in ppm (by volume). The
Henry’s Law constant for trichloroethylene is 0.9 kPa/(gmol/m3) and the
total air pressure in the soil is 1,013 kPa.
8.20 A groundwater has been contaminated with carbon tetrachloride at a
level of 250 ppm (by mass). Assuming equilibrium with the air in the soil
right above the groundwater table, calculate the concentration of carbon
tetrachloride in the air space. Give your answer in ppm (by volume). The
Henry’s Law constant for carbon tetrachloride is 3.0 kPa/(gmol/m3) and
the total air pressure in the soil is 1,013 kPa.
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Solid and Hazardous Waste Management
337
8.21 How much bleach (NaOCl) is needed to safely destroy the cyanide in a
5,000 L tank of cyanide solution containing 2,000 mg/L of CN– ions? Give
your answer in kg.
8.22 Rework Problem 8.21 but assume that you use Ca(OCl)2 to destroy the
cyanide.
8.23 Assume you must incinerate 2,000 kg/day of a waste that contains 15%
by weight of chlorobenzene (C6H5Cl). During the incineration process,
the chlorine atoms form HCl gas. Calculate the mass of HCl formed, kg/
day. If the incinerator exhausts a total of 40,000 kg/day of exhaust gases,
and the gases have an average molecular weight of 30, calculate the concentration of HCl in the exhaust gases, ppm (by volume).
8.24 Consider Example 8.5. If MSW comes in wet, much of its heat is “wasted”
in evaporating the water. Also, if most of the high energy-content wastes
(e.g., paper and plastics) are removed for recycling, the MSW has a much
lower heat content. Keeping all other variables the same as in Example
8.5, but changing the moisture content of the waste to 40%, and the HHV
to 2,000 Btu/lb, calculate how much heat (if any) will be left for steam to
the turbine/generator.
8.25 Chromium can be removed from metal-plating wastewater by first reducing Cr+6 to Cr+3, then precipitating out Cr(OH)3. One such method uses ferrous sulfate (FeSO4) as the reducing agent. The overall reaction is as follows:
Cr6+ + 3 FeSO4 + 12 NaOH →
Cr(OH)3↓ + 3 Fe(OH)3↓ + 3 SO42– + 12 Na+
a. Calculate the addition rate of FeSO4 (kg/day) required to remove 99%
of the chromium from a 100 L/min stream containing 200 mg/L of Cr+6.
b. Calculate the total amount of sludge produced [Cr(OH)3 and Fe(OH)3]
in kg/day (assume a 30% solids content in the final disposed sludge
material).
REFERENCES
Code of Federal Regulations, 40 CFR 261. “Title 40—Protection of the Environment, Part
261, Identification and Listing of Hazardous Waste.” Washington, DC: GPO.
Council on Environmental Quality. 1978. Environmental Quality—The Ninth Annual
Report of the Council on Environmental Quality. Washington, DC: CEQ.
Curran, Caroline. 2013. “New Hanover County to Solicit Proposals for Hauling Trash.”
Accessed April 2014. http://portcitydaily.com/2013/10/21/new-hanover-countyto-solicit-proposals-for-hauling-trash/
EHSO (Environmental Health & Safety Online). 2013. “Commercial Hazardous Waste
Landfills.” Accessed August 2013. http://www.ehso.com/cssepa/tsdflandfills.php
ENN (Environmental News Network). 2004. “EPA Projects Hazardous Waste Sites
Growing in Number and Cleanup Costs.” Accessed August 2013.
http://www.enn.com/top_stories/article/520
Fortuna, R. C., and D. J. Lennett. 1987. Hazardous Waste Regulation: The New Era. New
York: McGraw-Hill.
Cooper.book Page 338 Monday, June 23, 2014 9:58 AM
338
Chapter Eight
Freilich, Irvin M. 1992. “Causation Becomes a Factor under CERCLA.” Journal of the Air
& Waste Management Association, 42(10): 1274–1275, 1392.
Sampson County. 2014. “Sampson County Landfill.” Accessed April 2014.
http://www.sampsonnc.com/landfillwaste.asp
Seminole County. 2013. “Environmental Services—Seminole County Landfill.” Seminole County, FL. Accessed August 2013. http://www.seminolecountyfl.gov/
envsrvs/solidwaste/landfill.aspx
US EPA (Environmental Protection Agency). 1979. “The Love Canal Tragedy.” Accessed
August 2013. http://www.epa.gov/history/topics/lovecanal/01.html
US EPA. 1987. “The Hazardous Waste System.” Washington, DC: Office of Solid Waste
and Emergency Response.
US EPA. 2006. “Standards of Performance for New Stationary Sources and Emissions
Guidelines for Existing Sources: Large Municipal Waste Combustors; Final Rule.”
Federal Registrar 71(90): 27324–27327.
US EPA. 2012. “EPA Announces $69.3 Million to Clean Up Contaminated Sites and
Revitalize Communities.” EPA Press Release, May 25. Accessed June 2014.
yosemite.epa.gov/opa/admpress.nsf
US EPA. 2013a. “Municipal Solid Waste.” Accessed July 2013. http://www.epa.gov/
epawaste/nonhaz/municipal/index.htm
US EPA. 2013b. “Municipal Solid Waste Generation, Recycling, and Disposal in the
United States: Facts and Figures for 2011.” EPA 530-F-001. Washington, DC: EPA.
US EPA. 2013c. “Standard Volume-to-Weight Conversion Factors.” Accessed August
2013. http://www.epa.gov/osw/conserve/tools/recmeas/docs/guide_b.pdf
US EPA. 2013d. “The National Biennial RCRA Hazardous Waste Report (Based on 2011
Data).” Accessed August 2013. www.epa.gov/wastes/inforesources/data/br11/
national11.pdf
US EPA. 2013e. “National Priorities List (NPL).” Accessed August 2013.
http://www.epa.gov/superfund/sites/npl/
Washington State Department of Ecology. 1987. Solid Waste Landfill Design Manual. Publication 87-13. Olympia, WA: WSDE.
Waste Industries. 2010. “Proposal to Design, Finance, and Permit the Development and
Operation of a Long-Term Solid Waste Management Infrastructure System.” Presented to the city of Greensboro, NC, March 1.
Wentz, Charles A. 1989. Hazardous Waste Management. New York: McGraw-Hill.
Cooper.book Page 339 Monday, June 23, 2014 9:58 AM
CHAPTER
9
Other Important Topics
There are many important topics that have not been covered in the first
eight chapters. In an introductory book such as this, there always is a struggle
between the desire to include more information and the need to keep the size
of the book under control. In this book we wanted to present the fundamentals
of environmental engineering in a complete, clear, and concise way, but also to
include as much (as practical) of the material important to your education as an
environmental engineer. The additional topics in this chapter are important,
and each could easily be expanded into a full chapter in an introductory text, or
even into a separate text for a more advanced course! But the judgment was
made that these topics would be covered only briefly in order to make this
book manageable for a one-semester course.
9.1 Risk Assessment
Risk assessment is a quantifiable method for assisting decision makers
when dealing with situations or events that pose a danger to public health or
the environment. With regard to specific pollution events (either a large onetime release or a small but continuous release of low-level contamination), risk
assessment is an attempt to scientifically assess the probability of harm due to
that event. Most actions (or inactions) involve some degree of risk. A logical,
repeatable risk assessment process assists scientists, managers, and regulators
in quantifying the potential harm to human health and the environment when
trying to decide on a proposed project, or when weighing alternative plans.
The process can be used to assess accidental or routine workplace chemical
exposure, to evaluate chemical tolerance levels in food, to identify the need for
and extent of remediation at sites contaminated with hazardous wastes, to
compare various treatment process technologies, and to conduct comparative
risk studies.
Risk can be defined as the probability that a specific adverse outcome will
occur. Safety is the complement of risk, or the probability that the adverse
effect will NOT occur. Risk is expressed numerically, and is often quoted as a
339
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340
Chapter Nine
“chance” or “odds” that something will happen. For example, the US EPA
sometimes sets pollution limits in drinking water based on a 1-in-a-million
chance that ingestion of water containing that pollutant at that limit will result
in a person developing cancer.
Risk is often reported as a decimal fraction. For example, the Centers for
Disease Control (CDC) reports that more than 34,000 motorcyclists were killed
between 2001 and 2008, and in 2008, the risk was 1.74 deaths per 100,000 persons (CDC 2014a). But if we expand the time frame to a lifetime of motorcycle
riding, the lifetime risk of dying in a motorcycle crash is about 0.00108 or 108
per 100,000 (National Safety Council 2014)! The top ten causes of death in the
United States are listed in Table 9.1; the risk of dying from each cause is calculated in each case simply by dividing the number of deaths from that one cause
by the total number of deaths from all causes.
Risk to human health comes from accidents, natural disasters, disease, voluntary personal activities, involuntary chemical exposure, and others. Generally, risks taken during voluntary personal activities (such as driving under the
influence, smoking, overeating, snow skiing, parachuting, etc.) are freely
accepted and generally perceived by the individual to be much lower than they
really are. On the other hand, involuntary risks (risks that are imposed on people) are not willingly accepted and are perceived to be higher than they really
are. Such risks include living in a heavily industrialized area, living near a
large landfill or waste incinerator or nuclear power plant; eating contaminated
food; drinking contaminated water; or exposure to radon gas.
People are willing to tolerate relatively high levels of risk from voluntary
activities, but are not willing to accept substantially lower risks from involuntary activities. This sometimes results in seemingly outlandish public demands
for stringent environmental controls. Table 9.2 provides actual risk levels for
several activities and shows that actual risk levels for certain involuntary expo-
Table 9.1 Ten Most Common Causes of Death in the United States (2010 data)
Cause of death
Heart Disease
Cancer
COPD (pulmonary disease)
Stroke
Accidents
Alzheimer's
Diabetes
Nephritis (kidney disease)
Pneumonia and Influenza
Suicide
All other
Total
Number of deaths
Risk (fraction)
597,689
574,743
138,080
129,476
120,859
83,494
69,071
50,476
50,097
38,364
616,086
2,468,435
0.242
0.233
0.056
0.052
0.049
0.034
0.028
0.020
0.020
0.016
0.250
1.000
Source: Centers for Disease Control, www.cdc.gov/nchs/fastats/lcod.htm
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Table 9.2
341
Annual Risk of Death from Specific Activities
Activity
Smoking 10 cigarettes/day
Motor vehicle accidents
Drinking two beers/day (cirrhosis only)
Snow skiing
Skydiving
Eating peanut butter (aflatoxin)
Living near a nuclear power plant (cancer)
Drinking water with EPA limit of trichloroethane
Annual Risk of death
1.25 × 10–3
1.3 × 10–4
4 × 10–5
1.7 × 10–5
1.3 × 10–5
8 × 10–6
1 × 10–6
2 × 10–9
Sources: Rowe (1977); Wilson and Crouch (1987); and others.
sures are much lower than for various voluntary exposures. The involuntary
exposures are typically perceived as being objectionably high by most people.
Consider that we—as a nation—need a number of large commercial hazardous waste treatment and disposal facilities so we can safely handle the hazardous wastes that our society generates. Now consider this question: how
many people reading this sentence would like to have such a facility built in
their neighborhood or city? Such a development project carries benefits for a
large number of people widely scattered throughout the country, but imposes
involuntary risks on a relatively few people in one community. Some future
engineer or project manager in this class may one day be faced with the task of
trying to “sell” this kind of a project to a local community and its government
leaders. This future project manager or engineer should be warned, though. He
or she may know that the risks associated with the project are low compared
with other risks that people readily accept, but if he or she cites statistics like
the ones in Table 9.2 in the hopes of convincing people to accept the new project, it may well backfire and cause people to resist even more! When proposing
new projects that carry involuntary risk, it is better to acknowledge the risks
and communicate the benefits of the proposed project, rather than try to hide
the risks or compare the risks of the involuntary action to the readily-accepted
risks from everyday activities.
Since risk is a probability, we cannot apply risk numbers to any one individual. Some people may smoke cigarettes for 50 years and never develop lung
cancer; other people may die from lung cancer without having ever smoked.
However, if a quantified risk is applied to a large enough subset of the population, then meaningful insights can result as suggested in Example 9.1.
EXAMPLE 9.1
In the year 2012, there were 1,500 deaths in a city with a population of
120,000; 950 of them were from cancer. Does this number seem suspicious in
any way?
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Chapter Nine
SOLUTION
Based on the US statistics presented in Table 9.1, the expected number of cancer deaths in this town would be about 23% of the total number of deaths, or
about 350 (0.233 × 1,500). To have almost triple that number of cancer deaths
certainly raises suspicions that there may be some unusual or hidden exposure to carcinogenic chemicals or radiation from this city’s air, water supply,
or soil.
Risk Assessment Process
Interest in risk assessment began in the 1960s when the process was used to
weigh the benefits and negative side effects of certain pharmaceutical drugs.
The risk assessment process was first applied to hazardous waste site evaluations in the 1980s. Since that time, the process has evolved into a sophisticated
tool in hazardous waste management. Although the process has many limitations and uncertainties, its use has become widely accepted in setting environmental standards, as well as goals for cleaning up contaminated sites. This
latter use has important ramifications for costs—that is, how clean should a
contaminated site be made before it is acceptable?
The risk assessment process was recommended by the National Academy
of Science in the mid-1980s, and its use was soon adopted by the Environmental Protection Agency (US EPA 1989). Risk is a product of two elements: toxicity and exposure. Reduce either component and risk decreases as well. The risk
assessment process as practiced by the EPA is shown in Figure 9.1. The process
involves four steps:
1. Data collection and evaluation to identify potential hazards at the site
2. Toxicity (dose-response) assessment
3. Exposure assessment
4. Risk characterization
Steps (2) and (3) are shown in parallel in Figure 9.1, and can in fact be done
in parallel by different members of the same team. A fifth step—risk management—follows after the fourth, but is not truly part of the assessment process—it is part of the solution.
The data collection/evaluation step is sometimes called hazard identification. It is site-specific and must be undertaken for each different place where a
risk assessment is needed. In this step the team strives to determine which
chemical, physical, or biological agents are present; what their concentrations
are in the air, water, soil, and food in the site being evaluated; where the
sources of these contaminants are located; and how the substances are being
transported from each source and dispersed into the community. Such an
investigation requires sampling and testing with good quality control. Quality
control procedures are needed for the laboratory testing (including routine calibrations of instruments; the use of spikes, duplicates, and blanks; and the
proper safeguarding of the data). Quality control is also needed in the field
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Other Important Topics
Figure 9.1
Four-step risk
assessment process. (Adapted
from EPA 1989.)
343
Data Collection
and Evaluation
Site Data
Toxicity
Assessment
Exposure
Assessment
Qualitative
Pollutant releases
Quantitative
Exposed population
Risk
Characterization
Cancer Risk
sampling effort (proper sampling protocols, use of clean sample containers,
chain of custody records, etc.). The identification of a hazard does not create
risk in and of itself, but the presence of hazardous chemicals is a necessary condition for risk.
The toxicity (dose-response) assessment evaluates the relationship
between the level of exposure and the extent of injury. Response in this context
means the damaging effects of the identified chemicals on living organisms.
These may include acute (short-term) effects such as a skin rash, vomiting, or
breathing difficulties, and chronic (long-term) effects such as cancer, genetic
mutations, or birth defects. Chronic effects are generally classified as either carcinogenic or noncarcinogenic. Responses are typically evaluated by exposing
animals (such as minnows or mice) to various doses of agents; however, more
recent procedures have utilized bacteria, specific organ cells, and animal eggs
or embryos.
Dose is generally expressed as the mass intake of the chemical normalized
to the body weight of the exposed individual; typical units are mg/kg (mg of
chemical per kg of body weight). In many cases, individuals are exposed to
hazardous chemicals in small does over time, so often the dose is expressed as
an average daily dose, and has units of mg/kg-day. The dose is related to
exposure, but is not equal to it. The total amount of chemical absorbed into the
body is often less than the amount ingested, breathed, or contacted on the skin
because sometimes it is quickly removed (exhaled in breath, excreted in urine
or feces, or washed or brushed off the skin).
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Chapter Nine
In order to tell if a particular dose creates appreciable risk, some relationship between dose and response is required. The dose-response relationship is
often published as a plot of response vs. dose, based on a statistical analysis of
data from many studies. These curves attempt to explain quantitatively how, as
the dose increases, the mortality increases or the incidence of cancer increases.
One of the major problems is that because these data often are obtained from
animal studies, the measured effects occur at laboratory doses that may be
orders of magnitude larger than the environmental concentrations to which
humans are exposed. Significant extrapolations are necessary to get down to
the low environmental doses that
are observed in most situations,
and there is considerable uncerFigure 9.2 Carcinogenic dose-response curve.
tainty about real effects at such
low doses.
Slope = Cancer slope factor
Typical dose-response curves
for carcinogenic and noncarcinogenic agents are presented in FigX
X
ures 9.2 and 9.3, respectively. Note
that for carcinogenic agents, it is
X
assumed that a threshold does not
Response
X
exist; that is, a cancer risk is present at all doses. The slope of the
X
line is called the cancer slope factor (CSF) and is used in the calcuNo threshold,
lation of risk as explained below.
linear at low doses
For noncarcinogenic agents, it is
assumed that if the dose is below
Dose
the no observable adverse effect
level (NOAEL), then it will not
cause harmful effects.
Figure 9.3 Noncarcinogenic dose-response curve.
The CSF is developed by assuming a linear relationship between dose and response at low
dosages and is an upper boundary
estimate of risk. For noncarcinoX
X
gens, a reference dose (RfD) or a
reference concentration (RfC) is deX
termined from the noncarcinogenic
Response
X
dose-response curve using the NOAEL. The RfD can be thought of as
an estimate of a daily oral exposure
X
to the human population (including sensitive subgroups) that is
RfD NOAEL
likely to be without an appreciable
Threshold
risk of deleterious effects during a
X
lifetime. The RfC has a similar defiDose
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345
nition but for a continuous inhalation exposure. To calculate the RfD or RfC to
protect human health, the NOAEL often is divided by safety factors (typically
multiples of ten) to acknowledge the uncertainty associated with the extrapolation of response data from animals, the variability in animal and human response, and extrapolation from relatively high doses to low environmental
doses, among other factors. Typical CSFs, RfDs, and RfCs for many chemicals
are maintained in various databases for use during risk analysis. Table 9.3 gives
some examples of CSFs for various compounds and routes of exposure.
The exposure assessment identifies possible exposure pathways to the
affected individuals and calculates the actual or potential dose that an exposed
individual receives. The process measures or estimates the intensity, frequency,
and duration of human or ecological exposure to the agent or agents. The exposure assessment answers questions regarding the release routes from the
source to the site; attenuation along the transport pathways; concentrations in
the air, drinking water, food, soil, and dust; body burdens in the population;
and work, play, and consumption habits of the population. Exposure routes
Table 9.3
Toxicity Data for Selected Potential Carcinogens
Chemical
Arsenic
Benzene
Benzo(a)pyrene
Cadmium
Carbon tetrachloride
Chloroform
Chromium VI
DDT
1,1-Dichloroethylene
Dieldrin
Formaldehyde
Heptachlor
Hexachloroethane
Methylene chloride
Nickel and compounds
Polychlorinated biphenyls (PCBs)
2,3,7,8-TCDD (dioxin)
Tetrachloroethylene
1,1,1-Trichloroethane (1,1,1-TC)
Trichloroethylene (TCE)
2,4,6-Trichlorophenol
Toxaphene
Vinyl chloride
Source: US EPA IRIS database (1989).
