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Air
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Course S!:412D
Control of Gaseous and
Particulate Emissions
Self-instructional
Problem Workbook
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United States
Environmental Protection
Agency
A.r Pollution Training Institute
MD 20
Environmental Research Center
Researcn Trtangie Par*. NC 27711
EPA 450 2-84-C07
Sectemoer 1984
Air
APT!
Course SL412D
Control of Gaseous and
Particulate Emissions
Self-instructional
Problem Workbook
Written by:
Workbook produced by Louis Theodora
Solutions produced by Ui Young Choi
Production by:
Northrop Services, Inc.
P.O. Box 12313
Research Triangle Park, NC 27709
Under EPA Contract No.
68-02-3573
EPA Project Officer
R. E. Townsend
United States Environmental Protection Agency
Office of Air, Noise, and Radiation
Office of Air Quality Planning and Standards
Research Triangle Park, NC 27711
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Acknowledgement and Disclaimer
This workbook was produced as part of the requirements of EPA Purchase Order
ID6383NASA, The opinions, findings, and conclusions expressed are those of the author
and not necessarily those of the Environmental Protection Agency. Mention of company
or product names is not to be considered as an endorsement by the Environmental Pro-
tection Agency.
Note
EPA policv is to use SI (metric) units only or to list both the common British engineering
unit and its SI equivalent. However, British units are primarily used in this manual for
the convenience of the majority of the reading audience. Readers more familiar and at
ease with SI units are advised to refer to Appendix C of this workbook.
Author's Reminder
Plagerizing any or all of the enclosed material is viewed as unprofessional by the EPA's
Air Pollution Training Institute.
This government publication is not copyrighted; however, it does contain some
copyrighted material. Permission has been received by the author(s) and the APTI to use
this copyrighted material for its original intended purpose as described on the cover sheet
of this workbook. Any duplication of this material, in whole or in part, may constitute a
violation of the copyright laws; unauthorized use could result in criminal prosecution
and/or civil liabilities.
The recommended procedure for mass duplication is as follows. Permission to use this
material in toto may be obtained from APTI in RTP, NC provided the cover sheet is
retained in its present form. Permission to use part of this material may also be obtained
from APTI provided APTI and the author(s) are properly acknowledged.
Author's Comment
The reader should note (hat in converting iheoi igmal manuscript to tins final punted form. thr
t vpesetter lailed to include the author's recommended spacing lor the various steps provided m
(he solution 10 each problem. Spaces were ongmallv piovided at each step where a solution is
i ec jtin ed so that (lie i eader con Id pertorm t he ret] nested calcul at ion dn ect I v j n the manuscript.
I his will be c onected in a subsequent printing.
ii
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Foreword
The increasing complexity of air pollution control regulations, air pollution control
technology, and the diversity of educational qualifications among employees who conduct
the nation's air pollution abatement programs, combine to create challenge in providing
the training chat is essential to develop and maintain effective conduct of those programs.
This is a self-instructional problem workbook (SIPW). It attempts to meet the
challenge of effective instruction for students with diverse backgrounds, including non-
technical education, with an approach that lies between programmed instruction and the
conventional textbook.
This workbook is for use in EPA Courses 413, Control of Particulate Emissions, and
415, Control of Gaseous Emissions, and it addresses the objectives established for these
courses. The primary objective of each of these courses involves the first of the abilities
stated below. The other stated abilities are important for an abatement worker dealing
with che public and they are developed as a by-product of addressing the primary
objective.
1. The employee should be able to check the more basic elements of design in an
emission control system as reflected in an application for a permit to operate. The
purpose of the check is to judge whether a proposed system, when operating as it is
supposed to operate, will meet established regulatory standards.
2. An agency employee should be able to talk intelligently and on a professional level
about emission control with representatives of regulated industries. To do this, the
employee must first know the vocabulary and must have some understanding of the
control systems and how they operate.
3. The employee should be able to conduct or make constructive contribution to the
inspection of an emission control system for the purpose of deciding whether the
system is in compliance. This decision requires knowledge of how a system functions
and typical conditions which prohibit the system from properly functioning. The
knowledge required is essential in supporting the compliance decision both in a
written report and as a possible witness in a formal hearing.
Courses 413 and 415 are concerned primarily with problem solving, and this SIPW is
an adjunct to these two courses, illustrating many of the less complex problems typically
involved in review activity. This SIPW is divided into three parts: (1) the Nomenclature is
a "vocabulary": it defines terms commonly used in air pollution control; (2) the Basic
Operations section deals with calculations reflecting basic concepts from chemistry,
thermodynamics and physics and which appear in more than one type of control system
check; (3) the Problems section presents calculations linked directly with individual con-
trol processes discussed in EPA course material. These include combustion, absorption,
adsorption, electrostatic precipitation, fabric filtering, etc. The section concludes with
problems related to fans and economics.
These problems have been laid out in such a way as to develop the reader's technical
understanding of the subject and/or control devices in question. Throughout each of
these problems, emphasis has been placed on applications. A special feature of the book
in
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is the inclusion of a comment section (right side of page) with all problems and basic
operations. This material often contains more extensive information and details on key
topics.
It is hoped that this workbook provides material which makes some of the principles
governing the operation of emission control systems more understandable, and to which an
individual mav return for refreshment of knowledge and more efficiency than can be obtained
from review of a typical textbook. In addressing these problems, it is suggested that the workbook
user refer to the glossary to check the meaning of technical terms before the solution of the
problem is pursued. Please note that weauempt to stateeach step in a solution tn words as well as
reflecting the step in mathematical svmbology. This, combined with comments, frequenth
telling the "whv" of a step, mav facilitate the learning process for many.
As in most endeavors credit is also due to others who contributed to this product. The
present book has evolved from a host of sources, including: notes, homework problems,
and exam problems prepared by the author for a one-semester three-credit elective
graduate course at Manhattan College; "Selected Problems in Design of Air Pollution
Control Equipment." NTIS PB-246 363, NSF Grant GZ-3432 for which Dr. Theodore
served as author, technical editor, and project director; A. J. Buonicore and
L. Theodore's CRC Press text "Industrial Control Equipment for Gaseous Pollutants,"
Vol. I and II, 1975; EPA Course 415 Student Manual "Control of Gaseous Emissions,"
prepared by L. Theodore and A. J. Buonicore for Northrop Services under EPA Contract
No. 68-02-02374, 1978; and EPA Course 415 Student Workbook, authors unknown,
1968.
Last but not least, I believe that this modest work will help the majority of the
individuals working in the environmental field to obtain a reasonably complete
understanding of air pollution control equipment.
My sincere thanks go to Ann Kaptanis and Lucy Costa for their dedicated effort in
typing and aligning the manuscript. Thanks are also due Sonva Kreidenweis, Maritza
Montesinos, Steven Guptiil, Daniel Bourgeois, and John Zielinski for their assistance in
proofing the typed pages. The author is also deeply indebted to Ui Young Choi for her
contribution to this workbook. Fred Stinson, Science Administrator in the Office of
Academic Training at EPA, deserves special credit for conceiving the project. His direc-
tion and education of the author during the project period was a truly educational
experience; his patient handling of the author during this introduction to a brand new
world involving self-instructional training is most appreciated. Finally, thanks are
extended to Charles Pratt who maintained continuity in the project following Fred
Stinson s retirement.
This workbook is dedicated to the boys of Section X, including: Mustache John,
Marcus, Otto, Phil the Cabbie, Tony Leisure Suit, George, A1 the Salesman, DeMucci,
Uncle Ronnie, Bob S, but in particular to Big Tim and Nunzio.
Louis Theodore
Manhattan College
IV
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Table of Contents
Page
Foreword iij
Table of Contents
Nomenclature 1
Basic Operations 21
1. Specific Gravity and Density 23
2. Weight Fraction and Mole Fraction 24
3. Average Molecular Weight 26
4. Ideal Gas Law 28
5. Actual Volumetric Flow Rate and Standard Volumetric Flow Rate 30
6. Mass Flow Rate and Standard Volumetric Flow Rate 32
7. Collection Efficiency 33
8. Stoichiometry 34
9. Required Heat Rate 35
10. Gross Heating Value 36
11. Henrys Law 39
12. Discharge Velocity 41
13. Reynolds Number, Re 43
14. Pressure Terms 45
Problems 47
Particle Size Distribution —
Log-Normal Distribution 49
Andersen 2000 Sampler 52
Fluid-Particle Dynamics —
Particle Terminal Velocity 57
Stack Application 59
Gravity Settler —
Minimum Particle Size 61
Traveling Grate Stoker 62
Cyclone —
Cut Diameter and Overall Collection Efficiency 66
Plan Review 68
Electrostatic Precipitator —
Process Change
Collection Efficiency '3
Plan Review
Scrubber-
Design of a Venturi Scrubber
Overall Collection Efficiency 79
Plan Review 81
v
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Page
Baghouse —
Bag Selection 84
Cleaning Frequency 86
Bag Failure 88
Combustion —
Air Requirements 90
Thermal Afterburner Design 92
Plan Review of a Direct Flame Afterburner 96
Absorption —
Spray Tower 100
Packed Tower Plan Review 102
Tower Height and Diameter 106
Adsorption —
Working Capacity 110
Degreaser Ventilation Clean Up 112
Plan Review 114
Fans —
Erica ol Fan Wheel Size and Speed 120
Brake Horsepower Requirement 121
Economics —
Breakeven Operation 122
Annualized Installed, Operation, and Maintenance Costs 123
Solutions 125
Particle Size Distribution —
Log-Normal Distribution 127
Andersen 2000 Sampler 128
Fluid-Particle Dynamics-
Particle Terminal Velocity 131
Stack Application 133
Gravity Settler —
Minimum Particle Size 134
Traveling Grate Stoker 135
Cyclone —
Cut Diameter and Overall Collection Efficiency 137
Plan Review 138
Electrostatic Precipitator —
Process Change 139
Collection Efficiency 140
Plan Review 141
Scrubber —
Design of a Venturi Scrubber 142
Overall Collection Efficiency 143
Plan Review 144
VI
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Page
Baghouse
Bag Selection 146
Cleaning Frequency 147
Bag Failure 148
Combustion —
Air Requirements 149
Thermal Afterburner Design 150
Plan Review of a Direct-Flame Afterburner 152
Absorption -
Spray Tower 154
Packed Tower Plan Review 155
Tower Height and Diameter 156
Adsorption —
Working Capacity 157
Degreaser Ventilation Clean Up 158
Plan Review 159
Fans —
Effect of Fan Wheel and Speed 160
Brake Horsepower Requirement 161
Economics —
Breakeven Operation 162
Annualized Installed, Operation, and Maintenance Cost 163
Appendix 165
A. Gas Properties 167
B. SI Units 170
C. Tables of Specific Conversion Factors 176
vii
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Nomenclature
Absolute pressure
Absolute temperature
Absorber
Absorption
acfm
Activated carbon
Adhesion
Adsorber
A measure of pressure referred to a complete vacuum, or
zero pressure. Ideal gas law calculations employ absolute
pressure terms.
Ideal gas law calculations use absolute temperature values
expressed in either degrees Kelvin (K) or degrees Rankine
(°R). °R are associated with the Fahrenheit scale and K are
associated with the Centigrade scale,
°R = °F - 460
K = °C + 273
A control device for carrying on the process of absorption.
A process by which a liquid material is used to remove one
or more soluble gas (absorbate) components from a gaseous
mixture, usually without chemical reaction. Typical absor-
bents are: water, dilute basic or acidic solution, and lean
(low molecular weight) hydrocarbon oils.
A volumetric flow rate term with units of actual cubic feet
per minute. The actual flow rate is based on actual
operating (temperature and pressure) conditions. Predicting
performance and design calculations for control devices are
based on actual flow rates.
Holds the most promise for efficient treatment of gaseous
pollutants by adsorption. It has been used extensively in the
past for removal of organic compounds. Activated carbon is
prepared by first making a char. A material such as coal is
brought to a red heat but with insufficient supply of air to
sustain combustion. The char particle is then activated by
exposure to an oxidizing gas at high temperature. This gas
produces a porous structure in the char and thus creates a
large internal surface area. The size of the pore developed
during activation is of importance in adsorption.
Molecular attraction which holds the surfaces of two
substances in contact, such as a gaseous pollutant and a solid
adsorb ant.
A control device or piece of equipment for carrying on the
process of adsorption. The unit is usually a vessel or pipe
containing activated carbon or another adsorbent. Generally
it has a means of admitting and exhausting fluids plus
whatever other piping (connections) might be needed for the
operation of the unit, including desorption if involved.
1
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Adsorption
Aerodynamic diameter
(of a particle)
Aerosol
Afterburner
AICC
Air
A process by which a solid material is used to remove one or
more components from a liquid or gaseous stream, usually
without chemical reaction. The removal takes place through
adherence to the surface. Typical adsorbents are activated
carbon, molecular sieves, silica gel, and activated alumina.
The diameter of a sphere having the same resistance to
motion as a particle. In air pollution, it is defined as the
diameter of a sphere of unit density (g/cm3) having the same
falling speed in air as the panicle in question.
A two phase medium consisting of a gas (usually air) and
either particulates or small droplets of liquid.
An incinerator or combustion device employed to control air
pollutants by chemical reaction.
Annualized installed capital cost.
A gaseous mixture appearing in the atmosphere. The com-
position of air is:
Component
Volume or mole %
Nitrogen
78.09
Oxvgen
20.95
Argon
0.93
CO,, krypton, neon, helium.
H., xenon, ozone
0.03
Air-to-cloth ratio
Ambient air
Andersen sampler
Atmospheric pressure
(standard)
Atomic weight
The ratio of volumetric flow rate of contaminated air to the
total filtering area of the filter bags in a baghouse system.
Air-to-cloth ratio is equivalent to filtering velocity through
the bags in ft/min. The usual notation is A C or G C.
Any unconfined portion of the atmosphere: open air.
A device consisting of a series of stacked stages and collection
surfaces. It determines the particle size distribution of a gas
sample containing particulates.
The pressure of the air and the atmosphere at sea level.
Standard atmospheric pressure is usually expressed as 1.00
atm or 14.7 psia.
A number expressing the ratio of the weight of one atom to
that of another. Since the atomic weight is nothing more
than a relative weight, the numerical value must be obtained
with reference to some convenient standard. The modern
chemical atomic weight scale uses the oxygen atom as the
standard, giving it a weight of 16.000. The atomic weight of
hydrogen is 1.0, of sulfur 32, of iron 56, etc.
9
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Available heat i HA j The available heat (HA i at any temperature (T) is the gross
heating value (HVC) minus the amount of heat required to
take products of combustion to that temperature (EAH). See
gross heating value.
Baffle
Bag failure
Baghouse
Baghouse compartment
Bahco sampler
Breakthrough capacity
Brake horsepower
Bulk density
A deflector that changes the direction of flow or velocity of
gas or particulate matter. It is often used with centrifugal collec-
tors (cyclones) and gravity settling chambers.
Filter bags in a baghouse fail with time. There are three
basic failure mechanisms that can shorten the life of a bag;
these are related to abrasion, thermal degradation, and
chemical attack.
Also fabric filter. An air pollution control device used to
capture particulates by filtering gas streams tnrough large
fabric bags.
Baghouses are usually constructed in modular form. Each
module consists of one or more compartments. Each com-
partment can contain a few or several thousand bags.
A particle classifier used for measuring particle size distribu-
tion in a gas stream. Its working range is approximately 1 to
60 micrometers tor microns). It uses a combination ot ehuriation
and centrifugation to separate particles.
The adsorption capacity of a packed bed where traces of
pollutants begin to appear in the exit gas stream. The same
units as equilibrium capacity (ratio of the weight of the
adsorbate retained to the weight of adsorbent) are employed
for breakthrough capacity.
The fan power the user pays for. It is the power to operate the
unit. The term (gas) horsepower is used to describe the
power delivered by the fan to the polluted gas stream. The
efficiency of the fan is given by the ratio of the gas
horsepower to the brake horsepower.
Mass per unit volume. Different density terms are commonly
applied to the weight-volume characteristics of particle
masses. The true density refers to the material making up
the particle. For example, a particle of solid glass would
ordinarily have a "true" particle density of about 162 lb/ft3.
If the particle contained minute fissures, cracks, or pores,
these intraparticle voids would reduce the density to an
"apparent" value of perhaps 150 lb/ft3 or less. The "bulk"
density, obtained by weighing a vessel of given volume filled
with the glass particles, would probably be in the neighbor-
hood of 75 lb/ft3, depending on the interparticle voids.
3
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Buovancv
Carbon dioxide (CO,)
Carbon monoxide (CO)
Cascade impaccor
Catalvsis
Cacalvtic combustion
Catalytic converter
Centrifugal collector
Chlorine (Cl2)
Chromatography
Colburn chart
The (buoyant) force exerted on a panicle suspended in an
air stream. The magnitude of the force is given by the
weight of the fluid displaced by the particle. In air pollution
calculations, this force is generally neglected.
A compound formed by the complete combustion of carbon
and oxygen. A colorless, odorless gas normally part of
ambient air. Poisonous only at very high concentration.
Small concentrations in the atmosphere are desirable because
it helps regulate breathing.
Compound formed from carbon and oxygen. Generally
formed during incomplete combustion. Odorless, poisonous
in high concentrations.
Commonly used to measure particle size distributions of gas
streams. It consists of a series of stacked stages and collection
surfaces. Each stage consists of from one to as many as 400
precisely drilled jet orifices, identical in diameter in each
stage but decreasing in diameter with each succeeding stage.
The process of catalyzing or promoting a reaction by a
chemical agent that does not participate in the reaction. It is
a surface phenomenon like adsorption.
The combustion of (organic) compounds catalyticallv to
cause oxidation at a lower temperature than required for
thermal combustion. Generally employs a platinum or
vanadium catalyst.
Also catalytic combustion device, catalytic reactor, catalytic
incinerator, or oxidizer. An air pollution control device that
removes (organic) contaminants by reaction to carbon diox-
ide and water. The operating temperature is in the
600-900°F range.
Also see cyclone. A control device using centrifugal force to
remove particulates from a gas stream.
A chemical element generally in the form of a gas or dis-
solved in liquid. It is often converted to hydochloric acid,
which can create corrosion problems.
A method of analysis of separation based on adsorbing the
gases, vapors, or substances of a mixture and then driving
each off, one by one. This permits the separation of the mix-
ture into individual compounds. A detector is also employed.
See Number of overall gas transfer units, Noc. A graph pro-
viding a means for calculating Noc. Usually employed in air
pollution calculations for absorbers.
4
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Combustion
Combustion products
Condensate
Contact condenser
Contact power theory
Contaminant
Control agency
Control method
Critical diameter
Cumulative distribution
Also incineration. A general term used to describe burning
or rapid oxidation. It is accompanied by the release of
energy in the form of heat. It is a basic cause of air
pollution.
Chemical compounds in the form of gases, vapors, and par-
ticles produced by combustion action or a burning process.
Typical combustion gases contain products such as nitrogen,
oxygen (less than in normal air), water vapor, carbon diox-
ide, carbon monoxide, nitrogen oxide, sulfur oxide,
unburned fuel, carbon particles, etc.
Condensed vapor or the liquid portion of a mixture of gas
and liquid. Also a gas that has been cooled, causing some of
the vapor to be liquified.
A heat exchange unit where the coolant and vapor streams
are physically mixed.
A theory predicting the collection efficiency of a scrubber
system. The contact power theory assumes that the collection
efficiency of a scrubber is solely a function of the total
pressure loss for the unit.
An element or compound that is not desired and should be
removed. Generally a substance not normally existing in air,
water, or the fluid under consideration.
A governmental agency (city, county, State, or Federal)
which deals with pollution and its control.
Method of controlling and/or eliminating air pollutants. It
may involve process modification but generally refers to a
device specifically designed to reduce the substance from the
exhaust .
Also critical size. Particle diameters equal to or greater than
this size are collected with 100% efficiency.
A type of (particle) size distribution, generated by plotting
particle diameter versus cumulative percent. For log-normal
distributions, the particle diameter is plotted on logarithmic
coordinates and percent less than stated size (% LTSS) or
percent greater than stated size (% GTSS) is plotted on
probability (normal) coordinates. Thus, if 10% of the par-
ticles have a size equal to or less than 3.2 micrometers
(microns, n), the point (3.2 pL, 10%) would fall on a
cumulative distribution curve.
Cut diameter
The particle diameter collected at 50
% efficiency.
D
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Cut power rule
Cvclone
Cunningham correction
factor
A generalized method of calculating overall efficiency of a
wet scrubber. The method assumes the predominant collec-
tion mechanism is inertial impaction.
A control device widely used for collecting dust from
polluted air. It is a cylindrical or conical chamber, where the
dust-laden gas usually enters the chamber at the top; par-
ticles separate due to centrifugal forces and settle at the bot-
tom; the cleaner gas exits from another opening at the cop.
See Stokes' law. A correction factor applied to Stokes law
for particles less than 2.0 microns tor micrometers) in diameter.
Panicles below this size are affected bv collisions by air
molecules which alter the sen line; velocity ot the particle.
Darcv's law
Density
Desorption
Distillation
Drag coefficient, Cc
Drag force, Fc
An equation used to predict pressure drop across a filter. It
is given by:
Ap,= k,v.
Where: Ap, = pressure drop
kj = fabric resistance factor
v, = filtration velocity
Mass per unit volume. In engineering units the density is
expressed in pounds mass per cubic foot. In cgs, the density
has the units of gram per cubic centimeter for solids and
liquids, whereas for gases it is usually expressed as gram per
liter.
Also regeneration. Any process that accomplishes a partial or
complete separation of either an adsorbed substance from an
adsorbent or an absorbed substance from an absorbent.
A process employed for purifying liquids through boiling or
distillation.
See Drag force. A coefficient used to calculate the drag
force, F0, on a particle.
See Drag coefficient. The force a particle experiences
whenever there is relative motion between the particle and
gas stream. It is a resistive force and is given by:
Fo = irdf ev2C0/8
Where: v = relative velocity
d„ = particle diameter
q = fluid (gas) density
CD = drag coefficient
6
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Drift velocity
(migration velocity)
Dusts
Efficiency
Effluent
Electrostatic
precipitator
Elutriation
Emission factor
Energy balance
Enthalpy
Also particle migration velocity. Once a particle is
charged in an electrostatic precipitator, the particle will
migrate toward the collection electrode with a certain drift
velocity. It represents the collectability of the particles within
the confines of an electrostatic precipitator.
Powdered materials of earth or other matter. Usually formed
by crushing, grinding, combustion, or detonation.
The measure of the degree of performance of a control
device or process. Generally refers to the degree of purifica-
tion or removal of pollutant in air pollution control on a
mass basis.
A fluid, either gas or liquid, leaving a process, building, fac-
tory, stack, etc.
Also ESP. A unit used for separating dust particles
and/or mist from a polluted air stream. An electrostatic field
imparts a charge on the particles, which are then attracted
to a collecting electrode of an opposite charge. The polluted
air passes over a high voltage negative electrode, whose
voltage is of the order of 40,000-100,000 volts.
A particle separation technique that is accomplished by par-
ticles settling out of a gas stream that is moving in an
upward direction.
The relationship between the amount of pollution produced
and the amount of raw material processed. For example, an
emission factor for a blast furnace making iron would be the
number of pounds of particulate matter emitted per ton of
raw materials employed.
A concept based on a fundamental law of physical science
(conservation of energy) known as the first law of thermo-
dynamics which says that although energy may be trans-
formed it may not be destroyed. "Heat balance" is a loose
term applied to a special form of energy balance frequently
used in processes which are associated with gaseous pollutant
control by combustion.
Total heat content (Btu/lb or Btu/lb mole) above an arbi-
trarily established temperature base point. The base condi-
tion is often a temperature of 32 °F. Enthalpy may be
obtained directly from tables or calculated from the product
of the heat capacity of the gas multiplied by the number of
degrees F above the established temperature base.
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Entrainment
Entrainment separator
EPA
Equilibrium
Equilibrium capacity
Equilibrium line
Excess air
Exfiltration
Fabric filter
Fan efficiency
Carryover; for example, the carryover of liquid mist in air
after scrubbing. Mist eliminators are used to reduce
entrainment.
Also mist eliminator. A device employed in many scrubbers
and absorbers to control entrainment.
Environmental Protection Agency.
A state in which there is no tendency toward spontaneous
change. Equilibrium exists when the net interchange of
material between phases is zero. Thus, in the absorption
process, equilibrium exists when gas molecules leave a liquid
(desorption) at the same rate at which they enter
(absorption).
A term used in adsorption studies to describe the maximum
amount of pollutant retained by an adsorbent at equilibrium
and at a given operating condition. Common units are lb/ lb
adsorbent or lb/100 lb adsorbent.
See Equilibrium. An equilibrium line is a plot describing the
mole fraction of the solute (pollutant) in the gas phase in
equilibrium with the mole fraction of solute in the liquid
phase. This is often required in absorber calculations.
See also stoichiometric air. The percentage amount of air in
excess of stoichiometric air required for complete combustion
of all hydrocarbons.
The opposite of infiltration. The exhaust of gases from a
building or structure, due to wind velocity and thermal
effects, through defects in the structure and normal leakage
around openings.
Also see Baghouse. A bag type unit for removing particulate
matter from the air. It is similar to a large vacuum cleaner.
Efficiencies may he in excess of yy%. Various filter materials
used are: glass fibers, teflon, nylon and rouon.
See Brake horsepower also. The efficiency of a fan is given
by the ratio of the gas (air) horsepower to the brake
horsepower. Gas horsepower is the power delivered to the
moving gas. Brake horsepower is the power delivered to the
fan. The fan efficiency term accounts for irreversibilities in
fan operation; it must be taken into account in fan calcula-
tions since no fan is 100% efficient.
3
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Fan laws
Flame incineration
Flammabilitv range
Flare
Flooding
Flow regime
Flue gas
Fluoride
Fluorine (F)
Fly ash
Fume
Fan laws are used to predict how a change in one fan
variable can affect other fan variables. Fan laws are based on
the premise that if two fans are geometrically similar
(homologous) their performance curves are similar and the
ratio of fan variables can then be related through the fan
laws.
Also see Flare. Burning of combustible pollutants whose con-
centration is in the flammabilitv limit. Operating tempera-
tures are in the 2000-2500°F range.
Also explosive range. A concentration range of hydrocarbons
in air that will immediately ignite in the presence of a spark
or flame.
A (tall) stack for burning excess quantities of waste combusti-
ble gases. Typically present at oil refineries.
The excessive entrainment of liquid in a packed column or
tray tower. This leads directly to a condition of high pressure
drop.
A term applied to describe the characteristics of particle
motion in fluid flow. It includes:
1. Laminar (Stokes) flow; Reynolds number less than 2.0.
2. Transition (intermediate) flow; Reynolds number
greater than 2.0 but less than 500.
3. Turbulent (Newton) flow; Reynolds number greater
than 500.
The gas discharged from a stack after combustion. It can
include nitrogen oxides, carbon oxides, water vapor, sulfur
oxides, particles, and many other chemical pollutants.
A compound of fluorine with another chemical element.
A chemical element that is very reactive. It forms stable
compounds.
Noncombustible particles formed primarily during the com-
bustion of coal.
An aerosol or suspension of fine solid or liquid particles that
may be visible. It may combine with other gases, vapors, or
odors.
Gauge pressure
Geometric standard
deviation
A measure of pressure expressed as a quantity above atmos-
pheric pressure or some other reference pressure.
The degree to which numerical data tend to spread about an
average value is called the variation or dispersion of the data.
Various measures of dispersion or variation are available, the
most common being the (geometric) standard deviation.
9
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Grain
Grain loading
Gravity settling tank
Gross heating
value (HVC)
Heat balance
Heat capacity
Heat of combustion
Heel
A unit of mass equal to 65 milligrams or 1/7000 of a
pound.
The concentration at which particles are emitted from a
pollution source, e.g., grains/ft3. It occasionally is used to
describe particle mass flow rate, e.g., Ib/min.
Also gravity settler. Also settling chamber. The simplest of
the air control devices for removing particulates. The dust-
laden air enters through a pipe. Upon entry to the unit, the
linear velocity of the polluted air is decreased to a value
below 10 ft/sec. Due to gravity and this decrease in velocity,
particles are collected. Operating efficiencies are low.
The enthalpy change or heat released when a gas is stoichio-
metricallv combusted at 60 °F, with the final (flue) products
at 60°F and any water present in the liquid state. Also
referred to as the higher heating value, HHV. Usually
expressed as Btu/lb fuel or Btu/scf fuel.
