What Is Process

Design?

understanding, compmhangioa Of practical and

žimi.žažion.sa sense, abilily to do orzginaž and hard] ere
for a design engineer, they must be used in the
'Ypproceh to any design problem,
M".x S. Pelers, 1958

2.1 Introduction
In the the to plish goal. Throughout this text, we will the term
process design. to deicribù the GKn"1uence of steps from the
planning stage to the equipment specification stage of an air pollution
control project, The step* in the design sequence are the preliminary
problem definition, the final problem definition, and a series of
decision points that consist of" alternatives and their associated
subproblems. At each decision point, an engineer mt:st evaluate the
alternatives and choose the one that is the most technically and
economically feasible. The thorough evaluation of each alternative at
a decision point requires Che solution of ait of the subproblems
associated with that alternative.
We will illustrate the design sequence using the fo]lowing
hypothetical situation.
The Acme Corporatir.n produces for building indus• One
operatio:s iavc!oes the drying of c granular that cceEtai3ts
significant of fine dust, The exhaust from the dryer visible and
may exceed al-loužablc particulate emiË,sion
The preliminary problem in this situation is simply that the
exhaust is visible and company management is coneerned from both
76 Chapter Two What, Is Process Design?
acathetie and regulatory standpointa, The engineer assigned the
preliminary prob. lem initiates the design shown schematically in
Figure
75
Figure 2.1
78 Chapter Two What, Is Process Design?
Steps in the design process.
 Section 2.2 General 79
DesigïE Ccnsi¿eTBtions
Prelimina1T mat.c•riai balanceà and air flow measurements indicate
that the emission rates from the drya• do exceed allowable limits. The
eulated emission rates are verified hy lat:oratory analysis Of dryer exhaust,
samples. At this point, a decision is made that an emission control
must be on the dryer, problem definition establi*hes the

rate, inlet and exit particulnte loading, operating temperature end pres„ sure,
and proftw,ted variations in operating parameterse
At decision point I (DPI), alLernative control techniques are evalHated
that might include (1) electrostatic precipitations (2) filtration, and (3) wet
scrubbing, (These particulate control methods ave discussed in detail in
Chapters 5, 6, and respectively.) Each alternative technique will provide an
acccptable control level, but each has several subprob (SP) associated
with •The dust the dryer be thus precipitation would then he
ruled out because of safety consideratior;s, Filtration might impractical
because of the adhesion properties of tbc dust. Wet scrubbing might not be
acceptable in some areas bees-tuse of the problems associabed with
digpozal of the sludge.
If wet scrubbing (alternative 3) is selected at DPI* then LEP2
will involve the selection of a tyne af sctubber. Subproblems DP2
that lead to the selection a type of serubber will incluäe pressure drap
requirements, liqt(id recirculating rates, materials of construction,
and power rtquirementa.
An important decision point in every pollution control project is
the evaluation of methods for disposing of collected pollutants;
alternatives include recycling or reuse, incineration, and disoosal a
Igndfill. 'The decision point for disposal of oollutatitö has a maJor
impact on the overall economic feasibilitv of project. In the last few
years, much more attk•ntion been given to sustainabilily in
engineering work*. Indeed, Future engineers will have to produce
sustainable designs that are both flexible and robust using a smaller
set of avail„ able energy and material resources than ever before
(Davidson et ale 2007). Thus in the general design process pictured
in Figure 2.1, one the decision points and perhaps several Of the
subproblems will focus on how to make the project sustainable
The completion of decision points leads to the final design, from
which complete equipment specifications can be prepared and a
project cost. estimate can be developed. Construction, startupÞ and
evaluation complete the design process,
 80 Chapter Two What, Is Process Design?
2.2 General Design Considerations
Process Flow Sheets
A prtR.•ess flow sheet provides a graphic description of a process
and can vary in complexity from a simple block diagram to dètûiled
schematic Bhøwing instrumentation and stream operating conditicnsn
The degree of detail is determined by the etage of development of the
process and by the intended use of the flow sheet.
A flow sheet is a valüEble a;o! at every step of the design pru.•ess. In
the initial development state, a simple qualitative flow sheet is used in
the connectivity of the process. Connectivity refërs to bow individual
components of the process are joined and how the overall process connects
to the entire manufacturing facility, Ag an illustration, consider a project
for recovering a mixture of tDluene and ethyl dryer exhaust using a
fixed-bed carbon solvdnt* recovery sys(eme Figure 2.2 shows a sketch af
a preliminary flow sheet of fixed-hed system to serve as a starting point in
the projat. Figure 2.2 is qualitative and shows on)v the componenfs of any
fixed&d system.
As the process ik3 developed, the flow sheet. i* modified to include
additicntd requirements such as carbon drying hot air after steam
regenerationx process is complicated by the fact that ethyl acetate is stightty
soluble in water. If the recovered solvent. mixture is to be burned as fuel,
no further separation beyond decantation is necessary, However,
 Section 22 Generat Design Considerations 81
vent vent
steam

waste pump

Figure 2.2
Preiirniï'ery flow sheet tar a fixedmbad carbon adsorption/eoìvent
recovery system.
as shown the detailed flow sheet of Figure 2.3, a distillation step will
be necessary to return a dry solvent for reuse in the original process.
The detailed flow sheet can be used to prepare an equipment IiEt,
which is then used in the preparation of the preliminary cost estimate.
It is often desirable to include stream flow conditions on the detailed
flow sheet. This information can be included bv flagging particular
streams or by showing stream eonäiti0à:3 in a box below tbe flow
sheet. The lattznž' procedure makes it easier to compare streaD
conditionx

Material Balances and Energy Balances

Equipment selection and sizing require R complete knowledge
of a]] material and energy flow to and from each unit. The law of
conservat-ion of mass and energy, whic,b is the basis for material
and energy balance calculations, can be expressed as accumulation
input — output + net generation
 82 Chapter Two What, Is Process Design?
For steady-state operation, all operating parfitneters are time
independent, and ('2.1) becomes
U input — output + get generation (2.2)
In the majority of cases* Eq. (2.2) describes material and energy balances
pollution control equipment that is designed for ateady-state
operation, Some exceptions to this generalization encountered in the design
of dryen, and adsorbers owing to the heat
generation and/or pollutant nccumulation within the
The following steps are helpful in performing and energy balance
calculations.
Draw a sketch of the processv
2. Identify and labeE al} enteriBg and streams.
3. Label pertinent data on the sketch.
4. Draw a dashed envelope around that portion of the process
involved in the balance,
5. Select, a suitable basis for the calculation.

Exampie 2. ?
Exhausts two storage bins at a fibetqglass plant are combined
and passed through cyclone that* provides 95% particulate removal
on 8 weight basis. The following measurements were made.
Exhaust from bin A:
flow rate—3000 acfrn (actual ft8/min) dry air loading-—15
grús of Si02 per standard cubic foot (grfscf)
[assume stamiard P and T are: I and 77 OF)
 Section 22 Generat Design Considerations 83
vent vent organics steam

steam
I

steam

Material balance lb/hr average

Figure 2.3
Detailed flow sheet for fixed-bed carbon adsorption/so\vent recovery including
solvent drying by distillation,
(Adapted from Fair, 1969,)
84 Chapter Two What, Is Process Design?

dry solvent return

Water
o o o o
Strlpping
layer Bottoms stea

3172
107 424
635 o
21 428
3300 635
 Section 22 Generat Design Considerations
Stripper 85
m overhead
pressure—t6 K.ä)tmds per
square inch absolute (psia)
tempersture—90 OF Exhaust
from bin B: Stripper
tlnw TEite—2500 acfm dry eir bottoms
loading—IO grains of NE2C03
per scf pressure—17 psig 3554

Find the rate of solids discharge (collection of soüds) From the

cyclone, and the concentration of sot.ids Fernaining in t.hù air beir.g
ex.ha*-ksted from the
Solution
First, we draw a sketach of the process and label the operating
eonditions on the sketch. It is necessary to perform balance around
point C (the junction of streams A and B) to find the total solids in
the feed to the cyclone. We draw envelope around point C, and a
basiß of one hour of operation. Sketch for Example 2.1

Ail' balance:
QA pa + Qn pn = QFPF
where
Q = volumetric flow rate of air, in cubic feet per minute
(c.fgn) p density of air, lb/ftî
Solids balance:
QACSA + QnCsg QrC,sF MSF
where
Q = volumetric rate of air, in cubic feet per minute (cfm) C
= concentration of solids, in stream (grains/fta )
 86 Chapter Two What, Is Process Design?
= subscripts identifying each stream
mass flow rat* of solids in stream F
NOW that Q, C, and p mt.Rt have consistent
units. That is, Q enn be expressed in standard
cubic feet per minute (scfin)¥ C in
gr3inRfscf, and p in lb/scf; or Q can be
expressed in actual cubic feet per minute
(acfm), C in gr/acf, and p in lb/acf. Often,
particulate matter (PM) measurements are
reported as gr/scf, end regulations are written
iïž terms of gr/scf. So, we will first convert
the air flow rates to gefm.
QA = 3000 3188 scfm
16 psia (77 +460)R min .14.7
psi8 (90 + 460) R
acf 17 psia (77 +460)R
QA 2500— 2724 scfm
min 14.7 psia (110+460)R
Note that since the air flow rates have been converted tn scfrnp PA = PB
= air
QA + QB QFZ 5912 scfm
Also note that no air exits from the bottom of the cyclone (it all goes
out with stream E), Thus, QE = Qr.
NOWE we return to the solids búnce. The solids flow rate into the
cyclone is:
Mar 3186 scfm x 15 gr/scf + 2724 scfrn x 10 gr/scf 75,060 gr/min

75,060 gr 1 1b 60 min
= 643 lb/hr
min 7000 gr 1 hr
Since the cyclone iss 95% efficient, the solids collection rete is:

Mso 0.95 x 643 = 611 lbffir

The concentration of solids in the air exhausted from the cyclone is:
0.05 X 643 lb/hr Ihr 7000 gr 0.63 gr 5912
scthnin 60 min 1b
 Section 22 Generat Design Considerations 87
At this pc•int, we can list soma generaiizations governing the
solu•• tion of material and energv balance equations.
Por a dirùet solution of equations with n unknowns,
n. independent equations are required.
The one overafi balance plus C I component balances, where C the number of
campy.en•as,
3. If energy balance and material balance, are considered
simultaneously, an additional independent equation can be
for the overan enthalpy bàlane,e azotil.id the svsce:n.
With the cost of energy increasing significantly in the past few
years, pollution control engineers must work even harder to lower
energy consumption in control equipment. Minimization of enerw
use can be accomplished through the following:
1. Better equipment design leading to increased efficiencies
2, Better equipment selection for specific control applications
3. Optimization of equipment operation through frequent evaluation of
equipment performance
A basic understanding of energy fundam6ntals is a prerequisite for
good design. The following section presents B brief review of
enthalpy as applied to flow systems. For a detailed derivation of the
basic energy relationships, refer to any introductory
thermodynamics text.