Cancer Slope Factor
oral route
(mg/kg-day)–1
Cancer Slope Factor
inhalation route
(mg/kg-day)–1
1.75
2.9 × 10–2
11.5
—
0.13
6.1 × 10–3
—
0.34
0.58
30
—
3.4
1.4 × 10–2
7.5 × 10–3
—
7.7
1.56 × 10–5
5.1 × 10–2
—
1.1 × 10–2
1.1 × 10–2
1.1
2.3
50
2.9 × 10–2
6.11
6.1
—
8.1 × 10–2
41
—
1.16
—
4.5 × 10–2
—
—
1.4 × 10–2
1.19
—
—
1.0–3.3 × 10–3
—
1.3 × 10–2
1.1 × 10–2
1.1
0.295
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Chapter Nine
include ingestion, inhalation, and dermal absorption. Computer models are
available to predict the fate of chemical or biological agents in the environment
and determine exposures through all routes. To confirm the model results and
quantify the rate of uptake by an individual, field measurements are made
where possible. If data are unavailable, typical assumed exposure and uptake
rates are frequently used (see Table 9.4).
Table 9.4
Typical Exposure and Intake Rates for Risk Assessment
Variable
Value
Soil Ingestion Rate
200 mg/day (children 1 through 6 years old)
100 mg/day (age groups greater than 6 years old)
Exposure Duration
years at one residence (national 90th percentile):
30 years (adult)
6 years (child)
Body Weight
70 kg (adult); 10 kg (child)
Averaging Time
for noncarcinogens, exposure duration
for carcinogens, 70 years
Water Ingestion Rate
2 L/d (adult); 1 L/d (child, age 1–6)
Inhalation Rate
0.83 m3/hr (adult); 0.46 m3/hr (child)
From exposure data, the lifetime average daily dose (LADD) is calculated
as shown in Eq. (9.1). The LADD represents the average daily dose received
over a person’s lifetime, and typically is calculated assuming an average lifetime of 70 years.
LADD =
(C )( I )(EF )(ED)( AF )
( AT )(BW )
where:
LADD = lifetime average daily dose, mg/kg-day
C
= concentration, mg/L
I
= intake rate, L/day
EF
= exposure frequency, days/year
ED
= exposure duration, years
AF
= absorption factor, dimensionless
AT
= averaging time, days
BW
= body weight, kg
The use of this equation is demonstrated in Example 9.2.
(9.1)
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347
EXAMPLE 9.2
Calculate the LADD that an adult male receives from drinking his well water
that contains 10 µg/L of chloroform. Assume the individual is exposed for 30
years, drinks 2 L/day of water, 7 days/wk, 49 weeks per year (every year he
travels on vacation for 3 weeks), and weighs 70 kg. Assume 100% of the chloroform is absorbed.
SOLUTION:
LADD =
(10 µg/L )(2 L/day )(7 days/wk )( 49 wks/yr )(30 yrs) (1)
(1, 000 µg/mg )(70 kg ) (365 days/yr ¥ 70 yrs)
LADD = 1.15 ¥ 10 -4 mg/kg-day
Example 9.2 showed only the calculation of the LADD for the route of drinking
contaminated water. There may well be other routes whereby the person could
be exposed to the same chemical. All routes must be considered when doing
the risk assessment; the LADDs from all routes are additive.
The risk characterization step estimates the probability of adverse effects
occurring under the conditions identified during the exposure assessment. For
carcinogens, the LADD (developed during the exposure assessment) is multiplied by the CSF developed from dose-response curves to calculate the added
risk of cancer. Risks are additive for multiple carcinogenic contaminants. For
noncarcinogens, a hazard index (HI) is calculated by dividing the daily dose by
the RfD (developed from dose-response assessments). Where more than one
contaminant is present a hazard quotient (HQ) is determined by summing all
of the HIs. A value of HQ greater than one indicates an unacceptable risk. The
equations are:
Risk (cancer) = LADD × CSF
(9.2)
HI (noncancer) = LADD/RfD
(9.3)
and
where:
Risk = probability of occurrence, dimensionless
CSF = cancer slope factor, (mg/kg-day)–1
HI
= risk of noncancer effect, dimensionless
RfD = Reference dose, mg/kg-day
EXAMPLE 9.3
Calculate the added carcinogenic risk posed by the chloroform consumption
of Example 9.2.
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348
Chapter Nine
SOLUTION
From Table 9.3, the CSF for chloroform is 0.0061 (mg/kg-day)–1
Risk = Dose × Toxicity = LADD × CSF
Risk = (1.15 × 10–4 mg/kg-day) (0.0061 (mg/kg-day)–1)
Risk = 7.0 × 10–7
Risk = 0.7 (rounded up to 1) additional cancer per million people
In Example 9.3, a lifetime risk of an additional 1 cancer case per million
exposed individuals (risk = 0.000001) was the result of consuming contaminated water. This risk is in addition to the background risk of a person in the
US dying from all forms of cancer, which is 0.233 (as was shown in Table 9.1).
Chloroform also has a noncancer risk; the calculation of the hazard index is
demonstrated in Example 9.4.
EXAMPLE 9.4
Calculate the hazard index for the chloroform exposure calculated in Example
9.2 using a reference dose for chloroform of 0.010 mg/kg-day.
SOLUTION
HI =
LADD
RfD
HI =
1.15 ¥ 10 -4 mg/kg-day
0.010 mg/kg-day
HI = 0.0115
Since the HI is well below 1, there is very little noncarcinogenic risk present.
Risk Management
Risk management refers to making decisions using the results of the risk
assessment. The risk is evaluated in light of the cost of each potential action and
other factors such as the technical feasibility of reducing risk, the size of the population involved, the possibility of increasing risks during a particular response
action (for example, exposure to contaminated dust during soil excavation),
statutory mandates, and public concern. The particular decision may involve
setting the level of cleanup desired at an old hazardous waste site, establishing a
treatment standard for potable water, or giving an approval or denial to a proposed location for a new industrial plant. Generally, a carcinogenic risk of less
than 10–6 has been deemed acceptable by the EPA. As we learned in Chapter 5,
EPA has established maximum contaminant levels (MCLs) for a number of elements, ions, and compounds. MCLs are upper limits for safe drinking water,
and the actual numerical limits were based on a risk analysis that determined
levels below which there are no known or expected health risks (US EPA 2013).
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The risk assessment process is extremely conservative in nature and utilizes measurements that are uncertain. In addition, insufficient data or information gaps often exist in characterizing the potential risk of an agent,
resulting in the need for assumptions or “educated guesses.” Risk is generally
expressed as a single number because it is easier to present to the public in that
form. However, it would be more properly represented by a numerical range.
Monte Carlo simulations have been used to account for the uncertainties in the
process. This tool allows risk to be expressed as a probability distribution
rather than a single probability which can then be used to make a more
informed decision during risk management.
9.2 Energy Resources (with Emphasis on Nuclear Power and
Radioactive Wastes)
The United States (just like the rest of the world) needs energy to power its
economy and sustain its society. Our overall energy supply is heavily dependent on fossil fuels (just like the rest of the world). We use coal and natural gas
to generate electricity to power all parts of our economy (even our cell phones
and computers), we use gasoline and diesel to transport people and goods, we
use all three fuels (coal, oil, and gas) to run our factories and to heat our homes
and cook. In the United States, we are blessed with bountiful natural resources,
including energy resources, but we consume a lot each year. On average, from
2007–2012, we used about 100 quads per year (a quad is a unit of energy equal
to 1 quadrillion Btu or 1.0 (10)15 Btu). During that five-year period, burning fossil fuels provided about 83% of our total energy supply. All forms of renewable
energy (mostly hydro and biomass with a little bit of wind and solar) accounted
for only about 8% of our total supply, and nuclear power (to generate electricity) accounted for about 9%, according to the Energy Information Administration (US EIA 2012a). The 2011 breakdown of supply by source is shown in
Figure 9.4 on the following page, which also details our renewable sources.
The fossil fuels—coal, oil, and natural gas—provide the bulk of our energy
resources. Coal is mined in many parts of the world, and the United States has
enormous reserves of coal. Because it is hard to burn and expensive to clean up
the emissions, coal has been cheap historically. For that reason, it has been a
favorite of power companies, who have the financial resources to invest in
large-scale plants and expensive pollution control equipment. Crude oil is primarily used to make gasoline, diesel, jet fuel, and other fuels for mobile
sources, but it also is the raw material for many specialty chemicals and for
plastics. It must be explored for, found, drilled from underground, and transported to a refinery, where hundreds of products are then made from the oil.
Oil typically is the most expensive fossil fuel. Natural gas is primarily methane,
and is the cleanest and easiest fuel to burn. Recently, the technologies of
hydraulic fracturing (“fracking”) and horizontal well drilling have made natural gas supplies much more plentiful, which has driven down the price of gas.
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Figure 9.4 US energy supply by type, 2011. (Drawn from data from US Energy Information Administration, Monthly Energy Review, March 2012, preliminary 2011 data.)
Total
97.5 quadrillion Btu
Total
9 quadrillion Btu
Solar 1%
Geothermal 2%
Wind 13%
Biomass Waste 5%
Coal
20%
Petroleum
36%
Natural Gas
25%
Biofuels 21%
Renewable
Energy
9%
Nuclear
Electric
Power
8%
Biomass 48%
Wood 22%
Hydropower 35%
With the greater supply and lower price, natural gas has displaced coal in
many power plants in the last five years.
Two other categories of primary energy resources must be included to get
the full picture. Nuclear energy is almost exclusively used for generating electricity, and will be discussed in more detail later in this chapter. Renewable
energy refers to those forms of energy that are tied to the sun, gravity, the
earth’s rotation, or its internal heat, and include hydroelectric power, biomass,
tidal power, wind energy, solar power, and geothermal energy. These are all
termed renewable because they are continually being renewed by their fundamental power sources.
Energy demand comes from four main sectors of the economy—residential
(R), commercial (C), industrial (I), and transportation (T). In 2011, US demand
consumed 97.5 quads, and these four sectors used that energy in the following
percentages: R (22%), C (18.5%), I (31.4%), and T (27.8%). Each sector has its
own specific needs. For example, transportation requires liquid fuels, mainly
gasoline and diesel, both of which come primarily from oil. Residential and
commercial demand is satisfied primarily by electricity (although many homes
and small businesses also use natural gas or fuel oil for various needs). Industries use the most economical fuel available in their locality, and thus, collectively, they use all three types of fossil fuels as well as electricity. Energy supply
by source and its flow to the various final end users by sector of the economy is
mapped in Figure 9.5. In this figure, electricity generation is not shown separately, but total electricity use has been distributed to the end users.
Electricity generation is a special case because power companies are not
really end users of energy, although they do consume a lot of primary energy to
produce electrical energy for consumers. Power plants consume energy (e.g., fos-
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351
Figure 9.5 US energy supply and consumption by primary source and final enduse sector. (Adapted from “Annual Energy Review,” US EIA 2012.)
Exports
(10 Quads)
Coal (22)
Natural Gas (24)
Total
Domestic (15)
Supply
Oil
Imported (29)
(107 Quads)
Nuclear (8)
Renewables (9)
Coal (20)
Fossil
Fuels
(90)
Gas (24)
Residential
(22)
Commercial
(18)
Oil (36)
Nuclear (8)
Renewables (9)
Industrial
(30)
Total
Domestic
Consumption
(97 Quads)
Transportation
(27)
sil fuel, nuclear, hydro) and produce and sell electricity, typically with a fairly
low overall energy conversion efficiency. Electricity generation in the US still
relies mainly on coal, natural gas, nuclear, hydro, and biomass for its primary
energy sources. Despite recent large percentage increases in the use of wind and
solar energy, those two renewable sources supply only a tiny fraction of total
demand. US electricity generation in 2011 consumed 39.3 quads of primary
energy, supplied as follows: coal (46%), gas (20%), nuclear (21%), and renewables
(13%). As was seen in Figure 9.4, the majority of the renewable energy that produced electricity came from biomass, and most of the rest came from hydroelectric power. The net electricity produced was distributed to residential,
commercial, and industrial users primarily, with very little going to the transportation sector. Table 9.5 on the next page shows the recent history of electricity
generation in the United States along with the percentages supplied by each primary energy resource.
Reducing our Energy Use and Improving the Environment
As noted above, the United States (and every other developed nation) is
very dependent on fossil fuels. In 2010, world energy consumption was about
12,700 million tons of oil equivalent (Mtoe) or about 500 quads, with about 80%
of that being fossil fuels (International Energy Agency 2012). The extraction
and transport of each fossil fuel can have substantial environmental impacts,
and wastes from the production and use of each fuel must be dealt with properly. Coal mining around the world results in the deaths of hundreds of workers every year. The air pollution emissions from burning coal (ash, NOx, SO2,
mercury, and others) create unhealthy air (and undoubtedly contribute to thou-
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Table 9.5 Recent History of Electricity Generation and Use in the United States
Year
1990
1995
2000
2005
2010
Electricity Use (by Sector), billions of kwh
Residential
920
1,040
Commercial
840
950
Industrial
950
1,010
Total, billion kwh
2,710
3,000
1,190
1,160
1,060
3,410
2,020
1,280
1,020
4,320
1,940
1,330
970
4,240
Consumption of Primary Energy (by Source) to Generate Electricity, Quads
Coal
16.26
17.47
20.22
20.74
Gas
3.31
4.30
5.29
6.02
Petroleum
1.29
0.76
1.14
1.24
Nuclear
6.10
7.08
7.86
8.16
Renewables
3.52
3.75
3.43
3.41
Total, Quads
30.5
33.4
37.9
39.6
19.13
7.53
0.38
8.43
4.06
39.5
Source: US EIA (2012a), http://www.eia.gov/totalenergy/data/annual/pdf/aer.pdf
sands of deaths each year) in many countries around the world (especially
China). The ash and sulfur that is captured from coal-fired power plants creates
enormous masses of solid waste and sludges that must be disposed of. Oil production and transport has potential for massive oil spills, impacting large
swaths of the natural environment. Even natural gas (the cleanest fossil fuel)
has the potential for groundwater pollution via fracking (American Water
Works Association 2013). And, of course, all three fossil fuels create enormous
amounts of CO2 emissions, as illustrated in Example 9.5.
EXAMPLE 9.5
Given the data in the preceding paragraphs, estimate world CO2 emissions
from burning fossil fuels in 2012. Assume that all fossil fuels can be represented by an average fuel with the formula CH2.2 and an energy content of
15,000 Btu/lb.
SOLUTION
15
Fossil fuel use in 2012 = 500 quads ¥ 80% = 400 quads (= 400 (10 )
15
Mass of fossil fuels used = 400 (10 )
Fraction carbon in CH 2.2 fuel =
Btu ¥
Btu)
1 lb
13
= 2.67 (10 ) lb
15, 000 Btu
12 lb C
= 0.845
14.2 lb fuel
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13
Mass of CO 2 emitted = 2.67 (10 ) lb fuel ¥ 0.845 ¥
353
44 lb CO 2
12 lb C
13
= 8.27 (10 ) lbs of CO 2
or in more conventional units,
13
= 8.27 (10 ) lbs ¥
1 kg
1 gigatonne
O2
¥
= 37.6 Gt of CO
2.2 lbs
1012 kg
Reducing our dependence on fossil fuels has many benefits, but one of the
biggest is the reduction in the rate of CO2 emissions. There are several possibilities for reducing our current use of fossil fuels. The first and best is conservation
by consumers. Small actions by individuals such as turning off lights, adjusting
the thermostat, carpooling or taking transit to work, walking or riding a bike
for short errands, and others can add up to large savings in electricity and fuel
consumption. The typical fossil fuel steam electric plant is about 33% efficient,
meaning that it takes about 3 kwh of fuel energy to produce 1 kwh of electrical
energy. So saving 10 kwh of electricity not only saves you money, but if that 10
kwh was generated from coal, it saves about 30 pounds of CO2. Similarly, not
using 1 gallon of gasoline not only saves you money, but prevents about 20
pounds of CO2 emissions.
On the commercial and industrial scale, efficiency improvements in manufacturing processes, more energy-efficient appliances, and more efficient electric
power-generating technology (supercritical or combined cycle plants) could
save more than a gigatonne of CO2 per year. Of course, the more we can utilize
renewable energy such as solar and wind power, the better off we will be. However, we must recognize that solar and wind are both so small in the world
energy picture right now, that it will take years before they can replace a substantial amount of the fossil fuels. In the opinion of this author, nuclear power,
even with its particular issues, seems preferable to continued growth in fossil
fuel use (although recent decisions in several countries seem to indicate popular dissent with that opinion). There is much public opposition to nuclear
power; it is a contentious and emotional issue for many people. As engineers,
we must always strive to understand the facts behind the issues. Therefore, a
more detailed discussion of nuclear energy and nuclear power generation is
presented in the following section.
Nuclear Power
As was shown, nuclear energy provides a significant portion of the primary energy needed to generate electricity in the United States. In 2013, a total
of 30 countries utilized nuclear energy to produce electricity, and together provided about 12% of the world’s electricity demand (Nuclear Energy Institute
(NEI) 2013). In the United States, France, England, Hungary, Sweden, Japan,
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and others, nuclear fission reactors provide heat to boil water into steam and
produce electricity. In 2011, thirteen countries relied on nuclear energy to generate at least one-fourth of their total electricity; France provided almost threequarters of all its electricity with nuclear power (NEI 2013). In the United
States, 104 reactors were operated in 31 states in 2012; seven of those states
used nuclear energy to produce 40% or more of all their electricity: Vermont,
South Carolina, New Jersey, Illinois, Connecticut, New Hampshire, and Virginia (NEI 2013).
Nuclear plants have a significant advantage over fossil-fuel-fired plants—
they do not emit carbon dioxide. Without reviewing the details of nuclear reactions, or the details of the technology of nuclear reactors that generate electricity, let us simply say that in a nuclear plant, heat is generated as the uranium
fuel undergoes fission. That heat is transferred from the fuel rods to water
which is either boiled into steam directly or is used as a heat transfer fluid to
boil another stream of water. The steam produced then turns a turbine-generator to produce power as described in Chapter 3. Typically, the steam temperature in a nuclear plant is maintained a bit cooler than in a gas- or coal-fired
plant, so the overall thermal efficiency is not quite as high—perhaps 30–34%
versus 36–40% on average. But because there is no carbon being burned, CO2
emissions are nil. All in all, it is estimated that nuclear-generated electricity
avoided emitting 588 million metric tons of CO2 in the United States in 2013
(NEI 2014), and saved more than 185 million metric tons of coal and natural
gas. But nuclear plants, being less efficient, emit more heat into the environment. Examples 9.6 and 9.7 help put these numbers into perspective.