See Energy balance.
The amount of heat required to raise the temperature of
1 mole of substance by 1 degree (molar heat capacity) or the
amount of heat required to raise the temperature of 1 unit
mass of substance by 1 degree. Typical units of heat capacity
include:
Btu/lb-°F
Btu/lb mole-°F
cal/g- °C
cal/g mole-°C
The appropriate absolute temperature may also be used.
Also enthalpy of combustion. The heat (enthalpy) of reaction
resulting from the complete oxidation of a substance with
oxygen with both the reactants starting, and the products
ending at the same condition, usually 25 °C or 60°F and
1 atm. Since all combustion processes result in a decrease of
enthalpy of the system, energy in the form of heat is released
during the reaction. The adopted convention is for the values
of heat of combustion to be negative in sign.
A term used in adsorption calculations. It refers to the
amount of pollutant retained by the adsorbent after
regeneration (or desorbing).
10
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Henrv's law A mathematical equation describing the relationship between
the concentration of a component in a gas with its
equilibrium concentration in the liquid. Henry's law states
that the partial pressure of a solute (pollutant) in the solution
becomes proportional to its mole fraction in the limit of zero
concentration. The equation is usually expressed in the
following form:
or
y i = mx,
Where: p, = equilibrium partial pressure of solute i
X[ = equilibrium mole fraction of solute dissolved
in solution
y. = equilibrium mole fraction of solute in gas
stream
— Henry's law constant, pressure/mole fraction
m = Henry's law constant, dimensionless
Note that Henry's law assumes a linear relationship between
the two equilibrium concentrations. This greatly simplifies
calculations for absorber heights.
Hoc Height of a transfer unit based on the overall gas film coeffi-
cient. Hoc is primarily a function of the resistance (difficulty)
of mass transfer in absorption processes. The numerical value
of Hoc is usually based on field or pilot plant studies. The
height of packing required for a given operation is given by:
Z = HocNoc; Z, Hoc in feet
Nog is the number of overall gas transfer units.
Hood A canopy or collecting structure over a process area or piece
of equipment to collect fumes, gases, vapors, or odors, which
are then exhausted or treated in some manner.
Hydrocarbons Organic compounds of carbon and hydrogen. Many of these
are used for fuel. Sometimes the term hydrocarbon is used as
a general designation for organic compounds and includes
some substances that contain other elements such as
nitrogen, oxygen, chlorine, sulfur, etc.
Ideal gas An imaginary (or hypothetical) gas which exactlv obevs the
ideal gas law. No real gas obevs the ideal gas law exac t ly over all
ranges oi temperature and pressure, although the "lighter"
gases ihvdrogen. oxygen, air. etc.) at ambient conditions
approach ideal gas law behavior. The "heavier" gases such as
Miliar dioxide and hydrocarbons, particularly at high pressures
11
-------
and low temporalures. deviate considerable Iron) the ideal gas
law. Despite* these deviations, the ideal gas law i-> routinelv used
in all air pollution calculations.
Ideal gas law A law or equation describing the relationship among
pressure, volume, and temperature of an ideal gas. A genera]
form for any number of moles of an ideal gas is:
PV = nRT
Where: P = absolute pressure
V = total volume
n = number of moles
R = ideal gas constant
T = absolute temperature
Impingement separator A device used to remove particles from the air; a typical one
is made of closelv-spaced louvers (baffles), where the air
stream is caused to change directions, and the particles hit
and adhere to the plates.
See Combustion. Oxidation by burning.
A gas that does not react with other substances under ordinary
conditions
Air entering a building through cracks and windows due to
the velocity of wind. It can also be caused by stack action
around the building.
An isobar is a graphical representation (plot) of the amount
of vapor adsorbed versu.s temperature at a particular (constant;
pressure. Also used in absorption studies.
An isostere is a graphical representation of the log (or
natural log, In) of the partial pressure of the solute or adsor-
bate (pollutant) in the gas "phase versus the reciprocal of the
absolute temperature (1.0/T) for a fixed quantity of
adsorbed vapor.
An isotherm is a graphical representation of adsorbent
capacity versus the partial pressure of the adsorbate at a par-
ticular temperature. Also used in absorption studies.
Johnstone equation An equation used to describe collection efficiency in a ven-
turi scrubber. The equation is given by:
- * v?
¦q = 1 — e Op
Where: = collection efficiency
'Sf = inertiai impaction number (dimensionless)
k = empirical coefficient based on lab, pilot plant,
or field test data
Q//Qp = liquid-to-gas ratio
12
Incineration
Inert gas
Infiltration
Isobar
Isostere
Isotherm
-------
K. factor
Kelvin
Laminar flow
Lappie's method
Latent heat
LEL
Liquid-to-gas ratio
Loading
See Flow regime. A dimension less term used to determine the
proper llou- regime tor a settling particle in a gas. It i.s independ-
ent ot the panicle settling velocity and given bv:
K = d,(ge,ew3
Where: dP = particle diameter
qp = particle density
q = gas density
H = gas viscosity
g= local acceleration due to gravity
See Absolute temperature.
Low Reynolds number flow (usually <2100) in a tube, pipe,
stack, equipment, room, etc., that is characterized by the
absence of eddies, i.e., the flow is smooth. The Reynolds
number is given by:
Re =
Lvg
Where:
q = density of the fluid
v = velocity of the fluid
L = a linear dimension
fi = viscosity of the fluid
Re = Reynolds number idimensionless]
A method used to calculate cyclone efficiency for a given
particle size. dp. The cut diameter d„t.. must be known. The
describing equation is:
1
1 + ([dp]CIir/dp}2
Heat effect associated with a change of phase, e.g., from
liquid to vapor (vaporization), vapor to liquid
(condensation), or from liquid to solid (fusion), etc., at a
given temperature. Expressed usually as Btu/lb or Btu/lb
mole.
The lower explosive limit. It is defined as that concentration
of fuel (organic) below which combustion will not occur in
the presence of a flame or spark.
The term liquid-to-gas ratio can be applied to absorbers or
venturi scrubbers. With absorbers, it represents the ratio of
the molar flow ratio of the liquid to the gas. For venturi
scrubbers, it is the ratio of the liquid to gas in the device,
with units of gal/1000 acf or gpm/1000 aefm.
The concentration of pollutant, usually expressed in grains of
pollutant per cubic foot of contaminated gas stream.
13
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Log-normal
distribution
Mass velocity
Material balance
Mean particle diameter
Methane (CHJ
Micron
Migration velocity
Mist
Mole (mol)
Mole fraction
See Cumulative distribution also. Particle size distribution
when a linear relationship exists on a logarithmic/probability
(normal) plot of particle size/cumulative distribution.
This term is obtained by dividing the mass flow rate by the
cross-sectional area through which the mass is flowing. In
absorption practice the mass velocity is expressed in pounds
per second per square foot, pounds per hour per square foot,
or kilograms per second per square meter. An advantage of
using mass velocity is that it is independent of temperature
and pressure when the flow is steady (constant mass flow
rate) and the cross section is unchanged.
A concept based on a fundamental law of physical science
(conservation of mass) which says that matter cannot be
created or destroyed.
Also average particle diameter. In air pollution the mean is
always based on mass. Thus, if a particle size distribution has
a mean of 7.2 microns, 50% of the particles by mass have a
size equal to or less than 7.2 microns.
Lowest molecular weight hydrocarbon. The main constituent
of natural gas. It is occasionally called marsh gas.
10"" meter ions* millionth of a mrti-ri. Micrometer is now the
accepted technically correct unit although "micron" is still
primarilv used. The velontv which a panicle migrates (drifts)
toward the collec tion plate in an electrostatic precipitator.
The velocity which a particle migrates (drifts) toward the col-
lection plate in an electrostatic precipitator.
Liquid particles that are formed by condensation of vapor.
That quantity of a pure substance whose weight (any unit of
mass) is numerically equal to its molecular weight. In
engineering calculations the pound mole is most commonly
used. A pound mole of hydrogen (H2, molecular weight 2) is
2 pounds and of oxygen (02, molecular weight 32) 32
pounds. The measure applies to mixtures of gases, e.g., a
pound mole of air is approximately 29 pounds. A pound
mole of the gases just mentioned or of any gas or mixture
occupies the same volume under standard conditions at 60 °F
and 1 atm, i.e., 379 cu ft. This important fact, based on
Avogadro's principle, establishes the basis for calculation of
gas flow rates and other factors which are a part of control
equipment design,
A ratio nnplou-d in expressing concentrations of solutions
and mixtures. The mole fraction of any component of a mix-
ture or solution is defined as the number of moles of that
component divided by the sum of the number of moles of all
components.
14
-------
Molecular weight See atomic weight. The molecular weight is the sum of the
weight of the atoms which comprise the molecule; therefore,
the molecular weight of a substance would be equal to the
sum of the atomic weights of each atom in its molecule. For
example, a molecule of hydrogen (2 atoms, H2) has a
molecular weight of 2; a molecule of oxygen (2 atoms, 02)
has molecular weight of 32. Water (H20) has a molecular
weight of 18.
Monitoring Periodic or continuous sampling to determine the level of
pollution.
MTZ The mass transit*!" zone of an adsorbent bed where the concentra-
i ion gradient is present. It exists bet ween the location where the
concentration is saturated and a value approaching zero. MTZ
in activated carbon applications are in the 2-4 inch range, The
Ml Z is dependent on the adsorbent, packing size, bed depth, gas
velocity, temperature, and total pressure ot the gas stream.
A natural fuel containing primarily methane and hydrocar-
bons that occurs in certain geologic formations.
See Gross heating value. Similar to the higher heating value
except that the water produced by the combustion is not
condensed but retained as vapor at 60°F. Also referred to as
the lower heating value, LHV, Expressed in the same units
as HVC.
Air pollutants; the two primary forms are nitric oxide, NO,
and nitrogen dioxide, N02. NO is colorless but it is
photochemically converted to N02 which is one of the major
components of smog; it causes a reddish brown plume.
Number of transfer units based on the overall gas film coeffi-
cient. N0c is a measure of the ease and degree of separation
in absorption processes. It is employed in packed height
calculations for absorbers. The packed height, Z, is given by;
Z = Hog Not;
Where: Hoc = the height of an overall gas transfer unit
The degree to which light is obscured by a polluted gas
stream, object, or substance. A clean window has zero opac-
ity, a wall 100% opacity.
A term used to describe operating conditions within an
absorber. It generally appears on an equilibrium diagram as
a straight line for dilute solutions. The slope of this line is
given by the ratio of the liquid to gas molar flow rates.
Natural gas
Net heating value
CHV.v)
Nitrogen oxides (NO*)
Nqg
Opacity
Operating line
15
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Oxidation
Packed ower
PAN (peroxvacetyl
nitrate)
Partial pressure
Particle
Performance curve
ppm (parts per million)
Rankine
Reaction temperature
Relative humidity
Regeneration
Reynolds number (Re)
The conversion of a substance to another form by combina-
tion with oxygen. In rapid oxidation, heat is released. For
some types of oxidation, heat is required.
An absorption pollution control device that forces con-
taminated gas through a tower containing packing, while
liquid is introduced over the packing material. The
pollutants in the gas stream either dissolve or chemically react
with the liquid.
A pollutant created by the action of sunlight on hydrocar-
bons and nitrogen oxides in the air. An ingredient of smog.
The pressure of one gas component in a mixture of gaseous
components if it alone occupies the mixture \ container. If the
gas mixture behaves approximately as an ideal gas. she gas
m ixtim* exerts a total pressure in the container which in equal to
the Mini of the individual partial pressures.
Fine liquid or solid matter such as dust, smoke, mist, fumes,
or smog, found in the air or gaseous emissions.
A graph describing collection efficiency of a control device as
a function of particle size.
A measure of concentration. Generally refers to parts per
million by volume, although sometimes (when so specified) it
refers to parts per million by weight. If we have one cubic
foot of benzene vapor mixed with 1 million cubic feet of air,
the benzene concentration in air is 1 ppm. The same analysis
can be extended to ppb, parts per billion.
See Absolute temperature.
Temperature at which oxidation occurs in a combustion
system for gaseous pollutant control.
Represents the ratio of the actual amount of water vapor
present in the air to the amount which could exist at satura-
tion. When the radio or TV announcer talks about percent
humidity, he means percent relative humidity (% RH).
See desorption.
A number which reflects characteristics of fluid flow in a
tube or duct. Laminar (smooth) flow is usually encountered
at Reynolds number below 2100. Under ordinary conditions
of flow, the flow is turbulent at Reynolds number above
about 4000. Between 2100 and 4000 a transition region is
found, where the type of flow may be either laminar or
turbulent.
16
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I he spcrd at which a tan wheel <>r shaft rotates, with units of
te\ olutions per miuute.
Used in absorption and adsorption studies. In absorption, it
represents the maximum amount of pollutant retained by the
scrubbing liquid at a given operating condition. In adsorp-
tion, it is the maximum amount of pollutant retained by the
adsorbent at operating conditions.
Standard cubic feet per minute. It is the air flow rate at
standard pressure and temperature. Standard conditions in
air pollution control applications are usually 60 °F and 1
atm. Other standard or reference conditions may be used.
A general term typically used as a name for an air pollution
control device that uses a liquid to capture particulate
pollutants. The term absorber is most frequently used as a
name for the same process in capture of gaseous pollutants.
Heat, the addition or removal of which, results in a change
of temperature.
See Terminal settling velocity. In air pollution, it is the
velocity of a particle calculated by equating the external
force acting on the particle (gravitv, centrifugal, etc.) with
the drag force.
A substance, generally a liquid, used for dissolving purposes.
The most universal solvent is water. Generally, the term
"solvent" refers to organic liquids used to dissolve various
substances. The scrubbing liquid in absorption processes is
often referred to as the solvent.
A term used to describe the pollutant in absorber studies.
The recovery of volatile solvents that are present in air or a
gas stream.
The ratio of two densities — that of the substance of interest
to that of a reference substance. The reference substance is
normally water. The term is dimensionless.
The ratio of two heat capacities —that of the substance of
interest to that of a reference substance. The reference
substance is usually water. The term is dimensionless.
Liquid droplets created by mechanical disintegration; found
in absorbers and some particulate control devices.
A vertical duct or conduit that discharges exhaust »ascs into the
aniiospheic.
-------
Standard conditions
Stationary source
Stoichiometric
Stoichiometry
Stokes" law
Stripping
Superficial velocity
Sulfur dioxide (SO*)
Sulfur trioxide (S03)
Surface condenser
Terminal (settling)
velocity
A specified state, for air pollution problem solving in this
course, of temperature and pressure. Typical conditions
include: 1 atm and 32°F, and 1 atm and 60 °F. In engineer-
ing units, at 1 atm and 32 °F, one pound mole of any gas
will occupy 359 cubic feet; at 60°F, it will occupy 379 cubic
feet. Other standard conditions may also be employed.
A pollution location that is fixed rather than moving.
Term applied to a mixture of fuel and air containing
precisely the amount of air required for combustion of the
fuel. Thus, no oxygen remains in the flue gas following the
combustion process. Few combustion processes are designed
for operation with a stoichiometric mixture. Excess air is
usually provided, the amount varying with the process and
type of fuel, e.g., odor control with an afterburner fired by
natural gas would employ a minimum of excess air; a gar-
bage incinerator might be designed with introduction of
100% excess air.
Material balances involving chemical reactions.
See Flow regime. Used to describe panicle settling dynamics
in the laminar flow regime, i.e., Reynolds number less
than 2.0.
A physical operation by which the desired transfer of
material is from either a liquid to a gas phase or a solid to a
gas phase.
The velocity term used to describe velocity through packing
(absorber or adsorber) assuming the packing was not present.
The actual velocity through the packing is higher since the
gas can flow only through the void volume of the packing.
The gas produced by the combustion of sulfur in air or oxy-
gen. One of the most significant air polluting substances.
Emitted from stacks burning fuels containing sulfur.
The anhydride of sulfuric acid. An oxidized form of sulfur
dioxide.
A heat exchanger where the coolant is physically separated
from the vapor stream, usually bv tubular surfaces, and
the vapor is condensed.
The value of velocity of a body (particle) when all forces
(gravity, drag, buoyancy, etc.) acting on the body balance.
Thus, the sum of all the forces acting on the body is equal to
zero.
18
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Thermal incineration
Thermal pollution
Threshold
Throat velocity
Toluene (C7H8)
Traveling grate stoker
Tray column
Turbulence
UEL
Also see Combustion. The burning or rapid oxidation of con-
taminants. This type of process occurs at a high tempera-
ture. Contrasted with catalytic combustion, which takes
place at lower temperature, requires additional energy to
accomplish its objective.
The addition of large quantities of heat to air, water, or
land so that the resulting temperature increase may have
harmful effects.
A concentration term employed in odor studies; it is the
minimum concentration at which a substance can be
detected by whatever test method is employed. At the odor
threshold, the average person can just barely detect that an
odor exists. Higher concentrations can be stated in terms of
thresholds. Two thresholds means that the odor can be
reduced to the threshold level by diluting one part of it with
one part of odor-free air. One hundred thresholds means
that one must add 99 parts of odor-free air to bring it down
to the threshold concentration. In a general sense, a definite
odor is about 10 thresholds, a strong odor 100 thresholds,
and an overpowering odor 1000 thresholds.
Gas velocities through the venturi throat. The typical range
is 12,000 to 24.000 ft, mm.
A solvent used in paints and varnishes. It is similar to
benzene, but has a higher boiling point and is less toxic.
A tvpej ol boiler where coal is carried and burned.
Also plate column. Cylindrical tower housing perforated
plates, bubble caps, etc. It may be used for scrubbing
polluted air for gaseous (absorption) or particulate control.
Haphazard and irregular secondary motion associated with
eddies in a moving fluid. One of the three "T's" required for
efficient combustion: the remaining: time and temperature.
Irregular motion of a fluid, usually caused by high velocities.
In meteorology, it is a disturbance of the atmosphere caused
by irregular motion of wind.
Upper explosive limit. (See lower explosive limit.) It is
defined as that concentration of fuel (organic) above which
combustion will not occur in the presence of a spark or
flame.
Vacuum
A hypothetical condition denoting the complete absence of
matter. Also a condition of zero pressure.
19
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Vacuum pressure
Vapor
Vaporization
Vapor plumes
Vapor pressure
Velocity
Vent
Venturi scrubber
Viscosity
Volatile substance
VOC
Weight fraction
Weir
Working charge
A measure of pressure expressed as a quantity below atmos-
pheric pressure or some other reference pressure.
A substance in the gaseous state which is considered conden-
sible. For example, the steam in a steam/air mixture is a
condensible vapor. The same applies to a hvdrocarbon/air
mixture.
A change of phase of a substance from a liquid to a gas.
Discharge or exhaust gases that are visible because they con-
tain liquid droplets.
Also equilibrium pressure. It is the pressure at which the
liquid and vapor phases of a pure substance exist in
equilibrium at a given temperature.
The rait* at which a fluid is flowing in a given direct ion. Air
u'loriiv is normally stated in feet per second or tee! per minute.
An outlet or opening (or the equivalent) from a chamber,
room, process, piping system, tank, or other enclosure which
permits the exhaust of gases,
A scrubber using water to remove pollutants from the air,
which passes through a venturi, or small opening. The air is
accelerated to high velocities in the venturi (throat).
A measure of a fluid's resistance to flow. Viscosities vary
greatly, from materials like heavy lubricating oils to water.
The viscosity is also a strong function of temperature,
increasing with increasing temperature for gases and deer'easing
with decreasing temperature for liquids.
A chemical compound or mixture contained in a solid or
liquid which volatilizes or evaporates (due to the application
of heat) at or near room temperature.
Volatile organic compounds.
A term employed in expressing concentrations of solutions
and mixtures. The weight fraction of any component of a
mixture or solution is defined as the weight of that compo-
nent divided by the total weight of the mixture or solution.
A device for distributing, measuring, or controlling the flow
of a liquid. Such a device is used in some types of absorption
systems.
Also working capacity. A term employed in adsorber calcula-
tions. It refers to the net amount of pollutant adsorbed in a
cycle. It usually includes MTZ and heel effects. It is usually
expressed as lb/ lb adsorbent or lb/100 lb adsorbent.
20
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Basic Operations
21
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Basic Operation 1
Specific Gravity and Density
Given the specific gravity of a liquid deter-
mine its density.
Specific gravity is the ratio of two
densities, that of the substance of
interest to that of a reference substance.
The reference substance for solids and
liquids is normally water.
Technical Vocabulary
Specific gravity
Density
Data
Specific gravity of the liquid (methanol)
= 0.92
Density of reference substance (water® 60 °F)
= 62.4 lb/ft3
Solution
1. Write theequation defining specific
gravity.
Specific gravity Density of methanol
of methanol Density of water
The density of water is 62.4 lb/ft3 in
engineering units (English) and 1.0
g/'cm3 in cgs units.
2. Calculate the density of methanol in
engineering units by multiplying the
specific gravity by the density of water.
Density of methanol = (specific gravity of
methanol) (density
of water)
= (0.92)(62,4)
= 57.4 lb/ft3
The procedure is reversed if one is
interested in calculating specific gravity
from density data.
23
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Basic Operation 2
Weight Fraction and Mole Fraction
Given the weight of each gas in a gas mix-
ture determine the weight and mole frac-
tions of each component.
By definition:
Weight fraction = weight of A/total
weight
Mole fraction = moles of A/totai
moles
Moles of A = weight of A/molecular
weight of A
Technical Vocabulary
Atomic weight
Molecular weight
Weight fraction
Mole fraction
Data
The mixture contains 20 lb of 02, 2 lb of S02,
and 3 lb of S03.
Solution
1. Determine the molecular weight of each
component.
Molecular weight of 02 = 32
Molecular weight of S02 = 64
Molecular weight of SO, = 80
2. Calculate the weight fraction of each
component.
You need the atomic weight to deter-
mine the molecular weight, e.g., the
atomic weight of the oxygen atom (O)
is 16, and the molecular weight of the
oxygen molecule (02) is 32. The atomic
weight of sulfur is 32.
Compound
Weight
Weight fraction
0,
20
20/25 = 0.8
SO-
2
0,08
SO,
3
0.12
25
1.00
24
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Calculate the mole fraction of each
component
Compound
Weight
Molecular
weight
Moles
Mole
fraction •
0.
20
32
20 32 = 0.625
0,901
SO;
2
64
0.0313
0.045
SO,
3
SO
0.0373
0.054
0.6938
l.'JOO
Remember,
Moles = weight / molecular weight
The procedure can be extended to
calculate the average molecular weight
of a mixture (see basic operation 3).
-------
Basic Operation 3
Average Molecular Weight
Given a mixture of gases and the percent of
each gas in the mixture determine its
average molecular weight.
The concentration of gases in a stack or
mixture of gases is tvpicallv determined bv an
Orsat analysis. (CO-.. Oi and CO are
measured; X- is calculated.) The concentra-
tions are expressed on a mole (or volumei
percent basis. These values are on a drv basts,
i.e.. the amount of water is not considered.
Technical Vocabulary
Atomic weight
Molecular weight
Data
Identification of each gas and its mole per-
cent (the gas analysis) is given below:
79%
O, 5%
C02 10%
CO 6%
Solution
1. Determine the molecular weight of each
component.
You need an atomic table to determine
molecular weight, e.g., the atomic
weight of the nitrogen atom is 14, and
the molecular weight of the nitrogen
molecule (composed of two nitrogen
atoms) is 28.
I. Multiply the molecular weight by its mole
percent and complete the table given below.
Compound
Molecular
weight
Mole
fraction
Weight, lbs
N,
28
a. 79
28 :< 0.79-22.1
IX
VI
I) 05
1.6
CO,
44
0.10
4.4
CO
2S
0-06
1.7
1.00
In a gas, molar percent equals volume
percent and vice versa. Therefore, a
volume percent can be used to deter-
mine weight fraction as illustrated in
the table.
26
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Calculate the average molecular weight
of the gas mixture.
Average \
molecular) =22.1 + 1.6 — 4.4 -f- 1.7
y weight /
= 29.8
The sum of the weights in lbs represent
the average molecular weight. This is
true since the calculation above is
based on 1.0 mole of the °fas mixture.
O
-------
Basic Operation 4
Ideal Gas Law
Given the pressure, temperature, and
molecular weight of an ideal gas determine
its density.
An ideal gas is an imaginary gas which
exactly obeys certain simple laws (ideal
gas law). No real gas obeys the ideal
gas law exactly, although the "lighter"
gases (hydrogen, oxygen, air, etc.) at
ambient conditions approach ideal gas
law behavior. The "heavier" gases such
as sulfur dioxide and hydrocarbons,
particularly at high pressures and low
temperatures, deviate considerably from
the ideal gas law. Despite these devia-
tions, the ideal gas law is routinely used
in air pollution calculations.
Technical Vocabulary
Ideal sras
O
Ideal gas law
Data
Pressure =1.0 atm
Temperature = 60 °F
Molecular weight of gas = 29
Solution
1. Write the ideal gas law.
PV = nRT
Where: P = absolute pressure
V = volume
T = absolute temperature
n = number of moles
R = ideal gas law constant
Absolute temperature (°R or K) are
used in the ideal gas law.
°F - 460 = °R
°C + 273 = K
Absolute pressures must also be
employed.
28
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2. Rewrite the ideal gas law in terms of
density, g.
6 =
m
"v
nM
V
PM
RT
Where: M = molecular weight of gas
m = mass of gas
Remember that the density is defined
as the mass divided bv volume.
3. Calculate the density of the gas using the
appropriate value of R.
PM
Q =
RT
= (l)(29)/(0.73)(60 -r 460)
= 0.0764 lb/ft3
Typical values of R are given below:
R= 10.73 psia-ft3/lb mole-°R
= 1545 psqf-ft3/lb mole-°R
= 0.73 atm-ft3/lb mole-°R
= 555 mm Hg-ftVlb mole-°R
= 82.06 atm-cm3/g mole-K
= 1.986 cal/g mole-K
= 1.986 Btu/lb mole-°R
The choice of R is arbitrary, provided
consistent units are employed.
Since the molecular weight of the given
gas is 29, this calculated density mav be
assumed to apply to air.
29
-------
Basic Operation 5
Actual Volumetric Flow Rate and
Standard Volumetric Flow Rate
Given a standard volumetric flow rate
determine the actual volumetric flow rate.
The actual volumetric flow rate, usually
* in acfm (actual cubic feet per minute),
is the volumetric flow rate based on
actual operating conditions
(temperature and pressure of the
system). The standard volumetric flow
rate, usually in scfm (standard cubic
feet per minute), is the volumetric flow
rate based on standard conditions. The
standard conditions have to be
specified, for there are different sets of
standard conditions. For most applica-
tions in air pollution, standard condi-
tions are 60 °F and 1 atm or 32 °F and
1 atm.
Technical Vocabulary
Standard volumetric flow rate
Actual volumetric flow rate
scfm
acfm
Standard conditions
Data
Standard volumetric flow rate of a gas
stream = 2000 scfm
Standard conditions = 60 °F and 1 atm
Actual operating conditions = 700 °F and
1 atm
Solution
1. Combine Boyle's law and Charles' law to
relate the actual volumetric flow rate to
the standard volumetric flow rate.
Charles' law states that the volume of
an ideal gas is directly proportional to
the temperature at constant pressure.
Boyle's law states that the volume of an
ideal gas is inversely proportional to the
pressure at constant temperature.
30
-------
Where: Q,, = actual volumetric flow rate
Q, = standard volumetric flow rate
Ta = actual operating tempera-
ture, °R or K
T, = standard temperature, °R
or K
Ps = standard pressure, absolute
P„ = actual operating pressure,
absolute
Remember that absolute temperatures
and pressures are used in all ideal gas
law calculations.
If the pressure is constant, the following
equation is used.
2. Calculate the actual volumetric flow rate
in acfm.
= 2000(700 -f- 460)/(60 + 460)
= 4462 acfm
If it is desired to convert from acfm to
scfm, reverse the procedure and use the
following equation.
Actual flow rates must be used in order
to design or predict performance of
control equipment.
31
-------
Basic Operation 6
Mass Flow Rate and
Standard Volumetric Flow Rate
Given a mass flow race determine the stan-
dard volumetric flow rate.
Technical Vocabulary
Mass flow rate
Data
Mass flow rate of flue gas, m = 50 Ib/min
Average molecular weight of flue gas,
M = 29 lb/lb mole
Standard conditions = 60°F, 1 atm
Solution
1. How mam cubic feet of flue gas does one
lb mole of flue contain at these standard
conditions?