Enthalpy
Recall the thermodynamic definition of enthalpy H, where
(2.3)
Enthalpy is a physical property of the fluid and is very useful in
making an energy balance on a flowing stream of liquid or gas.
Enthalpy is a point function; that is, the enthalpy of a substance is a
function of the conditions at a point and not a function of the path to
that particular point. We do not use absolute enthalpies; rather, we
use a difference or change in enthalpy between a desired point and a
standard reference point or datum. The enthalpies of water and steam
can be obtained from standard steam tables, and the enthalpy of air
can be determined from standard psychrometric tables or charts, as
presented in Appendix B. At or near atmospheric pressure, air
behaves nearly ideally and its enthalpy is virtually independent of
pressure. Thus, air enthalpy change can be calculated by

AH = 11b -Ha (2.4)

 88 Chapter Two What, Is Process Design?
where
Cp specific heat at constant pressure, Btu/lbm
T = absolute temperature, R
For temperatures below 150 0C, it is sufficiently accurate to use

(2.5)
P(uvg) where Cpavg) average value
at (Tb + Ta)/2

Example 2.2
A waste heat boiler is a device that ofiz•n is used to recover heat
from a stream of hot gases before it exits a stack; it recovers the heat
 Section 2.3 Simple Analysis a Power Plant 89
Of Coal-I.'äred

by boiling water into steam, which is a useful commodity at most

industrial plants. Consider the diagram shown below. The hot air
stream is flowing at 60,000 acfm (at 1400 O F and I atm). CHIculate
the heat removed from the air when it is cooled from 1400 a F to 400
O
F. Also, calculate the rate of steam produced (in lb/hr), assuming
that the water comes in at 90 O F, and leaves as saturated steam at
212 O F.
steam

cooler
air

hot air

Solution
Using the density of air from Table B2 in Appendix B, the mass
flow rate of the air is:
0.0212 1b
M 60,000 acfm x 1272 lb/min

Using the enthalpies of air at the two temperatures from Table B. 7

in Appendix B, the heat removed from the air stream is: 1272 1b
(341.5-82.1) Btu
= Btu/min
min 1b
From Table B.8* Appendix B, the enthalpy of water at 90 O F is 58.0
Btu/lb and the enthalpy of steam at 212 O F is 1150 Btu/lb. Thus, in
this case, converting one pound of water at 90 O F to one pound of
steam at 212 O F requires 1150 — 58 or 1092 Btu. Assuming that all
the heat given up by the air is transferred into the water, the amount
of steam produced is:
90 Chapter Two What, Is Process Design?
330, 000 Btu/min 60 min
= 18,100 lbffir
1092 Btu/lb hr

23 Simple Analysis of a Coal-Fired Power

Plant
To illustrate the power of a combined material and energy balance
analysis, we now introduce the reader to pulverized coal (PC) power plants.
From an air pollution point of view, PC power plants are an excellent
choice for study because such plants produce a varietÿ of air
pollutants, all of which must be controlled, often using vastly
different control equipment. The most significant pollutants from
PC power plants include PM, S02, NOx, and mercury, Of course,
now that EPA has officially classified C02 as a pollutant, this gas
(which is emitted at a rate that is an order of magnitude greater than
all the other pol„ lutants combined) has moved to the forefront. In
later chapters of this book we will examine the design fundamentals
for equipment to trol each of these pollutants.
In a PC power coal is crushed and then pulverized into n fine
powder. It is then mixed with air and Glown_ into a large furnace/
boiler (usually just called the boiler) where it burns very quickly in
a large fireball. The heat released by the combustion of the coal is
absorbed by high-purity water that is circulating through steel tubes
that line the inside wails of the boiler. The water absorbs the heat
and boils into steam; the steam is transported through the tubes into
another section of the boiler where thc steam absorbs even more heat
and becomes superheated. The superheated steam is piped to a large
turbine which is coupled to a large generator. The steam expands
through the turbine, transforming its thermal and pressure energy
into mechanical energy by rotating the turbine at high speed (which
in turn rotates the generator, generating electricity), A simplified
process flow diagram (without the APC systems) is shown in Figure
2.4, und a straightforward but effective video explaining this process
can be found at http://www.opg.com/power/fossil/howitworks.asp.
The steam loses pressure and temperature while traversing the
turbine, and as the steam exits the turbine, it is condensed back into
water so that this high-purity water can be pumped back into the
boiler to begin the cycle again, The waste heat (given up while
condensing the steam that exits the turbine) is removed from the
process by 8 separate stream of (lower-purity) cooling water in large
 Section 2.3 Simple Analysis of a Coal-Fired Power Plant
heat exchangers, and eventually exhausted to the atmosphere via
massive cooling towers (see Figure 2.5).
Recall that a large quantity of air is used to blow the pulverized
coal into the boiler. The mixture of nitrogen, excess oxygen, and all
the gases produced by coal combustion is often called pue gas. This
hot gas stream exits the combustion zone and continues to travel
through the boiler, transferring its hcat to other streams. Because the
heat is very valuable (good money was paid for the coal), the plant
is designed to recover as much of the heat as possible. After boiling
the water into steam, the hot flue gas passes through the steam
superheater section, where more heat is transferred into the steam.
Then the gas flows through the economizer, where the incoming
boiler feed water is preheated before it flows into the, boiler. Finally,
the flue gas flows through the air-preheater where it heats up the
incoming air. Keep in mind that the cooled flue gas (it still is in the
350—400 OF range) carries all the pollutants that were created
during the combustion of the
92 Chapter Two What, Is Process Design?
 Section 2.3 Simple Analysis of a Coal-Fired Power Plant

Figure 2.5
Cooling towers at a coal„fired power plant.

coal, and therefore must now be routed to the necessary air pollution
control systems. The APC systems vary depending on the quantities
and types of pollutants and the local and federal reBIIations for the
final exhaust gasn
Despite attempts to recover and use as much heat as possible,
tram ditional coal-fired power plants are not highly efficient at
converting all the heat released by burning coal into electricity.
There is a thermodynamic limit to this conversion of heat into useful
work. The Carnot cycle (see Figure 2.6) is the classic idealized
thermodynamic heat engine, and Eq. (2.6) shows that the overall
thermal efficiency is dependent on the high temperature at which
heat is transformed into electricity, and the low tempcrature at which
heat is exhausted to the environment. Traditional stainless steel
tubes begin to get soft at about 1050 OF (about 840 K), which sets
the upper temperature limit, and heat is exhausted to cooling water
at about 85 OF (about 305 K), which sets the lower temperature limit.
Using Eq. (2.6), the Carnot efficiency limit is calculated to be about
64%,

(2.6)
89

where efficiency
94 Chapter Two What, Is Process Design?
W = useful work output by the engine heat input to the
engine at the high temperature,
QC — waste heat exhausted at the cold temperature, Tc
TH and Tc — — hot and cold temperature, K

Figure 2.6
Schematic diagram of a Carnot heat engine.

However, the theoretical limit of 64% is unrealistically high,

because the Carnot heat engine cycle is not applicable to the steam
electric power plant depicted in Figure 2.4. For this type of plant,
the Rankine cycle is more applicable (see Figure 2.7). The
combination of the lower theoretical efficiency of the Rankine
cycle* the realities of using water as the heat transfer fluid, the
limitations of using realworld equipment, and the need to use
significant amounts of energy to run the APC systems means that
even a reasonably modern coal-fired power plant will have an
operating thermal efficiency much lower than the Carnot limit. Over
the past 60 years, the industry has slowly made improvements, and
operating efficiencies have steadily crept higher—from the low 30%
range in 1950 to near 40% in 2008. This trend toward higher
efficiencies not only saves money but also reduces the amount of
greenhougc gases emitted per unit of electricity produced. In fact,
just in the past decade, research has produced break* throughs in
materials (exotic steel alloys) and technologies (supercritical steam
processes) that are pushing efficiencies above the 40% range, even
approaching 50%, This topic will be more fully explored in Chapter
22, carbon dioxide control.
To illustrate the use of combined material and energy balances,
we present Example which contains several subproblems.
 Section 2.3 Simple Analysis of a Coal-Fired Power Plant

Figure 2.7
Idealized schematic diagram of the Rankine cycle,

Example 2.3
A 750 MW power plant is burning coal that has the following
properbies (among many others): heating value = 23,500 kJ/kg,
carbon content = 60%, ash cnntent = 9%), sulfur content = 2.2%,
and mercury conten t 120 ppb. The plant an overall thermal
efficiency of 37.5%, Assume that 20% of the ash falls out in the
bottom of the furnace and the rest (the fly ash) goes out with the
gases. An ESP collects the PM, and wet scrubber controls S02. The
required air pollution control device efficiencies are as follows: the
F,SP is 99.4% efficient; the wet scrubber is 92% efficient.
Calculate (a) the ratc of heat emissions to the environment (kJ/
see), (b) the coal feed rate to the furnace (metric tong!day), (c) the
rate of ash emissions to the atmosphere (kg/day), (d) the rate of SOz
emissions to the atmosphere (kg/day), (e) the rate of Hg emissions
to the atmosphere (assuming 30% Ilg removal in the wet scrubber)
in gtday, and (9 the rate of C02 emissions (metric tons/day).
Solution
(a) First draw a simplified energy balance diagram.
96 Chapter Two What, Is Process Design?
91

QM (heat)

(electricity) 750
MW

The statement of thermal efficiency means that only 37.5% of the

input energy ig converted into electricity, so:
750 MW
2000 MW
0.375
The heat emitted to the environment is;
= (1-.375) x AMW 1250 MW
Converting to kJ/scc:

I
= 1250 MWxkJ / sec
(b) We calculate the coal input rate from the energy input rate
and thc hcnting valuc of the coal:
24 hr
Coal feed rate = 2000 MW
x
1 MW 1 day 1 kWh
1 kg 1 tonne
7353 tonneJday
23,500 kJ 1000 kg
TO calculate the Of PM emissions to the w e first
draw the following diagram showing the air and coal in and
the flue gas out:
ESP
 Section 2.3 Simple Analysis of a Coal-Fired Power Plant
coal
treated gases
air

collected fly ash

The ash that comes in with the coal is:

7353 tonne 1000 kg 0.09 kg ash
= 661,800 kg/day (ash in)
day
98 Chapter What Is Design?
Ptccc•ss

The fly leaving the furnace and escaping the ESP with the treated
gAse:-3 is:
661,800 kg
x (1-0020) 9176 kg/day (ash emitted)

(d) To calculate the rate cf SO} emissions to the atmosphere, we

continue the previous drawing to include the wet scrubber:
treated
oases
stack

coal

bottom ash coltected ny ash sludge

The sulfur that comes in with the coal is oxidized in the furnacvs to SOP
and the molecular weights of S and S02 are 32 64, respectivelyž so:
7853 1000 kg 0.022 kgs day tonne kg coal
64 kg S02
(1-.92) 25,880

(e) To calculate the rate of Hg effiissions to the atmosphere. we

simply caleulate a mass ba\anee on the mercury eotning in
tha coal. This estimate is conservative because some of
mercury may bc captured coincidentally with the fly ashy
and more bhan may be captured coincidentally by the s02
scrubber,
7353 tomle coal 1000 kg 10 e x 120 ppb x (I
— .30) = g Hg/dey day tonne
(f) To calculate the rate of COZ emissions to the atmosphert¥ we
simply calculate a bøEar;ce on the QRrbor.t coming in
with the coal.