EXAMPLE 9.6
(a) How much CO2 is avoided by one 1,000-MW nuclear plant with a thermal
efficiency of 32% compared with a 1,000-MW coal-fired plant that is 38%
efficient? Assume the coal plant emits CO2 at an average of 210 pounds of
CO2 per million Btu of heat input and assume it operates at 100% capacity
for 340 days/year. Give your answer in pounds per day and tonnes per year.
(b) How much more waste heat is discharged into the environment every day
from the nuclear plant? Give your answer in MW, in kJ/day, and in percent.
SOLUTION
(a) Heat input to coal plant = 1,000 MW/0.38 = 2,632 MW
7
CO 2 = 2, 632 MW ¥
1, 000 kw 3, 412 Btu 24 hr 210 lb CO 2 4.53 (10 ) lb
=
¥
¥
¥
day
MW
kwh
day
106 Btu
7
=
4.53 (10 ) lb
day
¥
340 days 1 tonne
6
= 7.0 (10 ) tonnes/yr
¥
yr
2, 205 lb
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355
(b) Nuclear plant: Waste heat = 1,000 MW/0.32 – 1,000 MW = 2,125 MW
Coal plant:
Waste heat = 2,632 MW (1 – 0.38) = 1,632 MW
Difference:
2,125 – 1,632 = 493 MW
493 MW ¥
1, 000 kw 1 kJ/s 86, 400 s
10
¥
¥
= 4.26 (10 ) kJ/day
MW
1 kw
day
493 MW
= 0.302 = 30%
1, 632 MW
EXAMPLE 9.7
Assume the average US passenger car is driven an average of 10,000 miles per
year and gets an average of 20 miles per gallon. Further assume that cars emit
about 20 pounds of CO2 per gallon of fuel burned. Calculate the number of
cars that would have to be taken off the road to avoid the same amount of CO2
emissions that were avoided by the one nuclear power plant of Example 9.6.
SOLUTION
CO 2 emissions per year per car =
10 , 000 mi 1 gal 20 lb CO 2 1 kg
¥
¥
¥
20 mi
gal
yr
2.2 lb
= 4 , 545 kg CO 2 /yr or about 4.55 tonnes/yr
6
Number of cars equiv. =
7.0 (10 ) tonnes/yr
4.55 tonnes/yr per car
= 1.5 million cars!
In addition to a direct CO2 emissions savings, nuclear power plants have
no air pollution emissions of mercury, particulate matter, NOx, or SO2, all of
which are problems associated with coal-fired plants. Furthermore, there are
no risks of coal-miner deaths or large-scale oil spills that can be attributed to
nuclear power plants. So why is there so much opposition to them? The answer
can be described in two words: fear and politics.
Nuclear plants have two very significant problems—(1) the risk of radiation
emissions, and (2) the disposal of radioactive wastes. The risk of radiation
release is present in two forms: (a) the high probability of very low-level releases
that occur on a steady basis, and (b) the very low probability of a very serious
accident happening at a nuclear power plant with a subsequent large-scale
release of radiation. Radiation is known to have damaging effects on living
organisms, and can cause or contribute to a variety of cancers in human beings.
Most people agree that the routine releases of radiation to the air or water (in the
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form of tritium, noble gases, carbon-14, and particles) from nuclear power plants
are very small and well within regulatory limits. These routine releases typically
result in annual exposures to people that are orders of magnitude below natural
background radiation, and well below the levels received from living in a house
with radon gas seepage, getting an x-ray, or taking a long plane flight.
Emergency releases, however, can be very significant. In the author’s lifetime, three major accidents have made headline news. The Three Mile Island
incident in the United States in 1979 included a partial melt-down of the core,
and resulted in some significant releases of radiation, but is not considered a
Level 7 catastrophe by international atomic agency standards. Two catastrophic (Level 7) accidents with large releases of radioactive materials have
occurred within the past 30 years: the runaway nuclear reaction at Chernobyl
in the USSR in 1986 and the earthquake and tsunami that destroyed the nuclear
reactors at Fukushima in Japan in 2011.
In April of 1986, the Chernobyl nuclear reactor #4 experienced an emergency that ultimately resulted in a reactor vessel rupture and a steam explosion. Subsequently, the graphite moderator of the reactor was exposed to the
air and caught fire, making matters much worse. The fire released a large
plume of highly radioactive material, which drifted for weeks over parts of the
Soviet Union and Europe. Over the next four years, more than 350,000 people
were evacuated from contaminated land areas in the Ukraine, Belarus, and
Russia. During the emergency as many as 200,000 emergency workers were
exposed to high doses of radiation, resulting in about 50 deaths shortly after
the accident. One recent estimate of total worldwide deaths from long-term
cancers attributed to this accident is 985,000 (Yablokov et al. 2009).
The Fukushima nuclear power plant on the east coast of Japan had six boiling water reactors, but in March 2011, only three of them were operating. A
major earthquake occurred off the coast and emergency shutdown procedures
were immediately initiated. Emergency generators (fueled by diesel oil) came
on to provide electricity to continue to pump cooling water through the reactors (even after a shutdown, the nuclear reactions continue to generate significant amounts of heat for days). The tsunami that followed the earthquake
inundated the plant and flooded the rooms in which the generators were
housed, shutting them down just minutes after they had started. With loss of
cooling water, the nuclear fuel rods overheated and melted in all three reactors.
To make matters worse, the nickel-alloy metal cladding around the fuel rods
got so hot that it reacted with water to form hydrogen gas, which eventually
exploded. Radiation was dispersed for miles around the plant, and whole
towns had to be evacuated. It is unclear at this time how many long-term cancer deaths will result from this disastrous accident.
The other problem associated with nuclear power plants involves disposal
of radioactive waste materials; how to handle nuclear waste is a major political
issue. Radiation is emitted spontaneously from the nuclei of radioactive elements. This radiation includes alpha and beta particles, gamma radiation, and
high-energy neutrons (see Table 9.6). All radiation causes damage to living tissue: alpha particles (helium nuclei) are the weakest, beta particles (free elec-
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Table 9.6
357
Four Basic Types of Radiation from Radionuclides
Name
Type
Description
Symbol
Properties
Alpha
Particle
Helium nucleus
4
2 He
Weak, easily shielded (skin or paper), only significant if inhaled or ingested
Beta
Particle
Electron
0
-1b
Medium strength, ionizing, can penetrate skin
but can be shielded by numerous materials
Gamma
Ray
Radiation
0
0g
Similar to x-rays, strongly ionizing, highly penetrating, need dense materials for shielding
Neutron
Particle
Neutron
1
0n
Highly penetrating, can induce radioactivity in
other materials, mainly occurs inside nuclear
reactors
trons) are stronger, and gamma rays are the most powerful. Each radioactive
element undergoes nuclear reactions, and each has its own characteristic emissions. Examples of some nuclear reactions are given in Table 9.7. Note the
example of a nuclear fission reaction. Fission reactions typically result in the
production of excess neutrons, which then go on to start new reactions. This
extremely fast set of chain reactions creates the self-propagating fission reactions in a power plant, and causes the explosion of a nuclear bomb.
Radioactive waste materials come from a variety of sources other than
nuclear power plants, including hospitals, universities, mining wastes, manufactured items, and contaminated garments. These low-level wastes (LLW)
may be slightly or weakly radioactive, and are short-lived (they decay quickly
and may remain radioactive for only hours or weeks). The large majority
(about 95%) of LLW wastes are classified as “Class A” (paper, rags, clothing)
and do not require special handling; these are normally disposed of in ordinary
landfills. Class B and C wastes (including used nuclear power plant filters and
Table 9.7
Examples of Nuclear Reactions
Type of Nuclear Reaction
Alpha emission (ejection of helium nucleus)
Beta emission (ejection of electron from nucleus)
Gamma emission
Nuclear fission
Example
226
222
4
88 Ra Æ 86 Rn + 2 He
14
14
6 C Æ 7N + e
238
1
239
0
92 U + 0 n Æ 92 U + 0 g
235
1
142
91
1
92 U + 0 n Æ 56 Ba + 36 Kr + 3 0 n
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Chapter Nine
solidified liquids) are more radioactive than Class A, but will decay to safe levels within 100 years. These are the regulated by the various states and generally
require the disposal facility to have a design life of 300 years.
High-level wastes (HLW) include transuranic wastes (transuranic elements
are those with atomic numbers higher than uranium) and spent fuel (SF) rods
from power plants; these materials are highly radioactive and can have either
short or very long half-lives. A radioactive half-life is the time it takes for half
the radioactive material to decay. For example, plutonium-239, a by-product of
the nuclear reactions of uranium-238 in a nuclear power plant, has a half-life of
24,000 years. This means that it will take 24,000 years for the mass of plutonium
in a sample of radioactive waste to drop to one-half its original value. Radioactive decay is a first-order process, and can be modeled in the same way as a
classic first-order chemical reaction.
Nt = N0 e–kt
(9.4)
However, it is more commonly modeled using the half-life equation:
Nt = N0 (½)t/t1/2
(9.5)
where:
Nt
= amount of radioactive material at time t from the starting time
N0 = amount of radioactive material at the starting time
t
= time
t1/2 = half life
EXAMPLE 9.8
Assume that some plutonium waste is 500 times more radioactive than a
“safe” level. After 1,000 years, what is its relative radioactivity compared to
the safe level?
SOLUTION
Setting the initial amount at 500 S where S is the “safe” level, from Eq. (9.5):
Nt = 500 S (1/2)1,000/24,000 = 0.9715 (500 S) = 486 S
So, even after 1,000 years, the radiation is still 486 times the safe level.
High-level wastes contain fission products and transuranic elements that
were generated from the US nuclear weapons program or from the processing
of spent nuclear fuel. After the uranium fuel rods have been depleted in
strength, they become spent fuel (SF), and require the same kind of careful isolation that HLW needs. Unfortunately, in the United States the current practice
for HLW and SF is to store them in above-ground and underground tanks
located at various power plants and other locations around the country. The
tanks are filled with water to dissipate the residual heat that is being continuously generated by the ongoing nuclear decay reactions.
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359
There are essentially only three options for the management of HLW and
SF: store it indefinitely at the point of generation, reprocess it to recover usable
fuel (which still leaves some residual waste), or isolate the waste in a national
deep underground repository. Reprocessing waste to chemically recover uranium and plutonium is being done in several countries, but is more expensive
than mining virgin uranium. Also, a highly radioactive liquid waste remains
which must be solidified for eventual disposal.
Radioactive waste disposal was called a political problem earlier. In the
United States, this is because the technical problems of long-term storage have
been solved, but the political problems of where to store it have not. The Yucca
Mountain nuclear waste repository was touted for more than 30 years by the
US government as the answer to our country’s nuclear waste disposal problem
(see Figure 9.6). Yucca Mountain is in Nevada, about 100 miles northwest of
Las Vegas, and its geology, isolation, and stability are ideal for using it as our
national repository. Comprehensive geologic studies and engineering conceptual designs were completed to prove its viability in this service, and its designation as the nation’s nuclear waste repository was approved by Congress in
2002. However, opponents of this logical solution never gave up fighting the
idea. The Obama administration withdrew its support for the project, and its
funding was cancelled by Congress in 2011. However, the project itself
remained in limbo. In August 2013, a federal appeals court in Washington, DC
ruled that the Nuclear Regulatory Commission can no longer delay a decision
on whether or not to issue a permit for the Yucca Mountain storage facility. As
of this writing, commercial HLW and SF are still being stored “temporarily” in
tanks at numerous locations around the country.
Figure 9.6
Schematic of Yucca Mountain facility for storage of high-level nuclear wastes.
Cutaway lifted to show
underground portion
Excavated rock
Access Ramp
Acces
s
Underground
storage area
Ramp
Railroad
spur
Receiving and
processing area
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The US government does operate one site for underground disposal of
HLW (along with certain kinds of LLW) generated from the research and production of nuclear weapons. It is called the waste isolation pilot plant (WIPP)
and is located about 26 miles east of Carlsbad, New Mexico. The Department of
Energy began studying this site in the early 1970s, but due to many political
battles and technical complications, the site did not open for full operations
until 1999. It now stores waste in salt deposits that sit about 2,000 feet below
the surface. Large rooms have been excavated from within a 3,000-foot thick
salt formation that has been geologically stable for more than 200 million years.
Salt formations behave with some plasticity, and cracks or holes seal themselves. The waste is stored in steel casks that are placed inside these rooms (see
Figure 9.7). In the late 1990s, after careful review and consideration of about
1,400 public comments, the EPA concluded that these wastes can be safely
stored for at least the next 10,000 years (US EPA 1998). The US is not the only
nuclear nation that has not settled on a final solution for safe disposal of its
nuclear waste. Despite its widespread use of nuclear power, France only
recently has started studying a proposed site for a deep underground waste
repository (World Nuclear News 2013).
Figure 9.7 Photograph of radioactive waste being stored at WIPP. (Department of Energy;
photo courtesy of the URS Corporation.)
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9.3 Indoor Air Quality
In Chapter 7, we focused on ambient (outdoor) air quality and industrial
air pollution control. However, in modern society, many people spend 90% or
more of their time inside: at their homes, at work, in stores, or inside vehicles
going to or from those places. Indoor air quality (IAQ) thus is very important
to people’s health. Indeed, there have been a number of instances of “sick
building” syndrome in the news over the past 30 years, resulting in a variety of
lawsuits against building owners and operators.
Indoor air pollutants include tobacco smoke, radon, formaldehyde, molds
and mildews, and others. The concentrations of pollutants to which people are
exposed can vary dramatically based on location, age of the building (but newer
isn’t always better), design and construction of the building, methods of ventilation, and whether control techniques are being used. This section presents
information on the major pollutants of concern—where they come from, how
they can accumulate in indoor air, and how we can predict and control IAQ.
Some Pollutants of Concern
Tobacco Smoke
Smoke from cigarettes, cigars, and pipes contains a number of pollutants
ranging from fine particulate matter and CO, to formaldehyde and nicotine, to
a variety of known or suspected carcinogens such as benzene and benzo-apyrene. Smoking is blamed for more than 5 million deaths per year, worldwide
(CDC 2014b). While smokers voluntarily choose to pollute their lungs, nonsmokers who share the same indoor space with smokers have no choice.
Smoke, in addition to circulating throughout the air within the home, restaurant, office, or shopping area, can deposit on surfaces (such as curtains) and the
smell can linger for a long time. Fortunately, in the US the public relations campaign against smoking appears to be successful, and the percentage of smokers
in the population is declining. There are now laws in most states that prohibit
smoking in public places. However, about 18% of all adults in the US are current smokers, and, on average, each day more than 3,200 persons younger than
age 18 smoke their first cigarette (CDC 2014b). In other countries, smoking is
still very prevalent, and growing.
Radon
Radon gas is a radioactive decay product of radium (which is itself a decay
product of uranium), and is found in rocks and soils in many parts of the country. Although radon is not chemically active, its decay products (polonium,
lead, and bismuth) are, and if inhaled radon gas decays while inside the lungs,
the decay products can lodge there where they emit more radiation. Radon gas
migrates upward through porous soils, and relatively large amounts of the gas
can enter homes and other buildings if they are in its migration pathway. The
gas enters through paths of least resistance (such as cracks in the concrete slab
under a house, through semi-porous basement walls or flooring, or loose joints
between floors and walls).
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Chapter Nine
The concentration of radon can be measured with a Geiger counter and
reported as pico-Curies per cubic meter (pCi/m3). A Curie (Ci) is a very large unit
of radioactivity, and so the pCi (10–12 Ci) is used most often. The radioactivity of
radon, combined with the chemical effects and radioactivity of its progeny, cause
several health effects, including lung cancer. The soil flux of radon varies widely
depending on location in the US; it has been reported between 0.1 and 100 pCi/
m2-s, with 1.0 being most typical. Many homes have indoor air concentrations in
the range of 1,500 pCi/m3 or 1.5 pCi/L. At this level, the risk of lung cancer is
about equal to that obtained by receiving 75 chest x-rays per year. (The EPA recommends remedial action to reduce radon levels in homes if the Ra concentration
reaches about 4 to 8 pCi/L.) It has been estimated that radon is responsible for
about 21,000 deaths per year from lung cancer in the United States (US EPA 2003).
Formaldehyde
Formaldehyde (HCHO) frequently is found inside buildings; it can come
from several sources. Formaldehyde is emitted from a variety of laminates and
glues, is found in some foam insulation, and in the “off-gases” from certain
kinds of furniture and shelving. The major sources of formaldehyde in new
mobile or manufactured homes (and often in schools or offices) are particle
board, plywood, and certain types of insulation. Fortunately, the emission rates
from those new products decrease over time, and within one year from installation, the rates are typically less than 20% of the original values. The principal
human health effect of formaldehyde is eye irritation, which occurs at very low
concentrations. The RfD for formaldehyde is 0.2 mg/kg-day, and the chronic
inhalation minimal risk level is 0.003 ppm in air (US EPA 2000).
Another source of HCHO is indoor combustion of fuels (HCHO is a relatively stable product of incomplete combustion of any carbonaceous fuel).
Sources include wood stoves and fireplaces, natural gas stoves and water heaters, or oil or coal furnaces with slight leaks. Also, cigarettes are known to emit
formaldehyde. Formaldehyde concentrations inside homes and offices range
from 0.02 to 0.3 ppm. In some mobile homes built in the 1970s, HCHO was
reported as high as 1 to 2 ppm (Godish 1985). Formaldehyde concentrations
have been shown to be substantially higher with increasing indoor air temperature, and/or increasing inside air humidity. As stated above, formaldehyde
emissions from newly installed items decrease substantially with time. Furthermore, the replacement of older, high-emitting products with modern ones
that do not emit formaldehyde has reduced overall exposure significantly. For
example, urea-formaldehyde foam insulation was widely used in the 1970s,
but now has been mostly replaced with other types of insulation.
Two other pollutants that come from combustion sources are CO and NOx.
Both are criteria pollutants and have been discussed previously. However, high
concentrations can be reached indoors if there is little or no ventilation for combustion products. It should be noted that some people die each year from CO
poisoning when CO accumulates indoors owing to the operation of a kerosene
space heater, a faulty furnace inside a poorly ventilated home, or a gasolinefueled portable electrical power generator near an open window of a home.
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363
Molds, Mildews, and Allergens
In buildings with central air conditioning, the inside walls of the ductwork
can become a breeding ground for mold, mildew, and bacteria. This problem
can be especially bad in warm humid climates, and has been exacerbated by
energy savings trends in building design during the last thirty years. Two popular energy saving steps are (1) to use a higher percentage of recirculated air
and a lower percentage of outside air, and (2) to keep the air moister when
using air conditioning. Inspections have found millions of spores per square
inch on the inside walls of ducts leading from the air conditioners in homes
and commercial buildings. People may be allergic to mold and mildew spores
and/or their gaseous metabolic products.
Ventilation and Infiltration
Ventilation is defined as fresh outside air coming in to replace inside air
that is being exhausted to the outside. This process is called simply air
exchange. Air exchange occurs in buildings in three ways: forced ventilation,
natural ventilation, and infiltration. Forced ventilation uses fans or blowers to
forcibly exchange the air, while natural ventilation permits natural air exchange
through open windows or doors. Infiltration refers to the natural air exchange
that occurs even when all windows and doors are closed. Air can leak into
buildings through numerous gaps and openings in the building envelope such
as the cracks around doors and windows, the gaps around pipes and electrical
conduits, kitchen and bathroom vent pipes, floor-wall joints, mortar joints, and
others. Ventilation and infiltration are depicted schematically in Figure 9.8.