PY = p.RT; R = 0.73 atm-ft3/ib mole-°R
V _ RT
n P
= (0.73)(60 + 460)/1.0
= 379 scf/lb mole
If the standard conditions are 60 °F and
1 atm, there are 379 standard cubic
feet of gas per lb mole of any ideal gas.
2. Calculate the standard volumetric flow
rate, Q,, in scfm.
Qj = (m/M)(379 scf/lb mole)
= (50/29)(379)
= 653 scfm
The calculational procedure is reversed
for determining molar (or mass) flow
rates from scfm data.
32
-------
Basic Operation 7
Collection Efficiency
Given inlet loading and outlet loading of a
control unit determine the collection
efficiency of the unit.
Collection efficiency is a measure of
degree of performance of a control
device; it specifically refers to degree of
removal of pollutant. Loading refers to
the concentration of pollutant, usually
in grains of pollutant per cubic feet of
contaminated gas stream.
Technical Vocabulary
Loading
Collection efficiency
Data
Inlet loading = 2 grains/ft3
Outlet loading=0.1 grains/ft3
Solution
1. Write an equation describing collection
efficiency in terms of inlet loading and
outlet loading,
7] = [(inlet loading - outlet loading)/(inlet
loading)](100)
2. Calculate the collection efficiency of the
control unit in percent.
[(inlet loading - outlet loading).- (inlet
loading)]( 100)
= [(2-0.1)/{2)]( 100)
= 95%
The collected amount of pollutant by
the control unit is the product of rj and
inlet loading. The amount discharged
to the atmosphere is given by the inlet
loading minus the amount collected.
Another term used to describe the
performance of high efficiency control
devices is penetration, Pt. It is
given by:
Pt = 1 - rj/ 100; fractional basis
Pt = 100-7); percent basis
So
o
-------
Basic Operation 8
Stoichiometry
Given a chemical reaction balance the equat-
ion anil determine the ratio of react-
• i!11> to products.
The chemical equation provides a
variety of qualitative and quantitative
information essential for the calculation
of the weight of reactants reacted and
products formed in a chemical process.
Technical Vocabulary
Stoichiometry
Reactant
Product
Data
The unbalanced reaction equation for the
(ombustion ol butane is shown below.
C4H,o + O2 — CO? + h2o
Solution
1. Balance the above equation.
C4H:o+ (13/2) 0,-4 C02~5 H,0
.Number of carbon atoms, in reactants =
Number of t arbon atoms. in products = 4
Number of oxygen atoms, in reactants =
Number of oxvgen atoms, in products = 13
Number of hvdrogen atoms, in reaitams =
Number ol hvdrogen atoms, in products = 10
2. Determine how many moles of C02 are
loaned from the combustion of 1 mole of
Moles dI CO-> = I
The balanced chemical equation must
have the same number of atoms of each
type in the reactants and products. For
example, given the balanced equation,
CH4 + 2 02 — C02 -r 2 HoO
Number of C's in the reactants —
Number of C's in the products — 1
Number of O's in the reactants =
Number of O's in the products = 4
Number of H's in the reactant =
Number of H's in the products = 4
Remember that the chemical equation
tells us in terms of moles (not mass) the
ratios among reactants and products.
34
-------
Basic Operation 9
TT*% * 1 __ __ T%
Required Heat Rate
Given the mass flow rate of a gas stream
and its heat capacity determine the required
heat rate to change the gas stream from one
temperature to another.
Technical Vocabulary
Mass flow rate
Heat capacity
Data
Mass flow rate of gas stream, m = 1200
lb/mm
Average heat capacitv of gas, C„ = 0.26
Btu/lb-°F
Initial temperature. T1=200°F
Final temperature, T,= 1200°F
Solution
1. Write the equation describing the
required heat rate, q.
• A T*
q = mC,AT
= mC,(T,-T,)
The required heat rate, q, can also be
expressed in terms of enthalpy.
q = niiAH
Where: AH = change in enthalpy
between Tj and T2)
Btu/lb
Note that if the enthalpy is given with
units of Btu/lb mole, employ the molar
flow rate instead of mass flow rate.
2. Calculate the required heat rate, q, in
Btu/min.
q = mCp(T2 - Tj)
= (1200)(0.26)(1200 - 200)
= 3.12 x 105 Btu/min
35
-------
Basic Operation 10
Gross Heating Value
Gi \en a gas mixture determine its gross (or
highen heating value
The gross heating value (HVC)
represents tlje enthalpy change or heat
released when a gas is stoichiometrically
combusted at 60°F, with the final (flue)
product at 60 °F and any water present
in the liquid state. Stoichiometric
combustion requires that no oxygen be
present in the Hue ga» following complete
combustion ol the hydrocarbons.
Technical Vocabulary
Gross heating value (HVC)
Combustion
Stoichiometry
Data
Composition (mole fraction) of the gas
mixture:
N2 0.0515
CH4 0.8111
C,H6 0.0967
C,H, 0.0351
QH10 0.0056
1.0000
Table 1 (Combustion constants).
36
-------
I al)li: I. Ooniliuslion cuiiMauls.
Ileal of combustion
For 100% total air
(iiiol/mol of combustible)
(fl'/ft* Of combustible)
For 100% total air
< 11)/11) of combustible)
Flammabiliiy
limits
Substance
Ft'/lb
(Uiu/fc')
(lltu/lb)
Kcquiii'il
for combustion
I'ltlC J)l Oil lid S
Rcquirc-il
for combustion
Flue products
(% l>y volume)
(iron
(high)
Nci
(low)
O ruu
(high)
Nci
(low)
o,
N.
Air
CO,
11,0
N.
o,
N,
Air
CO,
II,O
N,
l.*owcr
lJ|)|K.-r
Cm lion. C*
-
11.01)3
1-1.093
1.0
3 76
•1 76
10
3 76
2 66
8 86
1 1 53
3.66
8 86
i lydiogcn, i lt
0 0053
187 723
325
275
61.100
51.623
0 5
1 88
2 38
10
1 88
7.91
26 II
31 31
-
8.91
26 11
¦I 00
71 20
Oxygen. O,
0 0811.
II 81 68
2 0
3 0
13 18
3 73
12 39
16 12
2 93
1 80
12 39
3 00
12 50
l'io|>jnc, C,ll,
0 11%
8 365
251)0
2385
21.661
11) .1)11
5.0
18 82
23.82
3.11
1 0
18 82
3 63
12 07
15 70
2 99
1 68
12 07
2 12
9 35
n lllil anc. Ctl l,u
0.1582
6 321
3370
3113
21.308
11). 680
6 5
21-17
30.97
10
50
21 17
3 58
II 91
15 19
3 03
1.55
1191
1 86
8-11
Isolml JIIC, C *, 11,0
0.1582
6 321
331)3
3105
21.257
11).621)
6 5
21 17
30 97
10
5 0
21 17
3.58
II 91
15 19
3 03
1 55
1191
1 80
8 11
II IVlllJIlC. CJ 1,,
0 1901
5 252
•10 ll>
3701)
21.01) 1
ID.517
8 0
30 II
38 1 1
5 0
6 0
30 11
3 55
11 81
15 35
3 05
1 50
1181
U(i|)riu jiic, CJ 1,,
0 1901
5 252
•1008
3716
21.052
11). 178
8 0
30 11
38 II
5 0
6 II
30 II
3 55
I I 8 I
15 35
3 05
1 50
11 81
Nc<)| K ill JIIC. (ijll,,
0.1901
5.252
31)1)3
361)3
20.1)70
11).31)6
8 0
30 ||
38 1 1
5.0
6 0
30 11
3 55
II 8 I
15 35
3 05
1 50
1181
ll 1 lex .inc. C )fcl I,,
0 227-4
•1.31)8
•171)2
•1112
20.1)-10
11). 103
1) 5
35 76
15.26
6 0
7 0
35 76
3.53
II 71
15.27
3.06
1.16
11 71
118
7 10
< )lclill sciles
l:.lliylcnc. C,1 l4
0 07 Hi
13 112
11)1 1
1513
21.611
20.21)5
3 0
II 29
1 1 29
2 0
2 0
11.29
3 12
11 39
1181
3 J1
I 29
II 39
2 75
28 60
I'l ujjylcljc, ( .'.I 1.
0 11 10
9 007
233h
2186
21.011
11)61)1
15
16 1)1
21 11
3 0
3 0
16 91
3 12
1139
1181
3 11
I 29
1 1 39
2.00
1110
11 liiilcnc. O.I 1,
0 1 480
6.756
3081
2885
20.810
11). UN)
6 0
22 59
28 59
ill
10
22 59
3.12
1131*
II 81
3 11
I 29
11 39
1 75
9.70
ImiImH cue. ("til.
0 1 180
6 751)
301.8
2861)
20.730
11). 382
6 0
22 51)
28 59
•III
10
22,59
3 12
II 39
1181
3 11
1.29
11 39
n 1'clll cue. C,l ll0
0 1852
5 100
3831)
3586
20.712
11). 363
7.5
28.23
35 73
5(1
5 0
28 23
3 12
II 39
11 81
3 11
I 29
1 1.39
Aioiiij(i( sci ics
Hciwcdc. C'»l l»
0 2060
•1.852
3751
3(>0I
18,210
17.180
7 5
28 23
35 73
I) 0
3 0
28 23
3 07
10 22
13 311
3 38
0 69
10 22
110
7 III
1 utllclH", C,l I,
0 2 431
•1113
•1181
•1281
1 8.-1 III
17,620
9 0
33 88
12.88
7.0
10
33.88
3.IS
10 10
13 53
3 31
0.78
10 10
1 27
6.75
Xylcnc. (',1 l,o
0 2803
3 567
5230
•11)80
18.650
17,760
10 5
39 52
50.02
8 0
5 0
39 52
3 17
10 53
13.70
3 32
0.85
10 53
1.00
6.00
Miscellaneous gases
At clylcnc. ( ',1 1,
0 0697
11.311
I II)!)
1118
21.500
20,776
2.5
9.11
1191
2 0
II)
9 11
3 07
10 22
13 30
3 38
0 69
10 22
N.ij>ilulciic. C,ul 1,
0.338-4
2.1)55
5851
5651
17.21)8
16,708
12 0
15 17
57.17
10 I)
10
15 17
3 00
9 97
12.96
3 13
0 56
9 97
M. iliyl alcolml. CII.OH
0 08-11»
1 1 820
81)8
768
10,251)
9.078
15
5 65
7 15
10
2 0
5 65
I 50
1 98
6.18
1.37
I.IS
1 98
6 72
36 50
liliyl alcohol. 0,11,011
0 1211)
8.221
11)00
1 151
13.161
11,1)21)
3 0
11 29
11 29
2 0
3 0
11.29
2 08
6 93
9 02
1.92
1.17
6 93
3.28
18 95
Ammonia, Nil,
0 0156
21 1)1 1
111
365
1)668
8,001
0 75
2.82
3.57
SO,
15
3 32
III
1.69
6 10
SO,
1 59
5 51
15 50
27 00
Snlhii. S*
3.1)83
3.1)83
10
3 76
¦1 76
1 0
3 76
I 00
3 29
•I 29
2 00
3.29
1 lyilingcn sulllilc, t iZS
0 091 1
II) 1)711
ti 17
5116
7.100
6.5-15
15
5.65
7 15
1.0
III
5 65
111
1 69
6 10
l 88
0 53
1 69
1.30
15 50
.Slllltll llloxlllc. S( >,
0 1733
5 770
W.lll'l V,1|K1I, 11,0
0 0171)
21.017
All
0 07t>6
13 01)3
I • JSollllC
1.10
7 1,0
-------
Solution
1. Determine the gross heating value of each
component, HVG,. from Table 1.
Component
HVg, (Btu/scf)
0
ch4
1013
CjHs
1792
C,H»
2590
C«H|0
3370
2. Calculate the gross heating value of the
gas mixture, HVC( in Btu/sef.
HVC = (0.G51S)(0) + (0.811)( 1013)
- (0.0967)( 1792) + (0.0351 )(2590)
^(0.0056)(3370)
= 1105 Btu/scf
The net heating value (HVv) is similar
to HVg except the water is in the vapor
state. The net heating value (HVv) is
also known as the low heating value
and the gross heating value is known as
the high heating value.
n
HVr; = Ex,HVCl
i
Where: x, = mole fraction of i'*
component
38
-------
Basic Operation 11
Henry's Law
Given Henry's law constant and the partial
pressure ol a solute, determine the maximum
mole fraction of solute that can be dissolved
in solution.
Henry's law states that the partial
pressure of a solute in equilibrium in a
solution is proportional to its mole
fraction in the limit of zero concentra-
tion, In air pollution applications, the
solute refers to the pollutant.
Technical Vocabulary
Solute
Henry's law
Partial pressure
Mole fraction
Equilibrium
Data
Partial pressure oi hydrogen sn 1 tide. H-.S =
0.0! aim
Total pressure = 1 atm
Henry's law constant = 483 atm/mole
fraction
Solution
1. Write the equation describing Henry's
law.
p
Where: p«,j= partial pressure of H,S,
atm
.'Jf- Henry's law constant,
atm/mole fraction
x = mole fraction of H2S in
solution
For an ideal gas the partial pressure of a
component in a gas mixture is given by:
P«2j — y«,i-pr
Where: pr = total pressure
If Henry's law constant is given in
dimensionless form, the following equa-
tion is used.
y — mx
Where: y = mole fraction of solute in
gas stream
x = mole fraction of solute
dissolved in solution
m = Henry's law constant,
dimensionless
39
-------
2. Calculate the maximum mole fraction of
H2S that can be dissolved in solution.
*h2s = p
= 0.01/483
= 2.07 x 10"5
This maximum mole fraction represents
the amount of solute in the liquid in
equilibrium with its partial pressure.
Henry's law may be assumed to apply
for most dilute solutions. This law finds
widespread use in air pollution absorber
calculations since the concentration of
the pollutant in most industrial gas
streams is dilute.
40
-------
Basic Operation 12
Discharge Velocity
Given the exhaust gas volumetric flow rate
from a stack, calculate the discharge
velocity.
The stack discharge velocity is the
velocity at which the exhaust gas from
the process is discharged to the
atmosphere.
Technical Vocabulary
Mark
Suu'k uiM'har
-------
2. Calculate the cross-sectional area of the
stack, S, where D is the stack diameter.
S= ttD2/4
= (tt)(1.2)V(4)
= 1.131 ft2
3. Calculate the discharge velocity, v.
v = 0,/S
= (1461.5)/(l.131)
- 1292.2 ft/min
= 21.5 ft/sec
The velocity calculaied here is the bulk c
average velocity. The average velocitv can 1)
simplv calculated bv dividing theactual voli
metric gas flow rate bv the cross-section;'
area through which the gas flows
42
-------
Basic Operation 13
Reynolds Number, Re
Given the physical properties and velocity of
a gas stream through a circular duct
determine the Reynolds number of the gas
stream.
The Reynolds number provides infor-
mation on flow behavior. Laminar flow
is normallv encountered at a Revnolils
number below 2100 in a tube, but it
can persist up to Reynolds numbers of
several thousand under special condi-
tions. Under ordinary conditions of
flow, the flow is turbulent at a
Reynolds number above about 4000.
Between 2100 and 4000 a transition
region is found where the type of flow
may be either laminar or turbulent. By
definition, the Reynolds number is
dimensionless.
Technical Vocabulary
Reynolds number
Data
Velocity through the duct = 25 ft/sec
Duct diameter =1.5 ft
Gas viscosity = 1.16 x 10"5 Ib/ft-sec
Gas density = 0.075 lb/ft3
Solution
1. Write the definition of Reynolds number.
Re- L"e
M
Where: L = duct diameter
v = velocity
q = gas density
H = gas viscosity
43
-------
2. Calculate the Reynolds number of the gas
stream.
P-
= (1.5)(25)(0.075)/(1.16 x 1(H)
= 2.42 x 105
For most air pollution applications, one
can usually assume turbulent (high
Reynolds number) flow.
The flou- is turbulent.
44
-------
Basic Operation 14
Pressure Terms
Given the height of a column of liquid
whose top is open to the atmosphere deter-
mine the pressure (in various units) exerted
at the bottom of the column.
The pressure is defined as force per
unit area.
Technical Vocabulary
j
Atmospheric pressure
Absolute pressure
Gauge pressure
Vacuum
Data
Height of column = 2.493 ft
Liquid = mercury (Hg)
Density of mercury = 848.7 lb/ft3
Atmospheric pressure = 2116 lb//ft2 absolute
Solution
1. Write an equation describing the gauge
pressure in terms of the column height
and liquid density.
P = Qg h/g.-
Where: p = gauge pressure
q = liquid density
h = height of column
g = acceleration of gravity
gc = conversion constant
This equation is derived by a force
balance around the column, the details
of which are not given here. The gauge
pressure is a pressure term expressed as
a quantity above atmospheric pressure
(or some other reference pressure).
2. Calculate the pressure in lb/ft2 gauge,
p = egh/ge
= (848.7)(1)(2.493)
= 2116 lb//ft2 gauge
Note that g/g^ is equal to 1.0 lb, 11)
in engineering units.
45
-------
3. Determine the pressure in lb/ft2 absolute.
Absolute pressure = gauge pressure
-r atmospheric
pressure
= 2116 + 2116
= 4232 Ib/Zft2 absolute
By definition:
Absolute pressure = gauge pressure ¦+
atmospheric pressure. Remember that
the atmospheric pressure is the pressure
of the air and the atmospheric
surroundings while the absolute pressure
is a measure of pressure referred to a
complete vacuum, or zero pressure.
Expressed in various units, the standard
atmosphere is equal to:
1.00 Atmospheres (atm)
33.91 Feet of water (ft H20)
14.7 Pounds per square inch
absolute (psia)
29.92 Inches of mercury (in. Hg)
760.0 Millimeters of mercury
(mm Hg)
1.013 x 105 Newtons per square meter
(N/m2)
2116 Pounds per square foot
absolute (psfa)
4, Determine the pressure in psia.
1.0 psia = (4232 psfa)(l ft2/144 in.2)
= 29.4 psia
5. Determine the pressure ,p, in mm Hs*.
p =(29.4 psia)(760 mm Hg/14.7 psia)
= 1520 mm Hg
Remember that 760 mm Hg is equal to
14.7 psia which in turn is equal to
1.0 atm.
6. Determine the pressure in in. H,0.
p = (29,4/14,7)(33,91)(12)
= 813.34 in. H20
Many pressure terms in air pollution
are expressed in in. H20. Correspond-
ingly, pressure drops (or pressure
changes) across control equipment are
also expressed in in. H20.
46
-------
Problems
Readers should note chat the solution to each problem has been divided into a series of
steps. However, each step may contain a series of sub-steps which should be first solved
sequentially to answer each step. Sub-steps may also have a series of sub-steps, and these,
too, should be first solved sequentially in order to provide an answer to the sub-step in
question.
47
-------
Particle Size Distribution—Log-Normal Distribution
Situation
You have been requested to determine if a particle size distribution is log-normal.
Technical Vocabulary
Cumulative distribution
Log-normal distribution
Data
Particle size range, d„
(microns)
Distribution (*tg/m')
<0.62
25.5
0.62- 1.0
33.15
1.0 - 1.2
17.85
1.2 - 3.0
102.0
3.0 - 8.0
63.75
8.0 -10.0
5.1
>10.0
7.65
Total: 255.0
Solution
1. Generate a cumulative distribution curve. Cumulative distribution plots are
generated by plotting particle diameter
versus cumulative percent. For log-
normal distributions, plots of particle
diameter versus either percent less than
stated size (% LTSS) or percent greater
than stated size (% GTSS) produce
straight lines on log-probability (nor-
mal) coordinates.
a. Complete the table given below by
% of total = (mass/total rnass)(100)
The cumulative percent, based on
% GTSS, is the total percent of the
mass of particles greater than the upper
size for the range in question.
Complete the table given below by
calculating the percent of the total
mass and the cumulative percent
GTSS for each particle size range.
dp (microns)
% of local
Cumulative
% GTSS
<0.62
(25.5/255X100)= 10
100- 10 = 90
0.62- 1.0
13J.«/*«•*> ("to"! 1 ?
10.0
3
^ O
49
-------
b. Plot the cumulative distribution curve
on the blank log-probability graph
paper provided.
2. Is the distribution log-normal?
The cumulative distribution points are
plotted using the upper size particle
diameter of each range versus
cumulative % GTSS.
A linear (straight line) relationship on
log-normal coordinates indicates a log-
normal distribution.
50
-------
SUOJDIUI dplUZ£
51
-------
Particle Size Distribution—Andersen 2000 Sampler
Situation
Given Andersen 2000 sampler data, you have been requested to plot a cumulative
distribution curve on a log-probability paper and determine the mean particle diameter
and geometric standard deviation.
Technical Vocabulary
Andersen 2000 sampler
Cumulative distribution
Mean particle diameter
Aerodynamic diameter
Geometric standard deviation
Data
Volumetric flow rate, Q_=0.5 cfm
Figure 1 (Aerodynamic diameter versus flow rate through Andersen sampler for an
impaction efficiency of 95%)
Andersen 2000 sampler data:
Plate $
Tare weight (g)
Final weight (g)
0
20.48484
20.48628
1
21.38338
21.38394
9
21.92025
21.92066
3
21.55775
21.55817
4
11.40815
11.40854
5
11.61862
11.61961
6
11.76540
11.76664
7
20.99617
20.99737
back-up filter
0.20810
0.21156
52
-------
ID
H
?
h
:>
I 0
O.i)
U rt
II 7
(I li
Of>
0.!
(12
<} i
o l
i (i r> i) (i u h o io
•10 M) (lO 70 HO \H) 100
o :i o i o r, o i> ii h i o
:i o
Aerodynamic diameter, microns
Figure 1. Aerodynamic diameter vs. flow rate through Andersen sumpler
for an impaction efficiency of 95%.
-------
% by weight l('ss llian .suu;il size (% l.TSS)
Figure Z. Cuinuhwive
-------
Solution
1, Generate a cumulative distribution curve.
a. Complete the table given below bv
calculating the net weight, % of
total weight, and cumulative percent
for each plate. Use % LTSS.
Plate #
Net weight (mg)
%
Cumulative %
0
. c>!
, ,»"> <.-> U'sO?
Ci
5.6
S p."!,
2
O '
H i
"c.
j
4
3
6
, J O jAZ-
A
"I.Z
> C!
=7 "5
n >
7? ^
O* • I
back-up
filter
. 3 D\
i ' ^0
;1 2
5
-------
3. Determine the geometric standard devia-
tion, ac, if the distribution is approxi-
mately log-normal.
For a log-normal distribution:
Oc — .13/ Y50 = Y50/ Y 13 87 (for % LTSSj
Where: Y50 = mean diameter, 50%
point
Yj4.i3 = Particle diameter at
84.13 cumulative %
Y 15.87 ~ Particle diameter at
15.87 cumulative %
For % GTSS, use:
<7c = Y30/\ 84.13 = Y15 37/ Y50
56
-------
Fluid-Particle Dynamics—Particle Terminal Velocity
Situation
Three different sized fly ash particles settle through air. You are asked to calculate the
particle terminal velocity and determine how far each will fall in 30 seconds. Assume the
particles are spherical.
Technical Vocabulary
Cunningham correction factor
Stokes' law
Terminal velocity
Data
Flv ash particle diameters = 0.4, 40, 400 microns
Air temperature and pressure = 238 °F, 1 atm
Specific gravity of fly ash = 2.31
Solution
1. Determine the value for K. for each fly
ash particle size settling in air.
a. Calculate the particle density using
the specific gravity given.
Terminal velocity is a constant value of
velocity reached when all forces
(gravity, drag, buoyancy, etc.) acting
on a body balance. The sum of all the
forces is then equal to zero (no
acceleration). A dimensioniess constant
K determines the appropriate range of
the fluid-particle dynamic laws which
apply.
K = dP(gePe/y)!/3
Where: K. = a dimensioniess constant
which determines the
range of the fluid-particle
dynamic laws,
dp = particle diameter
g = gravity force
qp = particle density
q = fluid (gas) density
H = fluid (gas) viscosity
See basic operation 1.
0 I
-------
b. Determine the properties of the air.
c. Calculate the value of K for each
particle size.
See Appendix for the viscosity of air.
Calculate the density of air (see basic
operation 4).
A consistent set of units that will yield
a dimensionless K is: d(, in ft, g in
ft/sec2, qp in lb/'ft3, g in lb/ft3, and n
in lb/ft-sec.
2. Determine which fluid-particle dynamic
law applies for the above values of K.
The numerical value of K determines
the appropriate law.
K<3.3; Stokes' law range
3.30 43g° '9
For Newton's law range:
v= 1.74(gdpgp/g)°-3
4. Calculate how far the fly ash particles
will fall in 30 seconds.
Distance = (t)(v)
When particles approach sizes com-
parable to the mean free path of the
fluid molecules, the medium can no
longer be regarded as continuous since
particles can fall between the molecules
at a faster rate than predicted by
aerodynamic theory. To allow for this
"slip," Cunningham's correction factor
is introduced to Stokes' law.
v = (gdjg,,/18 n){C,)
Where: Q = Cunningham correction
factor = 1 -i-(2AX/dp)
A = 1.257 -f- 0.40 e_1 10 V2*
\ = mean free path of the
fluid molecules
(6.53 x 10~6 cm for
ambient air)
Check how the Cunningham correction
factor affects the terminal settling
velocity for the 0.4 micron particle.
The Cunningham correction factor is
usually applied to particles equal to or
smaller than 1 micron.
58
-------
Fluid-Particle Dynamics—Stack Application
Situation
You have been requested co determine the minimum distance downstream from a cement
dust emitting source that will be free of cement deposit. The source is equipped with a
cyclone.
Technical Vocabulary
Cyclone
Data
Particle size range of cement dust = 2.5-50.0 microns
Specific gravity of the cement dust = 1.96
Wind speed = 3.0 miles/'hr
The cyclone is located 150 ft above ground level. Assume ambient conditions are at 60 °F
and 1 atm. Neglect meteorological aspects.
Solution
1. What particle size should be considered The smallest particle will travel the
to find the minimum distance down- greatest horizontal distance,
stream from the source chat will be free
of the dust?
2. Determine the value of K for the
appropriate size of the dust. This
calculation is similar to that presented in
the previous problem.
K = dp(ge/»e/M2)l/3
Where: K = A dimensionless constant;
it determines the appro-
priate range of the fluid-
particle dynamic laws,
d, = particle diameter
g = gravity force
qp = particle density
q = gas density
H — gas viscosity
a. Calculate the particle density using
the specific gravity given.
b. Determine the properties of the gas
(assume air).
See basic operation 1.
See Appendix for the viscosity of air.
Calculate the density of air (see basic
operation 4).
59
-------
c. Calculate the value of K.
A consistent set of units that will yield a
dimensionless K is: d0 in ft, g in ft/sec2,
qp in lb/ft3, q in lb/ft3, and n in
lb/ ft-sec.
3. Determine which fluid-particle dynamic
law applies for the above value of K.
"b
K < 3.3; Stokes' law range
3.3
-------
Gravity Settler—Minimum Particle Size
Situation
A hydrochloric acid mist in air at 25 °C is to be collected in a gravity settler. You are
requested to calculate the smallest mist droplet (spherical in shape) that will be entirely
collected by the settler. Assume the acid concentration to be uniform through the inlet
cross section of the unit and Stokes' law applies.
Technical Vocabulary
«¦
Gravity settler
Stokes' law
Data
Dimensions of gravity settler = 30 ft wide, 20 ft high, 50 ft long
Actual volumetric flow rate of acidic gas = 50 ft3/ sec
Specific gravity of acid = 1.6
Viscosity of air = 0.0185 cp= 1.243 x 10~5lb/ft-sec
Density of air = 0.076 lb/ft3
Solution
1. Calculate the density of the acid mist.
See basic operation 1.
2. Calculate the minimum particle diameter
both in feet and microns assuming
Stokes' law applies.
For Stokes" law range:
Minimum dp = (18 ,«Q/'gg-BL)1''2
Where: fi = gas (fluid) viscosity
Q.= volumetric flow rate
g = gravity force
qp = particle or mist density
B = width of settler
L = length of settler
Note that the above equation can be
derived by substituting the terminal
settling velocity for Stokes' law into the
gravity settler equation v = Q/BL.