180 tonne C021dey

 Section Engineering Economics
2.4

Hong and Slatick (1994) have derived C02 emission factors (EPs)
for various coals in the United States; they range from 205 for
bituminous coal* to 216 for lignite, to 227 for anthracite coal, all in
lbs of C02 per million Btu. It is interesting to that a C02 EF can be
derived purely from material balance as denxs.nstrated in the next
example.

Example 2.4
Derive the C02 emission factor for the coal of Example 2.3,
Solution
Choose a basis of one millio= Btw
106 Btu kg coal 2.205 1b
1 million Btu 23,500 kJ 1 kg
44 1 C218 1b C02
0.948 Btu 1 1b coal 12 1b C I million Btu

2.4 Engineering Economics

problems
ally several viable solutions. As indicated earlier in this chapter? a
major decision point in the design process is the selection of the most
technically and economically {èasible sotution att.-ernative.
Although a detailed discussion of economie principles as applied to
decision ing is beyond the scope of this text, we will present
several basic proee. dures for compariÄ3g alternative investments
that are valuable tools for process engineers.

Optimizing Fixed Capital and Operating Costs

To develop comparison procedure for alternatives, we must, define
a base case. The base case is the alternative that will provide the
desired control resuP»s at least total cost. The base case should
represent a system operating under nearly optimum conditions. A
trade-off usuaiiy exists between capitol and operating cost,
illustrated by Figure 2.8 for a typical fabric filtero In this case, the
trade.cff is between fabric area (capital cost) and pressure drop
100 Chapter What Is Process Design?
(operating cost). Optimum conditions from an economic standpoint
would be in the region Of minimum total cost. The process design that
provides the lowest total cost may differ from the one that provides the
best operations—that is, the design thnt optimizes ease of operation
and stabilitv Of removal levels and minimizes maintenance. These
operational considerations might dictate a design somewhat removed
Two
Figure 2.8
Typical cost
relationships for $ablic
filters,

of
investment

from the economic optimum. 'Thus, it is obvious that process engino•rs

must be careful to dcfine which '{optimum" they are referring to.

Depreciation
All physical assets such as pollution control equipment decrease in
value with time owing tc deterior&Å-ion and/or obsolescence.
I.)eteriorstion is the result of physical wear and, in many cases, of corrosion.
Obsolescence be due to technological advances, or changes in control
requirementst either case, the end result is eventual equipment and
the necessity tör replacement capitaln Depreciation is recognized as a
legitimate operating cost and can be claimed as such in computing ineome
taxes, Keep in mind that depreciation is cost,
Depreciation can be ealculatzd by one of several methods;
CQEn• pany policy normally determines which method is used. All
methods require that a useful service life and salvage value be
est#nated each item of equipment. In the preparation of
 Section Engineering Economics
preliminary cast estimates, the usual practice is to ao-sume slr€ight-
line depreciation CRIcu„
lated by the following equation:
VR — Vs
(2.7)

d annual depreciation, dollars/year

VR = initial cost of equipment, dollars Vs
salvage or serap value, dollars
= service life, years
24

The asset value after a years in service is

found by

where
Va asset vaiuc,
dollars a time
in scrvice,
yeers
A further simplification often used in
preliminary ('cost estimates is to agsume a zero
salvage value and an average service life of 10 years, Under
these conditions, the, ar:nual depreciation
per year is 10% of the initial cost.
A new company might wish to depreciate an
asset rapidly during the first few years to
minimize the companys taxes during its early
development period. In this the declining-baiance method is often
used, this method* the annual depreci8tion is
a fLV.ed percentage of the asset value at the
beginning of the depreciation year. The asset
value at the end years of service is

where f = fixed tage factor, dimensionless

102 Chapter What Is Process Design?
The asset value at the end of the estimated
service Ma n is then
(2.10)
For depreciable assets placed in service
before 1981 e the IRS permits the use of
declining-balance factor equal to twice the
firstmyear straight-line rate (known as the double-declining-
balance method). The Economic Recovery Tax Act of
1981 stipu\ated that all depreciable property
in service in or later must be depreciated at
r&tes published in ACRS (Accelerated Cost
Recc,vevy System) tables, In 1986, the ACRS
was modified (and is now ag MACRS) hy the Tax
Reform ACL Of 1986 (TRA*86) to allow more
rapid depreciation Of assets, The modified depreciation
is similar to the double-decliningbalance method. The
intent of TRA-86 was to stimulate the purchase
of new capital equipment by industry, the
rules in force todav. Under MACRS, ail
property is classified into 7-, 10-, 15and
20myear service groups, with no selvage value.
The majority Of process equipment, including
pollution control equipment, fans in the 5-
year group. Depreciation ratf.* for equipment
plaen?d in seNicer 1986 for the 5-year
group were:
Year Rate

2 32%
3 19.2%

deprecinüon ill the first. and yecx,very yenrg

11.52%
5 11.52%
5.76%
'A. half.seffi is: allowed in the first n.•eovery years
For a detailed discussion of depreciation and effects, consult the text by
Steiner ( 1992). IRS publications 584 and 946 give. a list of R*SCL
 Section Engineering Economics
permissible depreciation rates. Humphreys and English (1992) provide
a gocd review ofACRS and MACRS stipulations,

Example 2.5
Companies A and B bought identical venturi ficrubbers in 2007
tbat $75,000 each, In both applications, the was esti„
mated to be 5 years with zero salvage value. The corporate income
tax rate for both companies was 50%. Company A used straightaline
depreciation and Company B used the MACRS method. How much
more money did Company B save over the first 3 years of service
based on its depreciation procedure?
Solution
Depreciation claimed by Company A:
$75,000 x3 =
$45,000
5
Depreciati0/ž claimed by Company B: The MACRS depreciation in the first three years totals 71.2%
of the initial cost. The deprEciatioz charged during those three years is

dB 000 0.712 $53,400

Since the corporate tax rate for each company is 50%.. Company B saved
0.50 x ($53,400 000) = $4200
As practical met.t,er, all companies the most advantageoug method ûf
depreciation allowed by law for tax purpcsesz whereas engineers typieally
use straightÄine depreciatiorž for evaluation of alternative cases.

Incremental Rate of Return on Investment

A
is the rate of return on the investment, ROI, defined by the
following:

ROI (2.11)
 Section Engineering Economics 104

ROI return investment, percent

P annual profit invegunent (income - dollars
total investment, dollars
Incremental ROI refers to the incremental return based an the
incremental profit and incremental investment between two
equipment alternatives. In this procedure, alternatives are comoared
with the acceptable alternative requiring the least investment as the
base case.
Often, when a company invest in pollution control eqtiip„ ment,
all the alternative solutions east money and none may generate any
true profit. However, faced with having to mnke to
comply with the law, the company must choose the best. alternative
among several, each of which solves the pollution probleza. In this
CüSee the concept of incremental ROE still be applied, with the
savings of one alternative compared with another being t.:teated as
"profit." It is interesting to note that, nowadays, certain pollution
con.

to invest in an altprnative that exceeds the requirements, and then

sell the credits on the open market!

Example 2.6
A company must purchase a cyclone to control dust from a fbundry
operation. The lowest bid on a cyclone that would meet ail control
requirements is for a carbon-steel cyclone with an installcd eost of $35,000.
The cyclone has a ser•åce of 5 years. A bid was received a
stairdess.steel cyclone that is guaranteed 10 years and would lower
maintenance costs by $1100 per year. The installed cost of the
cyclone is $60,000. Both cyclones are estimated to have
zero salvage value. If the company currently receives 8 j return before taxes
on all investments, which cyclone should be purchased?
Solution depreciation on carbon-
steel cyclone depreciation on
 Cost 105

5tRinleSE•stael cyclone $60,000

total yearly savings = $6000/year
with stainless-steel 10
cyclone incremental ($7000 — 6000) + $1100 = $2100 /
investment incremental year
ROI $60,000 $25,000
$35* 000 9100
$7000/year 25,000
5
l.)e$igzÑ

Since the incremental return doos not exceed company requirementRp the
higher bid should not be accepted. Note thRt, despite the fact that the
incremental return is not very high, FJVther c.otiâiiieration must be given
the stainless.steel cyclone because, presumably, a second car„ bon-steel
cyclone would need to be purchased after five years, perhaps making the
option of a stainless«3teel cyclone the better choice.

Comparison of Several Alternativesl

Where there are three or more alternatives to be c»mpared, the

I, Select the acceptüble unit with the lowest installed cost as the
base Cage and designate it ease 1 or the base case.
2. Designatc higher-cost units case 2, case 3, etc., in order of
increasing cost.
3, Calculate the incremental ROI between cases I and 2, If the
ROI is acceptable. then case 2 now becomes the base If the

is discarded,
4, Calculate the ROI between case 3 and the base case, which
no«.v may be either case 2 or case la If tae ROI is acceptable,
106 Chapter Two What Is Process
case 3 is the new base ease. If the ROI is unacceptable, case 3
is discarded.
5. Con•tainue this process until all cases have been evaluated,
This proeedure ensures that each increment of ins,Q9tment
will provide atl acceptable return.

Payout Period
A simple measure or orofitability is the length of time required
to recover the depreciable fixed capital investment for a project. This
period ofyears is called the payout period is defined by the follow*
ing equation, fixed capital investrnent payout period
(annual profit + annual depreciation)
The payout period can be adjusted to include interest costs by adding
such costs over the life of the project to the fixed capital investment,

