United States
Environmental Protection
Agency
Air Pollution Training Institute
MD20
Environmental Research Center
Research Triangle Park, NC 27711
EPA 450/2-81-006
May 1981
&EPA
Air
APTI
Course 415
Control of Gaseous
Emissions
Student Workbook
-------
United States
Environmental Protection
Agency
Air Pollution Training Institute
MD20
Environmental Research Center
Research Triangle Park, NC 27711
EPA 4 50/2-81-006
May 1981
Air
APTI
Course 415
Control of Gaseous
Emissions
Student Workbook
Northrop Services, Inc.
P.O. Box 12313
RtMarch Triangle P*rk. NC 27709
Under Contract No.
68-02-2374
EPA Project Officer
R. E. Townsend
United States Environmental Protection Agency
Office of Air, Noise, and Radiation
Office of Air Quality Planning andStandards
Research Triangle Park, NC 27711
-------
Notice
This is not an official policy and standards document. The opinions and selections
are those of the authors and not necessarily those of the Environmental Protection
Agency. Every attempt has been made to represent the present state of the art as well
as subject areas still under evaluation. Any mention of products or organizations
does not constitute endorsement by the United States Environmental Protection
Agency.
Availability
This document is issued by the Manpower and Technical Information Branch, Con-
trol Programs Development Division, Office of Air Quality Planning and Stan-
dards, USEPA. It was developed for use in training courses presented by the EPA
Air Pollution Training Institute and others receiving contractual or grant support
from the Institute. Other organizations are welcome to use the document.
This publication is available, free of charge, to schools or governmental air pollu-
tion control agencies intending to conduct a training course on the subject covered.
Submit a written request to the Air Pollution Training Institute, USEPA, MD-20,
Research Triangle Park, NC 27711.
Others may obtain copies, for a fee, from the National Technical Information Ser-
vice (NTIS), 5825 Port Royal Road, Springfield, VA 22161.
Sets of slides and films designed for use in the training course of which this publica-
tion is a part may be borrowed from the Air Pollution Training Institute upon writ-
ten request. The slides may be freely copied. Some films may be copied; others must
be purchased from the commercial distributor.
11
-------
Course Description
This course is a four-day lecture course dealing with the sources and control
methods associated with gaseous air pollutants. The course presents the fun-
damental concepts of the operation of gaseous emission control equipment for
stationary sources. The lessons include a description of equipment operation,
control equipment operating parameters that affect efficiency, and examples of
equipment types used in selected industrial applications. Lecture topics include a
discussion of the principles of absorption, adsorption, combustion, and conden-
sation. Problem sessions are held in which the student calculates equipment effi-
ciencies and determines if equipment design meets minimum engineering
specifications.
How to Use This Workbook
This workbook is to be used during the course offering. It consists of two parts:
the first contains a chapter corresponding to each of the fourteen lessons, and the
second contains five sets of problems.
Each chapter includes a listing of the lesson goal, the lesson objectives, any
special references in Part 1 that might be helpful to you, and several pages of
black-and-white line-art reproductions of selected lecture slides. The reproduc-
tions are intended to follow generally the slide presentations given in the lecture.
However, the instructor may on occasion change the order or present new
material not included in the workbook. It is therefore recommended that the stu-
dent take notes throughout the course and not rely on the graphic reproductions
as representing the total course content.
Part 2 of this workbook consists of five problem sets. Each set contains several
problems and any graphs or figures that you will need to work the problems.
in
-------
Table of Contents
Page
Part 1—Lessons
Chapter 1. Course Goal and Objectives 1-1
Chapter 2. Review of the Basics 2-1
Chapter 3. Concepts of Combustion as an Emission Control Method 3-1
Chapter 4. Stack Gas Afterburners 4-1
Chapter 5. Principles of Absorption 5-1
Chapter 6. Review of Absorption Design Principles 6-1
Chapter 7. Absorption Control Systems 7-1
Chapter 8. Adsorption Principles 8-1
Chapter 9. Adsorption Dynamics 9-1
Chapter 10. Adsorption Control Equipment 10-1
Chapter 11. Condensation Principles and Applications 11-1
Chapter 12. Reduction of SOa Emissions from Fossil-Fuel Burning Sources .. 12-1
Chapter 13. Control of Nitrogen Oxide Emissions
from Fossil-Fuel Burning Sources 13-1
Chapter 14. Exhaust Systems 14-1
Part 2—Problem Sets
Problem Set 1. Review of the Basics 15-1
Problem 1-1. Orsat Analysis 15-1
Problem 1-2. Partial Pressure of Gases 15-2
Problem 1-3. Properties of a Gas 15-3
Problem Set 2. Combustion , 16-1
Problem 2-1. Combustion of Gases 16-1
•Problem 2-2. Fuel Required to Incinerate Waste Gases 16-2
Problem 2-3. Design of Afterburner with Heat Recovery 16-4
Problem 2-4. Plan Review of a Direct-Flame Afterburner 16-5
Problem Set 3. Absorption 17-1
Problem 3-1. Equilibrium Diagram 17-1
Problem 3-2. Packed Tower for H2S Removal 17-5
•Problem 3-3. SO, Absorption by Water 17-6
Problem 3-4. SO, Absorption by Dilute Alkaline Solution 17-12
Problem 3-5. Permit Review of Ammonia Absorber 17-13
Problem 3-6. Spray Tower 17-14
Problem Set 4. Adsorption 18-1
Problem 4-1. Adsorption Working Capacity 18-1
Problem 4-2. Sizing an Adsorber 18-2
Problem 4-3. Benzene Adsorber Plan Review 18-3
Problem 4-4. Adsorption Plan Review—Gasoline Marketing 18-4
•Problem 4-5. Rotogravure Printing Adsorber 18-5
•The solution is included.
-------
Page
Problem Set 5. Condensation 19-1
•Problem 5-1. Contact Condenser 19-1
•Problem 5-2. Surface Condenser 19-4
Appendixes
Appendix A—Common SI Units 20-1
Appendix B—Conversion Factors 21-1
Appendix C—Constants and Useful Information 22-1
Figures
2-1 Two-bed heat recovery incinerator 16-4
2-2 Afterburner with heat recovery system 16-5
2-3 Combustion constants 16-7
2-4 Heat contents of various gases 16-8
2-5 Mean molal heat capacities of gases above 0°F 16-9
2-6 Available heats for some typical fuels (referred to 60°F) 16-10
3-1 Packing data 17-15
3-2 Ammonia-water absorption system 17-16
3-3 Colburn chart 17-17
3-4 Generalized flooding and pressure drop correlation 17-18
4-1 Adsorption isotherm for carbon tetrachloride 18-8
4-2 Adsorption isotherm for benzene 18-8
4-3 Adsorption isotherm for gasoline vapors 18-9
4-4 Adsorption isotherm for toluene 18-9
5-1 Contact condenser 19-2
5-2 Surface condenser 19-4
'The solution is included.
VI
-------
Nomenclature
a
A
AHP
BHP
C,
C,
c
C,
C
d,
D
D
E
f
F
AF
ge
G
G
Cm
H
H
H,
H,
Hv
HVC
HVN
HA
Hoc
h,
suiface area available for absorption per unit
volume of tower, ft2/fts
area, ft2
air horsepower, hp
brake horsepower, hp
specific heat, Btu/lb°F
average specific heat, Btu/lb °F
concentration
saturation capacity of adsorber (weight %)
empirical constant in BET equation
diameter, ft
bed depth (adsorber), ft
diffusivity, ftVsec
efficiency
percent of flooding
absorbed packing factor
free energy change, Btu/mol
gravitational constant, Ib ftVlb/ sec2
gas flow rate, Ib/min
superficial gas flow rate Ib/min ft2
gas molar flow rate, Ib mol/min
Henry's Law constant, atm/mol fraction
enthalpy, Btu/lb
heat of combustion, Btu/ft3 of fuel
sensible heat, Btu/lb
latent heat of vaporization, Btu/lb
gross heating value, Btu/ft3
net heating value, Btu/ft1
available heat, Btu/ft3
height of a transfer unit based on overall gas
phase, ft
height of a transfer unit based on overall
liquid phase, ft
system resistance, in. of water
individual mass transfer coefficient based on
liquid phase, Ib mol/hr ft2 aim
individual mass transfer coefficient based on
gas phase, Ib mol/hr ft2 atm
overall mass transfer coefficient based on
gas phase, Ib mol/hr ft1 atm
overall mass transfer coefficient based on
liquid phase, Ib mol/hr ft2 atm
SI
mVm3
m2
kW
kW
J/kg °C
J/kg °C
m
m
mVs
J/mol
kg m/N s2
kg/min
kg/min m2
kg mol/min
kPa/mol fraction
J/g
J/m3
J/g
J/g
J/m3
J/m3
J/m3
m
m
cm of water
kg mol/h m2 kPa
kg mol/h m3 kPa
kg mol/h m1 kPa
kg mol/h m2 kPa
vn
-------
L
L '
Lm
m
m
MW
M.,
N
OG
N
NX
P
Pt
p
P
Q
q
R
Re
T
AT
U
V
v
V.,
X
Y
Z
C
T
6
liquid flow rate, Ib/min
superficial liquid flow rate Ib/min ft2
liquid molar flow rate Ib mol/min
slope of a line
mass flow rate, Ibs/hr
molecular weight, Ib mol
mass of theoretical air per unit mass of fuel
combusted
number of transfer units based on overall gas
coefficient
number of transfer units based on overall liquid
coefficient
fan speed, rpm
mass flux, Ib mol/min ft2
total pressure, psia
barometric or atmospheric pressure, psia
partial pressure, psia
vapor pressure, psia
gage or static pressure, psia
power, hp
volumetric flow rate, cfm
heat rate, Btu/hr
ideal gas constant
Reynolds number
temperature, °F
mean temperature difference, °F
heat transfer coefficient, Btu/hr/ft2
volume, ft3
velocity, ft/sec
volume of theoretical air required to combust a
unit volume of fuel
mol fraction in liquid phase
mol fraction in gas phase
packed tower height, ft
density, lb/ft3
shearing stress, lb/ft2
kinematic viscosity, ftVsec
viscosity, centipoise
residence time, sec
ratio of specific gravity of water to solvent
flooding correlation ordinatc
kg/min
kg/min m2
kg mol/min
kg/h
g mol
kg mol/min m2
kPa
kPa
kPa
kPa
kPa
kW
mVmin
J/h
CC
°C
kJ/h/m2
m1
m/s
m
kg/m3
kPa/m2
mVs
mPa«s
s
vm
-------
Part 1
Lessons
-------
Chapter 1
Course Goal and Objectives
Course Goal
The goal of Course 415 is to present characteristics and operating features of
gaseous air pollutant emission control devices. Upon completion of the course,
you will be able to evaluate the gaseous emission control characteristics of emis-
sion control equipment shown in plans filed with a governmental air pollution
control agency for a permit to construct or modify an air pollution source. You
will also be able to evaluate the gaseous emission control characteristics of
operating air pollutant sources.
Course Objectives
At the end of this course, you should be able:
1. To use the ideal gas law, laws pertaining to temperature, pressure and
volume corrections, the properties of viscosity, specific heat and the
Reynolds number in gaseous emissions control calculations.
2. To calculate gross Meat value, available heat, and fuel requirements for
incineration.
3. To evaluate (given sets of design plans for flares, catalytic reactors, and
incinerators) in terms of equipment efficiency, problems which may impair
efficiency, and appropriateness of the control technique for the particular
source.
4. To evaluate (given sets of design plans for contact and surface condensers
in terms of vapor and odor control efficiency, problems which may impair
efficiency, and appropriateness of the control technique for the particular
source.
5. To calculate material balances and conditions of equilibrium for an
absorption system.
6. To evaluate (given plans for absorption systems) in terms of emission con-
trol efficiency, the suitability of absorption media, the effect of varying
physical conditions on the operation of the absorption system and the
appropriateness of the control technique for the particular source.
7. To describe the main methods used to control nitrogen oxide emissions
from combustion sources.
8. To describe the main methods used to control sulfur oxide emissions from
combustion sources.
1-1
-------
9. To evaluate (given sets of plans for adsorption control systems) in terms of
gaseous emissions removal efficiency, saturation capacity of the adsorption
bed, bed regeneration processes and appropriateness of the control tech-
nique for the particular source.
1-2
-------
Chapter 2
Review of the Basics
Lesson Goal
To explain the meaning of numerous symbols and basic concepts that are used
when performing gaseous emission control calculations.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. Define the following in general terms.
a. Pressure—gage, barometric, absolute
b. Temperature—Celsius, Farenheit, Kelvin, Rankin
c. Density
d. Ideal Gas Law
e. Molecular weight
f. Raoult's and Henry's laws
g. Gas viscosity
h. Reynolds Number
2. Calculate pressure, temperature, and volume changes of gases.
3. Calculate the Reynolds Number.
References
1. APTI Course 415 Student Manual.
2-1
-------
REVIEW OF
GAS
PROPERTIES
AND LAWS
PROPERTIES
OF GASES
Temperature
Pressure
Density
Molecular Weight
Viscosity
P»lnt »f H,0
C
r P
ID I
I
•%r;
2-2
-------
TEMPERATURE SCALES
1* 4U
Pttat •« M,0
mn MUM
BAROMETRIC OR
ATMOSPHERIC PRESSURE
M«M«r«d In HIM
•f m«rcury «r
GAGE PRESSURE
Tht difference
bttw««n »ytttm
pr«**ur« and
•tmo*ph«ric
prtttur*
ABSOLUTE
PRESSURE
P = barometric pressure
p.- gage pressure (positive or
negative value)
DENSITY
volume
2-3
-------
02 (one molecule)
turn of all atomic weight* « molecular
weight
• 32 amu
molecular
weight of
substance
___
Mmix ~
= molecular weight of gas
mixture
n = number of moles of each
component
M = molecular weight of each
component
Viscosity
mtermolecular
Cohesive
Forces
Momentum
Transfer
between the
Layers of Fluid
2-4
-------
VISCOSITY =
fluid retistance to flow
LowVbcMlty
HEATED
LIQUID =
lower
viscosity
Gas
Gas
KINEMATIC VISCOSITY
Q
V • kinematic viscosity
H m absolute viscosity
C - density
BEHAVIOR OF
GASES
• Meal Gas Law
• Van der Waal* Uw
• Dalton'sLaw
• Henry's Law
• Reynolds Number
2-5
-------
CHARLES BOYLES AVOGADRO'S
LAW LAW LAW
PocT Poc-J. Vocn
• constant • comtant • constant
prMturt temperature tempwatur*
•no ITIOMtt MM RtOffM ARQ pTVftMaTft
PV = nRT
n = number of moles
R - Ideal Gas Constant
— n~l mole
-
• T = standard tcmptratur*
3M.16K
P - standard prttture =
1 atm
v _ 1 (.08205)(298.16K)
1 atm
V = 24.46 Oters at EPA STP
VAN DER WAALS EQUATION
P - pressure
V - volume
R - ideal gas constant
T - temperature
a - constant
b -constant
D ALTON 'SLAW
— PA
~
YA = volume fraction of A
PA = partial pressure of A
1 = total pressure
2-6
-------
RAOULT'S LAW
PA=
PA = partial pressure of A
XA = mole fraction of A In solul
PA = vapor pressure of pure A
HENRY'S LAW
REYNOLDS NUMBER
Re
DvQ
i«- RtynoMt number Q - fatdtmtty
D - dtenwter of duct p - |M vtocwtty
¥ - (*•'
REYNOLDS NUMBER RANGE
PIPE FLOW
Re < 2000 Re > 2500
Laminar Flow Turbulent Flow
GASES
Properties Behavior
• Tamparatura • Waal CM
• Praatm • Van oar Waate Law
• Dtratty • Datton'* Law
• Vtoeoatty • Ray
2-7
-------
Chapter 3
Concepts of Combustion
as an Emission Control Method
Lesson Goal
To review the basic concepts of combustion and to illustrate some basic heat
calculations.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. Define a combustion reaction.