61
-------
Gravity Settler—Traveling Grate Stoker
Situation
A settling chamber is installed in a small heating plant which uses a traveling grate
stoker. You are requested to determine the overall collection efficiency of the settling
chamber given the operating conditions, chamber dimensions, and particle size distribu-
tion data.
Technical Vocabulary
Overall collection efficiency
Settling chamber
Terminal settling velocity
Traveling grate stoker
Operation and Design Data
Chamber width = 10.8 ft
Chamber height = 2.46 ft
Chamber length = 15.0 ft
Volumetric flow rate of contaminated air stream = 70.6 scfs
Flue gas temperature = 446 °F
Flue gas pressure = 1 atm
Particle concentration = 0.23 grains/scf
Particle specific gravity = 2.65
Standard conditions = 32 °F, 1 atm
Particle size distribution data of the inlet dust for the traveling grate stoker is shown in
Table 2.
62
-------
Table 2
Particle
size
range
(microns)
Average
particle
diameter
(microns)
Inlet
Grains/scf
Weight %
0-20
10
0.0062
2.7
20-30
25
0.0159
6.9
30-40
35
0.0216
9.4
40-50
45
0.0242
10.5
50-60
55
0.0242
10.5
60-70
65
0.0218
9.5
70-80
75
0.0161
7.0
80-94
85
0.0218
9.5
-94
-94
0.0782
34.0
0.2300
iOO.O
Assume chat the actual terminal settling velocity is one half of the Stokes' law velocity.
Solution
1. Plot the size efficiency curve for the
settling chamber.
a. Express the collection efficiency in
terms of the particle diameter, d„.
(1) Replace the terminal settling
velocity in the above equation
with Stokes' law.
(2) Determine the viscosity of the air
in lb/ft-sec.
(3) Determine the particle density in
lb/ft3.
You need the size efficiency curve to
calculate the outlet concentration for
each particle size (range). These outlet
concentrations are then used to
calculate the overall collection efficiency
of the settling chamber.
The collection efficiency for a settling
chamber can be expressed in terms of
the terminal settling velocity, volumetric
flow rate of contaminated stream, and
chamber dimensions:
¦q = v(BL/Qr
Where: -q = fractional collection
efficiency
v, — terminal settling velocity
B = chamber width
L = chamber length
Q_= volumetric flow rate of
the stream
Remember that the actual terminal
velocity is one half of the Stokes'
law range terminal velocity.
See Appendix.
See basic operation 1.
63
-------
(4) Determine the actual flow rate
in acfs.
(5) Express the collection efficiency in
terms of dp, with dp in ft. Also
express the collection efficiency in
terms of dP with dp in microns,
See basic operation 5.
Use the equation developed in l.a.(l)
above. Substitute values for qp, g, B, L.
p., and Q in consistent units. See
Appendix. Use the conversion factor for
ft to microns.
Steps b. and c. below are optional since the
size efficiency equation is given above in step
l.a.(5); you have the option of proceeding
directly to step 2.
b. Complete the following table in order
to generate a size efficiency curve.
d„
(jiin)
d'
(fim2)
7
(%>
93.3
8800
90
8100
80
6400
60
3600
40
1600
20
400
10
100
The particle diameter of 93.8 /mi is
obtained by setting the fractional effi-
ciency equal to 1.0 and solving the col-
lection efficiency equation above. The
values of d„ are chosen arbitrarily to
generate the size efficiency curve.
c. Plot the size efficiency curve for the
settling chamber using the table com-
pleted above (use Figure S).
The size efficiency curve is plotted as
efficiency versus particle diameter.
2. Determine the collection efficiencies for
each particle size from the size efficiency
curve completed above or from the size
efficiency equation.
Remember that percent efficiency is
obtained by multiplying fractional effi-
ciency by 100.
Average dP,
Utm)
Weight
fraction, w,
>7,
(%)
10
0.027
25
0.069
35
0.094
45
0.105
55
0.105
65
0.095
75
0.070
85
0.095
-r 94
0.340
1.000
3. Calculate the overall collection efficiency
for the settling chamber, -q.
rj = Lw.tj,
Where: 77, = efficiency for Ith size range
w, = weight fraction for i'* size
range
64
-------
% '•«U3IDlj]3
-------
Cyclone—Cut Diameter
and Overall Collection Efficiency
Situation
You are requested to determine the cut size diameter and overall collection efficiency of a
cyclone given the particle size distribution of a dust from a cement kiln.
Technical Vocabulary
Cut diameter
Overall collection efficiency
Average particle
size in range, d„
% wt
(microns)
1
3
5
20
10
15
20
20
30
16
40
10
50
6
60
3
> 60
7
Gas viscosity = 0.02 cp - | x.
Specific gravity of the particle = 2.9
Inlet gas velocity to cyclone = 50 ft/sec
Effective number of turns within cyclone = 5
Cyclone diameter = 10 ft
Cvclone inlet width = 2.5 ft
Solution
1. Calculate the cut diameter, [dp]cur.
The cut diameter is the particle
diameter collected at 50% efficiencv.
For cyclones:
m i _ r 9
Up I cut
|_27rn,v,(qp - q)_
Where: /*. = gas viscosity, Ib/ft-sec
Bc = cyclone inlet width, ft
n, = number of turns
v, = inlet gas velocity, ft/sec
Qp = particle density, lb/ ft3
q = gas density, lb/ft3
66
-------
a. Determine the value of qp- q.
b. Calculate the cut diameter using the
equation given.
2. Complete the following table using
Lapple's method.
d,
w,
d,/[d,L„,
VA%)
1
0,03
5
0.20
10
0.15
20
0.20
30
0,16
40
0.10
50
0.06
60
0.03
< 60
0.07
3. Calculate the overall collection efficiency.
Overall collection efficiency = 17 = Ew,tj(
Where: 17,-= efficiency for \'h size range
w, = weight fraction for i:" size
range
Since the particle density is much
greater than the gas density, (qp- q)
can be assumed to be qp. See basic
operation I,
Lapple's method provides the collection
efficiency as a function of the ratio of
particle diameter to cut diameter. Use
the equation
7J = (1.0)/[1.0 -r (dpc/dp)2]
(or Figure 4 provided in the next prob-
lem). For additional details on the
above equation, the reader mav refer to
L. Theodore and V. DePaola, "Predict-
ing Cyclone Efficiency," JAPCA, 30,
1132-3, 1980.
67
-------
Cyclone—Plan Review
Situation
As an air pollution control officer you have been asked to evaluate a permit application
to operate a cyclone as the only device on the ABC Stoneworks plant's gravel drier.
Technical Vocabulary
Cut diameter
Design and Operating Data
from Permit Application
Average particle diameter=7.5 microns
Total inlet loading to cyclone = 0.5 grains/ft3
Cyclone diameter = 2.0 ft
Inlet velocity =50 ft/sec
Specific gravity of the particle = 2.75
Number of turns = 4.5 turns
Operating temperature = 70°F
Viscosity of air at operating temperature = 1.21 X 10"5 Ib/ft-sec
The cyclone is a conventional one.
Air Pollution
Control Agency Criteria
Maximum total outlet loading = 0.1 grains/ft3
Figure 4 (Lapple's curve) provides the cyclone efficiency as a function of particle size
ratio.
Solution
1. Determine the collection efficiency of the
cvclone.
Utilize Lapple's method which provides
collection efficiency values from a graph
relating efficiency to the ratio of
average particle diameter to the cut
diameter. The cut diameter is the par-
ticle diameter collected at 50% effi-
ciency. See Figure 4 for graph or equa-
tion in previous problem.
68
-------
a. Calculate the cut diameter, [dp]£„,
'.1) Determine the inlet width of the
cvclone, Bc.
(2) Determine the value of qp - q.
(3) Calculate the cut diameter using
the equation given above.
b. Calculate the ratio of average particle
diameter to the cut diameter.
c. Determine the collection efficiency
utilizing Lapple's curve.
The equation for determining cut size
diameter is (see previous problem for
units and notation):
[dpjc,,, = [9 ,uBc/ 2;m,v\( qp - e)]1/2
Where: a = gas viscosity, Ib/'ft-sec
Bc = inlet width, ft
nr = number of turns
v. = inlet velocity, ft/sec
Qp = particle density, lb/ft3
q = gas density, lb/ft3
The permit application has established
this cyclone as conventional. The inlet
width of a conventional cyclone is lA
the cyclone diameter.
Since the particle density is much
greater than the gas density, (qp — q)
can be assumed as qp. Also see basic
operation 1,
See Figure 4.
2. Calculate the required collection effi-
ciency for the approval of the permit.
Input and output loadings are given.
See basic operation 7.
3. Would you approve the permit?
The collection efficiency of the cyclone
must be higher than the collection effi-
ciency required by the agency.
69
-------
10
!)
8
7
(i
<1
3
2
1 L—
(0 1)
9(1.0)
I'nniclc size ratio, (1,,/[(1,,],„,
Figure 1. Collection efficiency as a function of particle size ratio.
-------
Electrostatic Precipitator—Process Change
Situation
A horizontal parallel-plate electrostatic precipitator consists of a single duct 24 ft high
and 20 ft deep with an 11 inch plate-to-plate spacing. Given a collection efficiency at a
gas flow rate of 4200 acfm, you are required to determine the bulk velocity of the gas,
outlet loading, and drift velocity of this electrostatic precipitator. You are also requested
to calculate a revised collection efficiency if the flow rate and the plate spacing are
changed.
Technical Vocabulary
Deutsch-Anderson equation
Drift velocity
Data
Inlet loading = 2.82 grains/ft3
Collection efficiency at 4200 acfm = 88.2%
Increased (new) flow rate = 5400 acfm
New plate spacing =9 in.
Solution
1. Calculate the bulk flow velocity, v.
See basic operation 12.
\/- (SI
V
= Q/s
" " (24V £.
--V.% C4,4et
Where: Q_=gas volumetric flow rate
S = cross-sectional area
through which the gas
passes
2. Calculate the outlet loading.
Remember that
, 31.Z7U T/t1,
jj( fractional) =
/ inlet outlet N
\loading loading/
(inlet loading)
Therefore,
Outlet loading =
Z-S.2- 1 - . S&e.
(inlet loading)! 1 — tj)
3. Calculate the drift velocity.
Drift velocity is the velocity at which the
particle migrates toward the collection
electrode within the electrostatic
precipitator.
71
-------
The Deutsch-Anderson equation
describing the collection efficiency of an
electrostatic precipitator is:
r] = 1 - e^
Where: r\ = fractional collection
efficiency
A =collection surface area of
the plates
Q_= gas volumetric flow rate
w = drift velocity
a. Calculate the collection surface
Remember that the particles will be
area. A,
collected on both sides of the plate.
b. Calculate the drift velocity, w
Since the collection efficiency, gas flow
rate, and collection surface area are
now known, the drift velocity can easily
be found from the Deutsch-Anderson
equation.
4. Calculate the revised collection efficiency
Assume the drift velocity remains the
when the gas volumetric flow rate is
same.
increased to 5400 acfm.
5. Does the collection efficiency change
Note that the Deutsch-Anderson
with changed plate spacing?
equation does not contain a plate
spacing term.
72
-------
Electrostatic Precipitator—Collection Efficiency
Situation
You have been requested to calculate the collection efficiency of an electrostatic
precipitator containing three ducts with plates of a given size, assuming a uniform
distribution of particles. Also determine the collection efficiency if one duct is fed 50% of
the gas and the other passages 25% each.
Technical Vocabulary
Deutsch-Anderson equation
Drift velocitv
Operating and Design Data
Volumetric flow rate of contaminated gas = 4000 acfm
Operating (emprrature ami pressure = 200°C and I aim
Drift velocity = 0.40 ft/sec
Size of the plate = 12 ft long and 12 ft high
Plate-to-plate spacing=8 in.
Solution
1. What is the collection efficiency of the
electrostatic precipitator with a uniform
volumetric flow rate to each duct?
The Deutsch-Anderson equation is
assumed to apply if collection efficiency
is given by (see previous problem for
units and notation):
Where: rj = collection efficiency
A = collection surface area of
the plates
Q_= gas volumetric flow rate
w = drift velocity
a. Calculate the collection surface area Consider both sides of the plate,
per duct, A. z * 2
b. Calculate the collection efficiency of Remember that the volumetric flow
the electrostatic precipitator using the rate through a passage is one third of
D-A equation. ^ the total volumetric flow rate.
2. What is the collection efficiency of the
electrostatic precipitator if one duct is fed
50% of gas and the others 25% each.
73
-------
a. What is the collection efficiency of
the duct with 50% of gas, t?i?
(1) Calculate the volumetric flow rate
of gas through the duct in acfs.
(2) Calculate the collection efficiency The collection surface area per duct
of the duct with 50% of gas. remains the same.
b. What is the collection efficiency,
of the duct with 25% of gas flow?
(1) Calculate the volumetric flow rate
of gas through the duct in acfs.
(2) Calculate the collection efficiency
of the duct with 25% of gas.
c. Calculate the new overall collection The key equation becomes:
efficiency of the electrostatic = (O.SKtj,) + (2)(0.25)(tj2)
precipitator.
74
-------
Electrostatic Precipitator—Plan Review
Situation
Fractional efficiency curves describing the performance of a specific model of an elec-
trostatic precipitator have been compiled by a vendor. Although you do not possess these
curves, the cut diameter is known. The vendor claims that this particular model will per-
form with a given efficiency under your operating conditions. You are asked to verify this
claim and to make certain that the effluent loading does not exceed the standard set
by EPA.
Technical Vocabulary
Cut diameter
Deutscn-Anderson equation
Electrostatic precipitator
Operation and Design Data
Plate-to-plate spacing = 10 in.
Cut diameter = 0.9 microns -^0
Collection efficiency claimed by the vendor = 98%
Inlet loading = 14 grains/ft3
EPA standard for the outlet loading = 0.2 grains/ft3 (maximum)
Particle size distribution
Average particle
Weight range
size, < d„ >
(microns)
O
IO
O
oS
3.5
20- 40%
8.0
4-»
O
cr.
O
oS
13.0
60- 80%
19.0
80- 100%
45.0
Assume a Deutsch-Anderson equation of the form:
77 = 1 —
Where: 77 = collection efficiency
K. = empirical constant
d,, = particle diameter
/ 3
-------
Solution
1. Is the overall efficiency of the electro-
static precipitator equal to or greater
than 98%?
a. Determine the value of K by using the
given cut diameter.
- K ( ^ ^
.so- I- e
- ' A )
. r> - U
XL -- ,77
b. Complete the following table.
Weight
fraction, w,
< d„>
7.-
0.2
3.5
0.2
S.O
0.2
13.0
0.2
19.0
0.2
45.0
c. Calculate the overall collection
efficiency, rj.
d. Is the overall collection efficiency
greater than 98%?
Since the weight fractions are given,
you need collection efficiencies of each
particle size to calculate the overall col-
lection efficiency.
Remember that the cut diameter is the
particle diameter collected at 50% effi-
ciency. Since the cut diameter is
known, you can solve the D-A equation
directly for K.
Use the D-A equation to calculate the
collection efficiency.
?7 = — W,77,
Where: 77 = overall collection efficiency
w; = weight fraction of i""'1
particle size
7]i = collection efficiency of f*
particle size
Does the outlet loading meet EPA's
standard?
a. Calculate the outlet loading in
grams/ ft3. ^ i
Q.
b. Is the outlet loading less than
0.2 grains 'ft3' zl e «=>
Outlet loading = (1.0— T7)(inlet loading)
Remember thac 17 is the fractional effi-
ciency for the above equation.
3. Is the vendor's claim verified?
V-4
76
-------
Scrubber—Design of a Venturi Scrubber
Situation
You are required to calculate the throat area of a venturi scrubber to operate at a
specified collection efficiency.
Technical Vocabulary
Liquid to gas ratio
Throat area
Venturi scrubber
Data
Volumetric flow rate of process gas stream = 11,040 acfm (at 68 °F)
Density of dust = 187 lb/' ft3
Liquid to gas ratio = 2 gal/1000 ft3
Average particle size = 3.2 microns (1.05 x 10"5 ft)
Water droplet size = 48 microns (1.575 x 10"4 ft)
Scrubber coefficient, k = 0.14
Required collection efficiency = 98%
Viscosity of gas — 1.23 x 10~5 Ib/'ft-sec
Cunningham correction factor = 1.0
Solution
1. Calculate the inertial impaction param
eter, y, from Johnstone's equation.
Johnstone's equation describes the
collection efficiency of a venturi scrubber.
¦q — \ — e_ k(£U-'Qp)v?'
Where: r) = fractional collection
efficiency
k = correlation coefficient
whose value depends on
the system geometry and
operating conditions,
typically 0.1 to 0.2,
1000 acf/gal
Q//'Qp = liquid-to-gas ratio,
gal/'1000 acf
^ = Cgpvdp2/18 cU/x
the inertial impaction
parameter
77
-------
qp = particle density
v = gas velocity at venturi
throat, ft/sec
dp = particle diameter, ft
(!„ = droplet diameter, ft
ft —gas viscosity, lb/ft-sec
C = Cunningham correction
factor
2. From the calculated value of \p above,
Remember that
back calculate the gas velocity at the
i/' = CePvdp2/18 cL/1
venturi throat, v.
3. Calculate the throat area, S, using gas
velocity at the venturi throat, v.
S = Q/v
See basic operation 12.
78
-------
Scrubber—Overall Collection Efficiency
Situation
You are requested to calculate the overall collection efficiency of a venturi scrubber
which cleans a fly ash laden gas stream given the liquid-to-gas ratio, throat velocity, and
particle size distribution.
Technical Vocabulary
Liquid-to-gas ratio
Throat velocity
Venturi scrubber
Operating and Design Data
Liquid-to-gas ratio = 8.5 gal/1000 ft3
Throat velocity = 272 ft/sec
Panicle density of fly ash = 43.7 lb/ft3
Gas viscosity = 1.5 x 10"5 Ib/ft-sec
Particle size distribution data
d,, (microns)
Weight %
< 0.1
0.01
0.1 - 0.5
0.21
O
b<
I
0
0.73
1.0- 5.0
13.0
5.0-10.0
16.0
10.0- 15.0
12.0
15.0-20.0
8.0
>20.0
50.0
Use Johnstone's equation, with a k value of 0.2, to calculate the collection efficiency.
Neglect the Cunningham correction factor effect.
Use Johnstone's equation, as provided
in the previous problem.
— i — e — Qpk ?
Where: 77 = collection efficiency
k = correlation coefficient
whose value depends on
the system geometry and
operating conditions,
typically 0.1 to 0.2 gal/acf
Solution
1. What are the parameters used in
Johnstone's equation?
79
-------
a. Calculate the average droplet
diameter in ft.
(V 11 J
n,--
- 1
- O K19"
2. Complete the following table to cal-
culate the overall collection efficiency.
< d„>
< d„>
w,
(microns)
(ft)
(96)
1.
W:!?,
0.05
1.64 x 10"'
0.01
. 000^ ^
0.3
9.84 x 10-7
0.21
0.8
2.62 x 10"s
0.78
¦"jio S
3
9.84 x 10's
13.00
.cici9l
8
2.62 x 10"5
16.00
1 0
1 ^
13
4.27 x 10"'
12.00
1 0
It
18
5.91 x 10"5
8.00
1. 0
20
6.56 x 10"s
50.00
1 .0
Note that the average particle diameter
has been rounded to the nearest signifi-
cant figure.
3. Calculate the overall collection
efficiency, 77.
7] = Ew,77,
cjc7."7 IS
'80
-------
Scrubber—Plan Review
Situation
A vendor proposes to use a spray tower on a lime kiln operation to reduce the discharge
of solids to the atmosphere. The inlet loading is to be reduced in order to meet State
regulations. The vendor's design calls for a certain water pressure drop and gas pressure
drop across the tower. You are requested to determine whether this spray tower will meet
State regulations. If the spray tower does not meet State regulations, propose a set of
operating conditions that will meet the regulations.
Technical Vocabulary
Contact power theory
Spray tower
Operating and Design Data
Gas flow rate= 10,000 acfm
Water rate = 50 gal/min
Inlet loading = 5.0 grains/ft3
Maximum gas pressure drop across the unit= 15 in. H20
Maximum water pressure drop across the unit = 100 psi
Vendor's Design Data
Water pressure drop = 80 psi
Gas pressure drop across the lower =5.0 in. HiO
The state regulations require a maximum outlet loading of 0.05 grains/ft3. Assume that
contact power theory applies.
Solution
1. Calculate the collection efficiency based
on the design data given by the
vendor.
Contact power theory is an empirical
approach relating particulate collection
efficiency and pressure drop in wet
scrubber svMems. It assumes that paniculate
collect ion efficiency is a sole funct ion of t he
total pressure loss for the unit.
St = J>C+*L
.?G = 0.157 Ap
i = 0.583 Pl(Q//Q.g)
-------
Calculate the total pressure loss, &T-
(1) Calculate the contacting power
based on the gas stream energy
input, .lPG, in hp/1000 acfm.
(2) Calculate the contacting power
based on the liquid stream
energy input, SPL, in
hp/1000 acfm.
(3) Calculate the total pressure loss,
¦9?r, in hp/1000 acfm.
Calculate the number of transfer
units, Nr.
Where: ./>T = total pressure loss,
hp/1000 acfm
s/c = Contacting power
based on gas stream
energy input,
hp/1000 acfm
dp = pressure drop across the
scrubber, in. H.O
= contacting power based
on liquid stream energy
input, hp/1000 acfm
p£ = liquid inlet pressure, psi
Qj. = liquid feed rate, gal/ min
Qc = gas flow rate, ft3 mm
The scrubber collection efficiency is also
expressed as the number of transfer
units.
Nt = a .^Ta
= ln[l/(l — 7?)}
Where: Nr = number of transfer units
t? = fractional collection
efficiency
a,(3 = characteristic param-
eters for the type of
particulates being
collected
To calculate the total pressure loss, you
need the contacting power for both the
gas stream energy input and liquid
stream energy input.
Since the pressure drop across the
scrubber is given by the vendor, you
can calculate .lPG-
•JJ*c, — 0.157 dp
Since the liquid inlet pressure and
liquid to gas ratio are given, you can
calculate ,lPL.
= 0.583 Pi(Qj.! Q,g)
¦ .-/c, +.jPl
N, = a -A ;
The values of a and j3 for a lime kiln
operation are 1.47 and 1.05 respec-
tively. These coefficients have been
previously obtained from field test data.
-------
c. Calculate the collection efficiency
Nt = ln[ 1/(1 - r?)]
based on the design data given by
The number of transfer units, N., is
the vendor.
calculated above.
2. Calculate the collection efficiency
Since the inlet loading is known and the
n-qunrd In i rgulatiotiv
outlet loading is set by the regulation,
the collection efficiency can be
calculated readily. See basic
operation 7.
3. Does the spray tower meet the
The collection efficiency based on the
regulations?
design data given by the vendor should
be higher than the collection efficiency
required by the State regulations.
4, Assuming the spray tower does not meet
Note chat the calculational procedure is
the regulations, propose a set of
now reversed.
operating conditions that will meet the
regulations.
a. Calculate the total pressure loss, d^r,
To calculate you need N..
using the collection efficiency required
by the regulations in hp/1000 acfm.
(1) Calculate the number of transfer
x
ii
jr
*—>
»
1
units for the efficiency required
The collection efficiency required by
by the regulations.
the regulations is given above.
(2) Calculate the total pressure loss,
N, = a #T3
in hp/1000 acfm.
b. Calculate the contacting power based
= 0.157 Ap
on the gas stream energy input,
Ap of 15 in. H,0 is the maximum
using a -lp of 15 in. H:0.
value allowed by the design.
c. Calculate the contacting power based
I
II
on the liquid stream energy input, -JA-
d. Calculate Q^/Qp, in gal/acf, using a
Q^/Qp = '^,'0.583 pt
p£ of 100 psi.
A pL of 100 psi is employed since it is
the maximum value allowed in the
design.
e. Determine the new water flow rate,
Q/ = (Qj-/Qfi)( 10.000 acfm)
Q/ in gal/min.
f. What are the new set of operating
conditions that will meet the
regulations?
33
-------
Baghouse—Bag Selection
Situation
It is proposed to install a pulse-jet fabric filter system to clean an air stream containing
particulate pollutants. You are asked to select the most appropriate filter bag considering
the performance and cost.
Technical Vocabulary
Air-to-cloth ratio
Design and Operating Data
Volumetric flow rate of polluted air stream = 10,000 scfm (60°F. 1 atm)
Operating temperature = 250°F
Concentration of pollutants = 4 grains/ft3
Average air-to-cloth ratio = 2.5 cfm/'ft2 cloth
Collection efficiency requirement = 99%
Information given by filter bag manufacturers
Filter bag
A
B
C
D
Tensile strength
Excellent
Above
Fair
Excellent
average
Recommended
260
275
260
220
maximum
temperature, °F
Resistance factor
0.9
1.0
0.5
0.9
Cost per bag
S26.00
$38.00
S10.00
S20.00
Standard size
S " x 16'
10"x16'
1' x 16'
1' x20'
Note: No bag has an advantage from the standpoint of durability under the operating
conditions for which the bag was designed.
Solution
1. Eliminate from consideration bags which,
on the basis of given characteristics, are
unsatisfactory.
2. Determine comparative costs of the
remaining bags.
a. Establish the cost per bag.
Consider operating temperature and
bag tensile strength required for a
pulse-jet system.
Total cost for a type bag is the number
of bags times cost per bag. No one type
bag is more durable than the other.
This information is given.
84
-------
b. Determine number of bags, N, for
each tvpe.
(1) Calculate the total filtering
area, Ac.
- I
(a) Convert given flow rate to
acfm. i buSA.
(b) Establish filtering velocity, vf.
(c) Calculate the total filtering
area from the acfm and
filtering velocity determined
above.
Calculate the filtering area per
bag. A- i\.
u
n- ft.-
~~ ??.*•
A ~ Yt •
.*1
(3) Determine the number of bags
required, N. . Mi
/ is > 2
Determine the total cost for each
bag.
Q
C,3 "9
The number of bags required, N, is the
total filtering area required divided by
the filtering area per bag.
Total filtering area, Ac = Q/v/
Where: Q_= gas stream volumetric
flow rate, acfm
vf= filtering velocity, ft/min
See basic operation 5.
This is given. Air-to-cloth ratio
expressed in cfm/ft2 is the filtering
velocity given above as 2.5 cfm/ft2
cloth.
A = rDh
(bags are assumed to be cylindrical)
Where: D = bag diameter
h = bag length
Remember that N = Ac/A
Total cost = (N)(cost/bag)
3. Select the most appropriate filter bag
considering the performance and cost.
85
-------
Baghouse—Cleaning Frequency
Situation
You are requested to determine the number of filtering bags required and cleaning fre-
quency for a plant equipped with a fabric filter system. Operating and design data are
given.
Technical Vocabulary
Cleaning frequency
Fabric filter system
Operating and Design Data
Volumetric flow rate of the gas stream = 50,000 acfm
Dust concentration = 5.0 grains/'ft3
Efficiency of the fabric filter system = 98.0%
Filtration velocity = 10 ft/min
Diameter of filtering bag= 1.0 ft
Length of filtering bag = 15 ft
The system is designed to begin cleaning when the pressure drop reaches 8.0 inches of
water.
The pressure drop is given by:
dp = 0.2 V/
Where: -Id = pressure drop. in. H,0
v; = filtration velocity, ft/min
c = dust concentration, lb/'ft3
t = time since the bags were cleaned, min
Solution
1. What are the number of bags, N,
needed?
a. Calculate the total required surface
area of the bags, Ae>. in ft2.
To calculate N, you need the total
required surface area of the bags and
the surface area of each bag.
Ac Q/ vc
Where: Ac = total surface area of the
bags
Q_= volumetric flow rate
Vf — filtering velocity
36
-------
b. Calculate the surface area of each
A = TrDh
bag, A, in ft2.
Where: A = surface area of a bag
D = diameter of the bag
h = length of the bag
c. Calculate the number of bags, N,
N = Ac/A
required.
2. Calculate the required frequency of
Apr= 0.2 v.- 4- 5 cva
cleaning.
Since dp is given as 8.0 in. H,0, the
time since the bags were cleaned is
calculated by solving the above
equation.
87
-------
Baghouse—Bag Failure
Situation
An installed baghouse is presently treating a contaminated gas stream. Suddenly some of
the bags are broken. You are now requested to estimate the new outlet loading of this
baghouse system.