2.5 APC Equipment Cost Estimation

Company management must budget capital pollution control
equipment and its aperttion, It is essential that accurate cost estimates
are available to establish expenditure priorities that assist in
Section 25 APC Equipment Estimation

the overall deci*ion proeessn Certainly, a major restraint on all

equipment design is the overriding necessity to maintain a satiE"
factory corporate profit structure.
Factors BfTecting equipment costs include company policies,
local and federal government regulations, and national economic
conditionfi. Individual companies can establi•xh strict design
stand¼rds that kirnit equipment selection. Union ež3ntracts and
company agreements with fabricators can also indirectly affect costs.
Government regulations can have significant impact on costs through
depreciation and tax credit policies.
There are different kinds of cost estimalcs, classified by the
degree of accuracy required. *Three levels of arcurgcy that are
among those widely recognized in estimating capital cogts are (U.S.
EPA 2002):
Order of estimate for conceptual planning: Accuracy
± 50% or higher
2. Preliminary estimate far the initial go or no go decision on a project:
Accuracy ± 30%
 Cost 107
9, Final detailed eng:meering estimate for preparation of budget
requests: Accuracy ± 5%
It is noted that a highly acenrate estimate is often neither needed
nor desired. For example, in the early stages of the design process
(such as decision po:.nt 1 or 2; see Figure 2.1), when engineers are
making B rough cut to determine acceptable alternatives, rough cost
estimates are Süfficient. It would be foolish to s»enrl excess time and
money to develoo highly accurate cost estimates for each alternative,
At the early stages, the level of effort need oniy match the revel. of
accurttcy desired,
Making engineering estimates of capital costs requireg
information; in general, the more information one has. the more
accurate thc estimate. For order of magnitude estimates or
preliminary estimates, such information may consist of a rough
sketch of the process flow sheet, pre.\iminary sizes or capacities of
the main APC devices, rough estimates of utilities (electricity,
steatrŽR water, gas, wastewater, etc.) needed, and preliminary gizes
of auxiliary equipment sucb as ducts,
are cally
made from general equations based on type and size of equipment*
past costs of similar equipment or graphs based on past onst data, and
can be updated to the year of analysis by cost indexes and/or
extrepoh*tions. For final detailed estimatesE finished design is
needed along with bid* from vendors. In general, the information
provided in this text r.nn be used for prftliminary cost estimnÉes (±
30%), associated with a reasonably complete process design.
The UVS- EPA has spent decades studying the costs of air
pollution control equipment. Costs are important considerations in
the formula. tion of some regulations, or in the analysis of
alternatives to determine the standard control option for meeting
other regulations. Most of the
 108 Chapter Two What, Is Process Design?
cost equations and graphs presented in this text come from EPA studies and
reports, the most recent of which is the Air Pollution. Cont;ul Cost Manual
(2002). This manual has iDformation applienhle to a wide array of APC
equipmetlt that be found in a wide range industries, However, as out in the
large coal-fired power plants have significantly larger and more specialized
equipment needs, Also, the coai-fira power industry usually analyzes
and costs bìH)d guidance the Ekectric Power Research Institute
(F,PRI), such "levelized costing' to make conjparisons within ftcross the
industry. ÉPRI's 'Tech.ni£al Assessment GuØe (TAG) is published
Ëriodically and typically is available to members only, Therefore, the EPA
has also developed software to help estimate APC equipment costs spe•
cifically for coal-fired power plants, as discussed in the next panagrapha.
The U.S EPA bas developed a spreadsheet tool to help calculate the
APC-'nžlated costs fer c€Æl"fired power plant, The toot is called Coal Utile
ity Environmental Cost (CUECost), and is Excel that calculates rough-
order-of-magnitude (± 30%) cost estimates for the total installed eapital
cost (TIC) and the annualjZ2d operating costs (AOC) of a complete air
pollution system for a coal„fired power plant. The user inputs parameter
values for plant power coal type, % cxcess air, air moisture contentr and
required removal efficiencies, and CUEC(jst produces size and cost
estimates for thfY APC equipment needed to control particulate matter,
NOX and SOX, The workbook is set with five see. nario columns to easily
compare the results ofvarying inputs
The workbook and eccompanying documentation are avaitahle
for free download from the EPA's Technology Transfer Network,
Clean Air Technology Center CITN{CATC). The center's *Neb site
offers numerous Fact sheets, reports and other information dealing
'"ith APC including operating prmciplea and cost estimates
(http:// www.epaøgov/ttnfcatWprtbducts.EttmE), Downloadable
files for CUECost can be found by scrolling down to the "Software"
category. Learning to use CUECost requirey several hours cf self-
teaching, but since it frequently is referenced in academic
publications and industry* it is worthwhile for students and
practitioners alike te make the Another computer cost model
has been developed at Carneg{e Mellon University for the U.S.
Department of Energys National nnesgy Technology Laboratory.
This model is the Integrated Environmenta.! Control Model, which
the added benefit of using modern graphical user interface (GUI).
Additional information about IECM ant:t the free download can be
found at http://www.cmu.edu/epp!iecm]index.htrnl.
When available, cost data are •.:suaEfy several months to several years
out of date, and must: be adjustr:d for the effect of inflation. Cost data for
by-pes of air pollution control equipment are presented in later
chapters, ExtrapolHtion of equipment costs to a later date is accomplished
 Cost 109
by using one of several published cost indexes. A cost index is ratio of the
cast of an item or group at a specific time the cost of the
Section 2.5 APC Equiptnent. Estirnetion

item at some base tirrte in the paat. Indexes are

pzblisbed by tbe govern. [Bent for and matesaEs
(bath for retail and wholesale). The most familiar
index is the Consumer Price Index (CPD, which
provides a sure of inflation each year. In
equipmcnt cost estimation, ati equipment or process
cost index is n„qcd (n update equipment cost, as
follows:
present index
cost(2.13)
pregent past pagt ir.dex

Two indices frequently used in estimating

the cost of prccess and air pollution control
equipment shown in •Table 2.1, Brief
descriptions of the MarshaÙ-Swift and Chemical Engineering
cost indices nre given in a eur:ent issue of the
journal Chemical F.;ngineering.

Table 2.1 Equipment Cost Indexes

Year
1981 721,3 2970
1982 745.6 3f4,O
1983 760,8

789.6 3253
797 3184
3Z3.B
852.G 342,5

1990 35T6
930.6 361.3
1992 943.1 358.2
1993 3592
1994 993„4 328.1
 Chapter Two What Is Process Design?
1995 1027.5
1996
30.5
998 1061.9 389, S
1085.3 390.6
1080.6 392.6
2001 1093.9 3940
2002 11042 395,6
2003 1123.6 402.0
2004 1178.5 444.2
2005 12445
2006 1302.3 499.6
2007 13733 525.4
200B 1449.3 5753
2009 1486.6 521.9

iVhen conducting an engincering economic analYEis» planners

must congiàcr both capital and operating costs. They also must
account for the time value of money That is, one dollar today is
worth tnore than one doller in the future (after accounting for
inflation). Sometimes a cost analysis is conducted in "current'*
dollars, and other times in «conRta_nt't dollars. A current«doilar
analysis takes into account the expected effi*cts of
inflation by using an 88surned inflation ratÆ or interest
rate, and escalates costs into the future, For if coal costs $10.00/106
Btu in 2008 dollars and inflation is be 3% per year9 coal would cost
$13.4•1/10 in 2018, and $18.06 in 2028. Typically, current dollars
ure more understandable, egpecially when the analysig covers only
a short period of time.
analysis presents alt eogts in base year's dollars. the
above example, coal would still cost $10,00/10G Btu in 2018 and in
2028 (st&ted in constant 2008 dollars). Constant•dollar anelysifi is
sometimes preferred when making long-range projections, this text
wc will use currentmdollnr analysis.
 Cost 111
Equipment Costs
Process equipment costs can be correlated with size or capacity
by the following relationship.
capacity sizez costB costA
costA
capacity A size A
where capacityn equipment throughput, in standard units of cftn, gpm, etc.
sizeB = equipment sizee in standard units of ft2 Ù constants

The exponents b and b' vary from 0.5 to ana wilt avertWe
between and 0.7 for many types of equipment. Typical exponent
values for various types of air pollution control equipment are shown
in Table

Example 2.7
A carbon-steel cyclone designed tar a throughput of 5000 scfm
was purchased in 2003 for $15,000. Estimate the cost of a
carbon-steel cyclone purchased in 2007.

NVe use the (MS) equipment cost index from

Table 2.1 to extrapolate the cast of a 5000 scfrn unit to 2007. We
find the cost exponent for cyclones in Table 2.2
Section 2.6 Preliminary Fixed Cepit»} Estimates

M-S index in 2003 1123„6

M-S index in 2007 1378.3 Cost
exponent 065
065
• nooo
costi0ffr00 $15,000 = $28,768
5000

Due to estimation accuracy limitsp au such answers should be

rounded to no more than three significant figures; this ansgver
should be reported $28,300 (or pessibly $29r000). Using the
ChF,micel Engineering Index results in an estimate of $30,800 (or
$31.000).
 Chapter Two What Is Process Design?
Tab\e 2.2 Cost Exponents for Air Po$lut\on Contrai Equipment

EquipmentType

Size Rango Cost Exponent

5000-1

ScrubLvs tnžptngerne"t 5000-90,000 0.80 Gravity *ray 5000-200.000 062

CentFifugal0.76
Venturi. Energy 5000-100.000 crm 0.76
VenturiE Energy 5000-1 0.72
Electrostatc
Ptecíp;tators 062
Reverse PuEse

Adapted from Pekers Timmerhal.i%

2.6 Preliminary Fixed Capital Cost Estimates

for Poltution Control Projects
Cost analysesq for complete industrial plants have shown that a
preliminury Installed Cost ('TIC) estimate can be developed from gn
estimate of the Delivered Equipment Cost (DEC) for the major items
or equipment, The direct and indirect installation costs are estimated
as pcrcentAges of the DEC, and all costs ere added together to
produce the TIC. The DEC is a function of thc f.o.b. purchased
equip. ment cost (PEC) and some add-ons. Tho stands for
free•onboard, and refers to the cost of the equipment that the
manufacturer charges the customer, including any charges for
loading the equip-
10•i

ment onto a train or truck, but not including the shipping charges or
anything else. A similar approach can be used estimating the
installed cost of air pollution control equipment, Average
installedcost factors for pollution control cquipment ere given in
 Cost 113
Table 2.3. These factors are mcallt to serve as a general guide and
must he adjusted to fit specific installations,

Table 2.3 Average InstaRed„Cost Factors for Estimating Capital Cosb

Precipi„

Dlrect Costs tators ScrubbO$3 Filters Adsøtbers

1, Equipment
Costs a. Control
Device and

Equiprnent (fob,) AS Req'd-

u. Inslrumenis and {P)

0.03
d. Freight
Subtotal t .00 1.18 x p 2,
Costs
a. Foundations and
SÈ.1ÿPù<s 0.12 0 06
b ErecOcn
and
Hanclinta c.50 0.40 0050 0.14 020
0108 0,01 o,oa 0_04 o„oa

Piping 0005 0.02

0002 0.03 0.07 0.01 0102
tnsuiatan
f, Painlirtg g, Site 0.02 0.02 0.01 0.01
As Reg l d.
h. Faciiües and
As Req't:i
Roiidiags
(¥ DEC) 1.67 .62 1.72 1.30 I .44

Inditact
3, Indirecl Co•àfs
a, Enghærir,g and
0.20 o, 10 0.10
b Cons?Fuctirwì end
 Chapter Two What Is Process Design?
Fleld Expenses 0.20 0.10 0.05 c, Constmct.jcn 0.20
Fee 0010 0.10 10
d. Start„up 0.01 0.01 0.01 0.02 002
e, Pefformance Test 0.01 0.01 0,01 0.01
f Model Study
g. Contingencæs 0.03 0.03 0.03 0.03 0.03

Total DEC} 2.24 1.97 2.17 1,61 1.75

 Section 2.7 Annual Operating Cost Estimates

The delivered cost for a 1000 scfm fabric

fiitÆr is $22*000. The unit is to be insta!led outdoors
on an existing concrete pad. the, total fixed
capital investment for the gystem.
Solution
We use Table 2.3* omitting those items not
required, Based on the minimum information
available in the problem, we the costs for
foundation and supports and painting. The
total inBtalled cost (also celled the fixed
capital investment) is then
total = — 0.04 0.02) $46,420 or $46,400 (rounded)

2.7 Annual Operating Cost Estimates

We can prepare preliminary annual operating
cost estimates by using Table 2.4e The major
operating cost items for many installations are
labor, utilities, and depreciation. For many types of equipment,
utility costs will far exceed all other costs.
On the other hand, in the case of wet scrubbers
and spray (brers used to remove acid gases,
chemical reegent costs may equel or exceed
other operating •costs, For equipment requiring
component replacement at regular intervals, such carbon
replacement in adsorbers and catalyst
reolacement catalytic incinerators,
replacement cost on an annual basis is included
under direct operating costs. On major
replacement items, such as carbon nnd
catalysts that might be replacÆd at to 5„year
intervals, the

Table 2.4 Annual Equipment Operating Costs

 Chapter Two What Is Process Design?