2. List the 4 variables involved with combustion of any fuel.
a. Time
b. Temperature
c. Turbulence
d. Oxygen
3. Balance a simple chemical equation for a combustion reaction.
4. Describe the difference between yellow flame combustion and blue flame
combustion.
a. Fuel/air mixture
b. Type of fuel used
c. Burning characteristics
5. Write a balanced generalized equation for combustion of carbon fuels.
6. Calculate the amount of air needed for combustion of a unit amount of a
given fuel.
7. Define the terms commonly used in working combustion problems.
8. Calculate the available heat for a given fuel.
References
1. APTl Course 415 Student Manual.
3-1
-------
BASIC
PRINCIPLES
OF
COMBUSTION
COMBUSTION
REACTION
Fuel + O2 — CO2 + H2O + Energy
FACTORS
AFFECTING COMBUSTION
TEMPERATURE
minimum ignition
temperature
increase in temperature
accelerates reaction rate
TIME
• residence time
• 0- V/Q
3-2
-------
TURBULENCE
• mixing of air and fuel
OXYGEN
• completeness of combustion
depends upon amount of oxygen
• CH4 * O2 —C - 2H2O * Energy
• CH4 - 2O2 —CO2» H2O- Energy
FLAMH ABILITY/
EXPLOSIBILITY
• Upper Explosive Limit (UEL)
• Lower Explosive Limit (LEL)
MECHANISMS OF COMBUSTION
V..
Yellow
name
Blue
Flame
Water-Gas
Reaction
YELLOW FLAME COMBUSTION
•ir
uel (tolid/liquid)
3-3
-------
BLUE FLAME COMBUSTION
air
fuel (gas/vapor)
WATER-GAS REACTION
STOICHIOMETRY
• determination of oxygen
needed for complete
combustion of a fuel —
theoretical amount -
EXAMPLEOF A
STOICHIOMETRIC
CALCULATION
I4 •"• 2O 2 — COj - 2HjO - tnerg?
2 moles O, I mole air 9.57 mole* air
I mole fuel .209 note O,
Jle fuel
9.57
air 379«cf air 1
fuel
I Bible fuel
I
air 379 acf fuel
9.S7»cr air
I acf fuel
3-4
-------
GENERALIZED STOICHIOMCTRIC
EQUATION
C.N.S.O. *<•*!* *
— »CO, * '/j n,O - ISO, * Energy
EXCESS AIR
%O, - O.5 (%CO)
0.264 (K.IN ,) - [%O, -0.5 CfcCO)]
THERMOCHEMICAL
RELATIONS
• total heat present
• heats produced by
combustion
TOTAL HEAT PRESENT
Heat Content (H)
Sensible
Heat(Hs)
Latent Heat of
Vaporization
(Hv>
HEATS PRODUCED
BY COMBUSTION
Grow Heating Valve (HVQ)
Available
Heat(HA)
Heat Loss In
Exit flue Gas
3-5
-------
HEAT BALANCE
„ Afl*
fuel I
available
heal (q.)
i •ytterii boundary .
MEAT CONTENT (ENTHALPY)
Btu/lb
H = Cp(T-T0)
Where:
C,- specific heat. Btu/lb °F
T « temperature of ga>, °F
T. » reference temperatures. °F
ENTHALPY FROM A
TEMPERATURE CHANGE
Where:
,= initial temperature
j= final temperature
p= defined at T3 or T| .
respectively
AVERAGE SPECIFIC If EAT
H-Cp(T2-T,)
Where:
Cp « average specific heat
value between T, and T2
HEAT RATE REQUIRED
TO CHANGE THE TEMPERATURE
OF A GAS
Btu/hr
q-riiAH
Where:
m*mass flow rate. Ib/hr
(no phase change or
heat loss in the system)
3-6
-------
HEAT RATE REQUIRED
TO CHANGE THE TEMPERATURE
OF A GAS
q « ih[Cp (T2-T0)-Cp (T,-T0)]
or
3-7
-------
Chapter 4
Stack Gas Afterburners
Lesson Goal
To provide you with an understanding of the basic concepts of waste gas
incineration and heat recovery.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. List three methods used to incinerate gaseous air pollutants.
2. Identify the conditions necessary for efficient combustion of organic
materials in a direct flame incinerator—including the effect of gas condi-
tioning, temperature, and residence time.
3. Explain the operation of at least two types of direct flame
incinerators—explaining the purpose of baffles, mult5jets, and distributed
burners.
4. Identify the conditions necessary for the efficient combustion of organic
materials in catalytic incinerators—including the effect of catalysts,
temperature, and residence time.
5. Discuss the advantages and disadvantages associated with catalytic
incineration.
6. Discuss the problems inherent in using process or heating boilers for
incineration.
7. Diagram at least two types of methods used for heat recovery.
8. Discuss the advantages, problems, and limitations associated with primary
and secondary heat recovery.
9. Discuss how a choice would be made in applying a heat recovery
system—discussing relative costs.
References
1. National Technical Information Services (NTIS). 1972. Afterburner
Systems Study. PB-212560.
2. Environmental Protection Agency (EPA). 1977. Controlling Pollution
from the Manufacturing and Coating of Metal Products: Metal Coating
Air Pollution Control. Technology Transfer Publication.
EPA-625/3-77-009.
4-1
-------
INCINERATION
THERMAL OXIDATION
AFTERBURNING
* *oc*u 6o«*» *
Cototytle
DIRECT FLAME INCINERATION
|Fum*
100
«o
EFFECTS OF
TEMPERATURE AND
TIME ON RATE OF
POLLUTANT OXIDATION
400
MO «000 HOC «400 «*00 1MO MOO
Increasing T*«np*ratui»
TYPICAL AFTERDURNER
OPERATING TEMPERATURES
Coil Coating •
1200°-13008F
Point Dak* Ov«ns—1100°-1500° F
P*trol«um Refining - 1000°- 2000° F
Printing 1000°- 1300°F
Vir« Enameling 1000°-1400° F
4-2
-------
DESIGN RANGES
Temperature 1000 ° — 1500° F
Detention 0.3 - 0.5 seconds
Time
DESIGN PARAMETERS
Proems das Flaw fe|vvn, Concentration
THERMAL DESIGN FACTORS
Efficiency increases with>
operating temperature
detention time
Initial hydrocarbon concentration
flame/solvent contact
good mixing
CO removal
(at temperatures >1300°F)
DISCRETE BURNER
I rum*
BAFFLED MIXING DEVICES
Fum*
Fun*
4-3
-------
WSTWDUTID BURNERS
_ MMUMI&M
MULTIJET
BURNER
Fum*
CATALYTIC INCINERATION
Diffusion
Adsorption
PRINCIPLES OF
OPERATION
Reoction
Dcsorption
Diffusion and
Mixing
4-4
-------
ADVANTAGES OF
CATALYTIC INCINERATION
• lower operating temperatures
• lower ouxllary fuel needs
• tower construction materials cost
DISADVANTAGES OF
CATALYTIC INCINERATION
• participate fouling
• thermal aging
• catalyst poisoning
PLATINUM
CATALYST POISONS
Fast
P
Di
As
Sb
Hg
Slow
Zn
Pb
Sn
High T«mp.
( >1100°F)
Fe
Cu
DISADVANTAGES OF
CATALYTIC INCINERATION
• port iculate fouling
• thermal aging
• catalyst poisoning
• suppressants
PLATINUM
CATALYST SUPPRESSANTS
• Sulfur
• Halogens
action Is reversible
4-5
-------
PROCESS
DOILERS
CONDITIONS FOR INCINERATION USE
• sufficient residence time
• no dependency
• tew fuel/oxygen rot*
• unaltered flam* and radiation patterns
• non-fouling or add fumes
Contaminated Wastt Goi
Cotolyii. it any
AFTERDURNER CONFIGURATION
WITHOUT HEAT RECOVERY
To Atmotph*i» Cotoly...
Fu«l
Pi*n*oi«> Afivrbutn*' I
WITH MIMA AT HEAT MCOVEKY
To Aunotptor*
AFTEABUKNEK CONFKMMATION VITH
AND DWECT MCYCIE HCAT KECOVEAY
4-6
-------
Ptknory M»ot
Afl*rburn»f
i.octuj I AFTU6UANER CONFIGURATION
| VITH F1UMARY AND SECONDARY
MAT RECOVERY
ttcot R»cov*fy
RECUPERATIVE
HEAT
RECOVERY
DEVICE
REGENERATIVE
HEAT
RECOVERY
DEVICE
(R»«co)
TUOULAR HEAT EXCHANGER SYSTEMS
• Eff*ctiv»n*ss ratio, E
1. Stoft MX max.
2. Stof* 65% max.
3. Stag* 63% max
• Limitation
Easily foul*d
Structural failures
Chemical Imeraaleni with tumt eomooneoti
AFTEMUAHEK PKOCE5S HEAT KECOVEKY
10
/i«ao«F
,.«oei
I,
i.
NOU, Final Twnpwotw*
•lllu*9UB3M*F
Fum» ttr*om uMd to
^TSSff
.ioo
.
Afttfkurrwf Capacliyjdm
4-7
-------
CAPITAL COST OF INCINERATION
10 '
ANNUAL VARIABLE COST OF INCINERATION
1?»
J 170
I
•JO
**
TO
4S
Th»tmol
10 11 M JJ au
Hew. icfm • 10*
4-8
-------
Chapter 5
Principles of Absorption
Lesson Goal
To familiarize you with the basic concepts involved in the transfer of mass from
the gas phase to the liquid phase.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. Describe the following in general terms.
a. Solute
b. Carrier gas
c. Absorbent
2. Illustrate the steps involved in the absorption of a gaseous molecule.
3. Describe both the two-film and Higbee penetration theories.
4. Differentiate between liquid and gas phase controlling resistances.
5. Use Henry's and Dalton's Laws to draw an equilibrium diagram.
References
1. APTI Course 415 Student Manual.
2. Bethea, R. M. 1978. Air Pollution Control Technology, An Engineering
Analysis Point of View. NY., NY.: Van Nostrand Reinhold Co.
3. Buonicore, A. J., and Theodore, L. 1975. Industrial Control Equipment
for Caseous Pollutants, volume I. Cleveland, OH.: CRC Press Inc.
4. Treybal, R. E. 1968. Mass Transfer Operations. NY., NY.: McGraw Hill
Book Co.
5-1
-------
BASIC
ABSORPTION
SYSTEM
earner
moje
fraction
moles of solute
moles of , moles of
solute """ solvent
LJQUID PHASE
MOLE FRACTION
XA =
mole A
mole
A
mole
H2O
CAS PHASE
MOLE FRACTION
moles A
moles .moles
A + air
5-2
-------
DALTON'S LAW
V =
'tot
HENRY'S LAW
*=KX
DALTON'S HENRY'S
LAW LAW
Y = —
T p10,
where
x=
mole fraction in gas
mole fraction in liquid
04
•si
&
'*"
*la,
'Henry's Law
Constant
0 .02 .04 .06 .08 .19 .12 M .16
mole fraction of A in water
PRE-EQUILJBRIUM
5-3
-------
EQUILIBRIUM
^AG
AL
Vs,* «••• *• ••|i\li±i±2_r.
* 'l%VifeMjT37ipi
VUi^ltJSJV^1^
•%• •• •• ,• • • 11/I Ml I
I
MASSRUX
(INDIVIDUAL)
- CAL)
5-4
-------
MASS FLUX
(OVERALL)
NA=KC(PAG-P*)
NA=KL(c*-cAL)
% V«
J_
K<
J_
K.
k,
1
CAS PHASE UQUID PHASE
CONTROLLED CONTROLLED
• contaminant is
very soluble
in liquid
contaminant is
relatively insoluble
in liquid
• diffusion through • diffusion through
gas phase is rate Equid phase is
controlling step rate controlling step
HICBEE PENETRATION THEORY
gas { Equd
5-5
-------
kg = /RT VD/710
g
k, = 2
Where:
D = diffusivity
G = contact time
R = ideal gas constant
TYP1CAI ABSORPTION TRANSFER
COEFFICIENT PIOT
(from vendor literature)
100* , _ IWIAAA M. L. A.."^ Civwi Condifeoft'
• lowvr diimrtrr
n^ _,~^- . pirting h»«ht
te molf.
hr H Mm
10 •
concrfltutaon
• outtf t kqmd
concrnlubon
• liquid tvfnprrjlurr
500 10OO 3OOO
C*s HOM Hilt fc hr (1
5-6
-------
Chapter 6
Review of Absorption Design Principles
Lesson Goal
To familiarize you with the main criteria used in designing a packed tower
absorber.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. Perform material balance calculations for an absorber.