Technical Vocabulary
Bag failure
Baghouse compartment
Operation conditions of the system = 60 °F, 1 atm
Inlet loading = 4.0 grains/acf
Outlet loading before bag failure = 0.02 grains/acf
Volumetric flow rate of contaminated gas = 50,000 acfm
Number of compartments = 6
Number of bags per compartment = 100
Bag diameter = 6 inches
Pressure drop across the system = 6 in. HzO
Number of broken bags = 2 bags
Assume that all the contaminated gas emitted through the broken bags is the same as
chat passing through the tube sheet thimble.
Data
Solution
1. Calculate the efficiency and penetration
before the bag failure(s).
Penetration, Pt = 1 — 77
See basic operation 7.
The effect of bag failure on baghouse
efficiency is described by the following
equation.
/
Pt* = Pt + Ptc
0 = Q/(LD2VT+46O)
Where: Pt* = penetration after bag
failure
Pt = penetration before bag
failure
Ptc = penetration correction
term, contribution of
broken bags to Pt*
88
-------
Ap = pressure drop, in. HzO
0 = dimensionless parameter
0 = volumetric flow rate of
contaminated gas, acfrn
L = number of broken bap
D = bag diameter, inches
T = temperature, °F
For a detailed development of the
above equation, refer to "Effect of Bag
Failure on Baghouse Outlet^ Loading,"
Theodore and Reynolds, JAPCA,
August 1979, 870-2.
2. Calculate the dimensionless parameter, .
See above equation.
3. Calculate the penetration correction
factor, Ptc.
See above equation.
4, Calculate the penetration and efficiency
after the two bag failures.
Use the results of steps (1) and (3) to
calculate Pt*.
5, Calculate the new outlet loading after
the bag failures.
Relate inlet loading and new outlet
loading to the revised efficiency or
penetration.
6. What other system and/or operating
variables will affect the new outlet
loading? Discuss qualitatively.
Note that,
Pt* = Pt + Pt.
= Pt + 0.582v3p (LD2VT + 460)/Q,
89
-------
Combustion—Air Requirements
Situation
You are requested to calculate the amount of air required to completely combust a given
volumetric flow rate of natural gas and also determine the volumetric flow rate of com-
bustion products.
Technical Vocabulary
Combustion
Combustion products
Flue gas
Data
Composition of natural gas
Component
Mole fraction
N,
0.05
CH,
0.81
c,h6
0.10
C,H,
0.04
Volumetric flow rate of natural gas= 1.0 scfm
Solution
1. Write the balanced chemical equa-
tions for complete combustion.
Combustion with 100% theoretical (stoichio-
metric air insures that no oxygen will
remain in the flue gas with complete com-
bustion.
See basic operation 8.
Remember that the combustion pro-
ducts are C02, H20, and inert N2 from
both the air and natural gas.
2. Determine the scfm of 02 required from
the above balanced chemical equations.
A chemically balanced equation tells
us, in terms of moles (or volume), the
ratios among reactants and products.
3. Calculate scfm of air required.
The composition of air is 21% 02 and
79% N? by volume.
90
-------
4. Calculate total scfm of N2 in the flue gas.
Remember that the flue nitrogen is the
sum of the Ns in the natural gas and
the N2 in air accompanying the
required 02 for combustion.
5. Calculate scfm of CO* and H20 in the
flue gas. ^ ^
6. Calculate the total scfm of flue gas.
The total products of combustion (flue
gas) include C02) H20, and N2.
91
-------
Combustion—Thermal Afterburner Design
Situation
As an air pollution control engineer you have been requested to evaluate the gross
heating value of a natural gas of a given composition. You are also to determine the
available heat of the natural gas at a given temperature, the rate of auxiliary fuel
(natural gas) required to heat a known amount of contaminated air to a given
temperature, the dimensions of an afterburner treating the contaminated air stream, and
the residence time.
Technical Vocabulary
Available heat
Gross heating value
Latent heat of vaporization
Operating and Design Data
Natural gas composition (mole or volume fraction):
N,
0.0515
ch4
0.8111
CtH,
0.0967
CjH,
0.0S51
QH10
0.0056
1.0000
Linear gas velocity = 20 ft/sec
Length-to-diameter ratio of the afterburner = 2.0
Temperature of dry natural gas = 60°F
Volumetric flow rate of contaminated air,Q_= 5000 scfm (60 °F, 1 atm)
It is required to heat the contaminated air from 200 °F to 1200°F.
Table 1 (see basic operation 10)
Table 3
92
-------
Table 3. Enthalpies of combustion gases (Btu/lb mole).
°F
N,
Air
M.W. 28.97
CO,
HfO
32
0
0
0
0
60
194.9
194.6
243.1
224.2
77
312.2
312.7
392.2
360.5
100
473.3
472.7
597.9
545.3
200
1,170
1,170
1,527
1,353
300
1,868
1,870
2,509
2,171
400
2,570
2,576
3,537
3,001
500
3,277
3,289
4,607
3,842
600
3,991
4,010
5,714
4,700
700*
4,713
4,740
6,855
5,572
800
5,443
5,479
8,026
6,460
900
6,182
6,227
9,224
7,364
1,000
6,929
6,984
10,447
8,284
1,200
8,452
8,524
12,960
10,176
1,500
10,799
10,895
16,860
13,140
2,000
14,840
14,970
23,630
18,380
2,500
19,020
19,170
30,620
23,950
3,000
23,280
23,460
37,750
29,780
Source: Kobe, Kenneth, A., and Long, Ernest G.,
"Thermochemistry for the Petroleum Industry,"
Petroleum Refiner, Vol. 28, No. 11, November, 1949,
page 129, Table 9.
Solution
1. Determine the gross heating value of the
natural gas.
See basic operation 10.
2. What is the available heat of natural
gas at 1200°F?
a. Write the balanced chemical com-
bustion equations for each of the
four components of the natural gas.
Use one scf of natural gas as a basis.
The available heat (HAr) at any
temperature is the gross heating value
(HVC) minus the amount of heat
required to take products of combustion
to that temperature (EAH).
See basic operation 8.
For butane, we have
C4HI0 + (13/2)O,-4 CO, + 5 H20
For every scf of natural gas there is
0.0056 ft8 butane. Using one scf of
natural gas as a basis, the above equa-
tion becomes
0.0056 C4HIO + (0.0056)(13/2)0,-
(0.0056)(4) CO, + (0.0056X5) H,0
This simplifies to
0.0056 C«H10 +0.0346 O,-
0.0224 CO,+ 0.028 H^O
93
-------
b. Determine the number of standard
cubic feet for each of the following
components of combustion.
02; scf/scf of natural gas
C02; scf/scf of natural gas
H20; scf/scf of natural gas
N2; scf/scf of natural gas
c. How many cubic feet of products are
there per scf of natural gas burned
assuming that a stoichiometric
(theoretical) quantity of air (21% 02,
79% N2) is used?
d. Determine the amount of heat
required to take the products of
combustion from 60 °F to 1200 °F
(LAH).
(1) What are the changes in enthalpy
for each combustion product
going from 60 °F to 1200°F?
AHcq2 = Btu/scf natural gas
AH,v2 = „. Btu/scf natural gas
AHHjo(i) = -Btu/scf natural gas
(2) Calculate LAH.
e. Calculate the available heat of
natural gas at 1200 °F in Btu/scf
of natural gas.
Calculate the rate of auxiliary fuel
required to heat 5000 scfm of con-
taminated air from 200 °F to 1200°F.
a. Determine the enthalpy change of air
going from 200 °F to 1200 °F.
b. Calculate the heat rate, q, required to
heat 5000 scfm of air from 200 °F
to 1200°F in Btu/min.
c. Calculate the rate of natural gas
required to heat 5000 scfm of air from
200°F to 1200°F in scfm.
Remember that the chemical equation
tells us, in terms of moles (or volume),
the ratios among reactants and
products. Assuming the natural gas is
burned with air (21% 02 and 79%
N2), the total amount of nitrogen in
the products of combustion is the sum
of N2 in the natural gas (0.0515 scf)
and that amount accompanied by the
required 02, (79/21 Xscf of 02).
Total cubic feet of products per scf of
natural gas are the sum of scf C02/scf
natural gas, scf H20/scf natural gas,
and scf N2/scf natural gas.
The enthalpies of combustion gases are
given in Table 3 and the latent heat
of vaporization of water (AHX) at
60 °F is 1060 Btu/lb.
See Table 3 for enthalpies of combus-
tion gases. The changes in enthalpy in
Btu/scf natural gas can be calculated
by multiplying the change in enthalpy
in Btu/lb mole by scf of gas/scf of
natural gas and divided by 379 scf/lb
mole. See basic operation 6.
LAH = AHco2 + AH„2 + AHX + AHWjO(<)
Since the water present in the combus-
tion product is vapor, the latent heat of
vaporization should be included.
HA=HVg-LAH
To calculate the rate of natural gas
required, you need both the enthalpy
change of air and the available heat.
See Table 3.
See basic operations 6 and 9.
Natural gas rate = q/HA
94
-------
4. What are the dimensions of an after-
burner treating 5000 scfm of contamin-
ated air stream?
a. What is the total flue gas flow rate?
(1) Calculate the volumetric flow rate of
products of combustion in scfm.
b. Determine the diameter of the after-
burner in ft.
c. Determine the length of the afterburner
in ft.
d. Determine the residence time for the
gases in the afterburner in seconds.
To determine the dimensions of the
afterburner, you need the total flue gas
flow rate at 1200 °F.
The total flue gas flow rate is the sum
of the contaminated gas stream and the
flow rate of the combustion products.
The volumetric flow rate of the prod-
ucts of combustion is obtained by
multiplying the total cubic feet of prod-
ucts per scf of natural gas (previously
determined) by the natural gas rate
(also previously determined).
products of com-
process gas
S=*D«/4 = Q/v
Where: S = cross-sectional area of the
afterburner
D = diameter of the afterburner
Q_— total flue gas flow rate
v = linear gas velocity
Residence time = length/v
Typical residence times in an after-
burner are in the 0.1-0.5 second range.
(2) Calculate the total flue gas flow rate This includes both the
in scfm. bustion (flue) and the
stream.
(3) Determine the total flue eras flow rate See basic operation 5.
at 1200°F in acfm.
95
-------
Combustion—Plan Review
of a Direct Flame Afterburner
Situation
You must review plans for a permit to construct a direct flame afterburner serving a
lithographer. Review is for the purpose of judging whether the proposed system, when
operating as it is designed to operate, will meet emission standards. The permit appli-
cation provides operating and design data. Agency experience has established design
criteria which, if met in an operating system, typically ensure compliance with standards.
Technical Vocabulary
Available heat
Gross heating value
Heat of combustion
Operating and Design Data from
Permit Application
Application = lithography
Effluent exhaust volumetric flow rate = 7000 scfm
Exhaust temperature = 300 °F
Hydrocarbons in effluent air to afterburner (assume hydrocarbons to be toluene)
= 30 lbs/hr
Afterburner entry temperature of effluent = 738 °F
Afterburner heat loss= 10%
Afterburner dimensions = 4.2 ft in diameter, 14 ft in length
500 °F
To process
Effluent
from lithographei
Atmosphere
7000 scfm
300 °F
450 °F
Natural
gas
burner
Process
air
preheater
Direct
flame
afterburner
Afterburner
preheater
^ 1 Fan
1400°F
738 °F
Makeup air
Graphical representation
96
-------
Agency Design Criteria
Afterburner temperature = 1300-1500°F
Residence time = 0.3-0.5 seconds
Afterburner velocity = 20-40 ft/sec
Standard Data
Gross heating value of natural gas = 1059 Btu/scf of natural gas
Combustion products per cubic ft of natural gas burned = 11,5 scf
Available heat of natural gas at 1400 °F = 600 Btu/scf of natural gas
Molecular weight of toluene = 92
Specific heat of effluent gases at 738°F (above 0°F) = 7.12 Btu/lb mole- °F
Specific heat of effluent gases at 1400°F (above 0°F) = 7.38 Btu/lb mole-°F
Volume of air required to combust natural gas = 10.33 scf air/ scf natural gas
Solution
1. Does this application meet any of the
agency design criteria?
The permit application design tempera-
ture is already within agency criteria.
Therefore, you need be concerned only
with whether, under the conditions
given, the residence time and the after-
burner velocity will be within agency
design criteria.
2. Determine the fuel requirement for the
afterburner.
a. Calculate the total heat load (heating rate)
required to raise 7000 scfm of effluent
stream from 738°F to 1400°F,q, in
Btu min.
b. Calculate the actual heat load
required, accounting for a 10% heat
loss, in Btu/min.
c. Calculate the rate of natural gas
required to supply the actual heat
required to heat 7000 scfm of effluent
from 738 °F to 1400°F in scfm. ;7 / . 9
See basic operation 9.
Actual heat load = (1.1) q
/ Rate of \_/actual\ //available\
\ natural gas/ \ heat )/ v heat /
3. What is the total volumetric flow rate
through the afterburner, Qj-?
Once you determine the volumetric flow
rate, you can calculate the afterburner
velocity and residence time since the
dimensions of the afterburner are given.
The total volumetric flow rate is the
sum of the process effluent from the
lithographer plus the combustion
products of the natural gas.
97
-------
a. Calculate the volumetric flow rate of
the combustion products of the
natural gas, Qj, in scfm.
b. What is the volumetric flow rate of
the effluent stream from the litho-
grapher in scfm?
c. Calculate the volumetric flow rate
of air required to combust the natural
gas required, Q?, in scfm.
d. Calculate the total volumetric flow
rate through the afterburner, Q^,
in scfm.
e. Calculate Qr in acfm.
4. Does the afterburner velocity meet the
criteria?
a. Calculate the cross-sectional area of
the afterburner, S, in ft2.
b. Calculate the afterburner velocity, v,
in ft/sec.
c. Is the afterburner velocity within
agency criteria?
5. Does the residence time meet agency
criteria?
a. Calculate the residence time, t, in sec.
To calculate the volumetric flow rate
of combustion products of the natural
gas, you use the amount of natural gas
just calculated.
/11.5 scf of / N
/ rate of \(combustion/ scf of
Vnatural gas/^ products/natural gas
rate of VI0.33 scf/ scf of \
natural gas A of air/natural gas/
Qr = 7000 + Qj
Since primary air is employed in the
combustion of the natural gas, Q* need
not be subtracted from Qr. Primary air
is air external to the process effluent.
See basic operation 6.
To calculate the afterburner velocity
you need the cross-sectional area of the
afterburner and actual volumetric flow
rate through the afterburner.
S = 7rD2/4
v= Qj/S
- t = L/v
Where: L = afterburner length
98
-------
b. Is the residence time within agency
criteria?
Note that the above calculations have
assumed combustion of the natural gas
to take place with primary air, i.e., air
supplied external to the process efflu-
ent. In addition, the energy released
and volume of gases formed by the
combustion of the pollutant (toluene)
have been neglected. This is common
practice in combustion calculations. If
secondary air (from process effluent) is
employed for combustion of the natural
gas, a trial-and-error calculation is
required for a rigorous solution to this
problem. However, a simplified pro-
cedure for this calculation can be found
in AP-40, EPA's Air Pollution
Engineering Manual.
99
-------
Absorption—Spray Tower
Situation
A steel pickling operation emits 300 ppm HCI (hydrochloric acid) with peak values of
500 ppm 15% of the time. The air flow is a constant 25,000 acfm at 75 °F and 1 atm.
Only sketchy information was submitted with the scrubber permit application for a spray
tower. You are requested to determine if the spray unit is satisfactory.
Technical Vocabulary
Henry's law
Data
Emission limit = 25 ppm HCI
Maximum gas velocity allowed through the tower = 3 ft/sec
Number of sprays = 6
Diameter of the tower = 14 ft
The plans show a counter current water spray tower. For a very soluble gas (Henry's law
constant approximately zero), the number of transfer units (N0g) can be determined bv
the following equation;
Where: yx = concentration of inlet gas
y2 = concentration of outlet gas
In a spray tower, the number of transfer units (Noc) for the first (or top) spray will be
about 0.7. Each lower spray will have only about 60% of the Nog of the spray above it.
The final spray, if placed in the inlet duct, has a Nog of 0.5.
The spray sections of a tower are normally spaced at three foot intervals. The inlet duct
spray adds no height to the column.
Noc
Spray tower
NOG = ln(yl/y2)
Solution
!• Calculate the gas velocity through the
tower.
See basic operation 12.
V = Q/S
= Q/(tDV4)
2. Does the gas velocity meet the
requirement?
The maximum velocity is set at
3 ft/sec.
100
-------
3. Calculate the number of transfer units
required to meet the regulation.
Remember that,
Nog = ln(yi/y2)
Use the peak value for inlet gas
concentration.
4. Determine the total number of transfer
units provided by a tower with six spray
sections by completing the following
table.
Spray section
N oc
Top
0.7
2nd
3rd
4th
5th
Inlet
Total
Remember that each lower spray has
only 60% of the efficiency of the section
above it. This is due to back-mixing of
liquids and gases from adjacent
sections.
5. Calculate the outlet concentration of
gas.
Use the Nog calculated in 4. to obtain
Y*-
6. Does the spray tower meet the HC1
regulation?
Compare the Nog values in 3. and 4.,
or. compare the vs in 4. with the emis-
sion regulation.
101
-------
Absorption—Packed Tower Review
Situation
Pollution Unlimited, Inc. has submitted plans for a packed ammonia scrubber on an air
stream containing NHS. The operating and design data is given by Pollution Unlimited,
Inc. You remember approving plans for a nearly identical scrubber for Pollution
Unlimited, Inc. in 1978. After consulting your old Files you find all the conditions were
identical except for the gas flow rate. What is your recommendation?
Technical Vocabulary
Equilibrium
Henry's law
Hoc
Operating and Design Data Given
by Pollution Unlimited, Inc.
Tower diameter = 3.57 ft
Packed height of column = 8 ft
Gas and liquid temperature = 75 °F
Operating pressure = 1.0 atm
Ammonia free liquid flow rate (inlet) = 1000 lb/ft2-hr
Gas flow rate = 1575 acfm
Gas flow rate in 1978 plan = 1121 acfm
Inlet NHS gas composition = 2.0 mole %
Outlet NH, gas composition = 0.1 mole %
Air density = 0.0743 lb/ft3
Molecular weight of air = 29
Henry's law constant, m = 0.972
Molecular weight of water = 18
Figure 5 (packing "A" is used) and Figure 6
Emission regulation = 0.1% NH3
102
-------
4.0 |
~~0-Packing A ~~
Packing B
3.2 -—
G = gas flow lb/hr-ft*
2,8 --
2.4 --
2.0 —-
1.6 --
1.2 —-
0.8 - -
0.4 [- -
500 1000 1500 2000
Liquid rate, lb/hr-ft*
Hoc vs. liquid rate for ammonia-water absorption system.
O-Packing A
Packing B
G = gas flow lb/hr-ft1
0
Figure 5.
103
-------
ill! Ill
' I ¦ ¦ » il
1 1 II1 i
10 20 50 100 200
(y,-mx,)/(y,-nixt)
Figure 6. Colburn chart.
500 1000
Solution
1. What is the number of overall gas
transfer units, N0c?
The number of overall gas transfer
units, Nog, is used in calculating pack-
ing height requirements. It is a function
of the extent of the desired separation
and the magnitude of the driving force
through the column (the displacement
of the operating line from the
equilibrium line).
104
-------
a. Calculate the gas molar flow rate,
Gm, and liquid molar flow rate,
Lm, in lb mole/ft2-hr.
(1) Calculate the cross-sectional area
of the tower, S, in ft2.
(2) Calculate the gas molar flow
rate, Gm, in lb mole/ft2-hr.
(3) Calculate the liquid molar flow
rate, Lm, in lb mole/ft2-hr.
b. Calculate the value of mGm/Lm.
c. Calculate the value of
(yi — mx2)/(y2 — mx2).
d. Determine the value of N0c from
Colburn's chart (Figure 6). ^,4
2. What is the height of an overall gas
transfer unit, Hoc?
a. Calculate the gas mass velocity, G,
in lb/ft2-hr.
b. Determine the value of Hog from
Figure 5.
3. What is the required packed column
height, Z, in ft?
4. Compare the packed column height of
8 feet specified by Pollution Unlimited,
Inc. to the height calculated above.
What is your recommendation?
The values of Gm and Lm are needed to
use Colburn's chart (Figure 6) which
graphically predicts the value of Nog-
S = irD2/4
Gm= Q^/SM
Where: Q= volumetric flow rate of
gas stream
q = density of air
M = molecular weight of air
U = L/Mt
Where: L = liquid mass velocity,
lb/ft2-hr
= liquid molecular weight
The reciprocal of this term is defined as
the absorption factor, A.
(y! — mxj)/(y2 — mx2) is the abscissa of
Colburn's chart (Figure 6).
Where: yi = inlet gas mole fraction
y2 = outlet gas mole fraction
Xi = outlet liquid mole fraction
x2 = inlet liquid mole fraction
m = Henry's law constant
The height of an overall gas transfer
unit, Hoc, is also used in calculating
packing height requirements. Hog
values in air pollution are almost always
based on experience. Hog is a strong
function of the solvent viscosity and dif-
ficulty of separation, increasing with
increasing values of both.
G=eQ/S
Z = (Nog)(Hoc)
105
-------
Absorption—Tower Height and Diameter
Situation
A packed column is designed to absorb ammonia from a gas stream. Given the operating
conditions and type of packing, calculate the height of packing and column diameter.
Technical Vocabulary
Flooding
Mass velocity
Packing factor
Operating Data and Assumptions
Gas mass flow rate = 5000 lb/hr
NH3 concentration in inlet gas stream = 2.0 mole %
Scrubbing liquid = pure water
Packing type = 1-inch Raschig rings; packing factor, F = 160
Hoc of the column = 2.5 ft
Henry's law constant, m = 1.20
Density of gas (air) = 0.075 lb/ft3
Density of water =62.4 lb/ft3
Viscosity of water = 1.8 cp
Figure 7
The unit operates at 60% of the flooding gas mass velocity, the actual liquid flow rate is
25% more than the minimum, and 90% of ammonia is to be collected based on State
regulations.
106
-------
Flooding line
^i£5n.5)
W
QJ
5 0.05
° °0a08 (0 i n\
4: 0.02
tL.
0.005
inches
meters
— ' «» * ini-nvj «*IV
per meter per ft
of packing of packing)
1 I I I
0.002
101
0.01 0.02 0.05 0.1 0.2 0.5 1
(L/G)(c / Gi)o s
Figure 7. Generalized flooding and pressure drop correlation.
Pure water
Z = packed column height
D = column diameter
y( = inlet gas composition
)'z = outlet gas composition
x, = outlet liquid composition
x, = inlet liquid composition
Contaminated gas stream,
Xi
y, = 0.02
Graphical representation of the packed column
107
-------
Solution
1. What is the number of overall gas trans-
fer units, Nqc?
a. What is the equilibrium outlet liquid
composition, xu and the outlet gas
composition, y2, for 90% removal?
(1) Calculate the equilibrium outlet
concentration, x* at y] = 0.02.
(2) Calculate y2 for 90% removal.
b. Determine the minimum ratio of
molar liquid flow rate to molar gas
flow rate,, (Lm/Gm)m,„, by a material
balance.;
c. Calculate the actual ratio of molar
liquid flow rate to molar gas flow
rate, (Lm/Gm).
d. Calculate the value of
(y, - mx2)/(y2 - mxj).
e. Calculate the value of (mG„)/L„.
f. Determine Nog from Colburn's chart
using the values calculated in (d)
and (e).
2. Calculate the height of packing, Z.
Remember that the height of packing,
Z, is given by:
Z = (Hog)(N0g)
Since Hog is given, you only need Noc
to calculate Z. Hoc is a function of both
the liquid and gas flow rates; however,
it is usually available for most air pollution
applications.
Remember that you need the inlet and
outlet concentrations (mole fractions) of
both streams to use Colburn's chart.
According to Henry's law, xf=yi/m.
The equilibrium outlet liquid composi-
tion is needed to calculate the min-
imum Lm/Gm. See basic operation 11.
Since it is required to remove 90% of
NHS, there will be 10% of NHS
remaining in the outlet gas stream:
y2 = (0,1 Vi)/[(1 -y,)-i-(0.1 yi)3
Material balance around the packed
column:
G„(yi -yj) = Lm(xf- x2)
(Lm/Gm)mi„ = (yi - ys)/(xf- x2)
Remember that the actual liquid flow
rate is 25% more than the minimum
based on the given operating
conditions.
(Lm/'Gm) = 1.25 (Lm/Gm)min
Even though the individual values of G„
and L„ are not known, the ratio of the
two has been previously calculated.
108
-------
3. What is the diameter of the packed
column?
a. Calculate the flooding gas mass
velocity, Gf.
(1) Calculate the abscissa of Figure 7,
(L/G)(q/ol)0-5.
(2) Determine the value of the
ordinate at the flooding line using
the calculated value of the
abscissa.
(3) Solve the abscissa for the flooding"/
gas mass velocity, G/, in lb/ft2-sec.
b. Calculate the actual gas mass velocity,
GOCI, in lb/ft2*sec.
c. Calculate the diameter of the column
in ft.
The actual gas mass velocity must be
determined. To calculate the diameter
of the column, you need the flooding
gas mass velocity. Figure 7 is used to
determine the flooding gas mass veloc-
ity. The mass velocity is obtained by
dividing the mass flow rate by cross-
sectional area.
(L/G)(g/0t)o-5 =
Where: 18/29
(U,/G,)(18/29)(e/ei;
0.5
ratio of molecular
weight of water to air
L = liquid mass velocity,
lb/sec-ft2
G = gas mass velocity,
lb/sec-ft2
Note that the L and G terms in Figure
7 are based on mass and not moles.
Ordinate:
G2F^t0-2
Where:
Where:
Qi-Qgc •
F = packing factor
= 160 for 1-inch
Raschig rings
-¦P = ratio, density of water/
density of liquid
gc = 32.2 lb-ft/lb/-sec2
}Il = viscosity of liquid, cp
G= gas mass velocity,
lb/ft2-sec
Gacr — 0.6 Gf
S = m/ Gacr
m = mass flow rate of gas
stream = 5000 Ib/hr
S = 3-DV4
109
-------
Adsorption—Working Capacity
Situation
You are to calculate the working capacity of an adsorption bed given the saturation
(equilibrium) capacity, mass transfer zone, and heel.
Technical Vocabulary
Adsorption
Breakthrough capacity
Heel
MTZ
Saturation capacity
Working capacity
Data
Depth of adsorption bed = 3 ft
Saturation capacity =39%
MTZ = 4 in.
Heel = 2.5%
Solution
1. Calculate the breakthrough capacity.
The breakthrough capacity is the
capacity when traces of vapor first
begin to appear in the exit gas stream
from an adsorption bed.
„ 0.5 C,(MTZ) + C,(D - MTZ)
Cfl
D
Where: Cs = breakthrough capacity
C, = saturation capacity
MTZ = mass transfer zone
length
D = adsorption bed depth
2. Calculate the working capacity.
The working capacity is lower than the
saturation capacity or the breakthrough
capacity. It results from unrecoverable
solvent in the carbon and occasionally
from a lower packing factor than
laboratory specifications.
110
-------
Working /breakthrough} _ () /packing'
capacity \ capacity / \ factor ,
Neglect the packing factor for this
calculation.
Ill
-------
Adsorption—Degreaser Ventilation Clean Up
Situation
You are asked to determine the required height of adsorbent for an adsorber which treats
a degreaser ventilation stream contaminated with trichloroethylene (TCE) given the
following operating and design data.
Technical Vocabulary
Activated carbon
Adsorbent
Adsorber
Adsorption
Bulk density
Working capacity
Operating and Design Data
Volumetric flow rate of contaminated air stream = 10,000 scfm
Standard conditions = 60 °F, 1 atm
Operating temperature = 70 °F
Operating pressure = 20 psia
Adsorbent = activated carbon
Bulk density of activated carbon = 36 lb/ft3
Working capacity of activated carbon = 28 lb TCE/100 lb carbon
Inlet concentration of TCE = 2000 ppm (by volume)
Molecular weight of TCE= 131.5
The adsorption column cycle is set at 4 hrs in the adsorption mode, 2 hrs in heating and
desorbing, 1 hr in cooling, and 1 hr in stand-by.
The adsorber recovers 99.5% by weight of TCE.
A horizontal unit with an inside diameter of 6 ft and length of 15 ft is used.
Solution
1. What is the volume of activated carbon
required to treat the contaminated gas
stream?
Since the dimensions of the adsorber
are known, you only need the volume
of activated carbon to calculate the
height of adsorbent (activated carbon).