Excess outside air that enters a building must be heated or cooled to keep the
inside conditions comfortable, and infiltration accounts for most of this “excess”
Figure 9.8
Ventilation and
infiltration of air
into a building.
Forced
Ventilation
Natural
Ventilation
Infiltration
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364
Chapter Nine
energy use. Infiltration is much larger than one might think—by some estimates,
it accounts for 4–15% of total annual energy use in US buildings (Emmerich and
Persily 1998) or about 1–4% of total energy use in the United States.
Material Balance Models for Indoor Air Quality
The indoor concentrations of air contaminants can be predicted by simple
mathematical models—the key variables are emission rates and ventilation
rates. In some buildings, the air flow is very simple, and we can assume the
whole building acts like a single CSTR. For other situations, we might have to
model the building as many stirred tanks connected in series and parallel. For
the simple model of one well-mixed room, the material balance equation is
similar to those that we have seen in previous chapters:
VdCe
= QCa + E - QCe - kCeV
dt
(9.6)
where:
V = volume of the room, m3
Ce = concentration inside (and exiting) the room, µg/m3
Ca = concentration in the ambient air coming into the room, µg/m3
Q = ventilation rate, m3/hr
E = emission rate inside the room, µg/hr
k
= reaction rate constant (here assumed to be first-order destruction), hr–1
Equation (9.6) can be rewritten as
dCe Ê Q
QCa E
ˆ
+ Á + k ˜ Ce =
+
¯
dt Ë V
V
V
(9.7)
The inverse of the coefficient of Ce is called the characteristic time or time constant for this system, and is given the symbol, τ. The general solution to Eq.
(9.7) is:
È
Ê QCa E ˆ ˘ È -t/t ˘
Ê QCa E ˆ
Ce = ÍC0 - t Á
+ ˜˙ e
+t Á
+ ˜
Î
˚
Ë V
Ë V
V¯˚
V¯
Î
(9.8)
where:
C0 = initial concentration inside the room at time zero
τ
= characteristic time, (Q/V + k)–1
The quantity Q/V is given the symbol A and is called the air exchange rate; it
has units of room air changes per hour. In terms of A, τ is simply 1/(A + k). The
air changes per hour is a common measure of the degree of ventilation for a
building or room. Equation (9.8) also can be written as:
Ê QCa E ˆ
Ce = C0 e-t/t + t Á
+ ˜ 1 - e-t/t
Ë V
V¯
(
)
(9.9)
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365
where:
τ
= characteristic time, 1/(A + k)
The solution to the steady-state case (as t approaches infinity) is just:
Ê QCa E ˆ
Cess = t Á
+ ˜
Ë V
V¯
(9.10)
Ê ACa + E / V ˆ
Cess = Á
˜¯
Ë
A+k
(9.11)
which also can be written as
where A = Q/V, air changes per hour
Obviously, as the air exchange rate increases, and the other terms decrease in comparison, the indoor concentration approaches the outdoor ambient concentration.
EXAMPLE 9.9
Two friends go to a nightclub in China one evening to listen to a popular
band. As the room begins to fill up, more and more smokers light up. The
room fills quickly, and soon there are 600 people there, one-third of whom are
smoking at any given moment. Each cigarette emits 1.4 mg of formaldehyde,
and each smoker averages three cigarettes per hour. Formaldehyde decays to
carbon dioxide with a first-order rate constant of 0.40 per hour. The room
measures 20 m by 30 m by 5 m, the ventilation rate is 2.5 air changes per hour
for this event, and the outdoor concentration is zero.
(a) Assuming that the two friends stay there until the steady-state concentration is reached, what is the maximum concentration of formaldehyde to
which they are exposed?
(b) Assuming the initial HCHO concentration is zero, how long does it take to
reach 95% of the steady-state concentration?
SOLUTION
(a) At steady state, Eq. (9.11) applies, namely:
Ê ACa + E / V ˆ
Cess = Á
˜¯
Ë
A+k
The emissions rate is:
E = 0.333 ¥ 600 people ¥
1.4 mg 3 cig
¥
cig
hr
= 839 mg/hr of formaldeh
hyde emissions
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366
Chapter Nine
The room volume is 3,000 m3, and since Ca is zero, then
Cess =
=
E /V
A+k
or
839 / 3, 000
= 0.0964 mg/m 3
2.5 + 0.4
This concentration converts to 0.079 ppm, more than enough to cause serious eye irritation.
(b) To estimate the time to reach 95% of the steady-state concentration, we
must know how the emissions rate function behaves. A simple approach,
since the room fills quickly, is to assume that the emissions jump up immediately to their final level (a so-called step increase). For C0 = Ca = zero, Eq.
(9.9) reduces to:
Ce =
E /V
(A + k) (
(
1 - e-t/t
= Cess 1 - e-t/t
Setting
Ce
)
Ê
ˆ
1
Á recall t = 2.5 + 0.4 = 0.345 hr˜
(
)
Ë
¯
)
= 0.95, and solving for t, we find that t equals 1.03 hours.
Cess
The response of a first-order system (such as described in Example 9.9) to a
step increase forcing function is always an exponential response curve (see Figure 9.9). For such response curves, 63% of the final response is reached in a
time numerically equal to one time constant, 86% in two time constants, 95% in
three time constants, and so forth. Steady state is effectively achieved after 4
time constants. In Example 9.9, the time constant is 0.345 hours or about 20
minutes, so the time to reach 95% of the steady-state concentration is three time
Figure 9.9 Dimensionless response of a
first-order system to a step-increase forcing as a function of dimensionless time.
1.0
C et
C e ss
0.8
0.6
0.4
0.2
0
0
1
2
t
τ
3
4
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Other Important Topics
367
constants or about one hour, which confirms the answer to part (b) of Example
9.9. The unsteady-state problem of Example 9.9 can be solved easily using
numerical methods as illustrated in Example 9.10.
EXAMPLE 9.10
Solve Example 9.9 using numerical methods using a spreadsheet.
SOLUTION
We start by rewriting the mass balance [Eq. (9.6)] as a finite difference equation:
V
DCe
= QCa + E - QCe - kCeV
Dt
Defining Δt as 1 minute, and setting the Ce equal to C0 at time zero, we set up
a spreadsheet as shown in Figure 9.10 on the following page. Then simply
copy down any number of rows and watch the results until the calculated Ce
approaches a constant value.
Solutions to IAQ Problems
Review of Eq. (9.11) shows that the only viable approaches to solving IAQ
problems are to increase the air exchange rate A (by increasing the ventilation
rate Q) or to reduce the emission rate E. Increasing Q increases energy costs on
a continuing basis, and has been discouraged as a solution (after all, it is only a
“dilution” method). However, if we know exactly where the emissions are
originating, and if we are smart about where to place the fan, ventilation is
very effective. For example, to control cooking emissions, a local vent fan is
installed over the stove. It exhausts only a small volume of air compared with
the whole house, but that small volume may contain 90% of the emissions of
certain pollutants. In another example, if the problem is radon emissions entering through the basement of a house, the basement could be isolated and ventilated continuously as shown in Figure 9.11a on p. 369. In very cold climates, the
ventilation technique shown in Figure 9.11a results in high energy costs. A
more energy-efficient method to mitigate radon in basements is shown in Figure 9.11b, also on p. 369.
For other problems, such as molds and mildews, the solution may require
that a moisture leak be corrected, followed by cleaning and disinfection of the
contaminated area. Recently, smoking-related problems have been eased by
bans on the smoking of tobacco within the confines of public buildings. Each
IAQ problem is unique and requires careful investigation, analysis, design, and
implementation to solve the problem in a cost effective manner.
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368
Chapter Nine
Figure 9.10
Rows
Spreadsheet solution to Example 9.10.
Columns
C
B
D
E
F
G
3
4 Example Problem 9.10 Spreadsheet solution for smoky nightclub
5
0.4 1/hr
K=
6 Define system parameters
Czero =
0 mg/L
Also, Cin=0
7 Define initial conditions
0 minutes
t zero =
8
Q = 7500 cu m/hour
9
840 mg/hour
E=
10
V = 3000 cubic meters
11
1 minute
12 Define simulation params deltime =
13
14 Equation to be solved VdC/d t = QCin + E - QCout - kCoutV
dC/dt = (Q/V)Cin + E/V - (Q/V)Cout - kCout
or
15
16 Also, new estimate for Cout: Cnew = Cold + dC/dt*del t
Start in row 23 with Cold = Czero; calc dC/dt from above;
METHOD
multiply by del t & add result to Cold to get Cnew
18
copy Cnew into next row cell for Cold and repeat
19
20
With spreadsheets, a formula with the cell
21 Calculations and results
t,
min
C
old
dCdt
C
new
22
address as, for example, E10, when copied
23
0 0.0000 0.2800 0.0047 will change to follow the cell position. But
24
1 0.0047 0.2665 0.0091 with the cell address given as $E$10, the cell
25
2 0.0091 0.2536 0.0133
address will not be indexed.
26
3 0.0133 0.2413 0.0174
27
4 0.0174 0.2297 0.0212
28
5 0.0212 0.2186 0.0248
29
6 0.0248 0.2080 0.0283
30
7 0.0283 0.1979 0.0316
31
8 0.0316 0.1884 0.0347
32
9 0.0347 0.1793 0.0377 Note break here; rows 33–80 are hidden
81 58 0.0911 0.0158 0.0914
82 59 0.0914 0.0151 0.0916
83 60 0.0916 0.0143 0.0918 One hour
84 61 0.0918 0.0136 0.0921
85 62 0.0921 0.0130 0.0923 Note: rows 86–140 are hidden
141 118 0.0963 0.0008 0.0963
142 119 0.0963 0.0008 0.0963
143 120 0.0963 0.0007 0.0963 Two hours
144 121 0.0963 0.0007 0.0963
145 122 0.0963 0.0007 0.0963 Note: rows 146–200 are hidden
201 178 0.0965 0.0000 0.0965
202 179 0.0965 0.0000 0.0965
203 180 0.0965 0.0000 0.0965 Three hours
Cooper.book Page 369 Monday, June 23, 2014 9:58 AM
Other Important Topics
369
Figure 9.11a
One method of
mitigating radon
entry into the
living space of
a house.
(Cooper, Dietz, and
Reinhart 2000.)
Fan forces
outdoor air
into basement
Radon-laden air
exits through
windows
R adon seeps into basement
Figure 9.11b A more energy-efficient method to control radon. (Nadakavukaren 2011.)
outside fan
draws radon
away from house
jo
in
ts
ts
join
sealant
in
en
op
gs
cra
cks
sheet metal
covers exposed area
sealant
sump
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370
Chapter Nine
9.4 Noise Pollution
Noise pollution is a serious environmental problem and gets a lot of attention from agencies such as the Federal Highway Administration (FHWA) and
the Federal Aviation Authority (FAA). Exposure to high levels of noise has
been shown to have deleterious health effects, and may cause permanent hearing loss.
Noise can be defined as unwanted sound. People use four fundamental criteria for evaluating noise: loudness, frequency, duration, and subjectivity. Each
of these criteria is discussed below.
Loudness
Sound is a form of energy; it is propagated through air as a sinusoidal
wave of pressure fluctuations. The loudness or intensity of the sound is
directly related to the amplitude of that sound wave. The pressure fluctuations
cause the ear drum to move back and forth, stimulating the auditory nerves
and creating the sensation of sound. The human ear can sense pressure fluctuations as low as 2 × 10–5 Pa (the threshold of hearing) up to and above 63 Pa (the
threshold of pain). This represents a pressure change by a factor of more than 3
million! Figure 9.12 shows typical sound pressure levels for various community noises.
The six-order-of-magnitude range of pressure fluctuations cannot be conveniently represented by a linear scale. Thus, the decibel (dB) scale was developed to describe loudness. The decibel is defined by:
Ê P2 ˆ
SPL (dB ) = 10 log10 Á 2 ˜
Ë P0 ¯
(9.12)
where:
P0 = the reference pressure (2 × 10–5 Pa)
P = the sound pressure of concern (Pa)
The use of dB indicates that loudness is being reported as a sound pressure
level (SPL), which is different from sound pressure. While sound pressures can
be added linearly, SPLs add logarithmically. For example, if the sound pressure
is doubled, the increase in the SPL is only 3 dB. Computing the sum of several
SPLs in dB may be accomplished using the equation:
SPL total = 10 log 10 Â 10SPLi /10
(9.13)
EXAMPLE 9.11
A sound from a motorcycle 100 feet away from you is measured to be 70 dB at
your location. If two more identical motorcycles next to the first one are started
up, what is the resulting overall sound pressure level that you experience?
Cooper.book Page 371 Monday, June 23, 2014 9:58 AM
Other Important Topics
371
SOLUTION
We are adding sounds, so we can use Eq. (9.13):
SPL total = 10 log 10 Â 10SPLi /10
(
SPL total = 10 log 10 1070/10 + 1070/10 + 1070/10
)
= 10 log 10 ( 30, 000, 000 )
= 74.8 dB
A change of 10 dB is generally perceived to be a doubling or halving of the
sound that reaches the human ear. In outdoor situations, generally it takes a
change in SPL of greater than 3 dB to be noticeable, but changes in other components, such as the frequency or tone, can be perceived with no change in SPL.
Figure 9.12 Typical community noises and their sound pressure levels. (Federal Highway
Administration.)
Common Outdoor Noises
Sound
Pressure
(μPa)
Sound
Pressure
Level (dB)
Common Indoor Noises
6,324,556
110
Rock band at 5 m
2,000,000
100
Inside subway train (New York)
Food blender at 1 m
Garbage disposal at 1 m
Shouting at 1 m
Vacuum cleaner at 3 m
Normal speech at 1 m
Jet flyover at 300 m
Gas lawn mower at 1 m
Diesel truck at 15 m
632,456
90
Noisy urban daytime
200,000
80
63,246
70
20,000
60
Gas lawn mower at 30 m
Commercial area
Quiet urban daytime
6,325
50
Quiet urban nighttime
Quiet suburban nighttime
2,000
40
632
30
200
20
63
10
20
0
Quiet rural nighttime
Large business office
Dishwasher in next room
Small theatre, large conference
room (background)
Library
Bedroom at night
Concert hall (background)
Broadcast and recording studio
Threshold of hearing
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372
Chapter Nine
Frequency
Picturing sound as a wave, the frequency is the number of oscillations per
second, and has the unit of sec–1 or Hertz (Hz). The human ear can distinguish
among a large range of frequencies—extending from about 20 Hz to 20,000 Hz.
The different frequencies (rates of the pressure fluctuations) produce the different tones that we hear. For instance, a flute has a much higher frequency than a
bass guitar, and their tones are discerned as being quite different. A blend of
many different tones can make some sounds become beautiful music while
blends of other sounds simply become noise. The frequency times the wavelength equals the speed of sound. Although the speed of sound in air changes
slightly with temperature or altitude, a good approximation for the speed of
sound in air is 343 m/s or 1,100 ft/s.
fλ=S
(9.14)
where:
f = frequency (Hz)
S = speed of sound (m/s)
λ = wavelength (m)
EXAMPLE 9.12
What is the wavelength of a 550 Hz tone?
SOLUTION
Rearranging Eq. (9.14):
l=
S
f
l=
343 m/sec
550 Hz
l = 0.624 m
The human ear does not detect all frequencies equally well. Low frequencies
(less than 500 Hz) are not heard very well; neither are high frequencies (greater
than 10,000 Hz). When evaluating community noise, a broad approach is used. In
this approach, all frequency contributions are adjusted to approximate the way the
ear hears them, then the contributions are summed to a single number. The Aweighted scale mimics the way our ears respond to moderate sounds. We design
and build instruments to measure noise levels to account for the nonlinear
response of the ear at low and high frequencies based on the A-weighted scale.
Most regulations and evaluations applicable to community noise use the Aweighted scale and are reported as dB(A), decibels that have been A-weighted. It
should be noted that dB are the units and are separated from the weighting scheme
through the use of parentheses, so often we just see noise levels reported in dB.
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Other Important Topics
373
Duration
A Fourth-of-July fireworks show may be loud, but each burst only lasts for
a fraction of a second. Traffic noise may not be as intense, but the duration is
quite long. Quantifying how sound varies with time completes the description
of noise (along with loudness and frequency). Some of the more commonly
used community noise descriptors are: maximum sound level, [Lmax (t)], statistical sound levels [Lxx (t)], equivalent sound level [Leq (t)], and day/night level
[Ldn]. In each of these descriptors the capital L represents that each is a sound
pressure level with the units of dB. The (t) indicates each is given for a specific
period of time. By definition, Ldn is a 24-hour descriptor.
Lmax (t) represents the maximum noise level that occurs during a time period
equal to t. For example, 60 dB(A); Lmax (1 hr) indicates that the maximum sound
pressure level that occurs during a 1-hour period is 60 dB weighted by the Ascale. More description is possible with statistical descriptors (Lxx). The subscript
xx indicates the percent of time that the listed level is exceeded. For instance, a
reported sound level of 60 dB(A); L10 (1 hr), means that a sound pressure level of
60 dB on an A-weighted scale was exceeded 10 percent of the time in a one-hour
time period. The numeric value may be any fraction of the time, but L10 , L50 , and
L90 are most commonly used. L90 is the sound pressure level exceeded 90 percent
of the time and is commonly used as the value for the background level of noise.
The equivalent sound pressure level, Leq, is a single number descriptor that
represents the value of a non-varying tone that contains the same acoustic energy
as all the varying sounds that occur over the same time period. One might think
of Leq as an average acoustic energy descriptor. It should be noted that the average
energy is not an average of SPL over the time period because of the logarithmic
nature of the SPL. Leq has the advantage of allowing different noises that occur
during the same time period to be added. It has become the metric of choice for
many types of community noise description. Leq is mathematically described as:
Leq
È t2 10SPL(t )/10 ˘
Út
˙
= 10 log10 ÍÍ 1
˙
t2 - t1
ÍÎ
˙˚
(9.15)
The descriptor Ldn, the day/night level, takes into account that not only are
loudness, frequency, and duration important, but the time of day the sound
occurs is also important. Ldn consists of hourly Leq (A-weighted) values, energyaveraged over the entire 24-hour period, with a 10 dB (A-weighted) penalty
added to the sound level during the time period from 10 PM until 7 AM. Due to
the logarithmic nature of dB, the 10 dB penalty in effect requires a sound pressure during nighttime hours to be one-tenth the pressure during daytime hours
to be equivalent. Mathematically:
Ldn
Ld
Ln +10 ˆ ˘
È
Ê
Ê 1ˆ
Í
10
= 10 log10 Á ˜ Á 15 ¥ 10 + 9 ¥ 10 10 ˜ ˙
ÍË 24 ¯ Á
˜¯ ˙
Ë
Î
˚
where: Ld, Ln = SPL (Leq) in each day and night hour, respectively.