2. Draw an operating line to determine minimum L/C ratio.
3. Determine minimum tower diameter.
4. Calculate the number of transfer units for the special case of straight
equilibrium line.
5. Calculate the height of packing required to achieve the desired efficiency
for a given absorber.
References
1. APTI Course 415 Student Manual.
2. Bethea, R. M. 1978. Air Pollution Control Technology, An Engineering
Analysis Point of View. NY., NY.: Van Nostrand Reinhold Co.
3. Buonicore, A. J., and Theodore L. 1975. Industrial Control Equipment
for Gaseous Pollutants, volume I. Cleveland, OH.: CRC Press Inc.
4. Cheremisinoff, P. N., and Young, R. 1977. Air Pollution and Control
Design Handbook, part 2. NY., NY.: Marcel Dekker, Inc.
5. Environmental Protection Agency (EPA). 1973. Air Pollution Engineering
Manual. AP-40. 2nd ed. RTP, NC.
6. Treybal, R. E. 1968. Mass Transfer Operations. NY., NY.: McGraw Hill
Book Co.
7. Zenz, F. A. 1972. Designing Gas-Absorption Towers. Chem. Engr.
79:120-138.
6-1
-------
REVIEW OF
ABSORPTION
DESIGN
PRINCIPLES
FACTORS
AFFECTING SOLUBIUTY
• Temperature • Presiure
ABSORBERS
Packed Tower Hate Tower Spray Tower
OPERATING
CRITERIA
Inlet Liquid -,
•^-*
• competition •
Inlet Cat
• Now ulr
• cwnpotikon
• MfflpCfllU'f
DESIGN REVIEW
• Liquid flow (L/C ratio)
• Tower Diameter
• Height of Packing
6-2
-------
UQUID
ROW
RATE
(L/C ratio)
-
MATERIAL
BALANCE
Go,
•^
Y,
IN = OUT
Lm2X2= Cm2Y2 4-
Where:
Cm ~ g*s nuss flow rate (Ib moles min)
Lm m Equid mass flow rate (Ib moles /min)
Rewriting the equation, we get:
Cm(Y,-Y2) = Lm(X,-X2>
Y,-Y2 =lni(X1-X2)
Cm
SYSTEM OPERATING LINE
6-3
-------
EQUILIBRIUM DIAGRAM
.E V,
mote fraction in Bquid
MINIMUM OPERATING UNE
I
mole fraction in liquid
ACTUAL OPERATING LINE
Y2 - .003
EXAMPLE
PROBLEM 3.3
Xa - 0.0
Y, - .03
fQUUBRIUM DIACRAM SOj-HjO
6-4
-------
0 .0002 JXXW .0006 .0008
.0)
.02
sn
A
0 0002 XKHM .0006 .0006
PACKED
TOWER
DIAMETER
Packed Tower diameter
is dependent on:
• gas flow rate
• liquid flow rate
• tower internal design
LOADING
6-5
-------
£ FLOODING
GENERALIZED ROOWNC CORRELATION
FOR DUMPED PttCES
SUPERRQAL CAS MASS ROW RATE
G' =
C' expressed in Ib/min-ft2
CROSS ' SECTIONAL AREA
A =
1C'
Where:
( m operating percent of flooding
C * actual gas ma» flow rate (Ib min)
C' — wperficial gas mau flow rate (Ib min-fl }
TOWER DIAMETER
dt =
= 1.13 A
6-6
-------
EXAMPLE PROBLEM 33
(continued)
Oven:
«„,
434J
754
minulr
mnutr
3
C| - 62.4 b ft
Cg - .0732 Ib ft3
For HjO: - 10 H * -8 centipoise
CINERAUZED ROOWNC CORRELATION
FOft DUMPED PIECES
HEIGHT
OF
PACKING
PACKING
HEIGHT
Z - NTU x HTU
NIL • numtwt ol B»mlrr unlt^
HTU1 • htighl pff tundvt unn
PACKING HEIGHT
2 « NTU X HTU
Z - NOG x HOC
Z • NOI x HOI
CK; • ovcnll gis
Ol - ovcraliquid
6-7
-------
Chapter 7
Absorption Control Systems
Lesson Goal
To familiarize you with contaminent and absorbent characteristics that affect
mass transfer, and to describe the various types of absorption equipment used to
control gaseous pollutants.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. List at least four characteristics of the contaminent gas stream that affect
absorber performance.
2. List at least six characteristics considered desirable in an absorbent.
3. Describe the operation of the following types of absorbers.
a. Spray or baffle tower
b. Centrifugal scrubber
c. Venturi scrubber
d. Plate tower
e. Packed tower
References
1. APTI Course 415 Student Manual.
2. Bethea, R. M. 1978. Air Pollution Control Technology, An Engineering
Analysis Point of View. NY., NY.: Van Nostrand Reinhold Co.
3. Cheremisinoff, P. N., and Young, R. 1977. Air Pollution and Control
Design Handbook, part 2. NY., NY.: Marcel Dekker, Inc.
4. Environmental Protection Agency (EPA). 1973. Air Pollution Engineering
Manual. AP-40. 2nd ed. RTP, NC.
7-1
-------
*-
ABSORPTION
CONTROL
SYSTEMS
FUNCTIONS OF
AN ABSORBER
• removal of
contaminants
• pretreatment
CHARACTERISTICS OF
CONTAMINANTS
• concentration
— _t • temperature
• chemical properties
• solubility
• toucity
DESIRABLE CHARACTERISTICS
44 OF AN ABSORBENT
•
i 4 * k**1 &°'u^'ty'°' **
. j contaminant
4 |k • low volatility
• low vtscoiity
• chemical ttabiity
CONTAa METHODS
7-2
-------
TYPES OF ABSORBERS
• Spray Towers
• Cydonk Scrubbers
• Venturi Scrubbers
• Plate Towers
• Packed Towers
• Moving Bed Scrubbers
SPRAY
TOWER
BAFaE SPRAY
TOWER
CYCLONIC
SCRUBBER
VENTURI
SCRUBBER J
-------
SPRAY
VENTURI
SCRUBBER
WETTED
APPROACH
VENTURI
cydone
ADJUSTABLE
THROAT
VENTURI
VENTW-ROD SCRUBBER
PLATE
TOWER
7-4
-------
TYPES OF PLATES
Sieve Tray
Impingement Tray
TYPES OF PLATES
.
Bubble Cap
Roat Valve
PACKED TOWER
">*cto« (counlercurrent)
Iwdgas
lo*»in
PACKED TOWER
(cocurrent)
PACKED TOWER
(cross flow)
7-5
-------
PACKED TOWER
(cross flow-3 beds)
COMMON I j ,^V"l*u
TOWER ^ - Serf Saddle
PACKING toichi8Rin8 f^
MATERIALS fe£^ --'~~';
*t\^s- biUlok Saddle
' 7-T
Pafl Ring
Tefleretle
CONRGURATION OF PACKING
PIATE
TOWER
lower wt^h!
handkt Urg»
v*f atom bM«r
W PACKED
j^j^ TOWER
• lower Ap
• belWf innnaww
pntett»d to to»m>
aOODED
BED
ABSORBER
7-6
-------
R.UICXZED
ABSORBER
h«4|M
MKTEIJMINATOR
• prevents carryover of liquid droplets
• recovers scrubbing liquor
• performs additional scrubbing
MIST EUMINATORS
Wire Mesh *S" or Chevron Curve
REVIEW
Characteristic* of ConUminant ds Streams
Desirable Characteristics of an Absorbent
Types of Absorbers
• Spray • Plate
• Cydonk • Packed
• Venturi • Moving Bed
7-7
-------
Chapter 8
Adsorption Principles
Lesson Goal
To familiarize you with the basic principles of adsorption for the control of
gaseous emissions.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. Briefly define adsorption, adsorbate, and adsorbent.
2. List the three stages that occur during the mechanism of adsorption.
3. Describe the difference between physical attraction and chemical attraction
between adsorbent and adsorbate.
4. Name the major force, van der Waals, used in physical adsorption, and
describe the attraction of the gaseous molecules to the solid molecules.
5. List four types of materials used in adsorption and describe their features.
6. Name at least four factors that affect proper adsorption design.
7. Briefly define breakthrough, mass transfer zone, saturation, and
retentivity.
8. Name two types of adsorption systems.
9. Describe the adsorption process throughput cycle and the regeneration of
the adsorber bed.
References
1. APTI Course 415 Student Manual.
2. Bethea, R., Air Pollution Control Technology, 1978. NY., NY.: Van
Nostrand Reinhold Co.
3. Cerny, S., and Smirch, M. 1970. Active Carbon. NY., NY.: Elsevier
Publishing Corp.
4. Environmental Protection Agency (EPA). April 1973. Packing Sorption
Device System Study, EPA-R2-73-202. Washington, DC.
8-1
-------
- ADSORPTION
ADSORPTION
PHENOMENA
of charcoal
• • • general .
'. ; '- bulk area v
DIFFUSION
INTO
PORES ..V
8-2
-------
MONOLAYER
BUILDUP
HEAT OF
ADSORPTION
ADSORPTION
THEORY
VAN DER
WAAL
FORCE
molecule of
adsorbate
molecules in
adsorbent
wall
8-3
-------
CHARACTERISTICS
Chemlvorption
Phyclcal Absorption
.release* high heat
80-120 calorie* mole
• form* • chemical
compound
•deeorption it difficult
• ImpoMlble adtorbate
recovery
• release* low energy
40 calorie*/mole
• dipolar Interaction
• deaorptlon it ea*y
• «aty adcorbate
recovery
TYPES OF ADSORBENT
Polar
Non-polar
• silica gel
• activated oxides
•molecular sieves
• activated carbon
RETENTIVITY
Temperature
8-4
-------
c
if
Pressure
t,
DC
Surface Area /Pore Size
-
I
!
J
Molecular Weight of Solvent
ADSORPTION
SYSTEMS
nonregenerative
regenerative
.
!J
i1
trough Cyrv* —~>
c.
-Brrikpuini
• Vulvnir o( Illlurni ltc«lrd-
8-5
-------
2 BED
ADSORPTION
SYSTEM
vapor Inlet
2 BED
ADSORPTION
SYSTEM
vapor Inlet
mh«u»t
vcpor Inlet
REGENERATING A
BED A
Inlet ctcam ->
exhaust
METHODS
OF REGENERATION
• stripping
• thermal desorption
• vacuum desorption
8-6
-------
Chapter 9
Adsorption Dynamics
Lesson Goal
To explain, in terms of adsorption principles, the dynamic adsorption system and
the criteria for evaluating its effectiveness.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. Briefly define the following.
a. Isotherm
b. Isobar
c. Isotere
2. Distinguish between saturation and working capacities of the adsorber bed.
3. Explain how the MTZ, depth of bed, and gas velocity are determined for
adsorption systems.
4. List three factors affecting regeneration that should be considered when
reviewing the design of an adsorption system.
a. Time needed to regenerate
b. Amount of stream required
c. Drying and cooling systems provided
References
1. APTI Course 415 Student Manual.
2. Cheremisnoff, P. N., and Ellerbusch, F. 1978. Carbon Adsorption Hand-
book. Ann Arbor, MI. Ann Arbor Science Publishers, Inc.
3. Environmental Protection Agency (EPA). April 1973. Packing Sorption
Device System Study. Washington, DC. EPA-R2-73-202.
4. Parmele, C. S., O'Connell, W. L., and Basdehis, H. S. Vapor Phase
Adsorption Cuts Pollution, Recovers Solvent. Chem. Engr.: Dec. 31, 1979.
5. Stern, Arthur C. 1977. Air Pollution 3rd ed., volume IV NY., NY.:
Academic Press, Inc.
9-1
-------
ADSORPTION
DYNAMICS
ADSORBER OPERATION
ADSORPTION
EQUILIBRIUM CONCEPTS
• Isotherm
• Isobar
• Isostere
ISOTHERM
Vapor Pressure
TYPICAL ADSORPTION
ISOTHERM
(Ml •«' •'
Partial Prrnwr*.
9-2
-------
POLANYI EQUATION
LANGMUIR
(BET)
EQUATION
1 C-l P
^ v^c-p;
ISOBAR
e
<
mm Hg
Temperature
ADSORPTION
ISOBAR FOR H,S MOLECULAR SIEVE
- too
1000
1 mm
Hg
0 35 15
ISOSTERE
i;
Weight X
Reciprocal Temperature
9-3
-------
ADSORPTION
ISOSTERE
OF H,S ON
MOLECULAR
SIEVE
25 27 O 3 I 33 3537
RrriprocAl of Abtoiui* Trmpcraiur* 10
DYNAMIC
ADSORBER OPERATION
• MTZ
• saturation and working capacity
• depth of bed
• gas velocity through bed
MASS
TRANSFER
ZONE
MTZ depends on:
• type of adsorbent
• particle size of adsorbent
• gas velocity
• temperature
• pressure
• efficiency
MASS TRANSFER ZONE
Where
D, • twd depth
C, » breakthrough capacity of bed D,
C . - Mluratlon capacity
X - de«ree of saturation In the MTZ
9-4
-------
MASS TRANSFER ZONE
= Ci|V-C1D,L
Where:
C « breakthrough capacity
D - bed depth
ADSORBER BED CAPACITY
Saturation Breakthrough Working
Capacity Capacity Capacity
SATURATION
AND WORKING CAPACITY
Saturation
Capacity
weight of *d*orb*(r
weight of «dtorb*ni
Working _ wgighi of »d
Capacity
i of hc«l
weight of •dtortxm
Br*«kthrough
Capacity
«/!»' W •"> \ P.I....INH.W »*d _ "17 \
'"" I.I...H, Ml,l,«lh» \ 1 .,....,, /\l).nrh |n>(IAl
•rd UXIII
*5.«
DEPTH OF
? ADSORBENT
BED
9-5
-------
EFFECT OF GAS VELOCITY ON MTZ
Gas Velocity, feet/minute
GAS
VELOCITY
THROUGH
BED
0 IOO
Velocity. fe«t minute
REGENERATION
CONDITIONS
• types of regeneration .