Direct Operating Costs

Labor, haurs/year x cost per hour

15% of operating labcr
Maintenance. 5% Dt fixed capital investment (FCI)
Utilities
ElectricitYJ kWh/year x cost per kWh
Steam, 3b/year x cost per pound
Coolina waterl gallyaat cost gat
Indirect Operating Costs

Labor overhead, fiO% Of toW labor

106

replacement cost can be annualized by considering the total replace«

ment cost the pt?sent worth o? an ordinaFY annuity, The atmual cost
is then the annual payment of this annuity, which can be calculated
as follows:

AOC (2.15)
where
AOC = annualized operating dollars
= replacement cost Of dollars i = annual
discrete interest rate

The expression enclosed in brackets is referred to as the capital

recovery factor.

Problems
A textile finishing process involves drying 8 fabric that has been
treated with a volatile solvent, The wet fabric entering the dryer
contains 45% solvent, and the dried fabric contains 3% of the
 entering solvent. Solvent-free air enters the dryer at a rule of g
pounds per of solvent-free fabric. The solvent has a molecular
s*eight of 60,
Calculate the percentage of solvent in the dried fabric by weight.
b. Caieulate the concentration of solvent in the dryer exhaust in
mgÌm3 if the exhaust is at 160 and 1.0 atmosphere,
22 A company uses a 50% solution of NaOH to maintain the pH or
water circulated in one of its furnace exhaust scrubbers. The
cost of the solution is $200 per ton f.o.b. (free•an-board), The
suppJier says it would prefer to sell the company a 65% NaOH
solution Find the company says the water content is immaterial
long as the equivalent costs ere the same including freight. If
the freight is $15 per tan, what is the maximum price per ton
the company should pay for the 65% solution?
An exhaust stream from a reactor consists of 90% air and 10%
NE13 (by volume). The exhaust is treated in a packed scrubber
as shown in Figure P2.3. the rate of NH3 leaving the
vent stack, in pounds per day.
Problems 107

to vent
stack
15.0 75 0.3%
NHv dry basis

to waste treatment

Figure P2-3

2.4 Exhaust gas from a chemical process flows at a rate Qf 8,000

flCfm from point A (see Figure P2.4) at 250 OF and 1 atm
pressure. It flows through circular duet to a fan where the
oressure is boosted to 4 inches of water (gage) and then flows
to ineinere„ tor, Fuel gas is burned at a rate of 75 pounds per
hour and the temperature of the incinerúr exhaust gag is 1200
 Chapter Two What Is Process Design?
'F. Assuming both the process exhaust and the incinerator
exhaust streams have properties similar to air, calculatz: the
enthalpy that has been added by burning the fuel gas
(Btuffiour). Assume that vou can recover heat from the
incinerator exhaust by buying a heat exchanger to cool the gases
to 500 OF while recovering usefül heat. Assulning that such heat
is •worth $7.û(YmilIion Btu, how much rnoney could you save
by recovering that amount of heat ($/day)?

duct

Figure

A pr'N-•ess fnr a giags will rc•znove boron from the 1400 cF

furnace by eooüng the exhaust to 275 OF in evaporative coder
to condense B203. The cooled exhaust will then pass through a
rt?ve.rse-jet filtez• t'] remove suspended dust, including the
B20a particulates. Draw a preliminary flow sheet for the
process, indicating any au£liary equipment needed, List any
operating difficuEties you thiEk might arise with this system.
296 As an alternative Lo the process described in Problem 2„5, the
exhaust Will be treated in UOP-type wet scrubber. The (JOP
scrubber is a tray.type unit in which there is a layer of plastic
108 la
spheres on cach tray to improve gas.„liquid contact. The liquid from
the scrubber will be clarified in a settling basin and recirculated to the
seruhber. Sludge will be pumped from one end of the settling basin, Draw-
a tinw sheet fbr this process including pumps and other auxiliaries„ List any
problem areas you think may exist, '2.7 The delivered equipment cost for a
fixed-bed carbon adsorption system with 8 capacity of 5000 actin was
$100,000 in 1994. Estimate the installed cost of 8 Similar system with n
capacity of 7500 acfm in 2000.
The incinerator system shown in Figure P2,'f can be equipped with a
catl.dyst bed and the same degree hydrocarbon removal accomplished
at a teinperature of 750 T. The catalyst and instalIatjon will cost
$250,000, and will have a service life of 3 years. If the sa3vege value
of the catalyst is $6000» fuel costs $7.00 per million Btu* and the
company requires a ROI betbre taxes, sh0üId the catglyst be installed?
2.9 The exhaust gas from g. rotary kiln hazardous waste incinerator
must be cooled from 1200 cc to 300 oc before entering a spray
dryer for acid gas removal. The engineer in charge of this
projew.t investigated several alternatives to cool the exheust gas
and hes concluded that the two most practical solutions would
be either the installation of 8 waste heat boiler or the installation
of spray injection system in which '.vater is sprayed directly
into the exhanst duct upstream of the spray dryer. Based on the
fot„rowing data, Which option should be selected?
Waste Heat Boiler
Cost installed $2 million

Service life 10 years

Operating cost excluding depreciation $0.4 milli0ü/yr
Generated steam value $0.4 million/yr
Water Injection System
Cost installed $0.5 million
Service life 5 years
Operating cost excluding depreciation $0„15
millionfyr
Noil}: additional water vapor added to the exhaust g8S will
increase the size of the spray dryer by 50%, The original spray
dryer cost was estimated to be $1„0 million with service lith of
10 years.
 Chapter Two What Process Design?
2.10 A dryer similar to the one described in Problem 2.1 is
processing a nonwoven sheeting material that has just* passed
through a benzene wash to remove surface contaminants, The
dryer operating conditions are as foliows:
• sheeting dry weight 4.0 ounces (02) per square yard (yd)
sheetintl enters dryer at a rate of 150 yd2}ti3in
Problems 109

sheeting enters dryer with 50% benzene by weight and exits

with 1.8% benzene a "pure" air enters dryer nt 220 OF and 1 atm
exhaust gas leaves dryer at 180 and 1 atill* containing 5.0%
benzene (volume percent)
Since benzene i.3 a hazardouš air pollutant (HAP), the
company decided to install a direct-flame thermal oxidizer to
destroy the benzene. Insurance regulations require thet the
stream to the oxidizer (the dryer exhaust) contain no more than
25% of the lower explosive limit (LEI,) of any combustible
gases. 'The Llö,L of benzene is 1.4% by volume in air.
a. CEAlculate the additional air needed to dilute the dryer
exhausc prior to feeding it into the oxidizer (lb/min and
scfm).
In your opinion, where should the air be added to the
system—. in front of the dryer or after? Give reasons for your
answer.
c. If the oxidizer operates at 1600 OF, and the cost of buying
natural gas to heat air is $8.00/million Btu of heat added to
the air¥ calculate the additional operating cost incurred due
to 3dding this extra dilution ($fhour),
2.11 A gae containing equal parts of methane? ethane, and ammonia
flaws at a constant rate through e laboratory water-based
absorption unit, which absorbs 96% of the ammonia and retain*
it in the liquid. No methane or ethane is absorbed into the water,
and no water evaporates into Ete gas. Initially, there was exactly
5%00 kg of water in the absorber, and at the end of 4 hours oc
operation, the liquid mass is 525 kg. Calculate the molar flow
rate (mol/hr) of the gas strenrn coming intu the absorber, and
the mole fraction of ammonia in the exit gas stream.
2.12 A 600 MW coal-fired power plant has an overatt thermal
efficiency Of 38%. It is buraing coal that has a beating value of
12,000 Btuflb, ash center;t o? a sulfur content of and a C02
 emission factor of' 220 Btu. Calculate the heat emitted to the
environment (fSt.G./sec), the coal feed rate (tone/ day), the
degree (%) of sulfur dioxide control needed to meet emission
standard of O, 15 1b SOßfmillion Btu of heat input, and the
(_ù02 emission rate (metric tož:s$day).
2.13 The Four Corners power plant is located in New Mexico near
where four Btates meet: New Mexico, Arizona, Utah, and
rado. It is one of the largest cnalNfired plants i" the world, -with
a generating capacity of 2r040 MW Assume that the coal it
burns has a carbon content of 75% and a heeting value of
13*000 Btuflb. Also, assume that the plnnt has a thermal
efficiency of 38% and that it runs at an annual average tperating
rate of of capac„ ity. Calculate its emissions of carbon dioxide
(tonne;yr).
Is

2.14 A large PC plant burns coal very rapidly compared to our ordi
experience, For example, when cooking on a charcoal grill for a
backyard picnic, we typically burn 5 pounds of charcoal in about an
hour, Consider the Four Corners PC plant described in Problem 2.13,
Calculate how fast- it burns coal (lb/sec),
2.15 Given that a wet scrubber had a TIC of $800,000 in 2001,
estimate the 2008 TIC using the two cost indexes presented in
Table 2.1 a Cae a reasonable method to estimate the TIC the
year 2013.

CB11Negie Mellon Ur.i¥E-rsity. '*Integrated Envi.r..uunent.al Contrai Model."

Accessed September 2009 http://www.íecnžmorùine„ct•nnf
Danielson, J. A) Ed. Air Potìution Engineering Manuel (2nd ed.). EPA
Publication AP40, Research Park, NC, 1973,
Davidson, C. f., H. So Hendrickson. C, T., Bridges, M. W., Allenby, BF R.,
Crittenden, Je Chen, Williamsz Ea, Alien, D. T& * Murphy, C. F., and Austim
SR "Adding Sustainability to the Engineer'8 Toolbox: A Challenge for
Engineering Educators," Environmenta Science & Technolagy, 41, July 2007.
Ffiir, J, R, Mixed Solvent Recovery Purification, Case Study Washing. ton University Design Series, 1969.
W. * Ed. Engineering Mantü.z£ (6th ed.). New Yark:

Heingobn, R. J, and Kabe5J R. L. Sources and Control ofAir Ponztion.