112
-------
a. What is the mass of TCE to be
adsorbed during a 4 hr period?
(1) Calculate the actual volumetric See basic operation 5.
flow rate of contaminated gas
stream in acfh.
(2) Calculate the volumetric flow rate Qrcs = (yrczXQp)
of TCE in acfh.
(3) Calculate the mass flow rate of See basic operation 6.
TCE, m. in lb/hr. m = Qg = Q(PM/RT)
(4) Calculate the mass of TCE to be
adsorbed during the 4 hr period.
b. What is the volume of activated
carbon required in ft3?
Activated
carbon
required
TCE \ /( bulk Vads. cap.
adsorbed// \densitvAof carbon.
TCE adsorbed = (ih)(0.995)(4)
2. What is the height of adsorbent in ft?
Height >
of
adsorbent
actual
carbon
volume
carbon | /(6)(15)
113
-------
Adsorption—Plan Review
Situation
The permit to construct a solvent recovery plant which recovers acetone from a con-
taminated air stream with three adsorbers has been applied for and you are asked to
review the plans for the adsorption system. In reviewing these plans you should check the
time to run each adsorber to breakthrough and the amount of steam required to
regenerate the adsorbent.
Technical Vocabulary
Breakthrough capacity
MTZ (mass transfer zone)
Partial pressure
Regeneration
Saturation capacity
Vapor pressure
Information Given in the Permit Application
Volumetric flow rate of contaminated air stream = 30,000 acfm at 1 atm
Temperature of contaminated air stream = 20 °C
Molecular weight of acetone = 58
Composition of acetone in air stream = 0.15 mole %
Carbon type; charge per adsorber = B, 4x 6 mesh; 15,000 lbs
Heel = 2 wt. %
Packing factor (capacity correction due to different packing) = 3 wt. %
Adsorber dimensions = 10 ft by 24 ft
Adsorption bed depth = 1.96 ft
Vapor pressure of acetone at 20°C= 170 mm Hg
MTZ length = 2 inches
Figure 8 (% relative saturation)
Figure 9 (steam requirement)
Two of the three adsorbers are in the adsorption phase at all times. The plant operates
24 hours a day and 365 days a year. Plant steam is available at 5 psig.
114
-------
Dry
Solvent
+ air
, 1
| Adsorber
^1
V//////////,
1
1
I *(
1
|!
r
i
Adsorber
1 (
<8h
I
Acetone
i
Condenser
-------
ao vc
l.U -\n3BdE3 UOUK-iniE^
116
-------
2 ~
Carbon B
Carbon C
Carbon A
Working charge, %
Figure 9. lbs of steam/lbs of solvent vs. working charge.
Solution
1. What is the adsorption time per adsorber
bed?
a. What is the working capacity of the
activated carbon (B)?
The working capacity of the carbon is
lower than the saturation capacity or
the breakthrough capacity and results
from unrecoverable solvent in the car-
bon and from a lower packing factor
than laboratory specifications. Based on
previous experience, the unrecoverable
solvent, or heel, is 2 wt. % and the
lower capacity due to the lower packing
factor is estimated to be
3 wt. %.
117
-------
(1) Calculate the relative saturation of
acetone in the air stream.
(2) Determine the saturation capacity
of carbon B, C„ for acetone from
Figure 8 using the % relative
saturation calculated above.
(3) Calculate the breakthrough
capacity in wt. %.
(4) Calculate the working capacity
of the carbon B in wt. %.
b. Calculate the mass flow rate of
acetone, m.
(1) Calculate the volumetric flow
rate of acetone. O. in acfm.
(2) Calculate the mass flow rate, m,
in lb/min.
c. Calculate the adsorption time in
minutes for each adsorption bed
(time for breakthrough).
d. Calculate the regeneration time in
hours to complete the total cycle.
/partial press, of\
% relative \ acetone in air J
saturation / vapor press, of \
\acetone at 20°cJ
The partial pressure of acetone in air is
equal to the product of acetone mole
fraction and atmospheric pressure.
Break-
through = -
capacity
Where: C,
0.5 C,(MTZ) + C,(D — MTZ)
D
saturation capacity
MTZ = mass transfer zone
length
D = adsorber bed depth
The breakthrough capacity is the
capacity when the traces of vapor begin
to appear in the exit gas stream.
Working / breakthrough\ /packing^
capacity - \ capacity / \ factor j
n /mole fractionyvolumetric flow\
\ of acetone A rate of air /
See basic operation 6.
Adsorption
time
carbon ^/working
charge/\ charge
0.5 m
Note that two adsorbers are on stream
with each cleaning one half of the total
gas stream.
The regeneration time should be less
than one half of the time for
breakthrough.
118
-------
2. What is the amount of steam required?
a. Determine the steam requirement for
carbon B in lb steam/lb solvent
removed using the working charge
calculated above and Figure 9.
b. Calculate the lbs of steam required
for each adsorber during
regeneration.
This is commonly referred to as the
steam to solvent ratio.
Mass of / steam \/working\/carbon\
steam \ requirement^ charge Acharge/
119
-------
Fans—Effect of Fan Wheel and Speed
Situation
You are requested to determine the power requirement of fan "B" which is from the
same homologous series as fan "A." The speed (rpm), blade diameter (D), capacity
(acfm), and the power (bhp) of fan "A" are given. Fan "B" delivers a gas having the
same density as fan "A."
Technical Vocabulary
Brake horsepower (bhp)
Fan laws
rpm
Data
Fan "A"
Speed = 1622 rpm
Blade diameter = 46 in.
Gas delivered = 15,120 acfm
Power (brake) = 45.9 bhp
Fan "B"
Speed = 1590 rpm
Blade diameter = 50 in.
Solution
1. Calculate the power requirement of
There are many fan laws. One that
relates horsepower with fan wheel
(diameter) and speed (rpm) is given
below.
fan "B."
m/ m=(dB/dAnNB/NA)3
Where: SPA, = horsepower of fan
"A" and "B"
d*,dB = blade diameter of fan
"A" and "B"
NU,NS = speed (rpm) of fan
"A" and "B"
Remember that the brake horsepower
represents the power the user pays for.
120
-------
Fans—Brake Horsepower Requirement
Situation
You are requested to calculate the horsepower required for a fan treating a gas stream
through a packed tower. The pressure drop across the packing, pressure loss for the duct
work, elbows, valves, etc., and overall fan-motor efficiency are given.
Technical Vocabulary
Brake horsepower
Fan efficiency
Operating and Design Data
Volumetric flow rate of the gas stream = 6000 acfm
Pressure drop across the packing=6.0 in. H20
Pressure loss for the duct work, elbows, valves, etc., and expansion/contraction
= 4.0 in. H20
Overall fan-motor efficiency = 63%
Solution
1. Calculate the total pressure drop in
inches of H20.
The total pressure drop is the sum of
the pressure drop across the packing
and pressure loss due to the duct work,
elbows, valves, expansion/contraction,
etc.
2. Calculate the brake horsepower (bhp)
required to treat the gas stream in bhp.
The brake horsepower represents the
power the user pays for, i.e., it is the
power to operate the fan. The term gas
or air horsepower is used to describe the
power delivered to the gas.
Rrs Icp
horsepower = (0X^X1.575x 10-)
(bhp) ns
Where: Q= volumetric flow rate of
gas stream, acfm
Ap = pressure drop, in. H20
—fan efficiency, usually
in the 0.5-0.65 range
121
-------
Economics—Breakeven Operation
Situation
A plant emits 50,000 acfm of gas containing a dust at 2.0 grains/ft3, A particulate con-
trol device is employed for particle capture and the dust captured from the unit is worth
$0.01/lb of dust. You are requested to determine at what collection efficiency is the cost
of power equal to the value of the recovered material. Also determine the pressure drop
in inches of HaO at this condition.
Technical Vocabulary
Fan efficiency
Data
Overall fan efficiency = 55%
Electric power cost = 50.06/kw-hr
For this control device, assume that the collection efficiency is related to the system
pressure drop, Ap, through the following equation.
tj = Ap/(Ap + 5.0)
Where: dp = pressure drop, lb//ft2
t] = fractional collection efficiency.
Solution
1. Express the value of the dust collected
in terms of collection efficiency, rj.
Amount of dust collected = (Q)(inlet
loading)(ij)
Remember there are 7000 grains per
pound.
2. Express the value of the dust collected
in terms of pressure drop, Ap.
Remember that
7j = Ap/(Ap + 5.0)
3. Express the cost of power in terms of
pressure drop, Ap.
bhp = Q_x Ap/ r)(
Where: = fan efficiency
4. Set the cost of power equal to the value
of dust collected and solve for dp in
lb//ft2. Convert this pressure drop to
in. H20.
This represents breakeven operation.
Remember, there are 44,200 ft-lb, per
one kw-min.
To convert from lb//ft2 to in. H20,
divide by 5.2.
5. Calculate the collection efficiency using
the value of Ap calculated above.
Use the equation for rj in (2).
122
-------
Economics—Annualized Installed, Operation,
and Maintenance Costs
Situation
*
You are requested to determine capital, operating, and maintenance costs on an annual-
ized basis for a textile dye and finishing plant (with two coal fired stoker boilers) where a
baghouse is employed for particulate control. Operating, design, and economics factors
are given.
Technical Vocabulary
AICC (annualized installed capital cost)
Operating, Design, and Economic Factors
Exhaust volumetric flow from two boilers = 70,000 acfcn
Overall fan efficiency = 60%
Operating time = 6240 fir/vear
Surface area of each bag= 12.0 ft2
Bag type = Teflon® felt
Air-to-cloth ratio = 5.81 acfm/ft2
Total pressure drop across the system = 17.16 lb//'ft2
Cost of each bag =$75.00
Installed capital costs = S2.536/acfm
Cost of electrical energy = S0.03/kw-hr
\ early maintenance cost = S3000 plus vearlv cost to replace 25% of the bags
Salvage value = U
Interest rate (i) = 8%
Lifetime of baghouse (m) = 15 yrs
Annual installed capital cost = (installed capital cost)
i(l + i)"
(1 +i)"-l
Solution
1. What is the annual maintenance cost?
a. Calculate the number of bags, N.
b. Calculate the annual maintenance
cost in dollars/year.
N = Q/(air-to-cIoth ratio)(A)
Where: Q_= total exhaust volumetric
flow rate
A = surface area of a bag
Remember that the annual main-
tenance cost is 55000/year plus replac-
ing 25% of the bags each year.
123
-------
2. What is the annualized installed cost
(AICC)?
a. Calculate the installed capital cost
in dollars. ) 17
b. Calculate the AICC using the given
equation.
-7
o 1^0
Installed capital cost = (Q)($2.536/'acfm)
/lnstalled\
AICC = ( capital I
\ cost /
i(l + i)"1
_(1 + i)m- 1_
3. Calculate the operating cost in dollars
per year.
8
-------
Solutions
Prepared by: Ui Young Choi
125
-------
Particle Size Distribution—Log-Normal Distribution
1, See sub-steps.
d„ (microns)
% of total
Cum. % GTSS
<0.62
10
90
0.62- 1.0
13
77
1.0 - 1,2
7
70
1.2 - 3.0
40
30
3.0 - 8.0
25
5
8.0 -10.0
2
3
>10.0
3
0
b. The cumulative distribution curve is shown on Figure IS.
2. Since a straight line is obtained on log-normal coordinates, the particle size distribu-
tion is log-normal.
r*
Ihii IiiiiIiiii I'VJuumnli
0,01
98 99 99.8 99.9 99.99
% GTSS
Figure IS. Cumulative distribution curve.
127
-------
Particle Size Distribution—-Andersen 2000 Sampler
1. a. See sub-steps.
(1) Calculate the net weight and percent of total weight for each plate.
For plate #0; Net weight = final weight —tare weight
= 20.48628-20.48484
= 1.44 x 10"3 g
= 1.44 mg
Percent of total weight = (net weight/total net weight)(10G)
= (1.44/10.11)(100)
= 14.2%
The table given below shows the results of the above calculations for each plate.
Plate #
Net weight (mg)
% of total weight
0
1.44
14.2
1
0.56
5.5
2
0.41
4.1
3
0.42
4.2
4
0.39
3.9
5
0.99
9.8
6
1.24
12.3
7
1.20
11.9
Back-up-filter
3.46
34.2
Total
10.11
100.0
(2) Calculate the cumulative percent for each plate.
For plate #1: Cumulative % = cumulative % of plate #0 — % of total weight
of plate #0
= 100-14.2
= 85.8%
The table given below shows the cumulative percent for each plate.
Plate #
Cumulative %
0
100.0
1
85.8
2
80.3
3
76.2
4
72.0
5
68.1
6
58.3
7
46.0
Back-up-filter
34.1
128
-------
b. Read off the 95% aerodynamic diameter at Q= 0.5 cfm for each plate from
Figure 1.
Plate 9
95% aerodynamic
diameter Otm)
0
20.0
1
13.0
2
8.5
3
5.7
4
3.7
5
1.8
6
1-2
7
0.78
c. The cumulative distribution curve is shown on Figure 2S,
2. Since the mean particle diameter is the particle diameter corresponding to a
cumulative percent of 50%, read off the particle diameter at a cumulative percent of
50% from Figure 2S.
Mean particle diameter = Ys0 = 1.2 microns
3. See sub-steps.
a. The distribution approaches log-normal behavior. Read off the particle diameter at
84.13 cumulative % from Figure 2S.
Yg4.i3 = 17.0 microns
b. crc = Y84.i3/Y5()
= 17.0/1.2
= 14.2
129
-------
60
50
40
30
20
10
9
8
7
6
5
4
3
2
1
MHIH| I I I I I llll |1 111 | I I [I I M |IHIIIIII|II I I [111 I |IIII|I1M [III l|ll I I [lllllll 11)11 I I | I I [III m» nil [I [III III I I
tLi_L
11 n ii It hi I i i l i i ) i Inn
llllllll 11 till I llll
llll i Ilium n lit i i I I I till I linn
I I 11
Inn
0,1 0.2 1 2 5 10 20 30 40 50 60 70 80 90 95
% by weight less than stated size (% LTSS)
Figure 2S. Cumulative distribution curve.
98 99 99.899,9 99.
-------
Fluid-Particle Dynamics—Particle Terminal Velocity
1. See sub-steps.
a. Calculate the particle density (see basic operation 1).
6, = (2.31)(62.4)= 144.14 lb/ft3
b. Calculate the density of air (see basic operation 4).
6 = PM/RT = (l)(29)/(0.7302)(238 + 460) = 0.0569 lb/ft3
Viscosity of air (see Appendix).
#4 = 0.021 cp= 1.41 x 10"5 lb/ft-sec
c. Calculate K.
For dp = 0.4 micron;
K = [(0.4)/(25,400)( 12)][(32.2)( 144.14)(0.0569)7(1.41 x 10"5)2]1'3
= 0.0144
For dp = 40 microns;
K = [(40)/(25.400)(12)][(32.2)(144.14)(0.0569)/(1.41 x 10"5)2]1'3
= 1.44
For dp = 400 microns;
K = 14.4
2. Select appropriate law.
For dp = 0.4 microns; Stokes' law range
For dp = 40 microns; Stokes' law range
For dp = 400 microns; Intermediate law range
3. Calculate terminal velocity.
For dp = 0.4 micron;
v = gd|ep/18 n
= (32.2)[(0.4)/(25,400)(12)]2(144.14)/(18)(1.41 x 10"5)
= 3.15 X 10"s ft/sec
For dp = 40 microns;
v = gd*Qp/\8 n
= (32.2)[(40)/(25,400)(12)]2(144.14)/(18)(1.41 x 10"s)
= 0.315 ft/sec
For dp = 400 microns;
v=o.i53 go.»dpM*e<°-»/y»-»e0-"
= (0.153)(32.2)0-71 [(400)7(25,400)( 12)] 1-14(144.14)°-71/( 1.41 x 10"5)0 43(0.0569)0 29
= 8.76 ft/sec
131
-------
4. Calculate distance.
For dp = 40 microns;
Distance = (30)(Q, 315)
= 9.45 ft
For dp = 400 microns;
Distance = (30)(8.76)
= 262.8 ft
For dp = 0.4 microns;
K = 0.0144, Stokes' range
v= 3.15 x 10"s ft/sec (without Cunningham correction factor)
With the correction factor we have:
A = 1.257 + 0.40 exp[( - 1.10)(0.4 x 10"s)/(2)(6.53) x 10"3)]
= 1.2708
C= 1 +(2)(1.2708)(6.53 x 10"9)/(0.4 x 10"6)
= 1.415
Corrected v = (3.15 x 10"5)(1.415)
= 4.45 x 10"5 ft/sec
Distance = (30)(4.45 x 10"5)
= 1.335 xlO"3 ft
132
-------
Fluid-Particle Dynamics—Stack Application
1. A panicle diameter of 2.5 microns is used to calculate the minimum distance
downstream free of dust since the smallest panicle will travel the greatest horizontal
distance.
2. See.sub-steps.
a. Calculate the panicle density (see basic operation 1).
e, = (1.96)(62.4)
= 122.3 lb/ft3
b. Calculate the density of air (see basic operation 4).
_ pM
Qa'
RT
(1)(29)
(0.73)(60 + 460)
= 0.0764 lb/ft3
Viscosity of air, fi at 60°F = 1.22 x 10"5 lb/ft-sec (see Appendix),
c. Calculate K.
K = dP(gePeV)1/3
2.1
(25,400)(12)
= 0.104
(32.2)(122.3)(0.0764)
(1.22 x 10"5)2
1/3
3. Stokes' law range.
4. Calculate the terminal settling velocity.
v = gd*eP/18 ii
2.5
= (32.2)
(122.3)/(18)(1.22 x 10"5)
_(25,400)(12)_
= 1.21 x 10"3 ft/sec
5. Calcxilate the time for descent.
t = h/v
= 150/1.21 x 10-3
= 1.24 x 105 sec
6. Calculate the horizontal distance travelled.
Distance = (t)(u)
= (1.24 x 105)(3.0/3600)
= 103.3 miles
133
-------
Gravity Settler—Minimum Particle Size
1. Calculate acid mist density (see basic operation 1).
G, = (62.4X1.6) = 99.84 lb/ft3
2. Calculate minimum particle size.
Minimum dp = [(18)(1.243 x l(r5)(50)/(32.2)(99.84)(30)(50)]1/2
= 4.82 x 1Q~S ft
= (4.82 x 10"5 ft)(3.048 x 10s jxn/ft)
= 14.7 microns
134
-------
Gravity Settler—Traveling Grate Stoker
1. a. See sub-steps.
(1) Since the actual terminal settling velocity is assumed to be one half of the
Stokes' law velocity (according to the problem statement),
v = gd|eP/36 n
7? = gePBLd|/36 mQ.
(2) Viscosity of air at 446°F = 1.75 X 10~5 lb/ft-sec (see Appendix).
(3) Gp = (2.65)(62.4)
= 165.4 lb/ft3
(4) In order to calculate the collection efficiency of the system at the operating
conditions, the standard volumetric flow rate of contaminated air of 70.6 scfs is
converted to actual volumetric flow of 130 acfs.
(32 + 460)
= 130 acfs
(5) To convert d., from ft2 to /xm2, dp is divided by (304,800)2.
r] = vtBL/Q.
= ge„BLd2/36MQ.
= (32.2)(165.4)(10.8)(15)dg
(36X1.75 x 10"s)(130)(304,800)2
= 1.14X 10"4d2;dp in pm (microns)
b. Calculate the collection efficiency for each panicle size.
For a particle diameter of 10 microns: rj= 1.14 x 10"4df
= (1.14 x 10"4)(10)2
= l.lx 10"2
= 1.1%
The table below provides the collection efficiency for each particle size.
dp (microns)
n (%)
94
100.0
90
92.0
80
73.0
60
41.0
40
18.2
20
4.6
10
1.1
c. The size efficiency curve for the settling chamber is shown on Figure 3S.
135
-------
2. Read off the collection efficiency of each particle size from Figure 3S.
dp (microns)
Weight fraction (w,)
% (%)
10
0,027
1.1
25
0.069
7,1
35
0.094
14.0
45
0.105
23.0
55
0.105
34.0
65
0.095
48.0
75
0.070
64.0
85
0.095
83.0
+ 94
0.340
100.0
Total
1.000
-
3. Calculate the overall collection efficiency.
T}=Lw,7]i
= (0.027)(1.1) + (0.069){7,1) + (Q.094)(14.0) + (0.105)(23.0) + (0.105)(34.0) +
(0.095X48.0) + (0.070)(64.0) + (0.095)(86.0) + (0.340)(100.0)
= 59.0%
100
60 -
40 -
'20
100
120
Particle size, microns
Figure 3S. Size efficiency curve for settling chamber.
136
-------
Cyclone—Cut Diameter
and Overall Collection Efficiency
1. See sub-steps.
Calculate particle density (see basic operation 1).
a. Qp-Qt-Qp
= (2.9)(62.4)
= 180.96 lb/ft3
b. Calculate cut diameter.
[dp]€u, = [(9)(0.02)(6.72 x 10~4)(2.5)/(2tt)5(50)( 180.96)]0-5
= 3.26 x 10~5 ft
= 9.94 microns
2. Complete size efficiency table.
d,, (microns)
w,
dp/d^c
!?,(%)
1
0.03
0.10
0
0.0
5
0.20
0.5
20
4.0
10
0.15
1.0
50
7.5
20
0.20
2.0
80
16.0
30
0.16
3.0
90
14.4
40
0.10
4.0
93
9.3
50
0.06
5.0
95
5.7
60
0.03
6.0
98
2.94
>60
0.07
-
100
7.0
3. Determine overall collection efficiency.
Ew.-ij,- = 0 + 4 + 7.5 + 16 + 14.4 + 9.3 + 5.7 + 2.94 + 7
= 66.84%
137
-------
Cyclone—Plan Review
1. a. See sub-steps.
(1) Since the cyclone is a conventional one,
Bc = cyclone diameter/4
= 2.0/4
= 0.5 ft
(2) Since the particle density is much greater than the gas density.
Qp~~ Qg~ Qp
=(2.75)(62.4)
= 171.6 lb/ft3
(3) Calculate the cut diameter.
[dP]cu< = [9 liBc/2 xntv,(eP-g)]1/2
= [(9)( 1.21 x lG"5)(0.5)/2 ir(4.5)(50)(171.6)]1/2
= 1.5x10-* ft
= 4.57 microns
b. dp/[dp],„ = 7.5/4.57
= 1.64
c. Determine the collection efficiency from Lapple's curve (Figure 4).
17 = 72%
2. The required collection efficiency
= [(inlet loading — outlet loading)/(inlet loading)i(lQO)
= [(0.5 — 0.1)/(0.5)]( 100)
= 80%
3. Since the collection efficiency of the cyclone is lower than the collection efficiency
required by the agency, the permit should not be approved.
138
-------
Electrostatic Precipitator—Process Change
1. Calculate throughput velocity (see basic operation 12).
v= Q/S
= (4200)/(11 /12)(24)
= 191 ft/min
= 3.2 ft/sec
2. Determine outlet loading.
Outlet loading = (2.82)(1 -0.882)
= 0.333 grains/ft3
3. See sub-steps.
a. A = (2X24X20)
= 960 ft2
b. Determine drift velocity.
0.882 - 1 - exp[ - (960)(w)/(4200)]
Solving for w,
w = 9.36 ft/min
4. Revised efficiency with process change.
j? = 1 —
= 1 - exp[ - (960)(9.36)/(5400)|
= 0.812
= 81.2%
5. Does plate spacing affect D-A efficiency?
No.
139
-------
Electrostatic Precipitator—Collection Efficiency
1. a. Considering both sides of the plate,
A = (2)( 12)( 12)
= 288 ft2
b. See sub-steps.
(1) Remembering that the volumetric flow rate through a passage is one third of
the total volumetric flow rate,
Q,= (4000)/(3)(60)
= 22,22 acfs
(2) Calculate the collection efficiency using the D-A equation.
ij = 1 - e^'1"^
= 1 - exp[ - (288)(0.4)/ (22.22)]
= 0.9944
= 99.44%
2. a. See sub-steps.
(1) Calculate Q in acfs.
Qj= (4000)/(2)(60)
= 33.33 acfs
(2) Calculate the collection efficiency, remembering the collection surface area per
duct remains the same.
i?i = l- exp[ - (288)(0.4)/(33.33)]
= 0.9684
= 96.84%
b. See sub-steps.
(1) Calculate Q in acfs.
Q_= (4000)/(4)(60)
= 16.67 acfs
(2) Calculate the collection efficiency.
7jz = 1 - exp[ - (288)(0.4)/(16.67)]
= 0,9990
= 99.90%
c. Calculate the new overall collection efficiency.
7| = (0.5)(t?1) + (2)(0.25)(t)2)
= (0.5X96.84) + (2X0.25X99.90)
= 98.37%
140
-------
Electrostatic Precipitator—Plan Review
a. Since the cut diameter is the panicle diameter collected at 50% efficiency,
17 = 1 - e'Kdp
0.5 = 1 — exp[ — K(0.9)]
Solving for K,
K. = 0.77
b. Calculate the collection efficiency using the D-A equation.
For dp = 3.5; 17 = 1 - exp[ — (0.77)(3.5)]
= 0.9325
The table given below shows the collection efficiency for each particle size.
w,
< dp>
Vi
0.2
3.5
0.9325
0.2
8.0
0.9979
0.2
13.0
0.9999
0.2
19.0
0.9999
0.2
45.0
0.9999
c. Calculate the overall collection efficiency.
T? = Ew.tj,
= (0.2)(0.9325) + (0.2)(0.9979) + (0.2)(0.9999) + (0.2)(0.9999) + (0.2)(0.9999)
= 0.9861
= 98.61%
d. Yes.
a. Calculate the outlet loading.
Outlet loading = (1.0 - i7)(inlet loading)
= (1.0-0.9861)(14)
= 0.195 grains/ft3
b. Yes.
Yes.
141
-------
Scrubber—Design of a Venturi Scrubber
1. Calculate inertial impaction parameter.
rj — 1-e ~~ MQj/QpK?
0.98 = 1 - exp[ — (0.14)(2)V$]
Solving for
4/= 195.2
2. Calculate throat velocity using the inertial impaction number.
^ = QPvd*/18
v= 18 $ii/ Qpdj
= (18)(195,2)(1.575 x 10"*)(1.23 x 10-5)/(187)(1.05 x 10"5)z
= 330.2 ft/sec
5. Determine throat area (see basic operation 12).
A = Q/v
= (11,040)/(60)(330.2)
= 0.557 ft2
142
-------
Scrubber—Overall Collection Efficiency
1. a. do = (16,400/vc) + 1.45^J 5
= (16,400/272)+ 1.45 (8.5)1-5
= 96.23 microns
= 3.156 x 10"4 ft
b. \p= Q^vdl/18 c^u
= (43.7)(272)(d*)/'18(3.156 x 10~4)(1.5 x 10"5)
= 1.3945 x 10ud*
c. jj = 1 — 0—^(Q^//Qp)v^
rj,¦ = 1 - exp[ - (0.2)(8.5)a/L3945 x 10"d^, ]
= 1 -exp(- 6.348 x 105d„,)
2. Calculate the collection efficiency for each particle size.
For dp = 0.05 microns (1.64 x 10"7 ft);
y]i= 1 -exp[(-6.348 x 105)(1.64x 10"7)]
= 0.0989
w,07, = (0.01 )(0.0989)
= 9.89 x 10"4
The following table shows the results of the above calculation for each particle size.
(ft)
w ,• (%)
V.
W,77, (%)
1.64 x 10"'
0.01
0.0989
9.89 x lO'*
9.84 x lCT
0.21
0.4645
0.0975
2.62 x 10"'
0.78
0.8109
0.6325
9.84 x lO"8
13.00
0.9981
12.980
2.62 x lCT-
16.00
1.0000
16.000
4.27 x 10~4
12.00
1.0000
12.000
5.91 x lO"1
8.00
1.0000
8.000
6.56 x 10~s
50.00
1.0000
50.000
3. Calculate the overall collection efficiency.
t/ = Lw.-rji
= 9.89 x 10"4 + 0.0975 + 0.6325 + 12.980 + 16.00 + 12.00 + 8.00 + 50.00
= 99.71%
143
-------
Scrubber—Plan Review
1. a. See sub-steps.