• factors to be considered
Ways to Regenerate
Thermal Detorption at 100 C
Vacuum Detorption at 20 C
and 50 mm Hg
Gat Circulation at 130 C
Steam Stripping at 100 C
VlU 411 li.* >..i
I Percent of
Charge DitpeJIed
15
25
45
98
|
FACTORS
AFFECTING DESORPTION
• time available
• retentivity of adsorbent
• heat requirements of vessel
and adsorbent
• direction of steam flow
• hysteresis
9-6
-------
D«»orption/Y
4^ "Adsorption
HYSTERESIS
PHENOMENON
Adsorbate Pressure
Arrton*
1
- STEAM
" REQUIRE-
- MENTS
0 I 1 J « s »
Pound* o) St»m p»i Pound ol &olv*tii
ADDITIONAL FACTORS
• drying and cooling of bed
• heat of adsorption
• fire hazards
• attrition
9-7
-------
Chapter 10
Adsorption Control Equipment
Lesson Goal
To review factors affecting the adsorption process and to describe the operation
of various types of adsorbers.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. List five factors which affect adsorber performance.
a. Temperature
b. Pressure
c. Concentration
d. Gas velocity
e. Contaminants
2. Describe the operation of three types of adsorbers.
a. Fixed (regenerate and nonregenerable)
b. Rotary
c. Fluidized
3. List three source categories to which adsorbers could be applied for
gaseous emission control and list some typical installations related to each.
a. Bulk terminals
b. Solvent recovery
c. Odor control
References
1. APTI Course 415 Student Manual.
2. Environmental Protection Agency (EPA). April 1973. Packing Sorption
Device System Study. Washington, DC. EPA-R2-73-202.
3. Parmele, C. S., O'Connell, W. L., and Basdehis, H. S. Vapor Phase
Adsorption Cuts Pollution, Recovers Solvent, Chem. Engr.: Dec. 31, 1979.
4. Technical Bulletin, "Carbon Adsorption/Emission Control" Vic Manufac-
turing, Minneapolis, MN.
5. Technical Bulletin, "Purasiv HR for Hydrocarbon Recovery", Union Car-
bide, NY., NY.
10-1
-------
ADSORPTION
CONTROL EQUIPMENT
FACTORS AFFECTING ADSORPTION
• Adsorbat* - Gat Stream
• Concentration
• Temperature
• Pret»ure
• Velocity
• Contaminant*
• Adsorbent
• Retentfvtty
• Deactivation
• Mechanical Stability
-\
CONCENTRATION
Concentration
Concentration
TEMPERATURE
Temperature
120 r
PRESSURE
Pressure
10-2
-------
VELOCITY
N
"5
£.
I
Velocity
100 It
CONTAMINANTS
• paniculate matter
• entrained liquid
• high humidity
• corrosive materials
• high boiling point hydrocarbons
RETENTIVITY
• high selectivity
for contaminant
• easy regeneration
DEACTIVATION
MECHANICAL
STABILITY
10-3
-------
TYPES OF
ADSORPTION
SYSTEM
• Nonregenerable
• Regenerable
NONREGENERABLE
USES OF
NONREGENERABLE
ADSORBERS
• control of odor
• control of
trace contaminants
ADVANTAGES
DISADVANTAGES
low Ap
inexpensive
unit
replacement
cost
no recovery
credit
10-4
-------
REGENERABLE
ADSORBERS
Fixed Bed
REGENERABLE
Fixed Bed
USES OF FIXED BED
REGENERABLE
ADSORBERS
• Solvent Recovery
• Gasoline Marketing
• Odor Control
ADVANTAGES
(over nonregenerable)
• reuse of carbon
• potential recovery value
DISADVANTAGE
• Kile
carbon
10-5
-------
REGENERABLE
FJuidized B«d
USES OF FLUIDIZED BED
REGENERABLE ADSORBER
• Solvent Recovery
• degreasing
• printing
• surface coating
ADVANTAGES
DISADVANTAGE
• increased
efficiency of
regeneration
• eliminates idle
carbon
• smaller than
comparable fixed
bed
• high attrition of
carbon
ROTARYBED
rurmtmm riw »*« \7.• ...•...-
CtaMi In
•>:•»*&
10-6
-------
USES OF ROTARY BED
REGENERABLE ADSORBERS
• Odor Control
• Solvent Recovery
ADVANTAGES
DISADVANTAGE
• eliminates idle
carbon
• shorter than
fixed bed
• wear on moving
parts
TO ACHIEVE
HIGH EFFICIENCY
• allow for cooling.'drying cycle
• be able to handle varying
concentrations
• monitor the outlet concentration
• provide additional regeneration
capabilities
10-7
-------
Chapter 11
Condensation Principles
and Applications
Lesson Goal
To familiarize you with the basic types of condensers used for control of gaseous
air pollutants from industrial sources.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. List two types of condensers and briefly describe their operation.
2. Recognize some advantages/disadvantages of using either a surface or con-
tact condenser.
3. Recall three applications of condensers for control of gaseous emissions.
4. Recall the steps in reviewing permits for construction and operation of
condensers as air pollution control devices.
References
1. APTI Course 415 Student Manual.
2. Environmental Protection Agency (EPA). 1973. Air Pollution Engineering
Manual. AP-40, 2nd ed. RTP, NC.
3. Perry, John H. 1950. Chemical Engineering Handbook. New York:
McGraw-Hill Book Company Inc.
11-1
-------
CONDENSATION
PRINCIPLES AND
APPLICATIONS
TYPES OF
CONDENSERS
• Contact
• Surface
non-condematc
SPRAY TYPE
CONTACT
CONDENSER
condtnute
SHELL-AND-TUBE CONDENSER
•Mtof
eondcnul*
Surface Condensers
Contact Condensers
• less coolant required
• fes» compensate produced
• product easily recovered
• no separation problem
• simpler
• lest eipensive
• less maintenance
required
• separation problem
(coolant and pollutant)
11-2
-------
USES OF CONDENSERS
IN A CONTROL SYSTEM
• Vapor Reduction
• Final Control Device
• Odor Control
TYPICAL APPLICATIONS
OF CONDENSERS
• Rendering Plants
• Degreasing Operations
• Petro-chemical Industry
RENDERING PLANTS
• Odor Control
• cookers
• dryers
• grease processing
• raw materials
DECREASING OPERATIONS
Solvent Vapor Containment and Recovery
Ute condensation coils
PETRO-CHEMICAL INDUSTRY
• Vacuum Dfctilalion of Petroleum
• Vapor Recovery in Bulk Terminals
11-3
-------
SIZING OF A CONTACT CONDENSER
Calculate condensation
rale:
• balance heat
of vaporisation
• balance compensate
MibcooKng
SIZING OF A SURFACE CONDENSER
Calculate heat transfer
REVIEW OF A DESIGN PLAN
• Examine construction permit
• Review vendor literature
• Check process variables
• Confirm source compliance -
• Calculate outlet emissions
• Require source lest
REVIEW
• Types of Condensers
• Contact
• SurUcr
• Advantages of Each Type
• Typical Applications
• Rendering PUnis
• D«g'M»
-------
Chapter 12
Reduction of SO2 Emissions
from Fossil-Fuel Burning Sources
Lesson Goal
To familiarize you with the characteristics and uses of flue gas desulfurization
(FGD) systems.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. Recall the NSPS for sulfur oxide emissions from fossil-fuel-fired steam
generators (FFFSG).
2. List three types of nonregenerable FGD processes and give a brief descrip-
tion of each.
3. List three types of regenerable FGD systems and give a brief description of
each.
4. Describe the use of dry-scrubbing FGD systems with baghouses for
SO2 control.
5. List three processes or fuel modifications used to reduce SO, emissions.
References
1. Jones, D. G., Hargrove, O. W. Lime/Limestone Scrubber Operation and
Control. Presented at the 72nd annual meeting of the Air Pollution Con-
trol Association. Cincinnati, OH. June 24-29, 1979. 79-23.6.
2. Hollett, Grant T., Jr. Dry Removal of SO* Application to Industrial Coal
Fired Boilers. Presented at the 72nd annual meeting of the Air Pollution
Control Association, Cincinnati, OH, June 24-29, 1979. 79-23.1.
3. Hatfield, Dennis L., Harris, James L., Steiner, Thomas P., Silk, Charles
T. Six Years' Operation and Maintenance Experience with a Wellman-
Lord SOa Abatement System at a Sulfuric Acid Plant. Presented at the
72nd annual meeting of the Air Pollution Control Association. Cincinnati,
OH. June 24-29, 1979. 79-23.3.
4. Electric Power Research Institute (EPRI). 1980. Comparible Economics of
Advanced Regenerable Flue Gas Desulfurization Process. EPRI CO-1381.
5. Environmental Protection Agency (EPA). 1979. Definitive SO, Control
Process Evaluations: Limestone, Double Alkali, and Citrate FCD. EPA
600/7-79-177.
12-1
-------
REDUCTION OF SO2 EMISSIONS
From Combustion Sources
Hue Gas DesuHurization (FGD)
NSPS FOR SO2
FFFSC rated > 250 x 10* Btu/hr
or 73 megawatts
New Sources after August 1971
• Equid gaseous fuel — 0.8 fc 10' Btu
or 340 ng J
• solid fuel — 12 I) 10* Btu
or 520 ng j
New Sources after September 1978
• liquid gaseous fuel — 0.8 Ib 10' Btu
or 340 ng J
and 90"u scrubbing
(N bekm 02 Ib 10' Btu, or 86 ng J,
then no scrubbing required)
New Sources after September 1978
(continued)
• sofidfuet —
Ufo 10'Btu
or 520 ng |
and 90°o scrubbing
0.6 fc/10* Btu
or 260 ng J
and 70°o scrubbing
FGD SCRUBBING
S02
Wet
so2
Dry
12-2
-------
WET FCD SCRUBBING
Nonregenerable Regenerable
TYPICAL FCD SCRUBBER EQUIPMENT
• Spray Chambers
• Venturi Scrubbers
• Packed Towers
• Cross How Scrubbers
SIMPLE
SPRAY
CHAMBER
SPRAY
VENTURI
SCRUBBER
.^rrMI HflMUIor
liquor
PACKED
TOWER
12-3
-------
CROSS FLOW SCRUBBER
SO2 REDUCTIONS
• Wet Scrubbing—at least 90%
• Dry Scrubbing—at least 75-83°o
NONREGENERABLE
PROCESSES
• Lime Scrubbing
• Limestone Scrubbing
• Double Alkali Scrubbing
SCRUBBER
WASTE DISPOSAL
• Ponding
• Mine Disposal
UME
SCRUBBING
12-4
-------
UME SCRUBBING
PROCESS CHEMISTRY
Ci(OH)j
(Sorry
fcomOO)
UME SCRUBBING
RELATIVE COSTS
(Nonregenerable Procestes)
E
Urne
LIMESTONE
SCRUBBING
12-5
-------
UMESTONE SCRUBBING
PROCESS CHEMISTRY
(Surey)
UMESTONE SCRUBBING
RELATIVE COSTS
{Nonregenerable Processes)
IJme IJmeftone
DOUBLE
ALKALI
SCRUBBING
12-6
-------
DOUBLE ALKALI SCRUBBING
PROCESS CHEMISTRY
DOUBLE A1KALJ SCRUBBING
Regenerated
AbUbent
RELATIVE COSTS
(Nonregenruble Processes)
Lime Limestone
REGENERABLE
PROCESSES
• Wellman-Lord/Allied
Chemkal
• Gtrate
• Magnesium Oxide
12-7
-------
TYPICAL
REGENERATION
PRODUCTS
• Sulfur
• SuKuricAcid
• Gypsum Wallboard
WELLMAN-iORD'
AIDED CHEMICAL
PROCESS
WELLMAN4.ORD ALLIED CHEMICAL
PROCESS CHEMISTRY
WELLMAN^ORD/
ALLIED CHEMICAL PROCESS
12-8
-------
RELATIVE COSTS
(Regenerabte Proceuet)
C
WellnurUord
AIM Chemical
CITRATE
PROCESS
CITRATE PROCESS CHEMISTRY
*7T5\
[H2S03
R. Spent
Gtratt.
Sulfur
CITRATE PROCESS
Paniculate
Miner Removed
4^4
EsBTYr '•
Abvorbent
^
Reaction
12-9
-------
RELATIVE COSTS
w (Regenerate Processes)
WHbnin-iofd Otute
AIM
MAGNESIUM
OXIDE
PROCESS
MAGNESIUM OXIDE
PROCESS CHEMISTRY
>5*~rtX V > t ; ,
SO, ^ + **
-------
RELATIVE COSTS
(Regenerable Processes)
Wrilnun-lord Gtule Mignwum
AIM Chenneil Oud*
J.
\Ji
DRYFCD
SCRUBBING
TYPICAL DRY PROCESSES
• Spray Dryer with Baghouse 'ESP
• Dry Injection
• Alkali and Coal Combustion
SPRAY DRYER WITH BAGHOUSE
Alkali Spray/
SO
ALKALI SPRAYS
• Sodium Bicarbonate
• NahcoTrte
• Lime
12-11
-------
SOj REDUCTIONS
• Dry Scrubbing—75-85°o
(90°« possible)
• Estimated Cost—$80-120 1W
TYPICAL DRY PROCESSES
• Spray Diyer with Baghouse/ESP
• Dry Injection
• Aloi and Coal Combustion
OTHER SO2 REDUCTION TECHNIQUES
• Fuel Substitution
• Fuel Cleaning
• Coal Gasification
• Coal liquificabon
12-12
-------
Chapter 13
Control of Nitrogen Oxide Emissions
from Fossil-Fuel Burning Sources
Lesson Goal
To introduce you to practices and equipment used to reduce nitrogen oxide emis-
sions from combustion sources.
Lesson Objectives
Upon completion of this lesson, you should be able to:
1. List the NSPS for nitrogen oxide emissions from major combustion
sources.
2. Briefly describe the various combustion modifications used for reducing
nitrogen oxide emissions.
3. Briefly describe the operation of the Exxon Thermal DE-NO, process for
nitrogen oxide emission reduction.