Upper
 Chapter Two What Process Design?
Saddle River, NJ: 19994
Iloag, Do Slatiek, E. R. Diaxide Emigsion fur Coal" '
Information Administration, originally published in DOW ÕIA:z012i (94/Ql)E
Augast 1994. Aceeased September 2009 from http:{f doe ,
gov/cneaf/coaVquurterIy/c02_nrticIelc02.htmå,
Humphreys3 K. K. , and Enßlishp L. M, , Eds. Project and Cost Engineers' Hand"
boob (3rd ed b), New York: Marcel Dekker, 1993.
Neveríi, R. B. , Price, J. t.JtF and Engdahl, K. L. "Capitat and Opernting
Costs o?
Selected Air Pollutíon Control Systems—VI' Journal of the Air
Pollution Control Association. 28(12), December 19781
Perry, R. Il. Industrial Vertiilciien (15th ed.). Lansingp Ml: American Conference
of Gcvernmental Industrial Hygienists, 1978.
Peters, M. Se, and K, D.. Plan! Design and Economies for Chem¿cc! Engineers (4th
New York: 1991.
Steiner, H. M. Engi\žeering Principles. New Yor¥a McGraw-Hill, 1992
U.S. Environmental Protection Agency (EPA). "Coa[ Utility îuvironmental
(CUECost).?' Accessed September 2009 from http://txww.epa..gov/ttn/

. EPA Air Control Cost Manual, edet

Research Triangle park* NC, January 2002. Accessed September 2009
from bttp://www.epa.gov/ttn/cateipreductg,html.

Particulate Matter

The dust in smaller arose,

Than those whick fluid ¿oåigs
John Arbathnt?t, A Dusty Day*" c. 1725
 3.1 Introduction
Particulate matter (FM) eonstitutes a major class of air pollution.
PM (often called particulates) comes in a variety of shapes and sizes,
and can be either liquid droplets or dry dusts, with a wide range of
and chemical properties. Particulates are emitted from many
different sources, including both cornbtßtion and noncombus„ tion
processes in industry, mining and/or construction Activities, motor
vehicles, tmd refuse incineration. Natural sources of particu• lates
include voleanoes, forest fires, windstor:ns, pollen, ocean spra½ and
so forth. Some information on particulate causes, sources, and
effects presented in Chapter 1. In this chapteG we present infor„
a.bont certain characteristics or particles and partieu!ate
behaeior in fluids, with emphasis org charaeteristics that pertain to
the imaortant engineering task of separating and removing particles
from a stream of

3.2 Characteristics of Particles

Before attempting the design of any collection device€ we must
obtain infirmation about the particles, the gag stream, and the pro„
cess conditions. Important pa?¥ictlhite characteristics include size,
size distribr.\tionz shape, density, stickiness, corrosivity, reactivity,
and tox•icity. Gas stre»m characteristics of importance ate pressure,
tempera. ture, viscosity, humidity, chemical composition, end
flammabiiity. Process conditions include gas flow rate, particulate
loading (mass concentration of particles in the gas stream), removal
effic.ìoncy
 Chapter Three Particulate Matter

requirements, and allotcvable pressure drop. Many of thÚ devicae in

use today fov separating particulat2s from gas exploit the vast
difference in the physical sine of particles aad gas moleculex in the
next few tions, we will further explore this characteristic and its
implications.
One of the most characteristics of a suspension of
particles is the size distribution of the particles. Figure 3.1 illustrates
the wide range of sizes (five orders of magnitude) of common
particie dispersoids. tThe common unit of measurement for small
particles the micrometer (pm), often referred to as the micron,] As
can be expected with such a wide range. of sizesr one type of
collection device might be better suited then others for specific
particulatÆ dispersoid. Further. more, any given collection device is
typically more efficient in collecting larger particles and less
efficient in collecting smeller ones, Thus, to calculate the overall
collection efficiency of a device, it is imperative to have good
information on the size distribution of the particles.
The overall efficiency can be calculated on a basis of total
number of particles collected or total mass collected. Generallyp
regolßtions are written based on maE9 collection, and efficiencies
are calculated on a mass basis, as shown ia Eq, (3.1).

where overall collection efficiency (fraction)

= total mags input rate, g/s or equivalent total
rnHäs ?missicn rate, g/s equivalent particulate
loading in the inlet gas to the device,
= particulEtte loading in the e„ut gas stream, g/m3
When the particulate size distribution is known, and the
efficiency of the device is known as a function of pnrticle size, the
overall tion efficieMcv can be predicted us follows.
.jmj where
T)j efficiency Of collection for the9th size range
m mass fraction of particles in thejt,h size range
The mass collection efficiency can be vustly different from the
number efficiency beenuse large Rte mueb more massive than small
 Section Characteristics of Particles
particles (mass varieg with the cube of the diameter). This is
illustrated in Example 3.1.
 Chapter Three Particulate Matter

Example
 Section Characteristics of Particles
A particle dispersaid consists of 300 spherical particles: 100
particles, 100 10-pm particl££, and 100 100.pyn particles. A
device is efficient on the 1-prn particles, 50% efficient on the
particles, and 99% efficient on the IOO•pm particEes. AIL particles
have the same denäity, and the particles have unit mass. Calculate
the number efficiency and the mass efficiency of collection.
Solution
We construct the following table, Note that the mass of 8 sphere
increases with the cube of its diametnr, so thet the total mass of all
particles is virtually equal to that of the 100-pm particles, or
mass units.

Number of Mass Ot
d'.pm particles Collected

Particles Collected
010 10
10 050 50 50(1000)
100 99 99(106)
Totals: 159 99<10P

overall

For simnlicity, previous discussion has been based on an implicit

assumption that particles are spherical. In fact, most particles are not
spherical* and they can be quite irregularly shaped. When choosing
the most characteristic "diameter" Of an irregularly shaped object
with ono or two dimensions siB1ifieantly larger than the other, we
must remember that our objective is to removr: particles from a
flowing stream of gas, Therefore, our choice of a size dimension
should be based on the particles' behavior in the gas, and not some
microscopic visual examination. For example, a sheet of notebook
paper witl settle to the gî0ünd* hut ball of. crumpled
notehook pauer will fall rapidlv. Particle density is also important,
For example, a golf ball and a ping pong have approximatelv the
same diametƼ but behave quite differently when tossed into thc
airt
 Chapter Three Particulate Matter
The aerodynamic diameter of a specific particle is the diameter
of a sphere with unit density (density of water) that will settle still
air at the soma rote as the particle in question (Friedlander 2000).
The aerodynamic diameter wilt be defined mathematically later thig
chapter in Eqt (323), but for now it is sufficient to realize that the
32 115
aerodynamic diameter is the proper diameter to apply to particles*
whether discussing collection efficiencies or settling in the lungèA
The aerodynamic diameter automatically results from aerodynamic
clagsi* fiers such ag cascade impactors.
As mentioned earlier, we must have adequute information on the
size distribution of particulates in the exhRust stream to he
controlled. A good means of obtaining this informBtion is by use of
a cascade impactor, A impactor is a device that separates and sizes
suspended oarticuiBtes in a manner similar the way that nested
screens seDarate and size samples of sand. However* a cascade
impactor separates particles by their aerodynamic diameuzrs rather
than their physical diameters. Air with particles is drawn through a
series of steges that consist of' slots and impaction plates (see Figure
3.2) Each successive stage has narrower slots and closer plates so
that each successive stage captures incrcasingky smeller particles.
The masses of particles collected on all stages are then used t")
determine the size distribution of the particulate stream.
Particulate distributions are often skewed toward much higher
numbers in the smaller size ranges—even when considered on mass
basis, A typical (normalixed) plot of Inass fraction of particles versus
size is presented in Figure 3.3. A skewed distribution such Figure
is considerabiy less convenient to work with than a normal
(Gaussien) distribution. Fortunately* in many cases, if one plots
mess frac„ ticn versus the logarithm of the particle dinmeter, the
result is a log-normal distribution, as shown in Figure 3.4.
A log-normal distribution such as shown in Figure 3.4 can be
characterized by two parameters: the (geometric) moqn and the (geomet„

Air flow Figure 3.2

Section Characteristics of Particles

Schematic diagram of a cascade impactar.

130 Chapter Three Particulate Matter
Tic) standard deviation. In the case of a log-normal distribution, the
relationships between the mean and the standard deviation are log =
log(d50 ) + logog (3.3)
and log (d50 ) — log Ug (8.4)

d84,i diameter such that particles constituting 84.1% of thc

total mass of particles are smaller than this size dä0
geometric mean diameter diameter such that
particles constituting 15.9% of the total mass of particles
are smaller than this size geometric standard deviation
Equations (3.3) and (3.4) can be written together as

d59
50
d159
A representation even more convenient than Figure 3.4 is a plot
of the particulate distribution data on log-probability paper. By
definition, if the distribution plots linearly, it is log-normal, and
d50 and c; can be used to characterize it. The linear plot is more
convenient for interpolation when only a few actual data points are
available. Exam„ ple 3.2 illustrates these techniques.

Diameter
Figure 3.3
A skewed particle distribution.
3.2 117
 Section Characteristics of Particles

Log dp Figure 3.4

A fog-nonnai particle distribution.

Example 3.2
The following data were obtained from a
cascade impactor (a par• tieulate sampling
device that classifies the particles by size).
Show that the distribution is log„normal, and find d50 and

Size Range, pm
Mass, mg 4. 179.5 368 276 73.5 18.5 920
5

Solution
We prepare a table of size range versus
cumulative percent less than the stated size, as follows:

Mass Fraction In Cumulative Percent Less

Size Range,urn Size Range, Than Top Size
0.0049 0.5
2-5 0.195 20.0
5-9 0.400 60.0
9—15 90.0

98.0
132 Chapter Three Particulate Matter
A plot of these data is presented in Figure
3.5, from which values of d84.1, d50 and d 15'*
can be read as 13.5, 7.8, and 4.6 microns. Thus

0" 13.5/7.8 = 1.73, or 7.8/4.6 1.70. (Owing to inaccur»cies of reading Buch

plots, valuea rarely agree perfeetly,) Our final unswerg are db0 pm
1.72 avergge of our two calculated values)

Example 3.3
Assuming that Figure 3.5 is characteristic of the particulates,
predict the overall efficiency of a device that has the
efficiency versus size relationship:

Solution
From Figure 325, we determine the fraction of particles in each
size ranøe, We mùiply the fraction in each size range by the effi*
eiency G:.r that size* and sum the results to give uveraÆ efficiency
for the device on the particulate dispersoid.
 Section Characteristics of Particles
Cumulative Percent
Size

3,2
134 Chapter Three Particulate Matter

3.3 Particulate Behavior in Fluids

Particulates often must be separated from a fl"id as part of a
pollution control system, In water and wastæwater treatmentp
particles of sludge are removed primarily by sedimentation tanks
and/or filters. In air pollutiott control, garticies can removed by
gravity settlers, cea• trifi.igal settàezs, fabric filters, electrostatic
precipitators, or wet gcrub„ bers. In Hll of these devices, particles are
separated from the Burasounding fluid by the application of one or
more forces. These forces, which include gravitational, inertial,
cer*rifugal, and electrostatic forces, cause the particles to
 Section Characteristics of Particles
away from the direction of the mean fluid flow, toward the direction
of the net force. The particles must then be collected removed from
the system to prevent ultimate re-entrainmenŽ into the fluid. Thug*
design and operation of ticulate poîlution control equipnxent
require R basic understanding of the characteristics of particleg and
Of the dynamics of particles in fluidse