(1) &c = (0.157)Ap
=(0.157)(5.0)
= 0.785 hp/1000 acfm
(2) ^ = 0.583 Pl(Qj./Qp)
= (0.583)(80)( 50/10,000)
— 0.233 hp/1000 acfm
(3) &t = 3>c+3>l
= 0.785 + 0.233
= 1.018 hp/1000 acfm
b. N, = ct&r3
= (1.47)(1.018)10S
= 1.50
c. N, = ln[l/(l-i?)]
1.50 = ln[ 1/(1 — tj)]
Solving for 17,
¦q = 0.777
= 77.7%
2. Collection efficiency required by State regulations:
= [(inlet loading - outlet loading)/'(inlet loading)](100)
= [(5.0 — 0.05)/(5.0)]( 100)
= 99.0%
3. No.
4. a. See sub-steps.
(1) N, = ln[l/(l - ??)]
= ln[ 1/(1 - 0.99)]
= 4.605
(2) N, = a&Te
4.605 = (1.47)(^r)1 05
Solving for (?T>
3>T = 2.96 hp/1000 acfm
b. &g = 0.157 Ap
= (0.157)( 15)
= 2.355 hp/1000 acfm
c, 3>l=3>t-&c
= 2.96-2.355
= 0.605 hp/1000 acfm
144
-------
Qa/Qp = #/(0.583)(Pi)
=(0.605)/(0.583)(100)
= 0.0104
Q/ = (Q*/Qp)(10,000 acfm)
= (0.0104)(10,000 acfm)
= 104 gal/min
Q/ = 104 gal min
#=2.96 hp/1000 acfm
-------
Baghouse—Bag Selection
Bag D is eliminated since its recommended maximum temperature (220 °F) is below
the operating temperature of 250 °F. Bag C is also eliminated since a pulse-jet fabric
filter system requires the tensile strength of the bag to be at least above average.
a. Cost per bag is $26.00 for A and $38.00 for B.
b. See sub-steps.
(1) Calculate Af.
Since actual flow rate = (10,000)(250 + 460)/(60 + 460)
=13,654 acfm
and filtering velocity = 2.5 cfm/ftz cloth
= 2.5 ft/min
Ac = Qjvf
=(13.654)/(2.5)
= 5461.6 ft2
(2) For bag A; A = xDh
= (t)(8/12)(16)
= 33.5 ft2
For bag B; A = (tt)( 10/12)(16)
= 41.9 ft2
(3) For bag A; N = A,/A
= 5461.6/33.5
= 163
For bag B; N = 5461.6/41.9
= 130
c. For bag A; total cost = (N)(cost per bag)
= (163)(26.00)
= $4238
For bag B; total cost = (130)(38.00)
= $4940
. Since the total cost for bag A is less than bag B, select bag A.
146
-------
Baghouse—Cleaning Frequency
1. See sub-steps.
a. Ac = QNs
= 50,000/10
= 5000 ft2
b. A =
-------
Baghouse—Bag Failure
1. Calculate efficiency and penetration initially.
7/= (inlet loading —outlet loading)/(inlet loading)
= (4.0 —0.02)/(4.0)
= 0.995 = 99.5%
Pt= 1.0-7?
= 0.005
2. Calculate the bag failure parameter.
0 = Q/(LD2v'T + 460)
= 50,000/(2)(6)2V60 + 460
= 30.45
3. Determine penetration arising due to bag failure.
Pt = 0.582 y/Kp/
-------
Combustion—Air Requirements
1. Balance the combustion equations,
CH< + 2 02 —COz + 2 HjO
C2H6 + (7/2) Oz-2 C02 + 3 HzO
CsH8 + 5 02 —3 C02 + 4 H20
2. Determine oxygen requirements.
0.81 CH4 + 1.62 O#-0.81 CO, + 1.62 H20
0.1 C2H6 + 0.35 02 —0.2 C02 + 0.3 H20
0.04 CjHg + 0.2 0,-0.12 CO,+ 0.16 H20
scfm of 02= 1.62 + 0.35 + 0,2 = 2.17 scfm
3. Calculate air requirements.
scfm of air = (scfm of 02)(l.O scfm of air/0.21 scfm of 02)
= (2.17)(1/0.21)
= 10.33 scfm
4. Calculate nitrogen in flue gas.
scfm of N2 = 0.05 + (scfm of air)(0.79 scfm of N,/l .0 scfm of air)
= 0.05+ (10.33)(0.79)
= 8.21 scfm
Calculate water and carbon dioxide in flue gas.
CO2 = 0.81 +0.2 + 0.12 = 1.13 scfm
H20 = 1.62 + 0.3 + 0.16 = 2.08 scfm
j. Total scfm of combustion products = 1.13 + 2.08 + 8.21
= 11.42 scfm
149
-------
Combustion—Thermal Afterburner Design
1. From Table 1, the following gross heat of combustion values are obtained for each
component of the natural gas.
Component
He, (Btu/scf)
N«
0
CH,
1013
CjHs
1792
c3h8
2590
C4H,o
3370
Basis: 1.0 scf natural gas.
HVC of natural gas = En,HVCl-
= (0.0515)(0) + (0.8111)(1015) + (0.0967)(1792) + (0.0551)(2590)
+ (0.0056)(3370)
= 1105 Btu/scf of natural gas
2. See substeps.
a. 0.8111 CR, + 1.622 0,-0.8111 C02 + 1.6222 H20
0.0967 C2HS +0.3385 O2-0.1934 C02 + 0.2901 H20
0.0351 C3H8 +0.1755 0,-0.1053 C02 +0.1404 H20
0.0056 G,H,0 + 0.0364 02-0.0224 C02 +0.028 H20
b. For 02; 1.622 + 0.3385 + 0.1755 + 0.0364 = 2.172 scf/scf of natural gas
For C02; 0.8111 + 0.1934 + 0.1053 + 0.0224 = 1.132 scf/scf of natural gas
For H20; 1.6222 + 0.2901 + 0.1404 + 0.028 = 2.081 scf/scf of natural gas
For N2; 0.0515 + (79/21)(2.172) = 8.222 scf/scf of natural gas
c. The total cubic feet of combustion products per scf of natural gas burned are the
sum of scf C02/scf of natural gas, scf H20/scf of natural gas, and scf N2/scf of
natural gas.
Total cubic feet of combustion products = 1.132 + 2.081 + 8.222
= 11.435 scf of products/scf of natural gas
d. See sub-steps.
(1) From Table 3, the following values of enthalpies at 60 °F and 1200 °F are
obtained.
For C02; AHc0l = H at 1200°F-H at 60°F
'= 12,960-243.1
= 12,716.9 Btu/lb mole
Since there are 379 scf per lb mole of any ideal gas, and there are 1.132 scf of
C02/scf of natural gas,
150
-------
dHcOj = (12,716.9)(1.132)/(379)
= 38.0 Btu/scf of natural gas
For N2: AH.v2 = (8452 - 194.9)(8.222)/(379)
= 179.1 Btu/scf of natural gas
For H20 (g); AH„;0 = (10,176 - 224.2)(2.Q81)/(379)
= 54.6 Btu/scf of natural gas
(2) Since the water present in the combustion product is vapor,
AHX = (1060 Btu/lb)(18) lb/lb mole)(2.081 scf of H20/scf of natural gas)/
(379 scf/lb mole)
= 104.8 Btu/scf of natural gas
IAH = AHcOj + AH.v, + AHHjOU, + AH,
= 38.0+ 179.1 + 54.6+ 104.8
= 376.5 Btu/scf of natural gas
e. HA = HVC - EAH
= 1105-376.5
= 728.5 Btu/scf of natural gas
See sub-steps.
a. From Table 3,
AHair = H at 1200°F- H at 200°F
= 8524- 1170
= 7354 Btu/lb mole
b. q = (5000 scfm)(7354 Btu/ib mole)/(379 scf/lb mole)
= 97,018 Btu/min
c. Rate of natural gas required = q/HA
= 97,018/728.5
= 133.2 scfm
a. See sub-steps.
(1) Q_ of combustion products = (11.435 scf of products/scf of natural gas)
(133.2 scfm of natural gas)
= 1523 scfm
(2) Total flue gas flow rate = 5000 + 1523
= 6523 scfm
(3) Actual total flue gas flow rate = 6523 (1200 + 460)/(60 + 460)
= 20,823 acfm
= 347 acfs
b. D = (4 Q/vt)1'2
= [(4)(347)/(20)(x)]>"
= 4.7 ft
c. Since the length-to-diameter ratio is 2.0,
L = (2.0)(4.7)
= 9.4 ft
d. t = L/v
=(9.4)/(20)
= 0.47 sec
151
-------
Combustion—Plan Review of a
Direct-Flame Afterburner
1. The design temperature is already within agency criteria.
2. See sub-steps,
a. First, calculate the molar flow rate of effluent.
ft = (7000 scfm)/(379 scf/lb mole)
= 18.47 lb mole/min
The heat load can now be calculated.
q = ft[Cp2(T2-T6)-C,l(T1-T6)]
= 18.47[(7.38)(1400 - 0) - (7.12)(738 - 0)]
= 93,740 Btu/min
b. Actual heat load = (l.l)(q)
= (1.1)(93,740)
= 103,114 Btu/min
c. Rate of natural gas required = heat load/available heat
= (103,114)/(600)
= 171.9 scfm
3.. See sub-steps.
a. Qj = (rate of natural gas)(ll,5 scf of combustion products/scf of natural gas)
= (171.9)(11.5)
= 1976 scfm
b. Volumetric flow rate of the effluent = 7000 scfm
c. Qo = (rate of natural gas)(10.33 scf of air/scf of natural gas)
= 1776 scfm
d. Since primary air is employed in the combustion of the natural gas, Q, is not sub-
tracted from Qj.
Or = 7000 + Qi
= 7000+ 1976
= 8976 scfm
e. Q/ = (8976)(1400 + 460)/(60 + 460)
= 32,106 acfm
4. See sub-steps.
a. S = ttDV4
= ( tt)(4 .2)2/4
= 13.85 ft2
152
-------
b. v=Q7-/S
= (32,106)7(13.85)
— 2317 ft/min
= 38.6 ft/sec
c. Yes.
5. a. t = L/v
= 14/38.6
= 0.363 sec
b. Yes.
153
-------
Absorption—Spray Tower
1. Calculate gas throughput velocity (see basic operation 12).
y = Q/(7tD2/4)
= 25,000/[7t(14)2/4]
= 162.4 ft/min
= 2.7 ft/sec
2. Yes, since the gas velocity is less than 3 ft/sec.
3. Calculate the number of overall gas transfer units.
Nog = In (yi/yz)
= In (500/25)
= 3.0
4. Calculate transfer units for spray tower unit.
2nd spray = N0g = (0.7)(0.6) = 0.42
3rd spray = Noc = (0.42)(0.6) = 0.252
4th spray = N0c = (0.252)(0.6) = 0.1512
5th spray= Noc31 (0.1512)(0.6) = 0.0907
Inlet = Noc = 0.5 (given)
Total Noc = 0.7+ 0.42+ 0.252+ 0.1512+ 0.0907+ 0.5 = 2.114
Note that this value is below the required value of 3.0.
5. yt/y2 = exp (Noc)
= exp (2.114)
= 8.28
y2 = (500)/(8.2S)
= 60.4 ppm
6. Since y2 is greater than the required emission limit of 25 ppm, the spray unit is not
satisfactory.
154
-------
Absorption—Packed Tower Plan Review
1. a. See sub-steps.
(1) S = irD2/4
= (t)(3.57)V(4)
= 10.0 ft2
(2) Gm = Qe/SM
= (1575)(0.0743)/(10.0)(29)
= 0.404 lb mole/ft2-min
= 24.2 lb mole/ft2-hr
(3) U = L/M£
= (1000)/( 18)
= 55.6 lb mole/ft2-hr
b. rnGm/Lm = (0.972)(24.2/55.6)
= 0.423
c. (y, - mxj)/(y2 - mx,) = [0.02 - (0.972)(0)]/[0.001 - (0.972)(0)]
= 20.0
d. From Colburn's Chart (Figure 6),
Nog = 4.3
2. See sub-steps.
a. G = Qjg/S
= (1575)(0.0743)/10.0
= 11.7 lb/ft2-min
= 702 lb/ft2-hr
b. From Figure 5,
Hoc = 2.2 ft
3. Z = (Noc)(HoC)
= (4.3X2.2)
= 9.46 ft
4. Disapprove.
155
-------
Absorption—Tower Height and Diameter
1. a. See sub-steps.
(1) xf=y1/m
=(0.02)/(1.20)
= 0.0167
(2) ya = (0.1)y1/[(l-y1) + (0.1)yI]
= (0.1 )(0.02)/ [(1 - 0,02) + (0.1)(0.G2)]
= 0.00204
b. (Lm/Gm)m,„ = (yt-y2)/(xf-x2)
= (0.02 - 0.00204)/(0.0167 - 0)
= 1.08
c. UG, = (1.25)(U/GmU
= (1 -25)( 1.08)
= 1.35
d. (y, - mx2)/(y2 - rax,) = [(0.02) - (1.2)(0)]/[(0.00204) — (1.2)(0)] =9.80
e. (mG„)/ Lm =(1.2)/(1.35)
= 0.889
f. From Colburn's Chart,
Noc = 6.2
2. Z = (N0g)(Hog)
= (6.2)(2.5)
= 15.5 ft
3. a. See sub-steps.
(1) (L/G)(e/et)05 = (U/G,X18/29)(e/ei)0-5
= (1,35)(18/29)(0.075/62.4)0-5
= 0.0291
(2) From Figure 7,
(G*FM'*)/(QiQgc) = 0.19
(3) Gf=l(0.\9)(QLegc)/(F^nY<*
= [(0.19)(62»4)(0.075)(32.2)/(160)( 1)(1.8)0-2]1/2
= 0.409 lb/ft2-sec
b. Gact = (0.6)(0.409)
= 0.245 lb/'ft2-sec
= 884 lb/ft2-hr
c. D = [(4m)/(Ga<:,-r)]l/2
= [(4)(5000)/(884)(t)]"2
= 2.67 ft
156
-------
Adsorption—Working Capacity
1. Breakthrough capacity = [(0.5)(0.39)(4) + (0.39)(36 - 4)]/36
= 0.368
= 36.8%
2. Working capacity = 36.8 - 2.5
= 34.3%
157
-------
Adsorption—Degreaser Ventilation Clean Up
1. a. See sub-steps.
(1) Q^= 10,000[(70 + 460)/(60+460)][(14.7)/(20)]
= 7491 acfm
= 4.5 X 105 acfh
(2) Qjc£- = (yxczXQp)
= (2000 x 10~8)(4,5 x 10s)
= 900 acfh
(3) rf, = mrr£ =QTC£= Q(PM, RT)
= (20)(131.5)(900)/(10.7S)(70 + 460)
= 416.2 lb/'hr
(4) TCE adsorbed = (m)(0.995)(4)
= (416.2)(0.995)(4)
=1656.6 lbs
b. Activated carbon required = (TCE to be adsorbed) [(28 lb TCE adsorbed 1001 b
carbon) (bulk density)]
= (1656.6)( 100/28)/ (36)
= 164 ft3
2. Height of adsorbent = (activated carbon volume)/(cross-sectional area)
= (164)/(6)(15)
= 1.85 ft
158
-------
Adsorption—-Plan Review
1. a. See sub-steps.
(1) % relative saturation = [(mole fraction)(air pressure)/(vapor pressure of
acetone @20°C)](100)
= [(0.0015)( 7 60)/( 170)]( 100)
= 0.67%
(2) From Figure 8,
Saturation capacity = 23 wt. %
(3) Breakthrough capacity = [(0.5)(C,)(MTZ) + (C,)(D - MTZ)]/D
= [(0.5)(0.2 3)(2 /12) + (0.23)(1.96 — (2/12))]/1.96
= 0.22
= 22 wt. %
(4) Working capacity = (breakthrough capacity) - (heel) — (packing factor)
= 22-2-3
= 17 wt. %
b. See sub-steps.
(1) Q_=(mole fractionXQ,,,)
= (0.0015)(30,000)
= 45 acfm
(2) m = PMQ/RT
= (l)(58)(45)/(0.73)(68 + 460)
= 6.77 Ibs/min
c. Since two adsorbers are on stream with each cleaning 14 of the total gas stream,
Adsorption time = (carbon charge)(working capacity)/(0.5 m)
= (15,0G0)(0.17)/(0.5)(6.77)
= 753 min
d. Regeneration time =6 hrs 16 min
2. a. From Figure 9,
Steam requirement = 2.4 lb steam/lb solvent removed
b. Mass of steam = (steam requirementXworking capacity)(carbon charge)
= (2.4)(0.17)(15,000)
= 6120 lbs steam
159
-------
Fans—Effect of Fan Wheel and Speed
1. Calculate power requirement of fan "B."
PB/ ?A = (50/46)5( 1590/1622)3
= 1.429
Pfl = (1.429)(45.9) .
= 65.6 bhp
160
-------
Fans—Brake Horsepower Requirement
Total pressure drop = pressure drop across the packing + pressure drop due
duct work, etc.
= 6.0 + 4.0
= 10.0 in. HjO
Brake horsepower = (Q)(Ap)( 1.575 x 10~4)/>7/
= (6000)(10.0)(1.575x 10'4)/(0.63)
= 15 bhp
161
-------
Economics—Breakeven Operation
1. The value of dust collected = 50,000 (ftVmin) x 2 (grains/ft3) x (l/7000)(lb/grain) x 0.01
($/lb) x 7)
= 0.143 tj; $/min
2. The value of dust collected = 0.143[Ap/(Ap + 5,0)]; S/rnin.
3. Cost of power = dp (Ib/ft2) x 50,000 (ft3 /min x l/44,200)(kW-min/ft-lb/) x (1/0.55)
x 0.06($/kW-hr) x (l/60)(hr/min)
= 0.002 Ap; S/min
4. Calculate breakeven pressure drop. Set (2) equal to (3).
(0.143)Ap/(Ap + 5) = 0.002 Ap
Solving for dp, dp = 66.5 lb/ft2 = 12.8 in. H20
5. Calculate breakeven efficiency.
n = 66.5/(66.5 + 5) = 0.93 = 93.0%
162
-------
Economics—Annualized Installed, Operation,
and Maintenance Costs
1. a. N = Q/(air-to-cloth ratio)(A)
= (70,000)/(5.81)(12)
= 1004
b. Annual maintenance cost = $5000/year + cost of replacing 25% of the bags each year
= S5000 + (0.25)( 1004)(75.00)
= $23,825/year
2, a. Installed capital cost = (Q)($2.536/acfm)
= (70,000)(2.536)
= $177,520
b. AICC = (installed cost)
= (177,520)
i(l + i)m
(1 + i)m- 1
0.08(1 +0.08)"
(1+0.08)
15 .
1
=$20,740/year
Since 1 ft-lb/sec = 0.0013558 kW,
Operating cost = (Q)(Ap)(operating time)(30.03/kW-hr)/
-------
Appendix
Jmm
-------
Appendix A—Gas Properties
1. Chart of Viscosity of Air at 1.0 Atmosphere
2. Figure of Viscosity of Gases at 1.0 Atmosphere
3. Chart of Heat Capacity of Gases at 1.0 Atmosphere
Temperature Viscosity
°C °F (cp)
100 —I
-
r- 0.1
0.09
.
- 100
- 0.08
;
.
- 0.07
;
r 0.06
°~
r°
0.05
—
- 100
\ 0.04
©
o
111 1 1 1
f- 200
[• 0.03
200
r300 x
— 400
0.02
:
-500
-
300 -j
-600
-
400 -=
-700
-800
-
500 J
600 -
700-
800 -
900 -
1000 -
-900
-1000
-1100
-1200
-1300
-1400
-1500
-1600
—1700
F—1800
r 0.01
7 0.009
- 0.008
- 0.007
- 0.006
0.005
1. Viscosity of air at 1.0 atmosphere.
167
-------
0.036
0.034
0.032
0.030
Air
0.028
0.026
0.024
CO.
0.022
a.
HtO
0.018
CH.
0.016
0.014
0.012
0.010
0.008
0.006
0.004
0.002
100
200
300
400
500
600
700
0
Temperature, °F
2. Viscosity of gases at 1,0 atmosphere.
168
-------
C, = Heat capacity = specific heat
= Btu/lb-°F
= cal/g-°C
C u9*J F.
0-
iCO-
2Q0 -
3CQ-
*00-
500-
SOOr
roc-
800-
900-
1000-
HOO-
;200-
1300-
1400-
• 00
-200
-300
-400
-500
-600
-TOO
-800
-§00
'000
-iiOO
• 200
-1300
1*00
*500
-i600
1700
1800
i900
-2000
-2iOO
-2200
¦2300
- 2*00
-2500
3
48
5 s
0,0
0 M
too°
12
o
o
IS
l?i
®irc
00
8
O
i7Q
N9-
Got
ftana«-D«aC
1°
0 - 200
)5
»
200- 4O0
<6
•
400-1400
27
Air
0 -f*O0
12
Amrftoruo
0 * 6 OO
14
•
S00- '*00
18
Canton 0>o*id#
0 * 400
24
»
400-1400
26
Cocaon Monoid*
0 -r«00
32
O«0fin«
0 - 200
3*
•
200- 1400
3
£m«t«
0 - 200
9
«
200- GOG
$
¦
600-1400
4
£fhyl*n«
0 * 200
I
•
200- 6GO
13
•
SQ0-I4C0
178
CCGUfl
* *21 fCMCifn
0 - ?SO
S7C
0 - i50
17*
» -ZZtCHCJ fc)
• -ii3(CCi,*-CCt^)
0-150
J70
0-150
1
Nydrogtn
0 ¦ 600
z
•
600- 1400
35
0 -s40Q
30
• Chiorid«
0 -1400
20
" Ffyeridt
0 -J400
36
• Iodic**
0 M4O0
19
* Sut^id#
0 - 700
2i
TOO* 140Q
5
0 • 300
6
•
300 - 700
7
>
700*1400
25
NiffiC 0n4«
0 - 700
28
* m
700 " i4O0
26
0 *1400
23
0*yqtn
0 - 300
29
500-'400
33
Sulfur
300*1400
22
SuHur Oiqxtd#
0 * 400
31
« «
400*1400
i 7
WQfff
0 -14O0
2
O
7
o
80
a
19
o
01s
OI7
20
O
21
0 26,
»° o Co29
2S 7TJ8
29
32
O
33
O
MO
X
o
-4.0
-3.0
-2.0
-1.0
-0.9
-0.8
-07
-0.6
-0.5
I-<14
-0.3
-0.2
-0.1
-0.0S
-0.08
-0.07
-0.0«
-0.0S
Example: To
find the
specific heat
of air at
300 °C, lav a
scraight edge
across paper
from 300 °C
through the
circle for 27
and read off
specific heat
scale as 0.255
cal/g °C.
Source: From Chemical Engineers' Handbook by R. Perry and C. Chilton.
©1973 McGraw-Hill. Used with the permission of McGraw-Hill Book Company.
3. Heat capacity o£ gases at 1.0 atmosphere.
169
-------
Appendix B—SI Units
1. The Metric System
2. The SI System
3. Seven Base Units
4. Two Supplementary Units
5. SI Multiples and Prefixes
6. Conversion Constants (SI)
7. Selected Common Abbreviations
1. The Metric System
The need for a single worldwide coordinated measurement system was recognized over
300 years ago. Gabriel Mouton, Vicar of St. Paul in Lyons, proposed in 1670 a com-
prehensive decimal measurement system based on the length of one minute of arc of a
great circle of the earth. In 1671 Jean Picard, a French astronomer, proposed the length
of a pendulum beating seconds as the unit of length. (Such a pendulum would have been
fairly easily reproducible, thus facilitating the wide-spread distribution of uniform stan-
dards.) Other proposals were made, but over a century elapsed before any action was
taken.
In 1790, in the midst of the French Revolution, the National Assembly of France
requested the French Academy of Sciences to "deduce an invariable standard for all the
measures and all the weights." The Commission appointed by the Academy created a
system that was, at once, simple and scientific. The unit of length was to be a portion of
the earth's circumference. Measures for capacity (volume) and mass (weight) were to be
derived from the unit of length, thus relating the basic units of the system to each other
and to nature. Furthermore, the larger and smaller versions of each unit were to be
created by multiplying or dividing the basic units by 10 and its multiples. This feature
provided a great convenience to users of the system, by eliminating the need for such
calculating and dividing by 16 (to convert ounces to pounds) or by 12 (to convert inches
to feet). Similar calculations in the metric system could be performed simply by shifting
the decimal point. Thus, the metric system is a "base-10" or "decimal" system.
The Commission assigned the name "metre" (which we now spell meter) to the unit of
length. This name was derived from the Greek word "metron" meaning "a measure."
The physical standard representing the meter was to be constructed so that it would
equal one tenth-million of the distance from the north pole to the equator along the
meridian of the earth running near Dunkirk in France and Barcelona in Spain.
The metric unit of mass, called the "gram," was defined as the mass of one cubic cen-
timeter (a cube that is 1/100 of a meter on each side) of water at its temperature of max-
imum density. The cubic decimeter (a cube 1/10 of a meter on each side) was chosen as
the unit of fluid capacity. This measure was given the name "liter."
170
-------
Although the metric system was not accepted with enthusiasm at first, adoption by
other nations oeurred steadily after France made its use compulsory in 1840. The stan-
dardized character and decimal features of the metric system made it well suited to scien-
tific and engineering work. Consequently, it is not surprising that the rapid spread of the
system coincided with an age of rapid technological development. In the United States,
by Act of Congress in 1866, it was made "lawful throughout the United States of America
to employ the weights and measures of the metric system in all contracts, dealings, or
court proceedings."
By the late 1860's, even better metric standards were needed to keep pace with scien-
tific advances. In 1875, an international treaty, the "Treaty of the Meter," set up well-
defined metric standards for length and mass, and established permanent machinery to
recommend and adopt further refinements in the metric system. This treaty, known as
the Metric Convention, was signed by 17 countries, including the United States.
As a result of the Treaty, metric standards were constructed and distributed to each
nation that ratified the Convention. Since 1893, the internationally agreed-to metric stan-
dards have served as the fundamental weights and measures standards of the United
States.
By 1900 a total of 35 nations —including the major nations of continental Europe and
most of South America—had officially accepted the metric system. Today, with the
exception of the United States and a few small countries, the entire world is using
predominently the metric system or is committed to such use. In 1971 the Secretary of
Commerce, in transmitting to Congress the results of a 3-year study authorized by the
Metric Study Act of 1968, recommended that the U.S. change to predominant use of the
metric system through a coordinated national program.
The International Bureau of Weights and Measures located at Sevres, France, serves as
a permanent secretariat for the Metric Convention, coordinating the exchange of infor-
mation about the use and refinement of the metric system. As measurement science
develops more precise and easily reproducible ways of defining the measurement units,
the General Conference of Weights and Measures—the diplomatic organization made up
of adherents to the Convention — meets periodically to ratify improvements in the system
and the standards.
2. The SI System
In 1960, the General Conference adopted an extensive revision and simplification of the
system. The name "Le Systeme International d'Unites" (International System of Units),
with the international abbreviation SI, was adopted for this modernized metric system.
Further improvements in and additions to SI were made bv the General Conference in
1964, 1968, and 1971.
The basic units in the SI system are the kilogram (mass), meter (length), second (time),
Kelvin (temperature, ampere (electric current), candela (the unit of luminous intensity),
and radian (angular measure). All are commonly used by the engineer. The Celsius scale
of temperature (0°C-273.15 K) is commonly used with the absolute Kelvin scale. The
important derived units are the newton (SI unit of force), the joule (SI unit of energy),
the watt (SI unit of power), the pascal (SI unit of pressure), the hertz (unit of frequency).
There are a number of electrical units: coulomb (charge), farad (capacitance), henry
(inductance), volt (potential), and weber (magnetic flux). One of the major advantages of
171
-------
the metric system is that larger and smaller units are given in powers of ten. In the SI
system a further simplification is introduced by recommending only those units with
multipliers of 103. Thus for lengths in engineering, the micrometer (previously micron),
millimeter, and kilometer are recommended, and the centimeter is generally avoided. A
further simplification is that the decimal point may be substituted by a comma (as in
France, Germany, and South Africa), while the other number, before and after the
comma, will be separated by spaces between groups of three, i.e., one million dollars will
be $1 000 000,00." More details are provided below.