4. Briefly describe the -operation of the selective catalytic reduction (SCR)
process for nitrogen oxide reduction using ammonia.
5. Briefly describe the operation of the UOP-Shell process for NO, reduction.
References
1. Environmental Protection Agency (EPA). 1977. Proceedings of the Second
Stationary Source Combustion Symposium, volume II. Utility and Large
Industrial Boiler. EPA-600/7-77-073b.
2. Electric Power Research Institute (EPRI). 1979. Proceedings: Second NO,
Control Technology Seminar. EPRI/FP 1109-SR.
3. Mobley, J. David. Assessment of NO, Flue Gas Treatment Technology.
Paper presented at Symposium on Stationary Combustion NQ, Control,
USEPA and EPRI, October 6-9, 1980, Denver, CO.
13-1
-------
CONTROL OF NO*
EMISSIONS FROM
FOSSIL FUEL
BURNING SOURCES
METHODS OF REDUCTION
Combustion
Modifications
Flue Gas
Treatment
NSPS FOR NO*
FFFSC rated > 250 x 10' Btu hr
New Sources after August 1971
• gaseous fuel 0.2 Ib 10' Btu
• liquid fuel 0.3 Ib 10'Btu
• solid fuel 0.7 Ib 10» Btu
(except lignite)
NSPS FOR NO,
FFFSC rated at 73 megawatts
New Sources after August 1971
• gaseous fuel
• Squid fuel
• solid fuel
(except lignite)
Wng|
130 ng J
300 ngj
FFFSC rated > 330 * 10* Btu hr
New Sources after September 1978
• gaseous fuel OJ Ib 10' Btu
• Bquid fuel OJ Ib 10* Btu
• Mtbbituminous coil OJ Ib 10* Btu
• bituminous 'anthracite
coal, lignite 0.6 Ib 10* Blu
• KgnHe in slag top
furnace 0.8 b 10' Btu
13-2
-------
fFFSC rated at 73 megawatts
New Sources after September 1978
• gateous
-------
FLUE CAS RECIRCULATION
LOW NO> BURNER
tuf\ and
pnnur\ jir
FLUE CAS
ij TREATMENT
T
• Exxon Thermal De-NO^
• Selective Catalytic Reduction (SCR)
• UOP Shell Process
• Wet NO, /SO, Process
ammonia injection
EXXON
THERMAL
DE-NO*
THERMAL DE-NOX
PROCESS CHEMISTRY
4NH3 + 4NO + O2 — 4N2 + 6H2O
• Required Flue Gas Temperature — 950 °C
(H hydrogen injected — 700 °C)
13-4
-------
AMMONIA INJECTION SPRAY
MTO BOIUR
^
SELECTIVE CATALYTIC REDUCTION
(SCR)
MK> • 400 "C
!%«3ta£
jmmonu injection*
SCR PROCESS CHEMISTRY
4NH3 -(• 4NO + O2 S 4N2 + 6H2°
4NH,
lS'-c 3N2 + 6H2O
~-"v""
SHAPES OF
PARALLEL ROW
CATALYSTS
Ptullrl PUU-
o°o°o°c
TubuUr
frrrnrr
IJLJUJUU.
GOOOUUL
C«umk Honeycomb
UNIT CELL
OF A PARALLEL
aOW CATALYST
13-5
-------
TYPICAL
PARALLEL FLOW
CATALYTIC REACTOR
IT. ^Optimum Flue Gas Temperature
-c
REACTORS
fixed Bed Moving Bed Parallel Flow
UOP SHELL PROCESS
Simultaneous NO, SOt Reduction
NO, SO^ REDUCTION
CuO-f SO, 4 SO, — CuSO,
CATALYST REGENERATION
CuSO, 4- 2Hj — Cu4 SOj
Cu -«• !>O2 — CuO
13-6
-------
Relative Reduction and Cost
Thrmul SCR UOP Mfl
O*-NO,
NO, NO, NO, SOj NO, SOj
13-7
-------
Chapter 14
Exhaust Systems
Lesson Goal
To provide a review of the basic functions of the components of an exhaust
system.
Lesson Objective
Upon completion of this lesson, you should be able to:
1. Identify and locate the various pans of an exhaust system: hood, duct,
fan, and air pollution control equipment.
2. Define system efficiency as a function of the components of the exhaust
system.
3. Describe two types of hoods.
4. List at least three types of pressure losses that occur in a duct.
5. List two types of fans and briefly describe the characteristics of each.
6. Define and relate to each other the following terms pertaining to fans.
a. Air horsepower (AHP)
b. Brake horsepower (BHP)
c. Mechanical efficiency (£M)
7. Recognize the "Fan Laws" that are used for comparing similar fans.
References
1. APTI Course 415 Student Manual.
2. Environmental Protection Agency (EPA). 1973. Air Pollution Engineering
Manual. AP-40. 2nd ed. RTP, NC.
3. Baumerster, T. Fans. 1st ed. New York, NY.: McGraw-Hill Book Co.
14-1
-------
EXHAUST SYSTEMS
•xhoust dispersion
hood
source
twoim.ni
EXHAUST SYSTEM
SYSTEM
EFFICIENCY
E = E x E x E -
LES LH * LD x LAPCE
HOOD DESIGNS
i±J
\hopp»'/
Enclosed
Exterior
DESIRABLE HOOD
CHARACTERISTICS
• enclose process or source
if possible
• locate exterior hood in path
of exhoust
• with exterior hood, minimize
interference from cross drafts
14-2
-------
PRESSURE
MEASUREMENTS IN DUCT
oif flow
Totol
Pressure
Stotlc
Pressure
Velocity
Pressure
PRESSURE LOSSES
• inertia (velocity pressure)
• orifice
• straight run
• elbaw and branch entry
• contraction and expansion
DESIRABLE DUCT
CHARACTERISTICS
• minimize changes in
flow direction
• smooth duct surface
• ovoid abrupt expansions
AXIAL-
FLOW
FAN
CENTRIFUGAL
FAN
14-3
-------
PRESSURE
BLOWER
FAN
AIR HORSEPOWER
A un _ Q(cfm) x PT(in. water)
AMP ~ 6356
DRAKE
HORSEPOWER
DH- Q(cfm) x PT (in. woter)
6356 x fon mech. eff.
nooo
to
itock
control
•quipnwnt
tourc*
FORCED DRAFT
FAN (Dirty)
to
Slack
control
equipment
MMfC*
INDUCED DRAFT
FAN (Clean)
14-4
-------
FAN CHARACTERISTIC CURVES
s Velum* Flow
SYSTEM CURVE
Volume Flow
FAN CURVE VERSUS
SYSTEM CURVE
Fon
Performonce
Curve
Volume Flow
VARIABLES AFFECTING
FAN OPERATION
Air Volume (Q) and Horsepower (./)
• ton size (d)
• fan speed (N)
• gas density (C)
• system resistance (hf)
14-5
-------
FAN LAWS
are descriptions of the
relationships among the
variables
used to determine what
will happen (in terms of
fan performance) if a
given variable is changed
EXAMPLE PROBLEM
A fan is exhousting 12.000 efm of air
at 600°F (density = .0075 Ib/ft^). Fan
speed is 600 rpm and 13 horsepower
Is required. What would be the required
horsepower if oir ot 70°F (density =
.075 Ib/ft3) is pulled through the system?
SOLUTION
Using Fan Low 4c>
W* get
0.075
S 10 "r\ 0.0075
A = 26 hp
hood
EXHAUST
conrol SYSTEMS
equipment REVIEW
fon
stack
14-6
-------
Part 2
Problem Sets
-------
Problem Set 1
Review of Basics
Problem 1-1. Orsat Analysis
Problem Statement
From an Orsat analysis of a boiler's exhaust gas we get the following percentages
of components in the flue gas:
CO,
CO
(NOTE: Orsat gives % on a dry basis.)
Question
What is the molecular weight of this stack gas?
15-1
-------
Problem 1-2. Partial Pressure of Gases
Problem Statement
An air stream of 15,000 scfm contains 1% by volume water vapor and 1000 ppm
HaS.
Questions
1. What is the partial pressure of the water vapor and H2S?
2. If Henry's Law constant is 483 atm/mole fraction for H2S dissolved in
water, what is the maximum mole fraction of H:S that can be dissolved in
solution?
15-2
-------
Problem 1-3. Properties of a Gas
Problem Statement
Carbon dioxide gas (molecular weight = 44) flows through a duct that is three
meters in diameter. Assume the following to be true:
P* = l atm
p, = 0.1 atm, vacuum
T=150°F
R= (0-082 atnrQ(liters)
(g-mol)(K)
kinematic viscosity = 1.1 x 10~* mVs
velocity = —:——
s
Question
What are the values of the following?
a. density, Q
b. absolute viscosity, n
c. Reynolds Number, Re
15-3
-------
Problem Set 2
Combustion
Problem 2-1. Combustion of Gases
Problem Statement
Consider a gaseous fuel composed of:
N, = 5%
C,H,
C,H.
by volume.
Questions
1 . What is the volume of air required for complete combustion of 1 acfm of
the above fuel with 100% theoretical air?
2. What is the volume of the product of combustion of the fuel?
16-1
-------
Problem 2-2. Fuel Required
to Incinerate Waste Gases
Problem Statement
We have an exhaust air stream from a meat smokehouse that contains obnoxious
odors and fumes. The exhaust is 5000 acfm at 90°F and we want to incinerate
the fume at 1200°F.
Question
How much natural gas will be required if the gross heating value of the fuel is
1059 Btu/scf? NOTE: Use Figures 2-4 and 2-6 on pages 16-8 and 16-10,
respectively.
16-2
-------
Solution to Problem 2-2
Note: base all
calculations on
1 hour.
A. Must first find mass flow rate of air
m = (volume flow rate) x (density)
acfWlbmoleY 492 \( 29 Ib \/60
min/ \359 scf
21,6781b/hr
m= 5000
V-
0+90/\J
;
b mole/\
Note: AH values
are obtained from
Figure 2-4 on
page 16-8.
B. Heat required
Hat 1200°F = 288.5 Btu/lb
H at 90 °F is obtained by interpolating
Hat9°eF=(T^o)30=7-2Btu/lb
q = (21,678 Ib/hr) (288.5 -^--7.2 —}
\ Ib Ib /
= 6.098x10* Btu/hr
Note: can calcu-
late q another
way by using an
average specific
heat value.
C. =
Dt 1 1
0.26 -— ; average for air over the temp, range
q = (21 ,678 Ib/hr) o.26 (1200-90)
q = 6.256 x 106 '^- (more of an estimate)
hr
D. Heat available
from Figure 2-6 on page 16-10, HU for natural
gas with a He of 1059 Btu/cf at 1200°F
is 690 Btu/cf
Use heat from B.
£. Amount of natural gas needed
BtuV scf
5.098x10*
— R R^7 S
,690Btu/~8'837 h7
16-3
-------
Problem 2-3. Design of Afterburner
with Heat Recovery
Problem Statement
A two-bed heat recovery unit removes odors and fumes from a 10,000 acfm air
stream. The exhaust air enters at 200°F and is heated to 950°F in the preheater
before being combusted at 1400°F. The exit air leaves at 500°F.
Natural gas
Combustor
Bed on
preheat
To stack
Bed on
recovery
Contaminated air
Figure 2-1. Two-bed beat recovery incinerator.
Assume the following:
• C, = mean heat capacity of air = .26 Btu/lb°F
• HA » available heat of fuel « 950 Btu/scf
• velocity through combustion chamber = 20 ft/sec
" »5 ft3 of combustion products
ft3 of fuel
• there are no heat losses
• all calculations are to be based on a reference temperature of 60° F
• there is a minimum residence time of .3 sec
products of combustion
Questions
1. How much fuel is required?
2. What are the diameter and length of the combustion chamber?
3. What are the fuel savings by using heat recovery based on 5000 hrs/yr
operation at SI/1000 scf of gas?
16-4
-------
Problem 2-4. Plan Review
of a Direct-Flame Afterburner
Problem Statement
Plans have been submitted to your air pollution control agency for a permit to
construct a direct-flame afterburner serving a lithographer. As an engineer with
the agency, it is up to you to determine if the afterburner meets the following
equipment design standards required by your agency.
90Vo removal by:
a. afterburner temperature 1300-1500°F
b. residence time 0.3-0.5 seconds
c. velocity through afterburner 20-40 fps
The applicant has provided you with the following information needed to
evaluate the system:
300T
Effluent
from lithographer 7000 scfm 300°F
To process
•tm
Natural
I«s
burner
t
Direct
flame
afterburner
Afterburner
prebeater
738°F
Makeup air
Figure 2-2. Afterburner with beat recovery system.
Assume the following:
• afterburner dimensions are 4.2 ft diax 14 ft long
Btu
• gross heating value of natural gas-1059
ft3
hydrocarbon in effluent air to afterburner = 30 Ibs/hr (assume hydrocarbons
same as toluene)
heat loss from afterburner = 10%
molecular weight of toluene=92
LEL of toluene = 1.2%
refer to Figures 2-3, 2-5, and 2-6 on pages 16-7, 16-9, and 16-10, respectively
for natural gas assume 11.5 ft3 of the flue products per ft3 of gas burned and
10.3 ft1 of theoretical air required per ft3 of gas burned
16-5
-------
Questions
Is the concentration of toluene in effluent less than the required 25«7o of
the LEL?
How much fuel is required?
• use Cp values from Figure 2-5 on page 16-9
• use Figure 2-3 on page 16-7 for the heating value of toluene
• use Figure 2-6 on page 16-10 to compute the available heat of natural gas
Based on volumetric flow of the flue gases is the chamber size adequate?
• use Figure 2-3 for the products of combustion and theoretical air
required for burning of the toluene
16-6
-------
i
No. Subitonc*
1. Carbon*
2. Hydrooen
3 . O'yaen
4. Nitrogen (aim)
5. Corbon monoiidr
6. Corbon dionide
Paraffin teriet
7. Methane
8. (than*
9. Piapooe
10. n-luton*
II. liobuiane
12 . i»-P*nlon*
13, iiopenlane
UtJ*4Mu«lnHM
. ncopenrane
15. n-Heion*
Ote'in Mf i«i
16. flhyltne
17 . Prapylene
II. n-Bulene
19. liobulene
20. n-Peniene
Aromatic teriei
21 . lenient
22. toluene
23. Xylene
Ml(c*llaneou< gmei
24. Acetylene
75. Naphthalene
76. Methyl alcohol
?/.. Ethyl alcohol
78. Ammonia
29. Sulfur*
30. Hydrogen wl'iifc
31 Suite* d»onide
32.' Wote. Vapor
33. Air
Formula
C
H,
6,
CO
CO,
CHt
C,H,
C,H.