The Drag Force

Consider a particle in motion relative to a fluid, Either the particle
or the fluid or both can be moving relative to an absolute frame of
reference. The fluid exerts a drag force on the particle which acts to
oppese the relative velocity of the particle, The drag force can be
represented as
Fž., 12

where
= drag force, N
CD drag coefficient projected
area of particle, 2
PF density of fluid, relative velocity,
ügually, the drag
coefficient must he determined
experimentally
Of the regime
aa characterized by tho Reynolds numbers For a particle, the
Reynolds number is defined as

(3.7)

particle diameter,
U = fluid viscosity kg/m-s
Figure 3.6 is a plot of the drag coefficient as a function of the
Reynolds number for three simple shapes,
Sectİon PgrticulHte
3,3 Behwq•iar in F[uids
 Section Particulate Behavior in FluidR
The fallowing discussion and equations apply to spheres at R.e½
no}ds numbers less than I—the flow regitne as the Stokes regime.
Theodore and Buonieore (1988) present more detailed tions
that aw vnlid for Reynolds numbera up to 105 (that is, where other
models the drag force are applicable). For rigid spheres and Of the
tions of motion results in Stokes law:
(3.8)
SubstituthŽg Eqs. (3.8) and (3„7) into (3
k
) results in

Rquation (89) approximates the solid line in Figure 3.6 for

Reyfaolds numbers less than 1 (that is, in the StÆ)kes regime). For
Reynolds numbers greater thnn I, Figure, 3.6 or its statistical curve
fit (Theodore and Buonicore 1988) should be used to obtain the drag
coef ficient CD Ibr spheres. The statistical fit of the curve in Figure
3.6 is

tog CD
185237
2
-
(3.10)

where LRe log(Re)

EquatiQn (3, 10) prœzides good fit of the "spheres" curve in Figure
3.6 for Reynolds numbers greater than 1,0, but its complexity it
usoful only in computer
Another empirical approach recommended by Theodore and Buoni„ core
(1988) is to divide Figure 3.6 into three regions based on Reynolds number.
The general equation for dreg coefficient is then given by
(3.11)
where and are constants (given in Table 3.1)
There is a lower fimit for particle size in Stokes law. When the
particle is large relative to the mean free pt:th of the gasp the fluid
can be modeled as a continuum. However, the particle diameter is
of the same magnitude as the parbicle no longer "sense«" the guid
as a con:.ínuum, but as discrete molecules, The particle is able to
"slipn between the gas molecules and this slippage reduces the
effective drag on the particles predicted b" Stokes law. The
138 Chapter Three Particulate Matter
Cunningham corm rectüm factor corrects the Stokes drag
coeffieiel±t for this effect. The corsection factor is reported by
Wark, Warner, and Davis (1998) as

1.257+0.40 (3A2)
3.3 123

Table 321
Constan(s
Eq (3, 11)

'The mean free path can he obtained from the kinetic theory of
gases ag

0.499PJ8MW/rtRT
where moan free path, m
P nbso\ute Pa
R universal gas constant, 8314 J/kgmol„K
molecular weight, kg/kgrnoi
— absolute temperature, K
= absolute viscosiR.ÿE kgfm-s

The variation of the. Cunnizghazì factor with particle

size is shown in Table 3,2. For particles smaller than 1 micron, the
81ip correction factor is always significaat, hut rapidly approaches
1.0 ag particle size increases above 5 The corrected dreg coefficient
CD is the Stokes drag coefficient divided by the Cunningham
correcLion factor. Because the Reynolds number is almost always
small when C differs significantly from we can correct (3,9) as
follows:

co 24
C C(Re)

Table 3.2 Cunningham Correction Factor at atm and 25

 Section Particulate Behavior in FluidR
22,5
0.05 5.02
2.89
0.50 ,334
1.0 1.166
20 1.083
5.0 1.033

10,0 017
External Forces
For a relative motion to exist between a fluid and a freely
pended particleb at least one external force must exist. Considering
only one net external tarce, which is opposed by the drag force
Newton's second of motion can be written as

(3»15)
Fe net external force
M= of particle
For a spherical particle in the Stokes regime, Eq. (3.8) and the
prochlct of volume times density for a sphere can be substituted into
(3515) to vield

(3.16)
where pp = particle density
The grouping ppd"p/18p has dimension of time, and is n
tic time •r for Eq, (3*16) The charHcteristic time is
merely TC. This characteristic tixne is the distinguishing parameter
of particle seemingly different systems (with
different particle sizes and densities, and different fluids) will
behave in the same manner iff is the same for both systems. We can
rewrite Eq. (3.16) as

(3.17)
Equation (3.17) the basic different;el equation governing Lhe
motion of 8 particle i.n a fluid (in the Stokes regime), The term
Fe/Mp is the net external force per unit Eness npplied to the pmrtiele.
The net external force can be due to an e!cctroatat.ic field, a
centrifugal field, a gravitational field, and so forth. In the next
140 Chapter Three Particulate Matter
section, v,že will use Eq. (3.17) to derive a for the Stokes settling
velocity of a particle in quiescent air (that in pure gravitational field).

Gravitational Settling
If a particle is released into air with initial down„ ward
velocity of zero, and with only gravitational force nctir,g on the
partiele, the equation Of motion is

(3.18)
where g is the gravitational constant
33 125
The term (pp — pp)/pp corrects for the
buoyancy of the particle
in the fluid. For a (3.18) i solid particle in a
gas, this term is nearly unity Rnà
can often be ignored. The solution to
givenby
(3.19) where v: is the
terminal settling

(8.20)
18B
Note that has units of tižne, so that any consistent set of units can be
used for density, diameter, and viscesity.
From a plot of Eq. (3,19), we cnn easily
estimate the time when terminal velocity is
atbained (see Figure 3,7), From Figure 3.7, we
can see that after four characteristic times,
the particle's velocity is virtu„ ally equal
to its terminal velocitye
We often ignore the transient (acceleration) portion of a particle's
settling time. In Table 3.3, we have
calculated the terminal velocity and the
characteristic time for several part.icfes,
B«eaurse t' is small, the terminal velocity is attained in
a few milliseconds or legs, we arc justified in ignoring
the transient. portion of the settling time.
For a particle larger than 10—20 microns
settling at its terminal velocity, the
Reynolds number is too bif'h for the Stokes
regžme analy to be valid. For these larger
 Section Particulate Behavior in FluidR
particles, empirical Elean.s are required ta obtain
the settling velocity, that we must use either direct experimental
results or empi?ical models in place of EA. (3.9) in the
development of equations analogous to F„ü.
(3.20). For example, for 2 < Re < 500, the
terming[ velocity [using F,q, (3.11) is by
Theodore and B»onieore (1988) as
3.7
1
PF (3.21)

Figure 3.7
Dimensionless
parliCie velocity
versus dimensionless
time, Table 3.3
Settling Velocities
and Characteristic
Times for unit
3 Density
z Spheres in Air at 25
Q
C and •i etm

Also, for 500 < Re < 200,000, the terminal velocity is given by those
authors as

0.5 (3.22)
The terms in Fzqs. (3.21) and (3.22) must have a
consistent set of units; one set is {Vsec
for for dp. for p, Ysec2 forg, and Ibm/ft-sec for
N.
For quick reference, it is usually more
convenient (and often just accurate) to use
a chart (such as Figure 3.8) portraying actual
142 Chapter Three Particulate Matter
experimental settling velocities, Note that
the top and left-side axes must be used for
larger particles, and the bottom and right-
side axes are used for smaller particles,
In Section 32, we introduced the tarm
aerodynamic diameten We can now define this term
mathematicallv. The motion of a Stokes ticle depends
only on its value of t'. From the definition
of we can see that is the ratio of the particle properties (Cppdp)
to e fluid property (viscositv), Thus, systems with the same, value of
t' should exhibit the same aerodynamic behavior
regardless of their separate values of p, pp,
and d The aerodynamic diameter is defined as
the z: 1000 kg;m3) that has the game gettlivzg
velocity as the particle question. In Eq.
(3.23), any consistent set of units cen be
used; one such set is indicated.

181.1b': (3.23)

where da aerodynamic
diameter, m pas
viscositYÞ lcg/m„s
settling velocity, rn/s
density of water, kg/m3 g =
gravitational
acceleration, m/s
 Section Behevior in Fluids
3.3 ParticulAte

Spherical particle diametert dB pm

200 1000

810
Spherical particle diameterr ðp um
Figure 3.8
Terminal settling ve'ocity Of spherical paflicles in air at STP (particte
density given in g/cc).
{Adapted from Watk. •tind 1998.}

Collection of Particles by Impaction.

Interception, and Diffusion
a flowing fluid approaches a stationary object such as a
fabric fitter large wetær droplet, or metal plate, the fluid flow
streamlines diverge around that object Because of their inertia,
pa.rüt•tES in the flüi.d not foìžow streamlines exactly, but will tend
144 Chapter Three Matter
to continue in their original directions, If the particles bttve enough
inertia and are located close enough to the stationary object* they
collide with the object, and can be collected by This phenomenon
is depicted in Figure 3,9.
Paytic•uiatc

Figure
Conection of peelic',es on a stationary object,

Impaction of pmsticles occnrs when the center of mass of a

particle that is diverging from the fluid streamlines strikes stationary
abiect. Interception otcurs when the petrtiele's center cf mess
closeEy misses the abject, but, becauze of its finite size, the particle
strikes the object. Collection of particles by diffusion occurs when
small particulates (which are subject to random motion about the
mean path and would usually miss the object even considering their
finite size) hap. pen to "diffuse?' toward the object while passing
near it. Once striking the object by any of these means, particles are
collected only if there

forth) strong enough to hold them to the surface.

A simple means of explaining impaction is with the concept of
stopping distance, If a sphere in the Stokes regime is projected with
an initial velocity vo into a motionless fluid, its velocity as a function
of time (ignoring but dreg forces) is
(3.24)
The totRl distance traveled hy the particle before it comes to rest (the
stopping distance) is

('3.25)
 If the particle stops before striking the object, it then be swept
around the object by the altered fluid flow. Since is small, ,xstop is
also small. For instance, if a 1.0-pm particle with unit density is
projected 10 into it will stop after traveling 36 microns (less than
hBlf the diameter offi human hair).
Section 3.4 Overview of Particulate Control Equipment 129
Ali impaction parameter Nt can be defined ag the ratio of the
stop. ping diätnnce of a particle (based on upstream fluid velocity)
to the diameter of the stationary object, or

(3.26)
s
It is most Of the particles will it.tpaet the ObjecL. If N! is very small
p most of the particles will follow the fluid around the object?