(9.16)
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374
Chapter Nine
EXAMPLE 9.13
You have been supplied ten SPL measurements, equally spaced in time. Calculate the Leq.
Time
Sound Pressure Level (dB)
1
2
3
4
5
6
7
8
9
10
67.8
55.9
63.7
71.1
67.2
68.3
60.0
69.1
52.8
59.0
SOLUTION
Since the measurements are evenly spaced in time, we can use a variation of Eq.
(9.15). In this case, we will replace the integration with a simple summation.
SPL i
È
Ê 1ˆ
Leq = 10 log10 ÍÁ ˜ Â 10 10
ÍË n ¯
Î
˘
˙
˙
˚
È
Ê 6.78 + 10 5.59 + 106.37 + 107.11 +
ˆ˘
Ê 1 ˆ 10
˜˙
= 10 log10 ÍÁ ˜ Á
ÍË 10 ¯ Á 6.72
6.83
6.00
6.91
5.28
5.9 ˜ ˙
+ 10
+ 10
+ 10
+ 10
+ 10 ¯ ˙˚
Ë 10
ÍÎ
= 66.4 dB
Subjectivity
The three components of sound discussed above (loudness, frequency,
duration) are quantifiable. However, individuals have different responses to
various sounds, and the degree of pleasure or irritation felt when one hears a
certain sound is quite subjective. Rock music may be a pleasing sound to one
listener but only noise to another person. Transportation noise is a common
problem in urban areas, but the degree of annoyance caused by this noise is
highly subjective, and may vary considerably between day and night. For this
reason, evaluation criteria are usually based on attitudinal surveys.
Although a single very loud noise can result in an immediate hearing loss,
a single loud noise is not the typical problem with community noise. In modern society, noise tends to be more of a chronic problem, resulting in reduced
hearing ability after long-term exposure. In the short-term, annoyance or irrita-
Cooper.book Page 375 Monday, June 23, 2014 9:58 AM
Other Important Topics
375
tion is of more importance. Studies have shown that excessive noise interferes
with restful sleep, causes increased tension, inhibits communication ability,
and reduces the learning abilities of students.
Noise Attenuation
Noise levels decrease with increased distance from a source; this is called
geometric spreading. The attenuation due to geometric spreading may be
characterized by the type of source. If noise is emitted from a single location,
the source is referred to as a point source. An air conditioning unit, large industrial pump or blower, or a stationary idling truck is a point source. If there are
many similar point sources in a row (such as vehicles on a highway), the entire
grouping is considered a line source. Highway traffic could be considered to be
a set of moving point sources, but is more commonly modeled as a line source.
For a point source the sound energy spreads out over the surface of an ever
expanding sphere. The energy dissipates as the square of the distance from the
source (the radius of the sphere squared). The intensity and the root-meansquare SPL decrease proportionally to the inverse of the square of the distance
from the source (inverse-square law). Since the SPL is always defined at some
initial distance from the point source, the inverse-square law appears as follows:
Êr ˆ
DSPL (dB ) = 10 log 10 Á 1 ˜
Ër ¯
2
(9.17)
2
where:
ΔSPL (dB) = difference in SPL at the two locations
r1
= distance from the source to point 1
r2
= distance from the source to point 2
A common rule of thumb is that if we double the distance (from a point
source) we reduce the noise level by 6 dB. But for a line source, geometric attenuation is like spreading the energy over the surface area of a cylinder, which
results in decreases in the SPL that are inversely proportional to distance (not
distance squared). Using the same mathematical procedure as for a point
source, a line source sound decreases with distance as:
Êr ˆ
DSPL (dB ) = 10 log 10 Á 1 ˜
Ë r2 ¯
(9.18)
These calculations are demonstrated in Example 9.14.
EXAMPLE 9.14
What is the decrease in dB when the distance from a point source to a receiver
is doubled? What is the decrease in dB when the distance from a busy highway increases from 100 ft to 200 ft?
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376
Chapter Nine
SOLUTION
Point source: use Eq. (9.17):
Êr ˆ
DSPL (dB ) = 10 log 10 Á 1 ˜
Ë r2 ¯
Ê 1ˆ
DSPL (dB ) = 10 log 10 Á ˜
Ë 2¯
2
2
Ê 1ˆ
= 10 log 10 Á ˜
Ë 4¯
= -6 dB
Highway: use Eq. (9.18):
Êr ˆ
DSPL (dB ) = 10 log 10 Á 1 ˜
Ë r2 ¯
Ê 100 ˆ
DSPL (dB ) = 10 log 10 Á
Ë 200 ˜¯
Ê 1ˆ
= 10 log 10 Á ˜
Ë 2¯
= -3 dB
According to Example 9.14, it could be expected that the sound level
would decrease by 3 dB for each doubling of distance from a highway. However, in reality the highway is not in free space but is close to the ground. As a
result, the interaction of the sound wave with the ground surface causes excess
attenuation above what would be expected from just geometric spreading. The
excess attenuation effects are related to the type of soil, ground cover, and surface topography. Ground effects are difficult to predict. It has been determined
that a value of 3 dB per doubling of distance is more typical of hard surfaces
(like parking lots). More attenuation occurs for soft surfaces (like loose soil
with vegetative coverings). Instead of a 3-dB drop for each doubling of distance, a good rule of thumb for the decrease of SPL with distance from a line
source located in “soft” surroundings is 4.5 dBA.
Obstructions in the noise path may cause diffraction or reflection of the
sound, which reduces the noise transmitted beyond the obstruction. This is the
whole purpose of noise barriers, or more specifically, the large highway noise
walls that are often constructed between busy highways and residential neighborhoods. Properly designed noise walls are effective ways to significantly
reduce noise levels at residential areas near large highways.
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Other Important Topics
377
Legislation and Regulations
Federal legislation dealing with noise pollution started with the Housing
and Urban Development Act of 1965, continued with the Noise Control Act of
1972, and was strengthened by the Quiet Communities Act of 1978. This environmental legislation required noise pollution to be addressed in communities,
and led to the development of analysis methodologies and models to document and mitigate noise impacts.
Transportation facilities, as a category, are the most prevalent source of
noise pollution in communities. Busy airports or large highways that are in
close proximity to a community are often the subjects of much angst and discussion with regard to noise impacts. The Federal Highway Administration
(FHWA) has defined procedures that must be followed to predict the worst
hourly noise levels where human activity normally occurs. Noise Abatement
Criteria for various land uses are shown in Table 9.8. Regulations require that
when the Noise Abatement Criteria are approached or exceeded, noise mitigation must be considered. If mitigation is feasible (possible) and reasonable (cost
effective), then abatement measures must be implemented. Under the law,
abatement steps may not be taken if it is deemed infeasible or unreasonable
even though the criteria are exceeded. This leads to the requirement that each
project be documented and considered individually. If modeling shows that
existing noise levels near highways will be increased substantially due to a
highway improvement project, even though the Noise Abatement Criteria are
not exceeded, an abatement analysis must occur.
Table 9.8
Noise Abatement Criteria
One-hour sound levels (dBA)
Leq
L10
Lands on which serenity and quiet are of extraordinary significance
and serve an important public need, and where the preservation of
quiet is required for the area to continue to serve its intended purpose.
57 (ext)
60 (ext)
Picnic areas, recreation areas, playgrounds, active sports areas, residences, motels, hotels, schools, churches, libraries and hospitals
67 (ext)
70 (ext)
Other developed lands, properties, or activities
72 (ext)
75 (ext)
—
—
52 (int)
55 (int)
Description of Activity
Undeveloped lands
Residences, hotels, motels, public meetings rooms, schools, churches,
libraries, hospitals, and auditoriums
Note: exterior and interior sound levels are indicated by (ext) and (int), respectively.
Source: Code of Federal Regulations, Title 23 United States Code, Chapter 772 (2010).
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378
Chapter Nine
SUMMARY
This chapter has presented information on several important topics—risk
assessment, energy, indoor air quality, and noise pollution. Students who are
future engineers may well find themselves working in one or more of these
areas. Even if one doesn’t work directly within any of the branches of environmental engineering discussed in this chapter, having the background knowledge of these areas is important because it may well inform your work in
another area.
PROBLEMS
9.1. A groundwater is contaminated with 0.1 mg/L of trichloroethylene.
Determine the lifetime cancer risk for a 70-kg male associated with utilization of this groundwater as a potable supply for 30 years. From Table
9.3 the cancer slope factor for trichloroethylene is 0.011 (mg/kg-day)–1.
9.2. For the groundwater described in Problem 9.1, determine the removal
efficiency required to achieve compliance with the MCL (see Chapter 5)
of 0.005 mg/L. Assuming that treatment is provided to reduce the concentration to the MCL, determine the lifetime cancer risk for a 70-kg male
from drinking the treated water for 30 years.
9.3
A groundwater was contaminated by a spill, and benzene and vinyl chloride remain in the water, each at a concentration of 2.0 mg/L. The
groundwater is used as a potable water supply for a town of 75,000 people. Using the data presented in this chapter, calculate the LADD for a
male who weighs 80 kg and who lives in this town for 70 years, and
drinks only the groundwater with no removal of these chemicals. Repeat
for a female who weighs 55 kg and lives there for 70 years.
9.4
The MCLs for benzene and vinyl chloride in potable water are 0.005 mg/
L and 0.002 mg/L, respectively. The town of Problem 9.3 can treat the
groundwater to get the concentrations of those chemicals down to their
MCLs. Calculate the average additional risk of cancer to residents in a
town that uses the treated water exclusively. For this problem, assume
that half the population is males (80 kg) and half is females (55 kg), and
all live in this town for 70 years.
9.5
For the town of Problem 9.3, calculate the expected total number of additional cancers per year due to drinking the untreated water. Compare
your answer to that if the water is treated (Problem 9.4).
9.6
Starting from the classic first-order kinetic expression, Eq. (9.4), derive
Eq. (9.5).
9.7
A sample of a radioactive element had a mass of 35.0 g on Jan. 1, 2010. On
Jan. 1 2014, its mass is measured to be 25.0 g. Calculate the half-life.
9.8
Do some research on how much nuclear electric power was generated in
Japan in 2010. After the Fukushima catastrophe in 2011, there was talk in
Cooper.book Page 379 Monday, June 23, 2014 9:58 AM
Other Important Topics
379
Japan of shutting down all its nuclear power plants. Calculate the
increase in annual fossil fuel use that Japan would need to make up for
the lost energy if the country completely abandons nuclear power. Use
the same assumptions about fossil fuel composition and heating value as
in Example 9.5. What is the current situation in Japan with regard to its
use of nuclear power?
9.9
Calculate the time required (years) for a waste with a mixture of radioactive wastes with various half-lives to decay to the point that the overall
radioactivity level is less than 10% of its initial value. Assume equal parts
of each of three wastes, with half-lives of 5 years, 10 years, and 20 years,
respectively. Assume none of the decay products are radioactive. How
would your approach to this problem change if the third waste had a
half-life of 200 (instead of 20) years?
9.10 A sample of a radioactive element has a half-life of 180 days; what percent
of this sample will still be radioactive after 3 years? After 5 years?
9.11 An older home was built in an area that has a high radon flux (10 pCi/m2-s).
The concrete slab did not have a plastic vapor barrier installed, so assume
that 85% of the Ra flux passes through the slab and into the house. The
house has a slab area of 2,400 ft2 and an enclosed volume of 24,000 ft3 and
averages 1.2 ACH (air changes per hour). The half-life of radon is 3.82 days.
Assuming the ambient concentration is negligible, estimate the steady-state
indoor concentration of radon.
9.12 A homeowner likes to burn candles in her bedroom. The candles look and
smell nice, but they have a tiny bit of lead in their wicks. Assume the lead
emission rate from all candles burning simultaneously is 0.20 mg/min.
Assume the bedroom is 14 ft × 16 ft × 8 ft tall, and has 1.2 ACH. Calculate
the steady-state concentration of lead in the air that would be achieved if
candles are burned continuously. Give your answer in µg/m3.
9.13 A person with no knowledge of the potential for carbon monoxide poisoning (and no common sense) brings his charcoal grill into his apartment
when a sudden rain storm begins. The apartment has a volume of 7,000 ft3
and a ventilation rate of 0.5 ACH. The ambient CO concentration and the
initial indoor concentration of CO are both zero. After one hour, the CO
concentration is 80 mg/m3. Calculate the emission rate of CO from the charcoal grill in mg/minute. What would the concentration be after two hours?
9.14 Solve Problem 9.13 numerically, using spreadsheets, but assume that the
emission rate of CO is 1.0 gram per minute. Produce a figure to show how
the CO concentration in the apartment varies with time. How long will it
take before the CO concentration reaches the NAAQS of 40 mg/m3? How
long before it reaches the serious health effects level of 300 mg/m3?
9.15 Your college roommate likes to burn incense in the dorm room. The room
is 4 m × 4 m × 2.5 m, and has 1.5 ACH. If the incense emits PM at the rate
of 2.5 g/hr, calculate the concentration of PM in the room after 1 hour, in
µg/m3.
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380
Chapter Nine
9.16 If you measure a noise to be 72 dB(A) at 20 meters, what will be the resulting level at 100 meters if (a) the source is a point source, and (b) the source
is a line source?
9.17 A large industrial fan has been temporarily located 150 feet north of a
home. At the industry fenceline (50 feet), the fan noise is measured as 72
dB(A). What is the noise level at the home?
9.18 Noise from a highway is measured as 70 db(A) at 50 feet. A noise wall is
erected at 50 ft that absorbs or reflects 40% of the noise energy. What is the
noise level just behind the wall?
9.19 The home in Problem 9.17 is also located 200 feet from a highway. The
highway noise is measured at 50 ft as 70 db(A). What is the total noise
level at the home?
9.20 A large air compressor has been temporarily located 175 feet away from a
home which is also 100 feet away from a highway. The compressor noise
(at 50 feet) was measured and rated at 61.2 dB(A); Leq, while the roadway
noise (at 50 feet) was measured and rated at 66.1 dB(A); Leq. Calculate the
combined noise level at the home.
9.21 If you see a flash of lightning and 4.0 seconds later you hear the thunder,
how far away did the lightning bolt hit (km)?
9.22 A certain mobile home was renovated with new carpets and new laminated wood paneling. Shortly after the renovation, the formaldehyde
(HCHO) emissions were estimated to be 50.0 mg/hr. The air flow into
(and out of) the home was 115 m3/hr, and the inside volume of the home
was 85 m3. Formaldehyde is destroyed in air with a first-order rate constant of 0.4 hr–1. Starting from a basic material balance approach, derive
an equation to predict the steady-state room air concentration for this situation. Assuming the concentration of HCHO in the outside air is zero,
calculate the steady-state concentration of HCHO inside this mobile
home. Give your answer in µg/m3.
REFERENCES
American Water Works Association. 2013. “Water and Hydraulic Fracturing. White
paper from the AWWA. Accessed August 2013. www.awwa.org/fracturing
Code of Federal Regulations. 2010. “Title 23, Highways; Part 772, Procedures for Abatement of Highway Traffic Noise and Construction Noise.”
CDC (Centers for Disease Control). 2014a. “Motorcycle Crash-Related Data.” Accessed
April 2014. http://www.cdc.gov/features/dsmotorcyclesafety/index.html
CDC. 2014b. “Smoking and Tobacco Use.” Accessed April 2014. http://www.cdc.gov/
Tobacco/data_statistics/fact_sheets/fast_facts/index.htm
Cooper, C. D., J. D. Dietz, and D. R. Reinhart. 2000. Foundations of Environmental Engineering. Long Grove, IL: Waveland Press.
Emmerich, S. J., and A. K. Persily. 1998. “Energy Impacts of Infiltration and Ventilation
in U.S. Office Buildings Using Multizone Airflow Simulation.” Proceedings of IAQ
and Energy 98 Conference, New Orleans, LA, Oct. 22–27. Also available at http://
www.bfrl.nist.gov/IAQanalysis/docs/PB96-165782.pdf
Cooper.book Page 381 Monday, June 23, 2014 9:58 AM
Other Important Topics
381
IEA (International Energy Agency). 2012. Key World Energy Statistics-2012. Paris: IEA.
Nadakavukaren, A. 2011. Our Global Environment: A Health Perspective. 7th ed. Long
Grove, IL: Waveland Press.
National Safety Council (NSC). 2014. “The Odds of Dying.” Injury Facts, 2014 Edition.
Accessed April 2014. http://www.nsc.org/news_resources/
injury_and_death_statistics/Documents/2014-Injury-Facts-43.pdf
NEI (Nuclear Energy Institute). 2013. “World Statistics—Nuclear Energy Around the
World.” Accessed April 2014. http://www.nei.org/Knowledge-Center/
Nuclear-Statistics/World-Statistics
NEI. 2014. Emissions Avoided, State by State, 2013. Last revised April 2014. Accessed May
2014. http://www.nei.org/Knowledge-Center/Excel-Files/Emissions-Avoidedby-the-US-Nuclear-Industry-State
US EIA (Energy Information Administration). 2012a. “Annual Energy Review 2011.”
DOA/EIA-0384 (2011). Accessed July 2013. http://www/eia.gov/totalenergy/
data/annual/pdf/aer.pdf
US EIA. 2012b. “Monthly Energy Review.” (March 2012). Table 10.1.
US EPA (Environmental Protection Agency). 1989 (December). “Risk Assessment Guidance for Superfund, Vol. I—Human Health Evaluation Manual (Part A).” EPA/
540/1-89/002. Also available at http://www.epa.gov/oswer/riskassessment/
ragsa/pdf/rags_a.pdf
US EPA. 1998 (May). “EPA’s Final Certification Decision for the Waste Isolation Pilot
Plant.” EPA Fact Sheet 5, #402-98-002. Accessed July 2013. http://www.epa.gov/
plainlanguage/documents/5-fact.pdf
US EPA. 2000. “Formaldehyde—Hazard Summary.” Technology Transfer Network, Air
Toxics Website. Last revised January 2000. Accessed July 2013.
http://www.epa.gov/ttn/atw/hlthef/formalde.html
US EPA. 2003. “EPA Assessment of Risks from Radon in Homes.” EPA 402-R-03-003.
Office of Radiation and Indoor Air. Washington, DC: GPO.
US EPA. 2013. “Drinking Water Contaminants.” Last revised June 2013. Accessed July
2013. http://water.epa.gov/drink/contaminants/index.cfm#List
Wilson, W. R., and E. A. C. Crouch. 1987. “Risk Assessment and Comparisons: An Introduction.” Science, 236:267–270.
World Nuclear News. 2013. “Public Comment on French Waste Disposal.” Accessed
July 2013. http://www.world-nuclear-news.org/
WR_Public_comment_on_French_waste_disposal_16051311.html
Yablokov, A., V., Nesterenko, and A. Nesterenko. 2009. “Chernobyl: Consequences of
the Catastrophe for People and the Environment.” Annals of the New York Academy of
Sciences, 1181.