C«Hlfl
C«M|;
CfcH,,
CcH..
CSMM
C;H4
C|H|j
C.M,
C*HIO
C(»\
C,H,
C*H,0
:,H,
'HjOH
C,HSOH
NH,
»
l,S
so>
JT-J/
H,0
Molecular
u. • • .
Weight
12.01
2.016
37.000
78.016
78.01
44.01
16.041
30.067
44.0»?
58.111
58.118
77.144
72.144
77.144
86.169
28.0)1
47.07?
56.107
56.10?
70.178
71.107
97.13?
106.158
76.036
178.16?
37.041
46.067
17 .031
37.06
34.076
64 06
18.016
78.9
Ibper
Cu Ft
^
0.0053
0.0846
0.0/44
0.0/40
0.1170
0.0474
0.0803
0.1196
0.158?
0.158?
0.1904
0.1904
0.1904
0.7774
0.0746
O.IHO
0.1480
0.1480
0.185?
0.2060
0.7431
0.7803
0.0697
0.3384
0.0846
0.1716
0.0456
0.0911
1.1/33
).'o/66
CuFt
pef Ib
_
187.773
11.819
13.443
13.506
8.548
73.565
17.455
8.365
6.371
6.3?)
5.75?
5.25?
5.75?
4.398
13.41?
9.007
6.756
6.756
5.400
4.857
4.113
3.567
14.344
?.9i5
1 1 .870
8.721
71.914
10.979
5.7/0
I3.*063
SpG.
Air
1.0000
_
0.0696
1.1053
0.9718
0.96/7
1.578?
0.5543
1.0488
1.5617
7.0660
7.0665
7.48/7
7.487?
7.4A7?
7.9/04
0.9740
1.4504
1.9336
1 .9336
7.4190
7.6970
3.1/60
3.6618
0.9107
4.4708
I.IO'j?
I.5H90
0.5961
1.1898
7.7640
).67 l*j
1 .0000
Heat ol Combutlion
8lu pr< Cu (
CH.£',
mm
3?S
—
—
377
—
1013
1/9?
2590
33/0
3363
4016
4008
3993
4762
1614
7336
3084
3068
3836
3751
4484
5730
1499
5854
868
1600
441
647
~
-
Net
(low)
_
7/5
—
—
37?
—
913
1641
7385
3113
3105
3/09
3/16
3693
441?
1513
7186
7885
7A69
3')86
3601
4784
4980
1448
5654
/6R
1451
365
596
-
Blu per tb
(High)
14,093
61.100
—
—
4,347
-
73.879
77.370
71.661
71. JOB
71 ?5/
?l!iT9l
71.05?
70.9/0
70.940
71.644
71.041
70,840
70. /30
70.71?
11.710
18.440
18.650
71.500
I/. 798
10.759
13.161
9.668
3.983
/,im>
~
--
Net
(low)
14.093
51.673
—
—
4,347
—
71.520
70.43?
19.944
19.6(10
19,679
19.517
19,4/8
19.396
19.403
70.795
19.691
19.496
19,38?
19.363
17.480
17.670
17. /60
70.776
16. /OH
9,0/B
II.9J9
8.001
3.983
6.54'i
-
for 100% Total Alt
Mole< per Male ol Combmtrble
or
CuFl per Cu Ft of Comhuttible
Required lor Combuilion
Oj
1.0
0.5
_
_
0.5
—
7.0
3.5
5.0
6.S
6.5
8.0
8.0
8.0
9.5
3.0
4.5
6.0
6.0
7.5
7.5
9.0
10.5
7.5
12.0
1.5
3.0
0.75
1.0
1.5
-
N7
3./6
I.8R
_
_
1.88
—
7.53
13.18
18.8?
?4.4/
?4.47
30.11
30.11
30.11
35.76
11.79
16.94
??.59
77.59
78 .73
78.23
33.88
39.57
9.41
45.17
5.65
11.79
7.87
3.76
5.65
-
Air
4.76
7.38
_
_
7.38
—
9.53
16.68
73.8?
30.97
30.97
38.11
3D. II
38.11
45.76
14.79
71.44
78.59
79.59
3b./3
35.73
47.88
50.0?
11.91
57.17
/.I5
14.79
3.5/
4.76
7.15
-
Flue Product!
CO,
1.0
_
_
_
1.0
-
1.0
7.0
3.0
4.0
4.0
5.0
5.0
5.0
6.0
2.0
3.0
4.0
4.0
5.0
6.0
/.O
8.0
7.0
10.0
1.0
2.0
—
SP&
ro
-
H,O
_
1.0
_
_
—
—
2.0
3.0
4.0
5.0
5.0
6.0
6.0
6.0
7.0
2.0
3.0
4.0
4.0
5.0
3.0
4.0
5.0
1.0
4.0
2.0
3.0
1.5
1.0
_
—
NJ
3.76
1.88
_
_
1.88
—
7.53
13.18
18.82
74.47
24.4/
30.11
30.11
30.11
35.76
11.29
16.94
27.59
77.59
78.73
78.23
33.88
39.5?
9.41
45. 1/
5.65
11.79
3.3?
3.76
5.65
^^
—
For 100% Total AW
Ib par Ib ol CombuttibU
Required lor Combuttion
0,
7.66
7.94
_
_
0.57
-
3.99
3.73
3.63
3.58
3.58
3.55
3.55
3.55
3.53
3.4?
3.4?
3.4?
3.4?
3.4?
3.07
3.13
3.17
3.07
3.00
1.50
7.08
1.41
1.00
1.41
—
—
N,
8.86
26.41
_
_
1.90
—
13.78
12.39
12.07
11.91
11.91
11.11
11.81
11.81
11.74
11.39
11.39
11.39
11.39
11.39
10.27
10.40
10.53
10.27
9.97
4.98
6.93
4.69
3.29
4.69
^
—
Air
11.53
34.34
_
2.47
—
17.27
16.1?
15.70
15.49
15.49
15.35
15.35
15.35
15.77
14.11
14.81
14.11
14.81
14.81
13.30
13.53
13.70
13.30
17.96
.48
.07
.10
.79
.10
w
—
Flu* Product!
CO,
3.66
_
_
_
1.57
-
2.74
2.93
2.99
3.03
3.03
3.05
3.05
3.05
3.06
3.14
3.14
3.14
3.14
3.14
3.38
3.34
3.37
3.38
3.43
1.37
1.97
—
SOJL
7.00
1.88
_
—
H,O
_
8.94
_
_
_
—
2.25
1.80
.68
.SS
.55
.50
.50
.50
.46
.29
.79
.79
.79
.79
0.69
0.78
0.81
0.69
0.56
1.13
1.17
1.59
^
0.53
^
—
N.
8.86
76.41
_
_
1.90
—
13.78
12.39
12.07
11.91
11.91
11.81
11.81
11.81
11. 74
11.39
11.39
11.39
11.39
11.39
10.2?
10.40
10.53
10.22
9.97
4.98
6.93
5.51
3.29
4.69
_^
—
'Carbon and sulfur are considered as gases (or molal calculations only.
NOTE: This table is reprinted from Fuel Flue Gases, courtesy of American Gas Association. All gas volumes corrected to 60°F and 30 in. Hg dry.
-------
Temp
°F
60
100
200
300
400
500
600
700
800
900
1000
1200
1400
1600
1800
2000
2200
2400
2600
2800
3000
3200
3400
3600
Relative beat eootent (H) in Btu per pound (•( atmospheric pressure)
0,
0
8.8
30.9
53.3
76.2
99.4
123.1
147.2
171.7
196.6
221.7
272.5
324.3
377.3
430.7
484.0
539.3
594.4
649.0
702.8
758.6
816.4
873.4
931.0
N,
0
9.9
34.8
59.9
85.0
110.3
136.1
161.7
187.7
213.9
240.7
294.7
350.8
407.3
465.0
523.8
583.2
642.3
702.8
763.1
824.1
885.8
947.6
1010.3
Air
0
9.6
33.6
57.7
81.8
106.0
130.2
154.5
178.9
203.4
235.0
288.5
343.0
398.0
455.0
513.0
570.7
628.5
687.3
746.6
806.3
866.0
925.9
986.1
CO
0
10.0
34.9
59.9
85.0
110.6
136.3
162.4
188.7
215.6
242.7
297.8
354.3
407.5
465.3
523.8
583.3
643.0
703.2
771.3
832.6
894.0
956.0
1018.3
CO,
0
8.0
29.3
52.0
75.3
99.8
125.1
149.6
177.8
205.6
233.6
290.9
349.7
416.3
470.9
532.8
596.1
659.2
723.2
787.4
852.0
916.7
981.6
1047.3
SO,
0
5.9
21.4
37.5
54.4
71.8
89.8
108.2
127.0
146.1
165.5
205.1
245.4
286.4
327.8
369.1
411.1
452.7
495.2
557.5
580.0
622.5
665.0
707.5
H,
0
137
484
832
1182
1532
1882
2233
2584
2935
3291
4007
4729
5460
6198
6952
7717
8490
9272
10060
10870
11680
12510
13330
Ctt,
0
21.0
76.1
136.4
202.1
272.6
347.8
427.4
511.2
599.2
691.1
886.2
1094.1
1313.0
1542.6
t • •
* • •
...
• • •
...
• • •
H,O
0
• • •
1165
1212
1259
1307
1355
1404
1454
1505
1609
1717
1829
• • *
• * •
...
Figure 2*4. Heat contents of various gases.
Source: North American Combustion Handbook, North American Manufacturing Co., Cleveland, OH
1st ed. (1952).
16-8
-------
t m «F C, = Btu/flb
t
0
200
400
600
800
1000
1200
1400
1600
1800
2000
2200
2400
2600
2800
3000
3200
3400
3600
3800
4000
4200
4400
4600
4800
N,
6.94
6.96
6.98
7.02
7.08
7.15
7.23
7.31
7.39
7.46
7.53
7.60
7.66
7.72
7.78
7.83
7.87
7.92
7.96
8.00
8.04
8.07
8.10
8.13
8.16
0,
6.92
7.03
7.14
7.26
7.39
7.51
7.62
7.71
7.80
7.88
7.96
8.02
8.08
8.14
8.19
8.24
8.29
8.34
8.38
8.42
8.46
8.50
8.54
8.58
8.62
H,0
7.93
8.04
8.13
8.25
8.39
8.54
8.69
8.85
9.01
9.17
9.33
9.48
9.64
9.79
9.93
10.07
10.20
10.32
10.44
10.56
10.67
10.78
10.88
10.97
11.08
CO,
8.50
9.00
9.52
9.97
10.37
10.72
11.02
11.29
11.53
11.75
11.94
12.12
12.28
12.42
12.55
12.67
12.79
12.89
12.98
13.08
13.16
13.23
13.31
13.38
13.44
H,
6.86
6.89
6.93
6.95
6.97
t
6.98
7.01
7.03
7.07
7.10
7.15
7.20
7.24
7.28
7.33
7.38
7.43
7.48
7.53
7.57
7.62
7.66
7.70
7.75
7.79
CO
6.92
6.%
7.00
7.05
7.13
7.21
7.30
7.38
7.47
7.55
7.62
7.68
7,75
7.80
7.86
7.91
7.95
8.00
8.04
8.08
8.11
8.14
8.18
8.20
8.23
CH«
8.25
8.42
9.33
10.00
10.72
11.45
12.13
12.78
13.38
...
...
• ••
• •*
* • •
• • •
...
...
• . .
...
...
...
SO,
9.9
10.0
10.3
10.6
10.9
11.2
11.4
11.7
11.8
12.0
12.1
12.2
12.3
12.4
12.5
12.5
• • •
• • •
• • •
...
...
...
...
...
NH,
8.80
8.85
9.05
9.40
9.75
10.06
10.43
10.77
• • •
...
• • •
• • •
* • •
• * •
• * •
...
• • •
• • •
• ••
...
• • •
• • •
• • •
• • *
...
HO
6.92
6.%
7.01
7.05
7.10
7.15
7.19
7.24
7.29
7.33
7.38
7.43
7.47
7.52
7.57
7.61
* * •
• • •
...
...
• it
• ••
..*
...
"••
NO
7.1
7.2
7.2
7.3
7.3
7.4
7.5
7.6
7.7
7.7
7.8
7.8
7.9
8.0
8.0
8.1
...
...
...
...
• •*
• ..
• .*
• *•
• ••
AIR
6.94
6.97
7.01
7.07
7.15
7.23
7.31
7.39
7.48
7.55
7.62
7.69
7.75
7.81
7.86
7.92
7.%
8.01
8.05
8.09
8.13
8.16
8.19
8.22
8.26
Figure 2-5. Mean anolal heat capacities of gases above 0°F*.
•Williams, E. T. and Johnson, R. C. 1958. Stoichiometry for Chemical Engineers New York: McGraw-Hill
Book Company, p. 321.
-------
140,000
120,000
f 100,000
Tt
" 80,000
£
•
2 60,000
I
£ 40,000
•
3,000
2,400
T 1.800
| 1,200
600
Heavy fuel oil 14° API
152,000 Btu/gal
Light fuel oil 36.5° API
138,000 Btu/gal
Commercial butane
3210 Btu/cu ft
Commercial propane
2558 Btu/cu ft
Natural gas
1059 Btu/cu ft
I
I
I
I
300
2700
900 1500 2100
Flue gas exit temperature °F
Figure 2<6. Available beats for some typical fuels (referred to 60°F).
Source: North American Manufacturing. 1978. "North American Combustion Handbook" 2nd
edition. Cleveland, Ohio.
16-10
-------
Problem Set 3
Absorption
Problem 3-1. Equilibrium Diagram
Problem Statement
The following data are from McCabe, W. L., and Smith, J. C., 1956, "Unit
Operations of Chemical Engineering." McGraw Hill Book Co., NY., NY.