3.4 Overview of Particulate Control

Equipment
different, Of particulate
including mechanical separators (such as gravity settlers, or
cyclones), fabric filters, electrostatic precipitatot¾ and wet
scrubbers. In thia sec. tion, we will briefly introduce each ofthe
l.najor types of particulate con• trol equipment (each will be
discussed in detail in subsequent chapters).
A gravity settler is merely a large chiÌmber in which the gag
velocity is slowed, allowing particles to settle out by gravity. A
cyclone removes particles by causing the entire gas stream to flow
in a spiral pattern inside of tube, Owing to centrifugal the larger
particles move outward and collide with the wall of the tube. The
par. ticles slide down the wail and fall to the bottom of the cyclone,
where they are removed, The cleaned gas flows out of the top of the
cyclone.
A fabrie filter (baghouse) operntee on the same principle as a vncuum
Air carrying dust pnrticlea is forced through a cloth bag. As the air pasaes
through the fabrict the dust accumulates on the cloth, providing a cleaned
air The dust is periodically removed from the cloth by shnking or by
reversing the Rir flow,
An electrostatic precipitator applies electrical force to separate
particles from the gas stream, A high voltage drop is established
between electrodesp a-cd particles passing through the resulting
146 Chapter Three Matter
electrical field Bcquire a charge. chRrged particles attracted to the
through the devicz Periodically, the plates are cleaned by rapping to
shake off the layer of dust that accumulates. The dust is collected in
hoppers Bt the bottorn of the device.
A wet scrubber employs the principles of impaction and interw
ception of dust purticles by droplets of water, The larger, heavier
water droplets ere easily separated from the gas by grevity. The solid
particles then be independently separated from the water, or the
water cat: be otherwiae treated before re-use or discharge,
Although broad generalities are often best avoided, some general
Etatements might help tn put the various types of particulate control
equipment before embarking on detailed discussion of egch type in
the next few Mechanical collectors typi„
cally much less exprnaive than the others, but are usuaüy only
atelv efficient. They are much better for large particles than for fine
dusts, and are often used as precleaners for the more-efficient final
control devìee*, especially when the dust loading is high. Fabric
filters tend to have, very high efficiencies but axe costly, Fabric filters
usually are limited to dry, low-temperature conditions, but can handle
many different types of dusts, Electrostatic precipitators (ESPs) can
handle very large volumetric flow rates at low pressure drops, and
can achieve very high efficiencãea, However, ES£s are costly and are
relatively inflexible to changes in process operatmg conditions. Wet
scrub. bers can also achieve high efficiencies and have the major
advantage that some gaseous pollutants cun be. removed
simultaneously with the particulates. However, wet scrubbers can be
very costly to operate (owing to a high pressure drop), and they
produce a wet sludge that can present additional disposal problems,
All things consideredp there is no way to decide, a priori, whRt
the best system is. Each air pollution control probiem is unique and
demands an engineered solution, Thug, there is ample opportunity
design engineers to exercise their creativity, even when dealing with
problems and/or "off the shelf'"
The overall collection efficiency of a system composed of two or
more devices in series is not simply the sum nor the product of the
efficiencies of each device. Each device's efficiency is based on the
mass loading of particles entering that device, but the overall system
efficiency is básed on the total mass collected a fraction of the total
mB88 entering the first device.
'The simplest way to approach this problem is to define the pene„
tration of a device as thc mass fraction that is not collected (that is,
the fraction that penetrates through the device). Thus,
(327)
where Pt = penetration (fraction)
The ovoraîl penetration a system simply the product of' the
penetrations of all of the individual devices; that

(3.28)
whet"
Pto overall penetration
Pti penetration of device i
148 Chapter Three Particulate Matter
The overall collection efficiency of the system is simply I Pto, as
depicted in i'Ygu..re 3.10. In applying this approach, care must be used in
determining the efficiency of any downstream device because the larger,
easier-to»collect patt'icles will be removed by the upstream device.
Problems 131

Figure 3.10
Efficiencies and penetrations of a pafliculate control system Wilh
devices in series.

problems
3.1 A stream of gas tbwirsg at 3000 BCfin has a particulate loading
of
50 gr/ft3 (grains/cubic foot), What is the 8110wable exit loading if 95%
collection efficiency is required? What is the mass of partieles
collected, kg/day?
3.2 A coal with 6% ash find a heating velue Of 13*000 Btullb is
burned in a new power plant. Assume that 30% of the ash falls
out bottom ash in the furnace. Calculate the efficiency of an ESP
required to meet federal new source performance standards,
3.3 *'The following data were obtained from a cascade impactor:
 Size Range, um 4—8 8—16 30„50
Mass, mg 25 195 100 75 30 5
Is the distributiøn log-normal? If so, estimate and (160.
3.4 Consider 8 cyclone that has the following theoretical efficiency
versus particle size relationship:
Size Range, 0-2 2-4 4-7 7-10 10-15 15-25 25-40 40--60

Efficiency, % 10 25 45 70 85 95 97 98 100
Calculate the overall efficiency of this cyclone for the particle
distribution of Problem 3.3.
3.5 Consider an ESP that has the following theoretical efficiency
versus particle size relationship:
Size Range, 0-2 2-4 4-7 7-10 10-15 15-25 > 25

Efficiency, % 50 80 95 97 98 99 100
Calculate the overall efficiency of this ESP for the particle
distribution of Problem 3.3,
3.6 For the particle distribution given in Figure 3.5* calculate the
overall efficiency of the cvclonc of Problem 3.4.
3.7 For the particle distribution given in Figure 3.5, calculate the
overall efficiency of the ESP of Problem 3.5.
3.8 A particle distribution is log-normal with a d50 of 8.0 microns,
and a ag of 3.0. Plot this distribution and determine the mass
percent of particles below 2.0 microns,
3.9 A particle distribution is log•normal with e d50 of 6.0 microns
and a ag of 2.0. Plot thig distribution and determine the mass
perceñt of particles below 2.0 microns.
3.10 Calculate the Cunningham correction factor for a particle with a
0.03-micron diameter in air at I atm pressure and 150 ec.
3.11 Calculate the Cunningham correction factor for a particle with a
0.03-micron diameter in air at 1 atm pressure and O DC,
3A2 Assuming Stokes behnvior, calculate the terminal settling veloc„
ity in standard air for the following particles: (a) diameter = 20
specific gravity 2„0; (b) diameter = 5 p.m, specific gravity
0.8; and (c) diameter = 40 pm, specific gravity = 2.5.
150 Chapter Three Particulate Matter
3.13 Assuming Stokes behavior, calculato the terminal settling
velocity in standard air for the following particles: (R) diameter
10 pm, specific gravity = 1.2; and (b) diameter 4 specific gravity
2.0,
3.14 Check your result for the 40-pm particle of Problem 3.12, using
Eq. Explain any differences in the results.
3.15 A gas stream with particulate loading of 20.0 g/m3 is passed
through a 70%-efficient cyclone followed by a 95%-efficient
ESP. Calculate the overall efficiency of the system.
3.16 Particulate removal on certain gas stream must be 985% efficient
to kneet standards. If a 60%-efficient cyclone prccleaner is with
a wet scrubber, what is the required efficiency of the scrubber?
PTO ems

8.17 Calculate the overall efficiency of a system composed of the cyclone

of Problem 3.4 followed by the ESP of Probiem 3.5, operating on the
following particle distribution:
Size, 0-2 2-4 4-7 7=10 10- 15 15-25
Weight, % 6 23 30 18 13 8 2
Remember to consider that the size distribution of the remaining
particles will change after having passed through the cyclone,
and will be different when entering the ESP.
3.18 Assume that in the ambient air near industrial plant on a certain
dav, a sample of the suspended particulate matter has a size
distribution as follows:
Size Range, þrn 0—2 2—6 6—10 10—16 16—30

Mass, mg 15 20 45 40 20 10
A simple high-volume sampler yields a value of 80 pg/m3 for
total suspended particulates for this air. Predict what result 8
PM-IO sampler would yield.
3.19 A stream of air at 25 "C and 1 atm is laden with particulate matter
(unit density spheres) of only two sizes: 1 pm and 5 pm. The
stream flows through a pipe at an average velocity of 20 m/s.
The pipe exhausts into a water mist scrubber where the water
droplets average 100 um in diameter. Based only on a
comparison of their impaction numbers, what will be the ratio of
particles collected (5 pm to 1 pm)?
3.20 A gas stream flowing at 1000 efm with particulate loading of
 400 gr/ffß discharges from a certain industrial plant through an
80%-efficient cyclone. A recent law requires that the emissions
from this stack be limited to 10.0 lb/hr, and the company is
considering adding wet scrubber after the cyclone, What is the
required efficiency of the wet scrubber?
3.21 Dusty air at a fertilizer plant flows through a 70%-efficient
cyclone and then through an kSP. The inlet air to the cyclone has
a dust loading of 50 grains/cubic foot.
a. In order to meet R control standard of 98.5% a)llection
efficiency for the fertilizer plant as a whole, what is the
allowable concentration of dust (in grains.lcubic foot) in the
air that exits from the ESP?
b. The outlet air from the ESP actually contains dust at 0.50
grainslcubic root. Calculate the efficiency of the ESP,
322 A 5-micron droplet of water is being carried in standard Hir
toward a 500-micron drop of water at a relative velocity of 40
m/s. Will the two drops collide? Support your answer with
calculations.
 Chapter Three Particulate Matter

3.29 Based on Figure 3.5, what percentage of the mass of particles is in the size range 4 to 9 micrometers?
3.24 Estimate the terminal tRettling velocity of a unit*density sphere with a diameter of 2 falling through
standard air. First use Figure 3.8, then repeat using either Table 3.3 or one of the equations from the text,
Are there any significant differences? Now repeat the process for a 200 pm unitwdensity sphere. Explain
any significant, differences
3,25 A sample of dust v,'as taken aerodynamigellye dust is characterized as being Ing-normal with a geometric
mean of 6.0 pm. Also, 5% of the particles were larger 25 p.m. Estimate the PM-2-5 fraction (that the
mass percent of this dust that is less than 2.5 pm).
3.26 An air pollution control system at a plant eonsiBts of a cyclone prec]euner followed by ESP. The air
entering the cyclone has a PM concentration of Z5 The concentration in the gases exiting the ESP is 0.02
g/mó .
a, What ig the overnll collection efficiency of the system?
b. if it is known that the cyclone is 75% efficientë what is the lection cfficíency of the ESP?
3,27 Consider the system described above in Problem 3.26. If we can increase the cyclone efficiency to 85%, what would
be the new required ESP efficiency to maintain the same emissions (Or02 ym 3)?
3.28 A sample of dust is taken aerodynamically and is charecterized as being log-normal with a geometric
mean of 7,0 pm. It is known thet of the particles are larger than 30 Estimate the percent of particles that
fall into the PM-2.5 category.

References
Friedlander, Sm K. SmokF,» Dusl and Haze (2nd ed). New York: Oxford University Press, 2000„
Lappie, C. E. Stanford Research Institute 5(95),
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Industrial end Engineering Cltemistry, 32(5), 1940t
Theodore, L. and A. J. Air Pollution Control Equipment for uloles, Vol. l. Boett Raton, FL: CRC Press,
1988.
I-S Warner, C. F. , and Davis, W. T. Air Pollution—lis Origin end (3rd ed.). Upper Saddle River, NJ:
Prentice„HaùE 1998.