Cooper.book Page 382 Monday, June 23, 2014 9:58 AM
Cooper.book Page 383 Monday, June 23, 2014 9:58 AM
APPENDIX
A
Conversion Factors
How to use these tables:
1 unit from column A = table entry units from row B
Examples: 1 lbm = 453.6 g; 1 kg = 2.205 lbm
Table A.1
Mass
Row B
Column A
lbm
g
gr
kg
ton
tonne
(metric
ton)
g
gr
453.6
1.0
0.0648
1,000
9.07 (10)5
(10)6
7,000
15.43
1.0
1.54 (10)4
1.40 (10)7
1.54 (10)7
lbm
1.0
0.002205
0.000143
2.205
2,000
2,205
kg
ton
tonne
0.00050
1.10 (10)–6
7.14 (10)–8
0.0011
1.0
1.102
0.000454
1.0 (10)–6
6.48 (10)–8
0.001
0.907
1.0
µm
km
miles
10
3.05 (10)5
2.54 (10)4
1.0
1.0 (10)9
1.61 (10)9
0.001
3.05(10)–4
2.54 (10)–5
1.0 (10)–9
1.0
1.609
6.21 (10)–4
1.894 (10)–4
1.578 (10)–5
6.22 (10)–10
0.6215
1.0
0.4536
0.001
6.48 (10)–5
1.0
907
1,000
Note: 1 Gt = (10)9 tonnes = 1015 grams
Table A.2
Length
Row B
Column A
m
ft
in.
µm
km
miles
m
1.0
0.3048
0.0254
10–6
1,000
1,609
ft
3.281
1.0
0.0833
3.28 (10)–6
3,281
5,280
in.
39.37
12
1.0
3.94 (10)–5
3.94 (10)4
6.336 (10)4
383
6
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384
Appendix A
Table A.3
Area
Row B
ft2
m2
Ac
Ha
mi2
km2
1.00
10.76
43,560
1.076 (10)5
2.788 (10)7
1.076 (10)7
0.0929
1.00
4047
10,000
2.59 (10)6
1.0 (10)6
2.29 (10)–5
2.47 (10)–4
1.00
2.471
640
247.1
9.29 (10)–6
0.0001
0.4047
1.00
259
100
3.587 (10)–8
3.86 (10)–7
0.001562
0.003861
1.00
0.386
9.29 (10)–8
1.00 (10)–6
0.004047
0.01
2.59
1.00
Column A
ft2
m2
Acres (Ac)
Hectares (Ha)
mi2
km2
Table A.4
Volume
Row B
ft3
L
gal
m3
1.0
0.03531
0.1337
35.31
28.32
1.0
3.785
1,000
7.481
0.2642
1.0
264.2
0.02832
0.001
0.003785
1.0
Column A
ft3
L
gal
m3
Table A.5
Force
Row B
N
lbf
kg-m/s2
lbm-ft/s2
1.0
4.448
1.0
0.1383
0.2248
1.0
0.2248
0.03108
1.0
4.448
1.0
0.1383
7.232
32.17
7.232
1.0
Column A
N
lbf
kg-m/s2
lbm-ft/s2
Note: 1 dyne = 10.0 µN and 1 N = 1(kg-m)/s2
Table A.6
Pressure
Row B
Column A
atm
psi
mm Hg
in. H20
mbar
Pa
atm
psi
1.0
0.068
1.316 (10)–3
0.002458
9.87 (10)–4
9.87 (10)–6
14.70
1.0
0.0193
0.03614
0.0145
1.45 (10)–4
Note: 1 Pa = 1 N/m2
Pa (N/m2)
mm Hg
in. H2O
mbar
760
51.7
1.0
1.868
0.750
0.0075
406.8
27.67
0.535
1.0
0.4016
0.00402
1,013
101,325
68.9
6,891
1.333
133.3
2.49
249
1.0
100
0.01
1.0
Cooper.book Page 385 Monday, June 23, 2014 9:58 AM
Conversion Factors
Table A.7
385
Energy
Row B
Column A
Btu
kJ
cal
ft-lbf
kWh
liter-atm
Btu
kJ
1.0
0.948
0.00397
0.001285
3,412
0.0961
cal
1.055
1.0
0.004184
0.001356
3,600
0.01014
252
239
1.0
0.3239
8.60 (10)5
24.22
ft-lbf
kWh
liter-atm
–4
778
737.5
3.087
1.0
2.66 (10)6
74.74
2.93 (10)
2.778 (10)–4
1.163 (10)–6
3.766 (10)–7
1.0
2.82 (10)–5
10.41
98.62
0.0413
0.01338
3.55 (10)4
1.0
Note: 1 J = 1 N-m
Table A.8
Power
Row B
Column A
W
W
kW
ft-lbf /s
hp
Btu/hr
kW
1.0
1,000
1.356
745.5
0.293
ft-lbf/s
0.001
1.0
0.001356
0.7455
2.93 (10)–4
hp
0.737
737.6
1.0
550
0.216
0.00134
1.341
0.001818
1.0
3.93 (10)–4
Btu/hr
3.412
3,412
4.63
2,545
1.0
Note: 1 W = 1 J/s
Table A.9
Speed
Row B
Column A
ft/s
m/s
mi/hr
ft/min
Table A.10
ft/s
m/s
mi/hr
ft/min
1.0
3.281
1.467
0.01667
0.3048
1.0
0.447
0.00508
0.6818
2.237
1.0
0.01136
60.0
196.8
88.0
1.0
Viscosity
Row B
Column A
cp
g/cm-s
lbm/ft-hr
kg/m-hr
cp
1.0
100
0.413
0.277
g/cm-s
lbm/ft-hr
kg/m-hr
0.01
1.0
0.00413
0.00277
2.42
242
1.0
0.670
3.61
361
1.492
1.0
Cooper App B.fm Page 386 Thursday, March 3, 2016 10:15 AM
APPENDIX
Selected Physical and
Chemical Properties
• Selected properties of air
• Values of the Ideal Gas Law constant R
• Selected properties of liquid water
• Solubility constants of selected solids
• Ionization constants of selected acids and bases
• Henry’s Law constants for selected gases in water
• Henry’s Law constants for selected organic compounds at 25 °C
• Enthalpies of saturated steam and water
• Standard heats of combustion for various organic compounds
Table B.1
Selected Properties of Air
Approximate Composition
(2 gas approximation)
(3 gas approximation)
(4 gas approximation)
(% by moles or volume)
Molecular Weight of Dry Air
79% N2, 21% O2
78% N2, 21% O2, 1% Ar
78.09% N2, 20.94% O2,
0.93% Ar, 0.04% CO2
MW = 28.85*
MW = 28.96
MW = 28.97
* MW calculated as ΣyiMWi where yi = mole fraction of gas i; MWi = molecular weight of gas i
Density*
Temp, °F
77
150
300
500
1000
2000
3
lb/ft
0.0740
0.0650
0.0521
0.0412
0.0275
0.0161
Specific Heat (Cp)
Btu/lb-°F or cal/g-°C
kg/m3
1.185
1.041
0.834
0.660
0.440
0.258
0.240
0.240
0.241
0.242
0.246
0.260
*at standard pressure of 1.00 atm.
386
B
Cooper.book Page 387 Monday, June 23, 2014 9:58 AM
Selected Physical and Chemical Properties
Table B.2
387
Values of the Ideal Gas Law Constant R [R = (PV/nT)]
Units
R
(Pressure-Volume) / (Matter-Temperature)
0.08206
82.06
62.36
1.314
0.08314
998.9
0.7302
21.85
555.0
10.73
1545.
8.314
atm-L / gmol-K
atm-cm3 / gmol-K
mm Hg-L / gmol-K
atm-ft3 / Ibmol-K
bar-L / gmol-K
mm Hg-ft3 / Ibmol-K
atm-ft3 / Ibmol-R
in. Hg-ft3 / Ibmol-R
mm Hg-ft3 / Ibmol-R
psia-ft3 / Ibmol-R
psfa-ft3 / Ibmol-R
Pa-m3 / gmol-K
1.987
8314.
cal / gmol-K
J / kgmol-K
Note: 1 pascal (Pa) = 1 N/m2; 1 N-m = 1 joule (J)
Table B.3
Selected Properties of Liquid Water
Temperature °F
Density
Ibm/ft3
Viscosity
cp
40
50
60
70
80
90
100
200
62.43
62.42
62.37
62.30
62.22
62.11
62.00
60.13
1.55
1.31
1.13
0.98
0.86
0.76
0.68
0.30
Vapor
Pressure
psi
0.122
0.178
0.256
0.363
0.507
0.698
0.949
11.53
Specific Heat
Btu/lb-°F
or cal/g-°C
1.004
1.002
1.001
1.000
0.999
0.999
0.999
1.005
Adapted from Cooper, C. D., J. D. Dietz, and D. R. Reinhart, 2000. Foundations of Environmental Engineering.
Long Grove, IL: Waveland Press.
Cooper.book Page 388 Monday, June 23, 2014 9:58 AM
388
Appendix B
Table B.4
Solubility Constants of Selected Solids
Bromides
PbBr2
Hg2Br2
AgBr
4.6 × 10–6
1.3 × 10–22
5.0 × 10–13
Carbonates
BaCO3
CdCO3
CaCO3
CuCO3
FeCO3
PbCO3
MgCO3
MnCO3
Hg2CO3
NiCO3
Ag2CO3
SrCO3
ZnCO3
2.6 × 10–90
5.2 × 10–12
4.7 × 10–9
2.5 × 10–10
2.1 × 10–11
7.5 × 10–14
6.8 × 10–6
8.8 × 10–11
9.0 × 10–17
1.4 × 10–7
8.2 × 10–12
7.0 × 10–10
2.0 × 10–10
Fluorides
BaF2
CaF2
PbF2
MgF2
SrF2
1.8 × 10–7
3.9 × 10–11
4.0 × 10–8
5.2 × 10–11
7.9 × 10–10
Hydroxides
AI(OH)3
Ba(OH)2
Cd(OH)2
Ca(OH)2
Cr(OH)3
Co(OH)2
Co(OH)3
Cu(OH)2
Fe(OH)2
Fe(OH)3
Pb(OH) 2
Mg(OH) 2
Mn(OH)2
Hg(OH)2(HgO)
Ni(OH)2
Ag OH (Ag2O)
Sr(OH)2
Sn(OH)2
Zn(OH)2
1.3 × 10–33
5.0 × 10–3
5.3 × 10–15
5.0 × 10–6
6.7 × 10–31
2.5 × 10–16
2.5 × 10–43
1.6 × 10–19
4.9 × 10–17
6.0 × 10–38
1.4 × 10–20
8.9 × 10–12
2.0 × 10–13
3.0 × 10–26
5.5 × 10–16
2.0 × 10–8
3.2 × 10–4
3.0 × 10–27
4.5 × 10–17
Iodides
PbI2
Hg2I2
Agl
8.3 × 10–9
4.5 × 10–29
8.5 × 10–17
Chlorides
PbCl2
Hg2Cl2
AgCl
1.6 × 10–5
1.1 × 10–18
1.7 × 10–10
Chromates
BaCrO4
PbCrO4
Hg2CrO4
Ag2CrO4
SrCrO4
1.2 × 10–10
3.0 × 10–13
2.0 × 10–9
1.9 × 10–12
3.6 × 10–5
Phosphates
Ba3(PO4)2
Ca3(PO4)2
Pb3(PO4)2
Ag3PO4
Sr3(PO4)2
AlPO4
6.0 × 10–39
1.3 × 10–33
1.0 × 10–54
1.8 × 10–18
1.0 × 10–31
9.8 × 10–21
Sulfates
BaSO4
CaSO4
PbSO4
Ag2SO4
SrSO4
1.0 × 10–10
3.7 × 10–5
1.3 × 10–8
1.2 × 10–5
3.4 × 10–7
Sulfides
Bi2S3
CdS
CoS
CuS
FeS
PbS
MnS
HgS
NiS
Ag2S
SnS
ZnS
1.6 × 10–72
1.0 × 10–28
5.0 × 10–22
8.0 × 10–37
4.0 × 10–19
7.0 × 10–29
3.0 × 10–14
1.6 × 10–54
3.0 × 10–21
5.5 × 10–51
1.0 × 10–26
2.5 × 10–25
Miscellaneous
NaHCO3
KCIO4
K2[PtCl6]
AgC2H3O2
AgCN
AgCNS
1.2 × 10–3
8.9 × 10–3
1.4 × 10–6
2.3 × 10–3
1.6 × 10–14
1.0 × 10–12
Adapted from Cooper, C. D., J. D. Dietz, and D. R. Reinhart, 2000. Foundations of Environmental Engineering.
Long Grove, IL: Waveland Press; and other sources.
Cooper.book Page 389 Monday, June 23, 2014 9:58 AM
Selected Physical and Chemical Properties
Table B.5
389
Ionization Constants of Selected Acids and Bases
Monoprotic Acids
acetic
benzoic
chlorous
cyanic
formic
hydrazoic
hydrocyanic
hydrofluoric
hypobromous
hypochlorous
nitrous
Polyprotic Acids
arsenic
carbonic
hydrosulfuric
oxalic
phosphoric
phosphorous (diprotic)
sulfuric
sulfurous
Bases
ammonia
aniline
dimethylamine
hydrazine
methylamine
pyridine
trimethylamine
HC2H3O2  H+ + C2H3O2–
HC7H5O2  H+ + C7H5O2–
HCIO2  H+ + CIO2–
HOCN  H+ + OCN–
HCHO2  H+ + CHO2–
HN3  H+ + N3–
HCN  H+ + CN–
HF  H+ + F–
HOBr  H+ + OBr–
HOCI  H+ + OCI–
HNO2  H+ + NO2–
1.8 × 10–5
6.0 × 10–5
1.1 × 10–2
3.5 x 10–4
1.8 × 10–4
1.9 × 10–5
4.0 × 10–10
6.7 x 10–4
2.1 × 10–9
3.2 × 10–8
4.5 x 10–4
H3AsO4  H+ + H2AsO4–
H2AsO4–  H+ + HAsO4–2
HAsO4–2  H+ + AsO4–3
CO2 + H2O  H+ + HCO3–
HCO3–  H+ + CO3–2
H2S  H++ HS–
HS–  H+ + S–2
H2C2O4  H+ + HC2O4–
HC2O4–  H+ + C2O4–2
H3PO4  H+ + H2PO4–
H2PO4–  H+ + HPO4–2
HPO4–2  H+ + PO4–3
H3PO3–  H+ + H2PO3–
H2PO3–  H+ + PO3–2
H2SO4–  H+ + HSO4–
HSO4–  H+ + SO4–2
SO2 + H2O  H+ + HSO3–
HSO3–  H+ + SO3–2
K1 = 2.5 × 10–4
K1 = 5.6 × 10–8
K3 = 3.0 × 10–13
K1 = 4.5 × 10–7
K2 = 4.7 × 10–11
K1 = 1.1 × 10–7
K2 = 1.0 × 10–14
K1 = 5.9 × 10–2
K2 = 6.4 × 10–5
K1 = 7.5 × 10–3
K2 = 6.2 × 10–8
K3 = 3.6 × 10–13
K1 = 1.6 × 10–2
K2 = 7.0 × 10–7
strong
K2 = 1.3 × 10–2
K1 = 1.7 × 10–2
K2 = 5.6 × 10–8
NH3 + H2O  NH4+OH–
C6H5NH2 + H2O  C6H5NH3+ + OH–
(CH3)2NH + H2O  (CH3)2NH2+ + OH–
N2H4 + H2O  N2H5+ + OH–
CH3NH2 + H2O  CH3NH3+ + OH–
C5H5N + H2O  C5H5NH+ + OH–
(CH3)3N + H2O  (CH3)3NH+ + OH–
1.8 × 10–5
4.2 × 10–10
7.4 × 10–4
8.5 × 10–7
5.0 × 10–4
1.5 × 10–9
7.4 × 10–5
Adapted from Cooper, C. D., J. D. Dietz, and D. R. Reinhart, 2000. Foundations of Environmental Engineering.
Long Grove, IL: Waveland Press; and other sources.
Cooper.book Page 390 Monday, June 23, 2014 9:58 AM
Appendix B
390
Table B.6a
where:
Henry’s Law Constants for Selected Gases in Water
Pa = Ha x a
Pa = partial pressure of the solute a in the gas phasee, atm
x a = mole fraction of solute a in the liquid phase, mole fraction
Ha = Henry’s law constant, atm/mole fraction
Ha×10–4, atm/mole fraction
Temp °C
Air
CO2
CO
C2H6
H2
H2S
CH4
NO
N2
O2
0
10
20
30
40
50
60
4.32
5.49
6.64
7.71
8.70
9.46
10.1
0.0728
0.104
0.142
0.186
0.233
0.283
0.341
3.52
4.42
5.36
6.20
6.96
7.61
8.21
1.26
1.89
2.63
3.42
4.23
5.00
5.65
5.79
6.36
6.83
7.29
7.51
7.65
7.65
0.0268
0.0367
0.0483
0.0609
0.0745
0.0884
0.103
2.24
2.97
3.76
4.49
5.20
5.77
6.26
1.69
2.18
2.64
3.10
3.52
3.90
4.18
5.29
6.68
8.04
9.24
10.4
11.3
12.0
2.55
3.27
4.01
4.75
5.35
5.88
6.29
Note: To use this table, extract the table entry, then multiply by 104 to get the Ha . For example, the value of HH2 S at
20 °C is 483 atm/mole fraction.
Adapted from Cooper, C. D., and F. C. Alley, 2011. Air Pollution Control. 4th ed. Long Grove, IL: Waveland Press.
Table B.6b Henry’s Constants for Selected Organic Compounds at 25 °C for Henry’s Law
written as Pi = Hi Ci
where:
Pi = partial pressure of solute i in kilopascals (kPa)
Hi = Henry’s Constant in kPa/(gmol/m3)
Ci = concentration of solute i in gmol/m3
Compound
Acetone
Benzene
Carbon tetrachloride
Chlorobenzene
Chloroethane
Chloroethylene
Chloroform
Henry’s Constant
0.01
0.6
3.0
0.4
0.2
4.0
0.4
Compound
1,1,2,2-Tetrachloroethane
1,1,1-Trichloroethane
Trichloroethylene
Toluene
Vinyl chloride
Xylenes
Henry’s Constant
0.05
3.0
0.9
0.7
50
0.6
Adapted from Davis, M. L., and D. A. Cornwell, 1998. Introduction to Environmental Engineering. 3rd ed. Boston:
McGraw-Hill.
Cooper.book Page 391 Monday, June 23, 2014 9:58 AM
Selected Physical and Chemical Properties
Table B.7
391
Enthalpies of Saturated Steam and Water
Enthalpy, Btu/lbm
T
°F
Vap. Press
atm
32
40
50
60
70
80
90
100
110
120
130
140
150
160
170
0.0060
0.0083
0.0121
0.0174
0.0247
0.0345
0.0475
0.0646
0.0867
0.115
0.151
0.196
0.253
0.322
0.408
Sat.
Liq.
0
8.05
18.07
28.07
38.05
48.02
58.00
67.97
77.94
87.91
97.89
107.88
117.87
127.87
137.89
Enthalpy, Btu/lbm
ΔHv
Sat.
Vapor
T,
°F
Vap. Press
atm
Sat.
Liq.