Page 496.
Weigh! of SO, per
100 weights of H,O
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Partial pressure
SO,, mm Hg
of
45
87
137
182
225
273
340
376
The data are given at 30°C and 1 atm.
Questions
1. Plot the equilibrium curve for this SOj-air-water system on the following
graph paper.
2. Does Henry's Law apply?
17-1
-------
I
_j
. i
I
\ -
17-3
-------
Problem 3-2. Packed Tower for H2S Removal
Problem Statement
The Blackgold Refinery has submitted plans for an H2S scrubber. Hydrogen
sulfide is to be removed from a waste air discharge by scrubbing with a
triethanolamine-water solution in a packed tower at atmospheric pressure.
Specifications indicate that the gas flow rate is 10,000 acfm at 70°F and contains
15,000 ppm H2S. The inlet liquid is to be solute free. They must reduce the H2S
to 500 ppm (250 on SO2 basis).
The pilot plant data indicate that Henry's Law applies (m = 2.0) and the
Hoc =1.94 ft.
The company proposes to install a 5-ft diameter, 10-ft high packed tower.
Questions
1. What is the minimum L/C (gal/1000 cfm) that you would approve?
2. If they operate at 80.65 Ib moles of scrubbing liquid per minute, is their
tower adequate?
17-5
-------
Problem 3-3. SO, Absorption by Water
Problem Statement
An exhaust stream of 3000 acfm is known to contain 3% SO, by volume. The
plans must reduce the SO* content by 90% and plans to do this by scrubbing
with water.
Assume the following:
• Henry's Law constant = 42.7 mole fraction SO» in gas/mole fraction SO2
in liq (from Problem 3-1)
Cf=62.41b/ft3
C, = 0.0732 lb/ft3
packing is 2-inch Intalox saddle
refer to Figures 3-1 and 3-4 on pages 17-15 and 17-18, respectively
temperature 30°C
Questions
1. Draw the material balance.
2. What is the liquid requirement in gal/min at 1.5 times the minimum L/C
ratio?
3. What is the column diameter at 15% of flooding?
4. Determine the number of transfer units.
5. Based on an Hoc of 2.72 ft, what is the total height of the tower?
17-6
-------
Solution to Problem 3-3
Part 1—Material Balance
NOTE: Y = mole
fraction for the
gas. Since we are
starting with pure
water and no
recycle Xs = 0
L-?
Y,-.003
\
X,-?
Y.-.03
Exhaust gas flow rate - 3000 acfm
Part 2—Determining Liquid Requirement
We can compute U, minimum two ways
1. graphically by plotting the inlet and outlet
concentrations on an equilibrium diagram from
Problem 3-1, or
" 2. by using Henry's Law and the equation for a
straight operating line as follows:
NOTE: that we
are only con-
cerned with the
very dilute end of
our equilibrium
diagram from
from Problem 3-1
A. Using Henry's Law
we know at bottom of tower Yi and X, are in
equilibrium
,42 ? mole fraction SO, in gas exajnplc)
mole fraction SO2 m liq
Y.-X,//
X.-Y,///
X, = (.03)742.7 = .000703
moles of
mole HjO
17-7
-------
B. Minimum U,/Gm ratio— now we want to compute
the slope of this line, this will give us the minimum
U./G*.
.03-. 003
U/0,-38.4
Y.-Y,
X, - X2 .000703-0
Ib mole water
Ib mole air
C. Operating L^/Gm ratio—absorbers are operated at
some liquid rate above the minimum—since at the
min. we hit the equilibrium line and absorption
stops.
Typical situation would be at 1.5 times minimum.
Therefore,
(U,/GW) operating = 1.5 (U./G,,) minimum.
(U/GJ op = (1.5)(38.4) = 57.6
NOTE:
398 cf
Ib mole
is a conversion
factor corrected
for temp.
D. Operating liquid requirement
our gas flow into absorber is:
Ib mole air
min
our liquid required is:
(U./GJ operating = 57.6
u./n.6JbjnoleHjOW7iJ4
\ Ib mole air / \
Ib mole HaO
Ib mole air
Ib mole air
mm
U,-434.3 Ib moles HjO/min
in gallons per minute that is:
Ib mole
gal/min - 434.3
min
/ 1» »b \( ga
\lb mole/V8.34
gal
Ib
gal/min =937.34
liquid requirement = 937.34 gal/min.
17-8
-------
Part 3—Column Diameter Determination
NOTE: density
from Perry's
Handbook
A. Given information:
434.3 lbmoleH'° C£=62.41b/ft'
G.-7.54
nun
Ib mole air
min
.07321b/ft3
B. Convert Ib moles to Ib
7.54 Ib mole air ( J9jb\ 218.7 Ib/min
nun Mb mole/
L-434.3
mm
Vlb mole
7817.4 Ib/min
C. Compute the abscissa
(L/G)/i.r-<^«r=i.225
D. Use Figure 3-4 from the 415 Student Workbook
and read the ordinate from the flood line
e (ordinate)-.019
E. Gas flow rate at flooding conditions from graph
G"F*"" =ordina,c=e
for water <£ = 1 .0 and p = .8 centipoise
if we use 2 inch ceramic Intalox saddles
F=40, from Fig 3-1 415 Student Workbook
G'*(40)Q.O)(.8)01
(.0732)(62.4)(32.2)
G' = .271b/ft'sec
019
F. Gas flow at operating conditions
assume we operate at 75 Vo of flooding conditions
G operating = (f)(G/,w^)
Ib
G operating = (.75)(.27) = .20
ft2 sec
17-9
-------
G. Cross sectional area of column
mass flow rate (Ibs/sec)
A =
operating flow rate Ibs/sec ft2
min
/ mm \
lin) \60sec)
A= (218.7 Ibs/min) \60sec,
.20 Ib/sec ft2
A =18.23 ft2
H. Column diameter
d,
/4A\ *
\»/
d, = 1.13 (18.23 ft2)* =4.8 ft
.*. need a 5-ft diameter tower
Column diameter = 5 ft
Part 4—Number of Transfer Units
A. From previous sections
v A ,- - mole fraction SO2 in air
„ Xa^O m = 42.7 - — - - : - — — : — — -
mole fraction SO3 m hq
Lm - 434.4 Ib mole H,O/min
7 .54 Ib mole air /min
Y,
Ya
.03
.003
B. Computing NOG
[•/Y.-mX,
VT _«_L\Yi-mXj
mG,
U
mO.
U.
(42.7)(7.54)
434.4
f .03-0
.003-0
(1-.740 +
.74ll
1-.741
NOTE: can use
Colburn diagram
to get same solution
17-10
-------
C. Total height of packed section
given Hoc « 2.72 ft from pilot studies of
air-SO2-water system
Z - Noc x HOG - (4.65)(2.72)
Z= 12.65 ft of packing
D. Graphic solution:
17-11
-------
Problem 3-4. SO, Absorption
by Dilute Alkaline Solution
Problem Statement
Assume the conditions from Problem 3-3 (i.e., a 3000 acfm air stream containing
3% SO,), but this time reduce the sulfur content by 90% by absorbing with a
dilute sodium hydroxide solution. (NOTE: This is a very rapid reaction.)
Questions
NOTE: Use any additional information supplied or derived from Problem 3-3.
1. What would be the number of transfer units needed to accomplish this?
2. From pilot plant studies we know the new operating L/C ratio is 5.0
gal/1000 cf. What diameter tower is needed if the company operates at
15% of flooding?
17-12
-------
Problem 3-5. Permit Review
of Ammonia Absorber
Problem Statement
Pollution Unlimited, Inc. has submitted plans for a packed ammonia scrubber on
a 1575 cfm air stream containing 2
-------
Problem 3-6. Spray Tower
Problem Statement
A steel pickling operation emits HC1 fumes (hydrochloric acid) of 300 ppm
average with peak values of 400 ppm 15Vo of the time. The air flow is a constant
10,000 cfm at 75°F and 1 atm pressure. Regulations limit emissions to no more
than 50 ppm HC1 at any time. Only sketchy information was submitted with the
scrubber permit application. The plans show a 14-foot tall, 9-foot diameter
counter-current water spray tower.
Assume the following:
• Gas velocity through the tower is not to exceed 3 ft/sec.
• For HC1, Henry's Law does not apply because HC1 is very soluble.
• In a spray tower, the number of transfer units (Noe) for the first or top
spray will be about 0.7. Each lower spray will have only about 60% of the
(Noc) of the spray above it. This is due to the mixing of liquids with that
from upper sprays and to the back mixing of liquids and gases. The final
spray, if placed in the inlet duct, adds no height and has an Noc of .5.
• The spray sections of a tower are normally spaced at three foot intervals.
Questions
Is the spray unit satisfactory? You will need to determine:
1. actual velocity through tower
2. Noc needed
3. number of spray sections to meet NOG
4. height of required tower
5. whether or not the spray unit is satisfactory
17-14
-------
Picking
Raschig rings
(ceramic and
porcelain)
Raschig rings
(steel)
Berl saddles
(ceramic and
porcelain)
Intralox saddles
(ceramic)
Pall rings
(plastic)
Size
'/i
i
V/i
2
3
'/i x 1/32
1 x 1/32
2x1/16
!/4
'/i
1
2
>/«
!4
1
2
5/8
1
2
Weight
Ib/ft3
52
44
42
38
34
77
40
38
55
54
48
38
54
45
44
42
7
5.5
4.5
Surface
area, a
(ftVff
packing
volume)
114
58
36
28
19
128
63
31
274
155
79
32
300
190
78
36
104
63
31
Percent
void
fraction
65
70
72
75
77
84
92
92
63
64
68
75
75
78
77
79
87
90
92
Picking
factor
(F)
580
155
95
65
37
300
115
57
900
240
110
45
725
200
98
40
97
52
25
Figure 3*1. Packing data.
17-15
-------
4.0
3.6
3.2
2.8
2.4
}>••
*-:
= 1.6
1.2
0.8
0.4
0
O—IVz" Raschig rings
—1" Tellerettes
flow Ib/hr-ft'
I
I
1500
2000
500 1000
Liquid rate Ib/hr-ft2
Figure 3-2. Ammonia*water absorption system.
Source: Perry, J. H. 1973. Chemical Engineers Handbook, 5th ed. NY.. NY. McGraw Hill.
17-16
-------
I
z
I
j
5 10 20 SO 100 200
Y.-mX,
Y2 - mX3
Figure 3-3. Colburn chart.
500 1,000
17-17
-------
Q/
0.5
0.2
0.1
0.05
0.02
0.01
0.005
0.002
0.001
I I
I I T
per
of
packing
0.01 0.02 0.05 0.1 0.2 0.5 1 2
L
10
(Oft
(dimensiqnless)
Figure 3-4. Generalized flooding and pressure drop correlation.
17-18
-------
Problem Set 4
Adsorption
Problem 4-1. Adsorption Working Capacity
Problem Statement
Assume the following data:
• 10,000 cfm air stream at 77°F
• atmospheric pressure
• 2000 ppm toluene
• 3000 ppm carbon tetrachloride
• 95Vo removal objective
• isotherms—Figures 4-1 and 4-4 on pages 18-8 and 18-9 respectively
• MW carbon tetrachloride = 154
• MW toluene = 92
Questions
1. How many pounds per hour of organics must be removed?
2. Assuming stream regeneration at 212°F, what is the working capacity?
3. Using vacuum regeneration, what is the working capacity?
NOTE: For rough estimates of working capacity, we can use the difference
between the saturation capacity (read from the isotherm) at adsorbing partial
pressure and the saturation capacity at regenerating partial pressure.
18-1
-------
Problem 4-2. Sizing an Adsorber
Problem Statement
Assume the information given in Problem 4-1 and the following data:
• Bulk density of carbon = 30 lb/ft3
• 90 fpm velocity through bed
• 16 hr/day operation
• steam regeneration at 212°F
• depth of bed cannot exceed 4 feet
• need 3031 Ibs of carbon to remove 939 Ibs of solvent per hour
Questions
1. What is the appropriate cycle time (based on bed depth limitation)?
2. What is the volume of carbon needed?
3. What is the bed diameter?
4. What is the bed depth?
18-?
-------
Problem 4-3. Benzene Adsorber Plan Review
Problem Statement
A solvent recovery system was designed to recover benzene from an air stream.
The company has plans to increase production, which would result in a 75%
increase in benzene that must be controlled. You are given the following data on
the present system:
Existing System—Ctrbon I
Carbon density is 23 lbs/fts
Carbon size is 4 x 6 mesh
Gas velocity is 100 fpm
Bed area is 120 ft2
Bed depth is 24 inches
Concentration of benzene is 1316 ppm
Temperature is 26°C
Carbon charge is 5200 Ibs
Working capacity is 36 Ibs benzene/100 Ibs carbon
Residual capacity is 2.2%
Cycle time is 10 hrs adsorbing, 2 hrs steaming and drying
Vapor pressure of benzene at 26°C is 100 mm Hg
MTZ is 2 inches
MW is 78.11
Refer to Figure 4-2 on page 18-8, Adsorption isotherm for benzene (use
the line for .Carbon I).
Questions
1. What is the new saturation capacity?
2. What is the new breakthrough capacity?
3. What is the new working capacity?
4. What is the cycle time?
5. Can the present system handle the increased benzene load?
18-3
-------
Problem 4-4. Adsorption Plan
Review—Gasoline Marketing
Problem Statement
A bulk terminal that transfers gasoline to service stations has submitted plans to
install a carbon adsorption unit to control emissions from the filling of their tank
trucks. You are given the following information:
• Average daily throughput is 120,000 gal/day of gasoline
• Maximum pumping rate is 2000 gal/min
• Air flow to adsorber is 350 cfm
• Diameter of adsorber is 4 ft
• Height of carbon is 3 ft
• 2 beds, regeneration cycle time of Vi hour for each bed
• Emission factor is 5 lbs/1000 gal loaded
• Emission limit is .67 lbs/1000 gal loaded
• MW of gasoline is 68 Ib/lb mole
• Ideal gas constant, R = .732 *tm/*'
Ib mole °R
• T is 70°F
• Carbon density is 30 lb/ft3
• Vacuum regeneration
Refer to Figure 4-3 on page 18-9, Adsorption isotherm for gasoline vapors
NOTE: Rules of Thumb
1. Velocity through adsorber should be between 20-100 fpm
2. Working charge of carbon may be estimated by doubling the amount of
carbon at saturation capacity. (Remember, the saturation capacity is the
ideal amount of solvent the carbon can hold and is read from the adsorp-
tion isotherm.)