3. Seven Base Units
a. Length —meter (m)
The meter (common international spelling, metre) is defined as 1 650 763.73 wavelengths
in vacuum of the orange-red line of the spectrum of krypton-86. The SI unit of area is
the square meter (m2). The SI unit of volume is the cubic meter (m3). The liter (0.001
cubic meter), although not an SI unit, is commonly used to measure fluid volume.
b. Mass —kilogram (kg)
The standard for the unit of mass, the kilogram, is a cylinder of platinumiridium alloy
kept by the International Bureau of Weights and Measures at Paris. A duplicate in the
custody of the National Bureau of Standards serves as the mass standard for the United
States. This is the only base unit still defined by an artifact. The SI unit of force is the
newton (N). One newton is the force which, when applied to a 1 kilogram mass, will give
the kilogram mass an acceleration of I (meter per second) per second. 1 N = 1 kg* m/s2.
The SI unit for pressure is the pascal (Pa). 1 Pa= 1 N/m2. The SI unit for work and
energy of any kind is the joule (J). 1 J = 1 N*m. The SI unit for power of any kind is the
watt (W). 1 W=1 J/s.
c. Time—second (s)
The second is defined as the duration of 9 192 632 770 cycles of the radiation associated
with a specified transition of the cesium-133 atom. It is realized by tuning an oscillator to
the resonance frequency of cesium-133 atoms as they pass through a system of magnets
and a resonant cavity into a detector. The number of periods or cycles per second is
called frequency. The SI unit for frequency is the hertz (Hz). One hertz equals one cycle
per second. The SI unit for speed is the meter per second (m/s). The SI unit for
acceleration is the (meter per second) per second (m/s2).
d. Electric current — ampere (A)
The ampere is defined as that current which, if maintained in each of two long parallel
wires separated by one meter in free space, would produce a force between the two wires
(due to their magnetic fields) of 2 x 10"7 newton for each meter of length. The SI unit of
voltage is the (V). 1 V = 1 W/A. The SI unit of electrical resistance is the ohm (0).
1 Q = 1 V/A.
e. Temperature—Kelvin (K)
The Kelvin is defined as the fraction 1/273.16 of the thermodynamic temperature of the
triple point of water. The temperature 0 K is called "absolute zero." On the commonly
172
-------
used Celsius temperature scale, water freezes at about 0°C and boils at about 100°C. The
°C is defined as an interval of 1 K, and the Celsius temperature 0°C is defined as defined
as 273.15 K. 1.8 Fahrenheit scale degrees are equal to 1.0°C or 1.0 K; the Fahrenheit
scale uses 32°F as a temperature corresponding to 0°C.
f. Amount of substance —mole (mol)
The mole is the amount of substance of a system that contains as many elementary
entities as there are atoms in 0.012 kilogram of carbon 12. When the mole is used, the
elementary entities must be specified and may be atoms, molecules, ions, electrons, other
particles, or specified groups of such particles. The SI unit of concentration (of amount
of substance) is the mole per cubic meter (mol/m3).
g. Luminous intensity—candela (cd)
The candela is defined as the luminous intensity of 1/600 000 of a square meter of a
blackbody at the temperature of freezing platinum (2045 K). The SI unit of light flux is
the lumen (lm). A source having an intensity of 1 candela in all directions radiates a light
flux of 4 t lumens.
4. Two Supplementary Units
a. Plane angle—radian (rad)
The radian is the plane angle with its vertex at the center of a circle that is subtended by
an arc equal in length to the radius.
b. Solid angle—steradian (sr)
The steradian is the solid angle with its vertex at the center of a sphere that is subtended
by an area of the spherical surface equal to that of a square with sides equal in length to
the radius.
5. SI Multiples and prefixes.
Multiples and submultiples
Prefixes
Symbols
1 000 000 000 000 10'*
tera (ter'a)
T
1 000 000 000 10»
gig3 (j» 'ga>
G
1 000 000 10«
mega (meg'a)
M
1 000 10' .
kilo (kil 'o)
k
100 10*
hecto (hek 'to)
h
10 101
deka (dek' a)
da
Base unit 1 10°
0.1 10*'
deci (des 'i)
d
0.01 10"J
centi(sen 'ti)
c
0.001 10"'
milli (mil 'i)
m
0.000 001 10"8
'micro (mi 'kro)
u
0.000 000 001 10-'
nano (nan 'o)
n
0.000 000 000 001 10-'*
pico (pe 'ko)
P
0.000 000 000 000 001 10'15
femto (fern 'to)
f
0.000 000 000 000 000 001 10"1®
atto (at' to)
a
173
-------
6. Conversion constants (SI).
To convert from:
To:
Multiply by;
milligrams/ mJ
micrograms/ m3
1000
micrograms/ liter
1.0
ppm by volume (20°C)
(24.04/M)
ppm by weight
0.8347
lb/ ft1
62.43 x 10""
micrograms- m3
milligrams/ m1
0.001
micrograms/ liter
0.001
ppm by volume (20°C)
(0.02404/M)
ppm by weight
834.7 x 10"«
lb/ ft1
62.43 x 10""
micrograms* liter
milligrams/ mJ
1.0
micrograms/ mJ
1000
ppm by volume (20°C)
ppm by weight
0.8347
lb/ ft3
62.43 x 10"'
ppm by volume (20°C)
milligrams/ m3
(M/24.04)
micrograms/ m3
(M/0.02404)
micrograms/ liter
(M/24.04)
ppm by weight
t.M/28,8)
lb, ft3
(M/385.1 x 10*)
ppm by weight
milligrams/ m1
1.198
micrograms/ m3
1.198 x 10"»
micrograms/ liter
1.198
ppm by volume (20°C)
(28.8/M)
lb/ ft3 '
7 ,48 x 10*
lb - ft3
milligrams/ m3
16.018 x I0»
micrograms/ m3
16.018 x 10'
micrograms/ liter
16.018 x 10*
ppm by volume (20"C)
(385.1 x 10*/M)
ppm by weight
133.7 x 103
To convert from:
To:
Multiply by:
grams
milligrams/ m3
35.3145 x 10J
grams/ m1
35.314
micrograms/ m3
35.314 x 10*
micrograms/ ft3
1.0x10'
lb/1000 ft3
2.2046
grams/' mJ
milligrams/ m3
1000.0
grams/' ft3
0.02832
micrograms/ m3
1.0 x 10*
micrograms/ ft'
28.317 x 103
lb/ 1000 ft3
0.06243
micrograms/ ft3
milligrams/ m1
35.314 x 10"»
grams/ ft3
1.0 x 10"»
grams/ mJ
35.314 x 10"s
micrograms/ m3
35.314
lb/1000 ft3
2.2046 x 10"*
no. of particles/ ft1
no., m3
35.314
no. / L
35.314 x 10"'
no. .cm3
35.314 x 10"»
tons/ tni!
lb/ acre
3.125
lb/ 1000 ft"
0.07174
grams/m1
0.3503
kb/kjm1
350.3
milligrams, rnl
350.3
milligrams/cm*
0.0350S
grams/ft1
0.03254
lb
7000.0
micrometer
in.
3.837 x 10"
mm
1.0 x 10"3
174
-------
7. Selected common abbreviations.
AA
angstrom unit of length
abs.
absolute
amb.
ambient
app. mol. wt.
apparent molecular weight
acm.
atmospheric
at. wt,
atomic weight
b.p.
boiling point
bbl
barrel
Btu
British thermal unit
cal
calorie
cg
centigram
cm
centimeter
cgs system
centimeter-gram-second
system
conc.
concentrated, concentration
cc, cm3
cubic centimeter
cu ft, ft3
cubic feet
cfh
cubic feet per hour
cfm
cubic feet per minute
cfs
cubic feet per second
m3, M»
cubic meter
<3
degree
°C
degree Celsius, degree
Centigrade
op
degree Fahrenheit
K
degree Kelvin
°R
degree Reaumur, degree
Rankine
ft
foot
ft lb
foot pound
fpm
feet per minute
fps
feet per second
fps system
foot-pound-second system
f.p.
freezing point
gr
grain
g. gm
gram
hr
hour
in.
inch
kcal
kilocalorie
kg
kilogram
km
kilometer
liq.
liquid
L
liter
log
logarithm (common)
In
logarithm (natural)
m.p.
melting point
m, M
meter
/zm
micrometer (micron)
mks system
meter-kilogram-second
system
mph
miles per hour
mg
milligram
ml
milliliter
mm
millimeter
ttifi
millimicron
min
minute
mol. wt.
molecular weight
oz
ounce
PPb
parts per billion
pphm
parts per hundred million
ppm
parts per million
lb
pound
psi
pounds per square inch
psia
pounds per square inch
absolute
psig
pounds per square inch gage
rpm
revolutions per minute
sec
second
sp. gr.
specific gravity
sp, ht.
specific heat
sp. wt.
specific weight
sq
square
scf
standard cubic foot
STP
standard temperature and
pressure
temp.
temperature
wt
weight
175
-------
Appendix €—Tables of Specific Conversion Factors
1. Temperature
2. Pressure
3. Area
4. Volume
5. Flow
6. Weight
7. Concentration
8. Length
9. Emission rates
10. Velocity
L Temperature.
Desired units
"F
°C
°R
K
Degrees Fahrenheit
0.5555 x(°F-32)
°F + 460
0.5555 x (°F-32)-r 273
Degrees Centigrade
1.8°C + 32
1.8°C + 492
°C + 273
Degrees Rankine
°R - 460
0.555 x(°R-492)
0.5555 x {°R - 492) + 273
Degrees Kelvin
1.8(K- 273) + 32
K - 273
I.8(K-27S) + 492
176
-------
2. Pressure.
Desired
units
Given
units
gmm
dynes
#m
poundals
gmf
2
cm'
"Atmospheres"
0
cm-sec'
cm2
ft-sec2
ft2
ft2
in2
Emm
cm-see'*
1
1
6. 7197
X 10"2
6.7197
X 10"2
1.0197
X 10"3
2.0885
X 10"3
1.4504
X 10"5
9.8692
X 10"7
dynes
em<*
1
1
6.7197
X 10"2
6. 7197
X 10~2
1. 0197
X 10"3
2. 0885
X 10-3
1.4504
X 10"5
9. 8692
X 10"7
ft-sec2
14. 882
14. 882
1
1
1. 5175
X 1Q'2
3. 1081
X 10"2
2.1584
X 1Q"4
1. 4687
X 10~5
poundals
14. 882
14. 882
1
1
1.5175
X 10"2
3.1081
X 10"2
2. 1584
X 10"4
1. 4687
X 10"5
gmf
cm2
980. 665
980. 665
65.898
65. 898
1
2.0482
1. 4223
X 10"2
9.6784
X 10"4
#f
ft2
478. 80
478. 80
32. 174
32. 174
4. 8824
X 10"1
1
6.9444
X 10"3
4.7254
X 10-4
t{
in2
6. 8948
X 104
6. 8948
X 1G4
4.6331
X 1Q3
4.6331
X 103
70, 307
144.00
1
6.8046
X 10"2
"Atmospheres"
1. 0133
X 1G6
1. 0133
X 106
6. 8087
X 104
6. 8087
X 104
1. 0332
X 103
2.1162
X 103
14.696
1
To convert a value from a given unit to a desired unit, multiply the given value by the factor opposite the given units and
beneath the desired units.
-------
3. Area.
Dealred Units
Square
Inch
Square
Feet
Square
Yard
Square
Mile
Acre
Square
Centimeter
Square
Declmetei
| Square
Meter
Square
Ktlmeter
Square
Inch
1
6.9444
* 10*5
T7.1605
¦ 10-5
2.49
x 10-M
15. 94
* 10"*
6.452
6. 452
x 10-2
6.452
x I0"4
6.452
x 10-IP
Square
Foot
144
1
0.1U1
J. 587
x I0"B
2. 296
x lO"5
929.0341
929.0141
x 10-2
929.0341
x 10-4
929. 0341
x 10"10
Square
Yard
1296
9
1
3.228
* I0"7
2.066
x I0*4
83. 61
x lO2
83.61
83.61
x I0"2
83.61
x 10"8
a
Square
Mite
40.144
x 10*
2.788
* 10*
J. 098
* 1G6
1
640
2.589998
x lO*®
2.589998
x 10®
2.589998
x 10*
2.589998
i
S3
8
3
Acre
62.73
* 107 .
4. 3560
X 104
4840
15.625
* I0"4
1
4046.873
x I04
4046. 873
xlO*
4046. 873
4046.873
x 10"6
Square
Centimetei
15. 5*10-2
10.764
* 10""*
1.1960
x 10-*
3. 8610
x lO"*1
2. 471
x 10"®
I
1 x 10-2
1 x lO"4
1 x 10-10
Square
Decimeter
IS. 5
10.764
x10-2 .
1.1960
* 10"*
3.1610
x 10-9
2.471
x 10-6
I x 10Z
1
1 x 10-2
1 X 10"®
Square
Meter
IS. 5 x 10*
10.764
1.1960
3. 8610
x I0-*
2. 471
x tO"4
I x 104
lx 10*
1
1 x I0-*
Square
Kilometer
15.5 * 10°
10. 764
* Ml6
1.1960
x 10*
3.8610
x 10-'
2 471
x lO*
• 1 x 10»>
1 x 10H
1 x 106
1
To convert • value from • given unit • desired unit, multiply the given value fay the factor opposite the Riven units
and beneath the desired unit.
-------
4. Volume.
^s^Desired
Given>w Units
Units
Cubic
Yard
Cubic
Foot
Cubic
Inch
Cubic
Meter
Cubic
Decimeter
Cubic
Centimeter
Liter
Cubic
Yard
1
27
4.6656
4
X 10
0.764559
764.559
7.64559
x io5
764.54
Cubic
Foot
3.7037
x io"2
1
1728
2.8317
x io"2
28.317
2.8317
4
X 10
CO
CO
CO
Cubic
Inch
2. 143347
x io-5
5.78704
-4
X 10
1
1.63872
x io-5
1. 63872
x io"2
16.3872
1. 63868
x io"2
Cubic
Meter
1.30794
35.314445
6.1023
4
X 10
1
1000
1 x io6
999.973
Cubic
Decimete r
1.3079
x io"3
3. 5314
x io-2
61.023
0.001
1
1000
.99997
Cubic
Centimeter
1.3079
x io"6
3.5314
x io"5
6. 1023
x io-2
i x io"6
1 x io"3
1
9.99973
-4
X 10
Liter
1.3080
x io"3
3.5316
x io"2
61.025
1.000027
x io"3
1. 000027
1000.027
1
To convert a value from a given unit to a desired unit, multiply the given value by the factor opposite the given units
and beneath the desired units.
-------
5. Flow.
oo
o
Desired
Given
Units
M3
sec
min
hour
ft3
ft3
fti.
hour
L
sec
L
min
3
cm
3
cm
min
sec
min
see
jii
sec
1
60
3600
35.3144
21. 1887
X io2
12.7132
4
X 10
999,973
59.U98
x io3
1 X io6
6 X ID7
min
0.0167
I
60
0.5886
35. 3144
2i. 18a
X io2
16.667
989.973
16. 667
x io3
6
1 X 10
M3
hour
2.778
X id"5
16.667
X I0*3
i
08.90
x io"4
0.5866
35.3144
27.777
x io"2
16.667
2.777
x io2
1.666
4
X 10
J»i
sec
28.317
x io"3
1. 693
101.94
1
60
3600
28. 316
16.9896
X ID2
2.8317
4
X 10
1.699
X io6
jsi
min
4.7195
-4
X 10
28, 317
x io"3
1. 690
£6.667
xio"3
1
60
47. 193
x io"2
28.316
4.7195
X io2
2.8317
ji!
hour
7.8658
X 1Q*6
4.7 105
X io"4
28.317
x io"3
2.778
x io"4
16.667
xio"3
1
7.866
x io"3
0. 4719
78.658
4.7195
x io2
L
sec
1.000027
x io"3
6.00016
x io"2
3.6
35.316
xio"3
2.11886
127.138
1
60
1000.027
16.667
L
min
1.6667
x io"5
1.000027
x io"3
6.00016
-2
X 10
5.886
-4
X 10
35.316
X io"3
2.11886
1,6667
x io"2
1
16.867
1000.027
3
cm
sec
1 X io"6
6 X io"5
3.6 X 10"3
3.5314
X io"5
2. 1189
X io"3
1,271
x io"3
9.99973
-4
X 10
5,9998
x io"2
1
16.667
x io"3
3
cm
min
1,6667
x io"8
1 x io"6
6 X 10"5
5. 880
X io"7
0.3531
.4
X 10
2.11887
X io"3
5.9998
x io"2
9.D9973
-4
x 10
60
1
To convert a value from a given unit to a desired unit, multiply the given value by the factor opposite the given units and beneath the desired unit.
-------
6. Weight.
Deaired Unita
Micro-
gram
Mill I-
gram
gram
Kilo-
gram
(rain
Ounce
(avdp)
Pound
(xvdp»
Ton |
(U.S. short
Tonne
) (metric)
m
M
•H
s
c
•
>
<3
Micro-
gram
1
1 x HT4
1 x !0-b
1 x 10-1
15.4124
x 10"*
3.5274
x 10"8
2.2046
xlO-9
1.1023
x 10"»*
1 x 10-»*
Milli-
gram
I * 10'
I
1 x 10-'
1 x 10-*
15.4324
x 10-'
3. 5274
x 10-5
2.2046
x 10-6
1.1023
x 10-9
1 x 10-*
gram
1* 10°
I X 10 J
1
I x HP*
15.4324
3.5274
x 10-2
2.2046
x 10"'
1.1023
x 10-*
1 x 10-6
Kilogram
1 x 10*
1 x I0b
1 x !0J
1
15.4324
* »0J
35. 274
2.2046
1.1023
u W'
1 x 10-*
grain
64.799
* 10'
64.799
64.799
x 10-5
64.799
x 10-*
1
22.857
x 10"4
1.4286
x 10-4
7.143
x 10"8
64.799
x 10-9
Ounca
|xirdp)
28.349
x 106
28.349
x 10'
28.349
28.349
x 10"'
437.1
I
62. 5
x 10-'
3.125
x 10-5
28.349
x 10-6
Pound
(avdp)
453.59
x 10*
451. 59
x 103
451.59
453.59
x 10-'
7000
16
1
5 x 10-"*
453. 59
x 10-6
Ton
(U.S. ahoi
905.185
¦t) x 10?
907.185
x 106
907.185
* 10J
907.185
14 x 10b
3. 2
x 104
2000
1
0.90716:
Tonne
(metric)
I x !0»*
1 x 10*J
I * lo6
1 * 10'
1.543 *10* 3.5274
j x I04
2204.62
1.10211
1
-------
7. Concentration.
Desired
Units
*£
M s
-*4
MT
L
or
ft. 3
Ibf.
IT5
g rami
ft. 5
Ibt.
1000 ft. 3
grain*
ft. J
I
iooo
1. 00002?
9. 989
x 10-7
6. 243
xlO«
2.831?
x 10"®
6.243
x 10-5
4. 37
x 10-*
1 x IO"3
i
1. 000027
x 10-1
9. 989
x IO"1®
6.243
x IO-"
2.831?
x 10"®
6.243
x 10"®
4.3?
* 10-7
_flL
1.
.999973
f.99913
* 102
1
9. 988
x IO*7
6.242
x 10-®
2.8316
x 10"®
6.242
x 10-5
4. 37
* 10-4
+»
5
5
OB
11.5
1. 00115
x iO6
1. 00115
* Iq9
1.00118
* 10*
1
62.5
* 10"3
28.349
62.5
4. 3?5
x IO2
>
«r4
lbs,
1T3
I. 602
x 10*
1. 602
xlfllO
1. 602
* 10*
16
1
453.59
1 x IO3
7 * IO3
It ram ¦
ft. 5
3.511
x io<
5. 531
x IO7
3.531
x IO4
3. 5274
x 10-2
2.20462
x 10"3
1
2.2046
15.43
Ibi.
1000 ft. 3
1. 602
X 10*
1. 602
x 107
1.602
x IO4
1.6
x IO-2
1 x IO"1
453. 59
x 10-3
1
7
gralni
ft. 5
2. 288
x IO1
2. 288
x IO6
2.288
x 10*
2.2857
* 10*3
1.4286
x IO'4
b.4799
x IO"2
14.286
1
-------
8. Length.
oo
oo
^N>l^)esired
\ Units
Given
Units
Inch
Foot
Yard
Mile
Micron
Millimeter
Centimeter
Meter
Kilometer
Inch
1
83.33
x io"3
27.778
X 10
1. 578
x io*5
2.54
X io4
25.4
2. 54
2. 54
X 10~2
2.54
x io-5
Foot
12
1
3333
1.894
-4
X 10
30.48
X IO4
304.8
30.48
30.48
x io"2
30.48
x io"5
Yard
36
3
1
5.682
x io"4
91.44
4
X 10
914.4
91.44
91.44
X io~2
91.44
x io"5
Mile
6.3360
4
X 10
5280
1760
1
1.6094
X 109
1,6094
X 106
1.6094
X JO5
1.6094
X io3
1.6094
Micron
3.937
x io~5
32.808
-7
X 10
10.94
-7
X 10
62.137
x io"11
1
-3
1 X 10
i x io"4
i x io"6
ix io"9
Millimeter
3.937
x io~2
32.808
-4
X 10
10.94
-4
X io
62.137
x io*8
1 X io3
1
0. 1
I X io"3
1 x io"6
Centimeter
3.937
x io"1
32.808
_ 'i
X 10
10.84
-3
X 10
62.137
-7
X 10
1 X 104
10
1
1 X io"2
1 x io*5
Meter
39.37
32.808
x io"1
10.94
x io"1
62. 137
x io"5
1 x 106
1 X 103
1 X io2
1
I X io~3
Kilometer
3.937
X IO4
32.808
X 102
10.94
X io2
62. 11?
-?
X 10
1 X io9
1 X io6
1 X io5
1 X io3
1
To convert a value from a given unit to a desired unit, multiply the given value by the factor opposite the given units and beneath the desired unite.
-------
9. Emission rates.
Desired
units
Given
units
gms / sec
gms / min
kg/hr
kg/day
lbs/min
lbs/hr
lbs/day
tons/hr
tons/day
tons/yr
gms/sec
1.0
60. 0
3. 6
8. 640
X 10
1. 3228
X 10"'
7. 9367
1. 9048
X 102
3. 9683
X 10*3
9, 5240
X 10*2
3. 4763
X 10
gms/ mill
1.6667
X 10-2
1.0
6. 0
X 10"2
1. 4400
2. 2046
X 10-3
1. 3228
X io-i
3. 1747
6. 6139
X 10"5
I. 5873
X 10"3
5. 7938
X 10" 1
kg/hr
2. 7778
X 10" 1
16. 667
1. 0
2. 4000
X 10
3. 6744
X 10-2
2. 2046
5. 2911
X 10
1. 1023
X 10-3
2. 6456
X 10-2
9. 6563
kg/day
1. 1574
X 10"2
6. 9444
X 10"1
4. 1667
X 10'2
1.0
1. 5310
X 10-3
9. 1860
X 10"2
2. 2046
4. 5930
X 10"5
1. 1023
X 10"3
4. 0235
X 10"]
lbs/min
7. 5598
4. 5359
X 102
2. 7215
X 10
6. 5317
X 102
1. 0
60. 0
1. 44
X 103
3. 000
X 10"2
7. 2000
X 10"1
2. 6280
X 102
lbs/hr
1. 2600
X 10" 1
7. 5598
4. 5359
X io-i
1. 0886
X 10
1. 6667
X 10~2
1.0
24. 0
5. 0000
X 10*4
1. 2000
X 10-2
4. 3800
lbs/day
5. 2499
X 10*3
3. 1499
X 10"1
1. 8900
X 10"2
4. 5359
X 10" 1
6, 9444
X 10"4
4. 1667
X 10"2
1. 0
2. 0833
X 10"5
5.0000
X 10"4
1. 8250
x io"1
tons/hr
2. 5199
X 102
1. 5120
X 104
9. 0718
X 102
2. 1772
X 104
3. 3333
X 10
2. 0
X 103
4.8000
X 104
1.0
24.0
8. 7600
X 103
tons/day
1.0500
X 10
6. 2999
X 102
3. 7799
X 10
9. 0718
X 102
1. 3889
8. 3333
X 10
2.0
X 103
4. 1667
X 10"2
1. o
365. 0
tons/yr
2. 8766
X 10-2
1. 7260
I. 0356
X I0"1
2. 4854
3. 8052
X 10"3
2. 2831
X 10" 1
5. 4795
1.1416
X 10"4
2, 7397
X 10"3
1. 0
To convert a value from a given unit to a desired unit, multiply the given value by the factor opposite the given units and
beneath the desired units.
-------
10. Velocity.
Desired
units
Given
units
m/set:
ft/sec
ft/ min
km/hr
mi/hr
knots
mi/day
m/see
1. 0
3.2808
1.9685
X 102
3. 6
2. 2369
1. 9425
5.3687
X 10
ft/ sec
3. 04 80
X 10~ 1
1. 0
60
1. 0973
6.8182
X 10"1
5. 9209
X 10"1
1.6364
X 10
ft/min
5.0080
X 10" 3
1, 666?
X lO"2
1.0
1,8288
X 10"2
1. 1364
X 10'2
9.8681
X 10"3
2. 7273
X 10"1
km/hr
2.7778
X 10"1
9. 1134
X 10*1
5.4681
X 10
1. 0
e. 213?
X 10"1
5. 3959
X 10"1
1. 4913
X 10
mi / iir
4. 4707
X 10"1
1. 4667
88. 0
1. 6093
1.0
8. 6839
X 10"1
24
knots
5. 14 79
X 10"1
1. 6890
1. 0134
X 102
1. 8533
1.1516
1. 0
2. 7637
X 10
mi/day
1. 8627
X 10"2
g. nil
X 10~2
3. 6667
6.7056
X 10"2
4.1667
X 10"2
3. 6183
X 10-2
1. 0
To convert a value from a given unit to a desired unit, multiply the given value by the factor
opposite the given units and beneath the desired units.
-------
TECHNICAL REPORT DATA
(Please read Instructions on the reverse before completingj
1, REPORT NO. 2.
EPA 450/2-84-007
3. RECIPIENT'S ACCESSIOf#NO.
4. TITLE AND SUBTITLE
APTI Course SI:412D
Control of Gaseous and Particulate Emissions
Self-Instructional Problem Workbook
5. REPORT DATE
September 1984
6. PERFORMING ORGANIZATION CODI
7. AUTHORtSS
Louis Theodore
8. PERFORMING ORGANIZATION REPC
9. PERFORMING ORGANIZATION NAME AND AOOHESS
Northrop Services Inc.
P.O. Box 12313
Research Triangle Park, NC 27711
10. PROGRAM ELEMENT NO.
11. CONTRACT/GRANT NO.
68-02-3573
12. SPONSORING AGENCY NAME AND ADDRESS
U.S. Environmental Protection Agency
Manpower and Technical Information Branch
Research Triangle Park, NC 27711
13. TYPE OF REPORT ANO PERIOO CO'
Self-Instructional Wor:
14. SPONSORING AGENCY CODE
IS. SUPPLEMENTARY NOTES
EPA Project Officer for chis Self-Instructional Workbook is R. E. Townsend
EPA-ERC MD 20, Research Triangle Park, NC 27711
16. ABSTRACT • '
This self-instructional workbook is to be used in conjunction with EPA-APTI
Courses 413, Control of Particulate Emissions, and 415, Control of Gaseous
Emissions. This workbook illustrates example problems and solutions
dealing with agency review of air pollution control equipment. The workbook
is divided into four parts: 1. nomenclature, defining terms commonly used
in air pollution control; 2. basic operations, dealing with calculations
reflecting basic concepts from chemistry, thermodynamics, and physics and which
appear in more than one type of control system check; 3. problems, presenting
calculations linked directly with control processes which include combustion,
absorption, adsorption, electrostatic precipitation, and fabric filtering;
and 4. solutions, giving detailed solutions to the mathematical problems
of the workbook.
17. KEY WOROS AND DOCUMENT ANALYSIS
a. DESCRIPTORS
b.identifiers/open ended terms
c. COSATI Field,'G:'
Air Pollution Control
Particulate Emission Control
Gaseous Emission Control
Self-Training Manual
Self-Instructional
Problem Workbook
13B
51
68A
is. distribution statement Unlimited. Available
from: National Technical Information
Service, 5285 Port Royal Road,
Sorinqfield. VA 2?16l
19. SECURITY CLASS (This Report)
Unclassified
21. NO. OF PAGES
188
20. SECURITY CLASS (This page)
Unclassif ied
22, PRICE
EPA Form 2220-1 (t-73)
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