1,075.1
1,070.5
1,064.8
1,059.1
1,053.4
1,047.8
1,042.1
1,036.4
1,030.9
1,025.3
1,019.5
1,013.7
1,007.8
1,002.0
996.1
1,075.1
1,078.6
1,082.9
1,087.2
1,091.5
1,095.8
1,100.1
1,104.4
1,108.8
1,113.2
1,117.4
1,121.6
1,125.7
1,129.9
1,134.0
180
190
200
210
212
220
230
240
250
260
270
280
290
300
0.511
0.635
0.784
0.961
1.000
1.170
1.414
1.699
2.029
2.411
2.848
3.348
3.916
4.560
147.91
157.95
167.99
178.06
180.07
188.14
198.22
208.34
218.48
228.65
238.84
249.06
259.31
269.60
ΔHv
Sat.
Vapor
990.2 1,138.1
984.1 1,142.1
977.8 1,145.8
971.5 1,149.6
970.3 1,150.4
965.2 1,153.3
958.7 1,156.9
952.1 1,160.4
945.3 1,163.8
938.6 1,167.3
931.8 1,170.6
924.6 1,173.7
917.4 1,176.7
910.1 1,179.7
Adapted from Cooper, C. D., and F. C. Alley, 2011. Air Pollution Control: A Design Approach. 4th ed. Long Grove, IL:
Waveland Press.
Cooper.book Page 392 Monday, June 23, 2014 9:58 AM
392
Appendix B
Table B.8 Standard Heats of Combustion for Various Organic Compounds
[Products are H2O(g) and CO2(g) at 25 °C] (Lower Heating Values)
Compound
Formula
MW
ΔHc (kJ/kg)
ΔHc (Btu/lb)
n-Alkanes
Methane
Ethane
Propane
n-Butane
n-Pentane
n-Hexane
CH4
C2H6
C3H8
C4H10
C5H12
C6H14
16.0
30.1
44.1
58.1
72.2
86.2
50,150
47,440
46,350
45,730
45,320
45,090
21,560
20,390
19,920
19,660
19,480
19,380
1-Alkenes
Ethylene
Propylene
1-Butene
1-Pentene
1-Hexene
C2H4
C3H6
C4H8
C5H10
C6H12
28.1
42.1
56.1
70.1
84.2
47,080
45,760
45,300
45,020
44,780
20,240
19,670
19,470
19,350
19,250
Miscellaneous
Acetylene
Benzene
1,3-Butadiene
Cyclohexane
Ethylbenzene
Methylcyclohexane
Styrene
Toluene
Carbon monoxide
Hydrogen
Ethanol
Methanol
Water (liq. to gas)
C2H2
C6H6
C4H6
C6H12
C8H10
C7H14
C8H8
C7H8
CO
H2
C2H5OH
CH3OH
H2O
26.0
78.1
54.1
84.2
106.2
98.2
104.2
92.1
28.0
2.016
46.1
32.0
18.0
48,290
40,580
44,540
43,810
41,310
43,710
40,910
40,950
10,110
120,900
25,960
18,790
2,445
20,760
17,440
19,150
18,830
17,760
18,790
17,590
17,600
4,346
51,970
11,160
8,080
1,050
Adapted from Cooper, C. D., and F. C. Alley, 2011. Air Pollution Control: A Design Approach. 4th ed. Long
Grove, IL: Waveland Press.
Cooper.book Page 393 Monday, June 23, 2014 9:58 AM
APPENDIX
C
Answers to Odd-Numbered Problems
1.1
44 ft/s; 805 m/min; 8.06 (10)4 furlongs/fortnight
1.3
3.63 (10)4 tons/yr
1.5
(a) 1.07 (10)11 GJ; (b) 1.68 (10)10 boe; (c) 2.96 (10)13 kwh
1.7
1.37 (10)4 µm; 2.00 µg; 137 µm
1.9
9.0 billion in 2030 and 13.7 billion in 2060
1.11 20,000 wings/month; 640 kg of wings/month; 1,875 kg of chickens
1.13 8.94 (10)13 lb CO2/yr
1.15 9,900 ft3 or 74,060 gal
2.3
C7H4N3O6 + 5 O2 → 7 CO2 + 2 H2O + 1.5 N2; 70.8 g O2
2.5
Al(OH)3 will precipitate first at all pHs
2.7
(a) 12; (b) 2; (c) 4.8
2.9
1.36 (10)–5 gmol/L
2.11 0.0504 eq/L; 2,520 mg/L as CaCO3
2.13 C2H5OH + 3 O2 → 2 CO2 + 3 H2O; 106 L
2.15 0.0355
2.17 (a) KA = mg/L and k = mg/(L-min); (b) L/(mg-min); (c) min–1; (d) L/(mg
of B-min); (e) mg/(L-min)
2.19 855 min
2.21 111 hours; zero-order
2.23 3.41 hours
2.25 0.008 moles NaI; 3.13 mg NaI
2.27 2,780 L
2.29 2.9 (10)–6 mg/L
393
Cooper App C.fm Page 394 Thursday, March 3, 2016 10:16 AM
394
Appendix C
3.1
1,800 kg/day
3.3
QC = 55 L/s; CD = 628.5 mg/L; QE = 5 L/s; CE = 59,715 mg/L; QA′ = 150
L/s; CA′ = 2,210 mg/L
3.5
818.5 gmol/day
3.7
728 kg/day
3.9
Crainy = 18.2 mg/L; Cdry = 100 mg/L
3.11 687,500 L/day
3.13 6,897 gal of high-octane and 3,103 gal of straight-run
3.15 (a) 13,333 L; (b) 3,600 gmol B/day
3.17 0.042 gmol/L
3.19 In slurry stream: coal = 20.31, water = 20.31, total = 40.62; In separated
water stream: coal = 0.31, water = 15.31, total = 15.62; In separated coal
stream: coal = 20.0, water = 5.0, total = 25.0
3.21 748.9 L
3.23 27.26 mg/L
3.25 67.06 mg/L
3.27 C10 = 40.52 mg/L; Css = 42.78 mg/L
3.29 (a) 34.6 mg/L; (b) 44 oC
3.31 (a) 7.85 (10)6 kg coal/day; (b) 2.98 (10)4 kg SO2/day; (c) 4,536 kg ash/day;
(d) 5.62 (10)7 Btu/min; (e) 7.83 (10)7 lb water/day
4.1
0.20 per yr
4.3
13.2 cfs
4.5
33.6 cfs
4.7
5,445 ft3
4.9
9,840 m3
4.11 CBODu = 108 mg/L; NBODu = 72 mg/L; Total BODu = 180 mg/L
4.13 236 mg/L
4.15 466 mg/L
4.17 949 mg/L
4.19 570 mg/L
4.23 480 mg/L
4.25 DO = 5.50 mg/L; BOD = 7.46 mg/L
4.27 9.90 (10)8 kg/day; 687 m3/min
4.29 CBODu = 81.1 mg/L; NBODu = 64.9 mg/L; Total BODu = 146 mg/L
4.31 (a) BOD5 = 78 mg/L; (b) BODu = 109 mg/L
Cooper App C.fm Page 395 Thursday, March 3, 2016 10:16 AM
Answers to Odd-Numbered Problems
5.1
pH 7 is better
5.3
remove Fe and H2S; disinfection
5.5
to raise pH above 10 to precipitate Mg(OH)2
395
5.7
747 mg/L as CaCO3 ; yes softening is recommended
5.9
4,180 lb/day of lime; 1,740 lb/day of soda; 183 lb/day of carbon dioxide;
8,515 lb/day of sludge (dry basis)
5.11 21 ft long, 7.0 ft wide, and 7.0 ft deep; actual HRT = 46.7 min
5.13 width and length = 4.17 ft, height = 7 ft; HLR = 5.0 gpm/ft2
5.15 154 min
5.17 2 clarifiers, each with diameter = 70 ft, and depth = 16 ft
5.19 W = D = 1.2 m, L = 4.8 m
5.21 85.5 ft
5.23 50,000 ft2 ; 2.63 MGD
5.25 Diameter = 70 ft, depth = 16 ft; OFR = 650 gpd/ft2; HRT = 4.0 hours; water
depth = 14.5 ft; tank wall depth = 16 ft
6.1
4.6%
6.3
1,736 gal
6.5
not satisfactory
6.7
0.184 lb/(cap-day) for each
6.9
6.9 days
6.11 309,000 L/day
6.13 160,200 ft3
6.15 calculate 1.06 days, but select 5 days
6.17 10,700 ft3
6.19 (b) 300,000 L/day; (c) 2,123 kg/day; (d) 4.77 (10)6 L/day; (e) 0.59 day–1; (f)
3.77 days
6.21 (b) 757,000 L/day; (c) 4,554 kg/day; (d) 0.41; (e) 2,277 mg/L
6.23 2 clarifiers, D = 95 ft; or 3 clarifiers D = 75 ft; or 4 clarifiers D = 65 ft
7.1
187.5 ppm
7.3
12 µg/m3
7.5
0.50 ppm
7.7
98.25%
7.9
7.4 tons/day
7.11 5,543 lb CO2/year for small car and 12,330 lb CO2/year for large SUV
7.13 480 kg SO2/day
Cooper.book Page 396 Monday, June 23, 2014 9:58 AM
396
Appendix C
7.15 65 oC; no
7.17 3.6 (10)8 tons/year
7.19 $40,800/year
7.21 3.14 (10)5 Btu/min
7.23 2.78 (10)3 m3/min; 12.2%
7.25 121 µg/m3; 110 µg/m3; 56.4 µg/m3
7.27 3,832 tonnes/yr
8.1
226 acres
8.3
44.2 years
8.5
8.02 (10)7 kwh/yr; 4,457 households
8.7
31.5 miles
8.9
for 20-yd3 trucks: 1,104 hrs/wk and 28 trucks; for 30-yd3 trucks: 1,067
hrs/wk and 28 trucks; the 20-yd3 trucks are cheaper and so are preferred.
8.11 14.1%
8.13 1.61 (10)8 lb; 5.35 (10)6 ft3
8.15 35.1 kg/day SO2; 43.9 kg/day NaOH; 37.7 kg/day dry solids; 94.3 kg/
day wet sludge
8.17 5.8 years
8.19 1,017 ppm
8.21 71.6 kg
8.23 97.3 kg/day; 2,000 ppm
8.25 (a) 250 kg/day FeSO4; (b) 775 kg/day wet sludge
9.1
1.35 (10)–5
9.3
males 0.05 mg/(kg-day) for each chemical; females 0.073 mg/(kg-day) for
each chemical
9.5
153 (untreated); 0.16 (treated)
9.7
8.24 years
9.9
41.25 years
9.11 235 pCi/ft3
9.13 336 mg/min; 129 mg/m3 after 2 hours
9.15 3.24 (10)4 µg/m3
9.17 62.5 dB(A)
9.19 66.3 dB(A)
9.21 1.37 km
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Index
Acid deposition, 255, 257, 259
Acid-base chemistry, 52–57
acid-base reactions, 45
ionization constants, 389
pH of a strong or weak
acid, 54–57
the pH system, 53
weak and strong acids
and bases, 53–54
Acidity, defined, 63
Activated carbon adsorption,
331
Activated sludge process,
215, 223–227
design guidelines for activated sludge system,
227
material balance
approach, 227–234
See also Sludge
Advanced wastewater treatment, 214
Aeration/air stripping
decontamination of
groundwater, 187–188,
331–333
surface, 231
Aeration tanks/biological
reactors
activated sludge process
and, 223–225
biomass growth in, 225
material balance diagram
for, 230
reactor balances, 229–230
Aerobic/anaerobic bacteria,
223
Agribusiness, industrial
wastewaters from, 211
Air
exchange/ventilation and
infiltration, 363–364
selected properties of, 386
stable and unstable meteorology, 286
Air pollutants
accumulation of, 253
atmospheric dispersion
of, 286–298
box model for concentration, 287–290
carbon monoxide (CO), 256
causes, sources, and
effects of, 258
emissions/rates of
removal, 253
Gaussian dispersions
model for concentration, 290–297
hazardous (HAPs),
255–256
indoor, 361–363
legislative/regulatory
standards for, 265–267
nitrogen oxides (NOx), 255
ozone and other oxidants, 257
particulate matter, 255
plumes from stacks,
290–291
397
sulfur oxides (SOx),
255–256
US emissions estimates of
major pollutants, 259
volatile organic compounds (VOCs), 256
Air pollution
control of
gaseous emissions,
276–281
mobile sources,
281–286
NOx, 281
particulate matter,
269–276
residuals from, 298
SO2, 278–280
stationary sources,
269–281
traffic management
techniques for, 285
VOCs, 277–278, 281
from fossil fuels, 351–352
global issues, 257–265
indoor, 361–367. See also
Indoor air quality
legislative and regulatory
standards for, 265–267
primary and secondary
pollutants, 255
standards, 265–266
transportation control
measures, 285
Air resource management
(ARM), 254
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Index
Air resources, 253–302
air pollutants, 255–257
management of, 253–254
Air stripping of contaminated groundwater,
331–332
Alkalinity, 63, 192–193
Alkaline scrubbing, 278–281
Allergens, 363
Alum, 172–173, 175, 177, 181
Ambient air quality standards (AAQSs), 265–266
Ammonia oxidation, 148–149
Ammonia-nitrogen, 186
Anaerobic digestion, 243–244
Aqueous solutions, 34, 53
Aquifers, 138–139
Arrhenius equation, 76
Arrhenius, Svante, 263
Artesian wells, 139
Atmospheric stability
classifications, 290–295
defined, 286
dispersion modeling,
287–297
Attenuation, noise, 375–376
Average daily dose, 343
Average global temperature,
263–264
Avogadro’s Number, 34
A-weighted decibel scale,
372
Backwashing, 199
Bacteria, aerobic/anaerobic,
223
Bagasse, 23
Baghouses, 272–273
Bar screen/bar rack, 216
Basic material balance equation, 87–89
Batch bioassay analysis
(BOD test), 151, 153–154
Batch reactor, 97
Biochemical oxygen demand,
defined, 147–148. See also
BOD
Biofiltration, 281
Biological nutrient removal
(tertiary treatment), 213
Biological reactors/aeration
tanks. See Aeration
tanks/biological reactors
Biomass
activated sludge system,
224, 227–232
defined, 225
electricity generation
from, 351
mixed liquor volatile suspended solids
(MLVSS), 230
production/specific
growth rate, 237
specific growth rate,
236–237
steady-state mass balance,
228
total discharge rate, 229
Biosolids, gasification/incineration of, 244–246
Bioreactor landfills, 313
BOD
BOD5, 151
carbonaceous (CBOD), 148
kinetics, 149–161
nitrogenous (NBOD), 148
progression curve, 150
testing, 151–154
ultimate (BODu), 147–150,
153, 160
Box model of pollutant concentration, 287–290
Breakpoint chlorination, 185
Brownfields, 331
Buffered solutions, 57
Buffering systems
carbonate system, 61–67
Cancer slope factor, 344–345
Carbon adsorption, 277
Carbon dioxide (CO2), atmospheric, 14–15, 255–256,
261, 263, 354
Carbonaceous BOD (CBOD),
148
Carbonate system, 61–67
equilibrium in surface
water, 62
hardness (CH), 188–193
Carcinogenic dose-response
curve, 344
Carcinogens, toxicity data
for, 345
Carson, Rachel, 18
Characteristic wastes, 322–323
Chemical equations, balancing, 45–50
Chemical equilibrium, 51–52
Chemical oxygen demand
(COD) test, 147, 154
Chemical reaction
kinetics, 67–78
stoichiometry of, 44–51
Chemistry
definition of, 32
reporting concentrations
in solutions, 33–38
Chernobyl nuclear disaster,
356
Chlorination, 213, 240
Chlorine residual, 185–186
Chlorofluorocarbons (CFCs),
20, 260
Civilization, growth and
environmental impact
of, 7–15
Clarifier(s)
balances, 231–234
hydraulic loading rate
(HLR), 225
primary, 215, 221–222
secondary, 223–224, 231,
240
solids loading rate (SLR),
225
Climate change
global, 260–265
sustainable development
and, 23
CO2. See Carbon dioxide
Coagulation/flocculation,
173, 176–179
Coal mining, 349, 351
Coal-fired power plants, 109
Colloidal dispersion, 33
Combined chlorine residual,
185
Compactor trucks, 306,
308–309
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Comprehensive Emergency
Response, Compensation, and Liability Act
(CERCLA), 330–331
Concentration expressions,
common, 38
Concentrations in gaseous
mixtures, 38
Cone of depression, 140
Confined aquifers, 139
Conservation of mass, 84–85
Continuity equation, 85
Continuous flow stirred tank
reactor, 97
Conversion factors, 383–385
Corrosivity, 322
Criteria pollutants, 255
Critical point, 157
Crude oil, 349
CSTR reactor, 98
Cyclones, 269–270
Data collection/evaluation,
342
Death
annual risk of, for specific
activities, 341
ten most common causes
of, 340
Dechlorination, 213, 241
Decibel (dB) scale, 370
Denitrification, 214
Density (concentration of
total mass per unit of
volume), 34
Design storm, 131
Detention time/hydraulic
residence time (HRT),
175, 221
Direct-flame thermal oxidizer, sectional view, 279
Disinfection
by-products, 169
of municipal discharge,
213, 240–241
primary and secondary,
184–187
Dispersion parameters,
curve fit equations and
constants for, 295
Dissolved oxygen
deficit, 156
rate processes, 141–143
in rivers and lakes, 143,
154–161
sag calculations/curves,
158–159
solubility/saturation values, 142
Domestic wastewater treatment
composition of untreated
wastewater, 209
generalized flow diagram of, 215
unit operations/processes of, 214–227
Dominguez, R., 131
Dose-response curves, 343–344
Drainfield, 242
Drinkable water. See Potable
water
Drinking water standards,
primary/secondary,
168–171
Dry basis (calculation of), 182
Effluent concentration limits, wastewater treatment, 213–214
Electricity
generation, 350–352
nuclear powered, 354, 356
Electrostatic precipitators,
270–271
Elementary reactions, 70–75
Emissions (air pollutants),
253, 354
Emissions inventory (EI), 254
Endocrine disruptors, 145
Energy
balances, 105–111
consumption, global, 12
real-world conversion
processes, 107–108
reducing use of, 351–353
various forms of, 103–105
Energy resources
categories of demand,
350–351
399
categories of supply,
350–351
conservation, 351–353
electricity generation,
351–352
fossil fuels, 349–352
nuclear power, 350–351,
353–360
renewable energy, 350–351
Engineers
environmental, 1–4
ethics, 4
problem solving abilities
of, 5–6
professional engineer
(PE), 3–4
Engines and fuels, characteristics of, 283–285
Enthalpy, 105–106, 391
Environmental agencies, 25
Environmental engineering
branches of, 2
credentials, 3
definition of, 1–3
Environmental ethic, 18–20, 29
Environmental health and
safety (EHS) managers, 3
Environmental laws/legislation, 25–28
Environmental Protection
Agency (EPA), 25
Environmental protection,
rise of, 15–17
Environmental regulations, 25
Equalization tanks, 216–220
Equilibrium concepts, 51–52
Equilibrium constant, 51–52
Equivalent weight, 36
Equivalents and normality, 36
Euler’s Method, 111
Eutrophication, 144
Evapotranspiration, 129
Ex situ/in situ site remediation techniques, 331
Exhaust emissions, effects of
air/fuel ratio on, 284
Exposure and toxicity, risk
elements of, 342
Exposure assessment,
345–346
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400
Index
F:M ratio (organic loading),
226
Filtration
air, 272
water, 173, 183–184
Finite difference method,
111–112, 114–116
First-order reactions, 71–72, 74
Fish kills, 16–17
Fission reactions, 357
Fixed-bed carbon adsorption system, process
flow diagram for, 277
Flocculation/coagulation,
173, 176–179
Floodplain, 132
Flow
of energy and materials in
power plants, 109–111
of energy/energy balances, 103–108
of materials/mass balances, 84–102
steady-state flow processes without chemical reaction, 89–94
unsteady-state flow processes, 95–96
Flue gas desulfurization
(FGD), 278, 280
Formaldehyde, 362
Fossil fuels
as energy resource,
349–352
power plants run on,
109–110
reducing our dependence
on, 353
worldwide combustion
of, 262–263
Fracking, 349, 352
Free chlorine residual, 185
Fuels
and engines, characteristics of, 283–285
nominal energy content
of, 103
See also Fossil fuels
Fukushima nuclear disaster,
356
Fundamentals of Engineering (FE) exam, 3
Gas behavior
ideal gas law, 39–40
partial pressure and partial volume, 40–41
Raoult’s Law/Henry’s
Law, 42–44
Gaseous solutions, 34
Gasification of waste activated sludge, 245–246
Gasoline-powered vehicles,
fuel economy standards/air pollution
emission limits for, 283
Gaussian dispersion equation, 292
Gaussian dispersion model,
290–297
Global climate change
(global warming), 22,
260–265, 353