Questions
1. What is the velocity through the adsorber?
2. How much carbon is required?
3. What is the bed depth?
4. Will the unit be in compliance?
18-4
-------
Problem 4-5. Rotogravure Printing Adsorber
Problem Statement
A printing company must reduce the amount of toluene they emit from their
Rotograve printing operation. The company comes in with some preliminary
information on installing a carbon adsorption system. You are given the fol-
lowing information:
Air flow is 20,000 cfm
They operate at 25% of LEL for toluene in the exit air
LEL for toluene is 1.2%
Toluene MW is 92.1 Ib/lb mole
Carbon density is 30 lb/ft3
Working charge is 30% of saturation capacity
Regeneration is just under 1 hour
Temperature is 77 °F
Maximum velocity through adsorber is 100 fpm
Refer to Figure 4-4 on page 18-9, Adsorption isotherm for toluene.
Questions
Determine the minimum size of adsorber that you would approve:
1. Diameter of the adsorber:
2. Square feet of carbon face area:
3. Depth of bed:
18-5
-------
Solution to Problem 4-5
(NOTE: all amounts are based on 1 hour because desorption takes only 1 hour)
NOTE: can point
out again that
saturation capacity
is idea] amount.
A. amount of carbon needed
1st calculate amount of toluene emitted per hour
(20,000 cfm)(25%)(1.2%)=*60 cfm of toluene
mole\/492°R 92.1 lb
/,n scf \ / w . w
\ mSr \359~s7f A53TR
/60 min\
\~hT"]
8461b/hr
in order to compute saturation capacity from Figure 4-4
on page 18-9, we need partial pressure of toluene
saturation capacity = 38% from Figure 4-4
working capacity is 30% of saturation
"•*»» toluene
(38%)(30%) =
or
100 lb carbon
amount of carbon needed for 1 hour cycle
/846 lb tolueneV 100 lb carbon \, j
\ hr All.41btolueneA
= 7,421 Ibs carbon/hr
Amount of carbon = 7,421 lb
B. volume occupied by the carbon
(7,421 Ibs carbon)(-?^-) = 247 ft3
>30 lb'
C. cross-sectional area of bed
20,000 (S-lf- =200
\m
i
100ft
based on 100 fpm limiting
NOTE: this requirement could be met by a horizontal
flow bed 10 ft wide and 20 ft long.
18-6
-------
D. depth of carbon bed
__ volume of carbon _ 247 ft3
cross sectional area 200 ft2
Depth of carbon bed-1.24 ft
1.24ft
Note that these calculations are based on minimum acceptable design condi-
tions using only two beds. In practical situation may opt for three bed
systems each handling only 10,000 cfm. This will decrease the diameter (cost)
of each individual unit and allow for more carbon length to ensure against
breakthrough.
18-7
-------
0.0001
0001
0.01
Partial pressure, psia
Figure 4-1. Adsorption isotherm for carbon tetrachloride.
f
r
.
*
\
100
26° C
10
1
0.1
Carbon J_
1
Percent relative saturation
Figure 4-2. Adsorption isotherm for benzene.
K)
18-8
-------
100
10
0,1
0.001
0.01 o.l
Psniil pressure, psia
Figure 4-3. Adsorption isotherm for (tioline vapors.
10
0.0001
0.001
0.01
Pmrtitl pressure, psii
Figure 4-4. Adsorption Isotherm for toluene.
0.1
18-9
-------
Problem Set 5
Condensation
Problem 5-1. Contact Condenser
Problem Statement
In an oil refinery, a stream of light hydrocarbons is to be condensed by a direct
contact condenser as shown in Figure 5-1. The light hydrocarbon stream is
essentially benzene. From Perry's Chemical Engineering Handbook you found
that for benzene: the boiling point is 175°F, the latent heat of vaporization (Hv)
is 160 Btu/lb, and the specific heat (C,) is 0.45 Btu/lb°F. Water is used as the
coolant at 60°F and C,= 1.0.
Questions
1. For a benzene mass flow rate (m) of 10,000 Ibs/hr, how much coolant (W)
is required if condensate temperature (T<) can be no higher than 100°F.
2. How much benzene is lost in the water if the solubility of benzene at
100°F is 0.05 lb/100 Ib of water.
19-1
-------
Water
Vapor
Recoverable
product
7*7)' ^ Noncondensibles
Waste water
Figure 5-1. Contact condenser.
19-2
-------
Solution to Problem 5-1
Note: there are
8.34 Ib/gal HaO
A. To calculate amount of water must set up a heat
balance
IN - OUT
. . . . heat required
heat required coo, heat lied
to condense + ,_ * u i;__
vapors
by water
to outlet
temperature
m x Hv+m x C, x (Tw - Te)« WC,(T« - Tw)
10,000 Ib w 160 Btu ^ 10,000 Ib „ .45 Btu
hr
Ib
hr
Btu
lb°F
x (175-100°F)«Wxl.0^(100-60)
160 xlCTBtu/hr-i-33.8 xlO4
Btu
40 Btu
hr Ib
W = 48,450 lb/hr-97 gal/min
Amount of coolant required*97 gal/min
B. Benzene lost in water
benzene _ .05 Ib benzene 48,450 Ib water
lost " 100 Ib water hr
Benzene lost in cooling water - 24 Ib/hr
>241b/hr
19-3
-------
Problem 5-2. Surface Condenser
Problem Statement
The surface condenser shown below is used to condense the hydrocarbon vapors
for the same conditions as in Problem 5-1.
V«por
Pore water
100°F
Pure condensate
Figure 5-2. Surface condenser.
The overall heat transfer coefficient (U) from Perry's Chemical Engineering
Handbook is 110 Btu/hr/ft3.
°F
Question
What is the surface area of the tubes required for the surface condenser?
19-4
-------
Solution to Problem 5-2
Note: °F is a
temperature
measurement
F° is a temperature
difference
To calculate surface area of tubes, use equation
A. Must first calculate the mean temperature change
175°F
60°F
i30er
100°F
(175-60)-(130-100)
fn(175-*°\
\i3o-ioo;
AT.
(115)-(30)
tn
115
30
62.2 F°
B. The heat required is
q = heat required to + heat required to
condense vapors subcool
q=rhH.+mC,(Tc,-Tca)
q- flO,000 *Wl<9Btu
\ hr/ \ hr
, 0,000
hr
q * 160 x 104 Btu/hr + 20.25 x 104 Btu/hr
q* 180.25 x 10* Btu/hr
19-5
-------
C. Surface area
A_ Q
UAT.,
A= 180.25 xlO4 Btu/hr
/110Btu/hr/ft2\ (62 2
= 263 ft2
Surface area of the condenser tubes = 263 ft2
19-6
-------
Appendix A
Common SI units
Qttithy (1)
length
area
volume
speed or velocity
acceleration
rotational frequency
mass (5)
density
force
movement of force (6)
pressure (or vacuum)
stress
viscosity (dynamic)
viscosity (kinematic)
energy, work, or
quantity of heat
power, or heat flow
rate
temperature, or tem-
perature interval
aOBM COBUBOB ••ItS
kilometer
meter
centimeter
millimeter
micrometer
aquare kilometer
square hectometer
square meter
square centimeter
square millimeter
cubic meter
cubic decimeter
cubic centimeter
meter per second
kilometer per hour (4)
meter per second squared
revolution per second
revolution per minute (4)
megagram
kilogram
gram
milligram
kilogram per cubic meter
kilonewton
Dcwton
newton meter
kilopascal
megapascal
niillip4ttCft] second (7)
square millimeter per
second (8)
joule (9)
kilowatt hour (10)
kilowatt
watt
kdvin
degree Celsius (11)
Symbol
km
m
cm
nun
km'
hm'
m'
cm'
nun*
m'
dm'
cm'
m/s
km/h
m/s*
r/s
r/min
Mg
kg
g
mg
kg/m'
kN
N
N.m
kPi
MPa
mPa*s
mm'/s
J
kW.h
kW
W
K
•c
•LQvnrvtBt Synbol
hectare (2) ha
liter (3) L
milliliter (3) mL
metric ton t
gram per liter g/L
kilowanhour kWh
NOTES
(1) Any measurable prop-
eny (such as length, area,
temperature) is called a
quantity. Listed in same
sequence as ISO 1000 and
ISO 31, except plane
angle.
(2) For land or water area
only.
(3) To be used only for
fluids (both gases and
liquids), and for dry ingre-
dients in recipes, or for
volumetric capacities. Do
not use any prefix with
liter except milti.
(4) The symbols for
minute, hour, and day are
min, h, and d,
respectively.
(5) Commonly called
weight,
(6) Torque or bending
movement.
(7) 1 mPa>s«l cP(cen-
tipoise, which is obsolete).
(g) 1 mm'/s -lcSt(cen-
tistokes, which is
obsolete).
(9) The unit-multiples
kitojoule (kJ) and mega-
joule (MJ) are also com-
monly used.
(10) To be abandoned
eventually. 1 kW.h-3.6
MJ.
(11) The degree mark * is
always used in *C to avoid
confusion with coulomb
(Q, but never with K for
kdvin.
Source: The American National Metric Council, Metric Editorial Guide, 3rd ed., January 1978.
20-1
-------
Appendix B
Conversion Factors
Length
1 inch = 2.54 cm
1 m = 3.048 ft
1 ft =.305m
Mass
1 lb = 453.6g
1 Ib « 7000 grains
1 kg = 2.21b
Pressure
1 atm=.101, 325 Pa
- 760 mm Hg (O'C)
>=14.7 psia
Force
1 N-l kgm/s2
1 N = 0.225 lb,
Energy
1 cal = 4.184 J
1 J = 9.48xlO-4Btu
1 Btu = 252.2 cal
Kinematic viscosity
1 mVS = 10* stokes
1 mVS = 3.875 104 ftVhr
Power
1 W=l J/S
1 W = 3.414 Btu/hr
1 W = 1.341xlO-3hp
1 hp = 33,479 Btu/hr
Area
1 m* = 10.764 ft2
1 cm2 = .155 in2
1 m2= 1.196 yd2
Volume
1 m3 = 35.31 ft3
1 cm3 = 0.061 in3
1 ms = 264gal(US)
1 ft3-28.317 L
1 barrel (oil) = 42 gal
1 ft3 = 7.48 gal
Density
1 kg/m3 = .06241b/ft3
Dynamic viscosity
1 Pa s(Nm/s)= 1000 cemipoise
1 cp = .000672 Ib/fts
Volume flow
1 m3/s = 35.3ft3/s
1 m'/min = 35.3 ftVmin
1 scfm = 1.7 mVh
1 gpm = 0.227 mVh
Velocity
1 m/s = 3.28ft/s
1 m/s= 196.85 ft/min
1 mi/hr = 0.447 m/s
Geometry
area of circle -rr2
circumference of circle=2 rr
surface area of sphere = 4 rr2
volume of sphere = 4/3 xr3
21-1
-------
Appendix C
Constants and Useful Information
Gas constants
R = 0.0821 atm liter/g mol K
= 83.14x 10* g cmVs2 g mol K
= 8.314Nm/gmol K
= 0.7302 atm ftVlb mol °R
1.987 g cal/g mol K or Btu/lb mol *R
Acceleration of gravity
g = 32.17 ft/sa = 980.7 cm/s1
Newton's conversion constant
gc=32.17 (Ib mass)(ft)/(lb force)(s2)
1 Ib mol = 359 ft3 of ideal gas at STP (32°F and 14.7 psia)
1 g mol = 22.4 L of ideal gas at STP (0°C and 760 mm Hg)
C, for water* 1 Btu/lb°R- 1 cal/g °C (at 20°C and 1 atm)
C, for air* .26 Btu/lb°R* .26 cal/g °C
viscosity of water 0*) = 1 cp = 0.01 g/cms (at 20°C and 1 atm)
density of air = 1.29 kg/m3 = 7.49 x l
-------
TECHNICAL REPORT DATA
(Please read Imamcrions on the-revene before completing!
NO
EPA 450/2-81-006
3. RECIPIENT'S ACCESSION NO.
4 TITLE AND SUBTITLE
APTI Course 415
Control of Gaseous Emissions
Student Workbook
6. REPORT DATE
June 1981
6. PERFORMING ORGANIZATION CODE
7 AUTMOR(S)
G.T. Joseph, D.S,
8. PERFORMING ORGANIZATION REPORT NO,
Beachler
9 PERFORMING OR'ANIZATION NAME AND ADDRESS
Northrop Services, Inc.
P.O. Box 12313
Research Triangle Park, NC 27709
10. PROGRAM ELEMENT NO.
B18A2C
11. CONTRACT/GRANT NO.
68-02-2374
12, SPONSORING AGENCY NAME AND ADDRESS
U.S. Environmental Protection Agency
Manpower and Technical Information Branch
Air Pollution Training Institute
Research Triangle Park, NC 27711
13. TYPE OF REPORT AND PERIOD COVERED
Student Workbook
14. SPONSORING AGENCY CODE
EPA-OANR-OAQPS
15. SUPPLEMENTARY NOTES
Project Officer for this publication is R.E. Townsend, MD-17, EPA-ERC, RTP, NC
This Workbook is designed for student use in APTI Course 415,
"Control of Gaseous Emissions." Its chapters correspond to each of the
14 lessons and 5 problem sets in the course. Each chapter contains lesson
goals and objectives, references, and printed versions of selected course
slides.
The Workbook is intended for use in conjunction with the Instructor's
Guide (EPA 450/2-81-004) and Student Manual (EPA 450/2-81-005) for APTI
Course 415.
7.
KEY WORDS AND DOCUMENT ANALYSIS
DESCRIPTORS
6.IDENTIFIERS/OPEN ENDED TERMS C. COSATI Field/Group
Air Pollution Training
Control of Gaseous Emissions
Training Course
Student Workbook
13B
51
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------- |