Green engineering, 23–24
Green roofs, 24
Greenhouse effect, 261, 263
Grit chamber, 216
Groundwater
aeration for removal of
hydrogen sulfide,
187–188
contaminated, cleanup of,
330–331, 333
fracking pollution, 352
hydrology, 138–141
lime-soda softening,
188–197
protection of, 141
schematic diagram of
resources, 139
site remediation, 328–334
sources, 168
treatment, process flow
diagrams, 174
Hardness, 64, 188–190
carbonate (CH), 191
non-carbonate (NCH),
191
Hazard identification, 342
Hazardous air pollutants
(HAPs), 255–256
Hazardous wastes, 320–327
characteristics, 322–323
“cradle-to-grave”
approach to, 321, 323
generators of, 323–325
improper disposal of, 328,
330
incineration of, 327–328
landfilling of, 325
laws and regulations on,
321–323
listed, 322
site remediation, 328–334
treatment, storage, and
disposal of, 325–327
Head of water, 139
Henry’s Constant, 42, 390
Henry’s Law, 42–44, 332, 390
Hoist trucks vs. compactor
trucks, 309–310
Horizontal dispersion coefficient, 293
Human civilization, growth
and environmental
impact of, 7–15
Hydraulic conductivity, 140
Hydraulic fracturing (“fracking”), 349, 352
Hydraulic gradient, 139
Hydraulic loading rate
(HLR), 184, 226
Hydraulic residence time
(HRT), 175, 178, 221, 225
Hydrochlorofluorocarbons
(HCFCs), 20, 260
Hydroelectric power, 351
Hydrogen sulfide, removal
from groundwater,
187–188
Hydrographs, input and output, 136–138
Hydrologic cycle, 128–129
Hyetograph, 136
Ideal gas law, 39–40, 387
Ignitability, 322
In situ/ex situ remediation
techniques, 331
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Index
Incineration
of digested sludge/biosolids, 244–245
of hazardous wastes, 327
of municipal solid wastes,
315–320
vapor incineration, 278
Indoor air pollutants
formaldehyde, 362
mold, mildews, and allergens, 363
radon gas, 361–362
tobacco smoke, 361
Indoor air quality, 361–367
material balance models
for, 364–367
pollutants of concern,
361–363
solutions to IAQ problems, 367, 369
Industrial wastewaters,
211–214
discharge standards for, 214
toxicity bioassay testing,
212
Infiltration
groundwater, 128–129, 138
and ventilation, 363–364
wastewater systems,
208–209
Inflow (into wastewater systems), 209
Inhibitory model for substrate utilization, 235
Intrinsic attenuation, 332
Inverse-square law, 375
Inversion layer, 286–288
Ion product, IP, 58
Iterative solution methodology for nonlinear algebraic equations, 116–120
Kinetics
biological, of wastewater
treatment plants,
234–242
BOD kinetics, 149–161
chemical reaction, 67–78
elementary reactions, 70–75
first-order reactions, 72, 235
second-order reactions,
72, 235
temperature effects on,
76–78
variable-order reactions,
75, 235
zero-order reactions, 71,
235
Landfills
bioreactor, 313
disposal of municipal
solid wastes, 310–315
for hazardous waste, 325
methane gas collection/
recovery, 311, 313
preliminary landfill
design, 313–315
sanitary landfills, 310–313
Leachate, 310, 313
Leopold, Aldo, 18
Licensure (PE), 3
Lifetime average daily dose
(LADD), 346
Lime-soda softening of
groundwater, 188–197
Limestone scrubbing/flue
gas desulfurization,
278–280
schematic process flow
diagram for, 280
Liquid water, selected properties of, 387
Loading
definition of, 208
hydraulic rate vs. solids
rate, 226
organic, 226
Love Canal, 19, 328, 330
Marsh, George, 18
Mass
balances, 84–87
conservation of, 84–85
flow of materials/mass
balances, 84–102
flow rates, 85–91
and mole concentration
expressions, 34–35
See also Biomass
401
Material balances
approach, 84, 253, 261
basic balance equation, 87
basis and boundaries,
87–88
with chemical reaction, 96
without chemical reaction, 89–100
clarifier balances, 231–234
continuity equation, 85
indoor air quality balances, 364
mixing point, 89, 94, 155,
159
reactor balances, 229–231
reactor models, 97–100
recycle, 91–92
separator tank, 91
splitter point, 90
steady-state flow, 89, 193,
198, 287
use of, 84
WWTP system balances,
228–229
Materials Recovery Facility
(MRF), 306–307
Maximum Contaminant
Level (MCL), 170, 348
Maximum Contaminant
Level Goal (MCLG), 170
Membrane processes, 173,
197–199
Mercury poisoning, 17
Metathesis reaction, 44
Meteorology and atmospheric stability, 290–295
Methane, 311, 313, 349
Microbiological decomposition of organic materials in water, 146–149
engineered systems,
147–148
natural systems, 146
Mixed liquor suspended solids (MLSS), 225
Mixed liquor volatile suspended solids (MLVSS),
225
Mixing point, 89, 94, 155, 159
Mixing zone, 155
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Index
Molarity, 35–37
Molds/mildews, 363
Mole, 34
Mole fraction, 35
Mole/volume and mass/volume expressions, 35–38
Monod equation, 235
Montreal Protocol, 20
Multimedia filters, 184
Municipal solid wastes
collection of, 306, 308, 310
composition of, by
weight, 304, 306
discards, 305–306
generation rate, 304
incineration of, 316
landfill disposal of,
310–315
quantities and composition of, 303–304
recycling, 304–305
thermal destruction of,
315–320
transfer stations for, 308
Municipal wastewater,
205–241
biological kinetics, 234–238
characteristics, 208–209
discharge standards for,
212–214
disinfection, 240-241
primary/secondary treatment of, 214–234
secondary clarifier, 240–241
water reclamation, 206–207
National Ambient Air Quality Standards, 265–266
National Environmental Policy Act of 1970, 25
National Pollutant Discharge
Elimination System
(NPDES), 212
National Priority List of contaminated sites, 330
Natural gas, 349–350
New Source Performance
Standards (NSPS), 265,
268
Nitrification, 214
Nitrogen oxides (NOx), 255
Nitrogenous BOD (NBOD),
148
No observable adverse effect
level (NOAEL), 344–345
Noise abatement criteria, 377
Noise attenuation, point
sources vs. line sources
of, 375–376
Noise pollution, 370–377
duration, 373–374
frequency, 371–372
legislation and regulations, 377
loudness, 370–371
noise attenuation, 375–376
subjectivity of, 374–375
Non-aqueous phase liquid
(NAPL) contamination,
332
Non-carbonate hardness
(NCH), 191
Non-carcinogenic doseresponse curve, 344
Nonlinear algebraic equations, iterative solution
of, 116–120
Normality, 36–37
Nuclear disasters, 356
Nuclear energy, 350, 353–360
Nuclear plants
advantages of, 354–355
disadvantages of, 355–360
Nuclear reactions, examples
of, 357
Nuclear waste, 356–360
Numerical methods, 111–118,
136–137, 367
Nutrients, 131, 142, 144, 214,
224, 238–239
Observed yield coefficient, 229
On-road vehicles, global statistics for, 282
Organic loading/food-tomicroorganism ratio,
225–226
Overflow rate (OFR), 180, 221
Oxidation numbers, 48–49
Oxygen, dissolved. See Dissolved oxygen
Oxygen sag curves, 157–158
Ozone
ground-level, as secondary pollutant, 255
photochemical formation
of, 257
stratospheric, depletion
of, 20, 259–260
Partial pressure/partial volume, 40–41
Particle collectors, typical
efficiencies for, 275
Particulate matter, 255,
269–275
Permeate, 198
Pesticides, 144
pH system, 53–57
Pharmaceuticals, 145
Piezometric/hydraulic head
of water, 139
Pinchot, Gifford, 18
Plug flow reactor model,
97–98, 100
Pollution prevention, 20–22
Pollution Prevention Act of
1990, 211
Population growth
environmental impacts of,
7–15
global, 13
implications of, 10–12
mathematics of, 8–9
resource consumption
and pollution generation, 12–15
urbanization and, 8
Potable water
disease transmission
linked to, 16
drinking water standards, 168–170
processes to produce,
172–174
sludge disposal, 200
source water characteristics, 166–168
treatment, 166–204. See
also Water treatment
unit operations
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Index
Power and energy, various
forms of, 103–105
Power plants, flow of energy
and materials in, 109–111
Precipitation
acid deposition, 255, 257,
259
defined, 129
design storm, 131
intensity, duration, and
frequency relationships, 132–133
maximum recorded
events of, 132
quantities, 131–134
time of concentration, 135
Primary clarifiers, 215,
221–223
Primary Drinking Water
Standards, 170–171
Primary/secondary disinfection, 184
Process engineering, 83–127
combined material and
energy balances,
108–111
definition of, 83–84
flow of energy/energy
balances, 103–108
flow of materials/mass
balances, 84–102
numerical methods,
111–120
Professional Engineering
(PE) license, 3
Radiation
emissions, 355, 357
radioactive half-life, 358
types of, 357
See also Nuclear energy;
Nuclear plants
Radioactive wastes, 355–360
Radionuclides, types of radiation from, 357
Radon gas, 361–362, 369
Rainfall. See Precipitation
Raoult’s Law, 42
Rapid mix, 175–176, 178
Rapid sand filtration, 183–184
Reactivity (hazardous
wastes), 322
Reactor balances, 229–231
Reactor models
batch reactor, 97
CSTR reactor, 98
plug flow reactor, 98–100
Recarbonation, 192–193
Recharge, 139
Reclaimed water, 207
Recycle ratio (R), 225–226
Recycling municipal solid
wastes, 304–305
Reduction-oxidation (redox)
reactions, 45–51
Renewable energy, 350–351,
353
Resource Conservation and
Recovery Act (RCRA) of
1986, 321
Resource consumption rates,
13, 349–351
Risk assessment, 339, 343–349
process of, 342–348
typical exposure/intake
rates for, 346
Risk characterization, 347
Risk management, 348
Roosevelt, Franklin D., 18
Roosevelt, Theodore, 18
Runoff
coefficients, 135
defined, 129
detention ponds, 136
hydrograph, 136
nutrients, 144
Rational Equation for estimation of runoff values, 134
stormwater management, 134–138
surface water pollution
from, 131
Sand County Almanac, A
(Leopold), 18
Sanitary landfills, 310–313
Secondary clarifiers, 223–227,
231, 240
Secondary Drinking Water
Standards, 170
403
Secondary treatment, 213–215
Secondary/primary disinfection, 184
Second-order reactions, 71–72
Sedimentation, 173, 179–183
Selected physical and chemical properties, 386–392
Selective catalytic reduction
(SCR), 281
Separator tanks, 91
Septic tanks, 241–242
Settling tanks, design/monitoring of, 180–181
Sewerage systems, early, 16
Shaker baghouse, cutaway
view of, 273
Silent Spring (Carson), 18
Sludge
activated sludge process
for WWTPs, 215,
223–227
design guidelines for,
227
material balance
approach, 227–234
definition of, 181
digested, incineration of,
244–245
treatment process, 147
waste activated, gasification of, 245–246
wastewater, treatment and
disposal of, 243–247
water treatment plants,
181–182, 192, 200
Smeaton, John, 15
Smog, 257
Snow, John, 16
Softening (water)
defined, 189
lime-soda softening of
groundwater, 188–197
Soil vapor extraction, 331–333
Solid waste management,
municipal, 303–310
Solids
biomass, 215, 225, 232,
dissolved (TDS), 142, 167,
170
dry basis, 182, 243–244
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Index
mixed liquor suspended
solids (MLSS), 225, 230
mixed liquor volatile suspended solids
(MLVSS), 225, 230
suspended (TSS), 91–93,
142–143, 167, 181, 207,
217, 240
Solids loading rate (SLR), 226
Solids residence time (SRT),
225
Solubility
of dissolved oxygen,
141–142
of solids, 57–61, 388
Solubility product constant, 57
Solutions, 32–44
dilute, 35–38
mass and mole concentration expressions of, 34
reporting solute concentrations, 33–38
Solvent, 33
Sound pressure level (SPL),
370–371
Specific biomass growth rate,
236–237
Specific gravity, 34
Specific substrate utilization
rate, 235
Spent fuel (SF), 358
Splitter point, 90
Spray tower scrubber, cutaway view of, 275
Stabilization (of water), 193
Standard heats of combustion, 392
Steady-state flow processes
without chemical reaction, 89–94
Stoichiometry of chemical
reactions, 44–51
Storage of water in hydrologic cycle, 129
Storm frequency/return
period, 132
Storm intensity/duration,
131–132
Stormwater management,
131, 134–138
Stratospheric ozone depletion, 20, 259–260
Substrate, 149–150, 154, 225,
228, 237
Substrate removal. See Activated sludge process
Substrate utilization, 235
Sulfur dioxide, limestonebased scrubbing system, 280
Sulfur oxides (SOx), 255–256
Superfund sites, 330–331
Surface aerators, 231
Surface loading rate, 180
Surface water
quality, 141–146, 211
dissolved oxygen
level, 141–143
pollutants affecting,
142–146
taste and odor, 143
turbidity and color,
143
self-cleaning capacity in
natural environments,
211
sources, 167–168
treatment process flow
diagrams, 172
Surficial/unconfined aquifers, 138
Suspension, 32
Sustainable development,
22–23
Sustainable industry example, sugarcane, 23
Synthesis gas (syngas), 245
Temperature, average global,
264
Temperature correction factor model, 76
Temperature effects on reaction kinetics, 76–78
Tertiary treatment (biological
nutrient removal),
213–214
Thermal desorption, 331
Thermal oxidation/vapor
incineration, 278
Thermal waste destruction,
315–320
Three Mile Island, 356
Tobacco smoke, 361
Total biomass discharge rate,
229
Total dissolved solids (TDS),
142
Total organic carbon (TOC),
148
Total suspended solids (TSS),
142
Toxic Characteristic Leaching Procedure (TCLP)
test, 322
Toxic metals, 144–145
Toxicity
acute (short-term) effects
vs. chronic (long-term)
effects, 343
data for selected potential carcinogens, 345
defined, 322
dose-response assessment, 343
and exposure, risk elements of, 342
Transfer stations (for MSW),
308
Transuranic radioactive
wastes, 358
Turbidity, 143
US energy supply and consumption, 351
Ultimate BOD (BODu),
147–150, 153, 160
Unconfined aquifers, 138
Unit conversion factors
(Appendix A), 383–385
Unsteady-state processes,
100–102, 364–367
flow processes, 95–96
simulation of, 111–116,
136–138
Urbanization, and world
population growth, 8
Vapor incineration/thermal
oxidation, 278
Cooper Index.fm Page 405 Tuesday, July 1, 2014 3:53 PM
Index
Vapor pressure of a liquid, 42
Variable-order reactions, 75,
235
Velocity gradient, 175
Ventilation and infiltration,
363–364
Vertical dispersion coefficient, 294
Volatile organic compounds
(VOCs), 255–256
Volatile suspended solids
(VSS), 225, 230
Volume per volume concentration expressions, 38
Wasteload allocation, 159
Wastewater(s)
calculating the stoichiometric nutrient requirements for, 239–240
flow of water from environment to users and
back, 205
flow rates and loadings
of, 208
industrial, 211–214
municipal. See Municipal
wastewater
pollution prevention/
waste minimization
efforts, 211
treatment, 205–252
advanced (nutrient
removal), effluent
standards for, 214
characterization of,
208
criteria for, 212
domestic, unit operations/processes
of, 214–227
in early 1900s, 16
material balance
approach,
227–234
primary/secondary
treatment, 215
schematic process
flow diagram of,
215
South Water Reclamation Facility,
Orange County,
FL, 206
standards for,
211–214
treatment plants
activated sludge process for, 223–227
biological kinetics of,
234–242
publicly owned, 206,
213–215, 217, 221,
225
sludge treatment and
disposal, 243–247
Water chemistry, 51–67
acid-base chemistry, 52–57
carbonate system, 61–67
equilibrium concepts,
51–52
pH system, 53–57
solubility product, 57–61
Water flux rate, 199
Water pollutants, 142–146
excess heat, 145–146
in municipal wastewater,
207
nutrients (nitrogen and
phosphorous), 144
oxygen-demanding
wastes, 146–149
pesticides, 144
pharmaceuticals and
endocrine disruptors,
145
stormwater runoff, 131
total suspended solids/
total dissolved solids,
142–143
toxic metals, 144–145
Water Pollution Control Act
Amendments of 1972
(PL 92-500), 211
405
Water quality, surface,
141–146. See also Surface
water quality
level of dissolved oxygen
and, 141–143
pollutants affecting,
142–146
taste and odor, 143
turbidity and color, 143
Water quantity, 128–141
Water resources, 128–165
Water treatment unit operations
aeration for removal of
hydrogen sulfide,
187–188
coagulation/flocculation,
176–179
disinfection, 184–187
lime-soda softening of
groundwater, 188–197
membrane processes,
197–199
rapid mixing, 175–176, 178
rapid sand filtration,
183–184
reverse-osmosis, 173–174,
198
sedimentation, 179–183
treatment and disposal of
residuals, 200
Weather, impact of global
warming on, 264–265
Wet scrubbers, 274–278
Weir loading rate (WLR),
181, 222
WIPP (waste isolation pilot
plant), 360
World energy consumption, 12
World population growth, 8,
13
Yucca Mountain nuclear
waste repository, 359
Zero-order reactions, 71
Zone settling, 240