Gilbert masters's Introduction to Environmental Engineering and Science part-1

Introduction to
ENVIRONMENTAL
ENGINEERING
AND SCIENCE
Third Edition
GILBERT M. MASTERS I,
AT-':;l.Tn-':;1 1 D "):;'1 A

When you can measure what you are speaking about, and express it in numbers,
you know something about it; but when you cannot measure it, when you
cannot express it in numbers, your knowledge is of a meagre and unsatisfactory
kind; it may be the beginning of knowledge, but you have scarcely, in your
thoughts, advanced to the stage of science.
- William Thomson, Lord Kelvin (1891)
xii Preface
lib
M,
I
M,
The rapid expansion of the environmental field, and the increa ing awarenes
that broad mterdlsclp]mary perspectives are needed ro successfully tackle environ-
~ental problems, makes It particularly difficult (but also importam) to write a text
t at ISup-to-dare, readIly accessibleto a broad range of students, technically sound
and sufficiently succmct so as not to overwhelm the wallet or motivation of the
reader. To theI degree that this text fulfillsthesegoals, it is due ro the intentional and
urunrennona mpur from If'
leagues and students In a ticulgroup a very open, mtelligent, and dedicated col-
and Eduardo Saez of'the Upamcuar, wfeAwouldlike to thank Professors Bob Arnold
mvefSltya mona for rhei itical i .
oversight At Stanford U' . D R err cnnca mpur and consisrenr
. mversIfY r. oyal K p d ided i . .
and assistance with the chapte ' I' a peru proVI ed invaluahlr- insights
rs on air qua tty and li h
publisher would also like to thank th f II' c rrnarec ange. The authors and
Loyola Marymount University B e a ~wmg rev.ewers.Joseph C. Reichenberger,
Raymond W. Regan, Sr., Penn' Sta:~Ctni~ Rmmann, Anzona State University,
Montgomery Brion, University of Kentuckersuy, Keith Stolzenbach, UCLA, Gail
Umversity, Ronald R. Cohen C I d S Gtegory G. WIlbur, Oklahoma State
Berkeley, John Novak, Virgini~ T~cohraadA
C
001 of Mines, David L. Sedlak, UC
M ' , an very Demo d U' , f .
osr Importantly both autho ff . nc, mversiry 0 Michigan.
K h ' rs a er special th k '
aren, w 0 served as ghost~write di . an s to our WIves, Mary and
rhi ff Thei rs, e itors motlVat d .IS e art. elf support enco ' Drs an , too often victims of
I . ,uragement and . h '
camp enon, ' patIence ave been essential ro its
Mass and Energy
Transferw,
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1.1 Introduction
1.2 Units of Measurement
1.3 Materials Balance
1.4 Energy Fundamentals
Problems
v
Gil Masters
Stanford University
Wendell Ela
University of Arizona
r
1.1 I Introduction
CHAPTER
1
Although most of the chapters in this book focus on specific environmental prob-
lems, such as pollution in surface waters or degradation of air quality, there are a
number of important concepts that find application throughout the study of envi-
ronmental engineering and science.
This chapter begins with a section on units of measurement. Engineers need to be
familiar with both the American units of feet, pounds, hours, and degrees Fahrenheit
as well as the more recommended International System of units. Both are used in the
practice of environmental engineering and both will be used throughout this book.
Next, two fundamental topics, which should be familiar from the study of ele-
mentary physics, are presented: the law of conservation of mass and the law of con-
servation of energy. These laws tell us that within any environmental system, we
theoretically should be able to account for the flow of energy and materials into, and
out of, that system. The law of conservation of mass, besides providing an important
tool for quantitatively tracking pollutants as they disperse in the environment,
reminds us that pollutants have to go somewhere, and that we should be wary of
approaches that merely transport them from one medium to another.

2 Chapter 1 Mass and EnergyTransfer
Libr
Mas
In
Mas
, In a similar way, the law of conservation of energy i al 0 an nnal a ounr-
mg tool with special envrronmenrn] implications, When coupled with Other thermo-
dr~~lcl P~Inclples, It WIllbe useful in a number of application in ludtng rhe rod"
o goa c irnate change, thermal pollution, and the di per ion of air polluranlS. '
1.2 Units of Measurement 3
TABLE 1.2
Common Prefixes
Quantity Prefix Symbol
10 15
femro f
10-12
pICO P
10-9
nano n
10-6
mICro J.L
10-3
milli m
10-2
centi c
10-1
deci d
10 deka da
102
hecro h
103
kilo k
10' mega M
109
glga G
1012
tera T
1015
peta p
1018
exa E
1021
zetta Z
1024
yotta y
Ir
I~
1
Wei
T
6
In the United States environm I ' ,
the U,S, Customary System (US~~a q~a~tItles are measured and reported in both
is important to be familiar with b) ~n I t ~ InternatIonal System of Units ( IJ so II
although the U.S. system wil! b ot
d
" n tms book, preference i given 10 I units,
, fIe use In some CI -r- b .
sion actors between the 51a d uses rcumstances, I a le 1.1 II t conver-
'II n systems f f
WI be encountered. A more e t ' b or some 0 me mo r ba ic units roar
"U fie x enSIveta Ie of co ' , , .
se u auversian Factors." nversions 1 given In Appendix A1
In the study of environmental en inee ' "
extremely large quantities and gl nng, It IS common ro encounrer both
roxi b extreme y small Th
OXICsu stance may be meas d i ones, e concentration of some
, ure In pans per b'W (
COUntrys rate of energy use rna be I ron ppb), for example, whereas a
(lerawans). To describe quantitie; th measured In lhousands of billions of warts
to have a system of prefixes thaI at may take On such exrreme values it is useful
f accompa h ' ,
pre ixes are presented in Table 1 2 ny t e urnrs, Some of the most important
Ofte " h ..n, It IS t e <oncent .
interesr U' h ratIon of some sub "
b d' SIng t e metric syslem i ' h srance In air Or warer thar is of
ase On mass (II n eit er medi
mol) hi h usua y mg Or g) velum ( IlIum, concentrarions may be
,W IC can le d ' e usua y 1 3 II
thar one I a to some confusion It 01 m ), or number (usua y
(6.02 X 1O~0 e of any substance ha; A~ay dbe ,helpfUl ro recall from chemistry
moleCUles/mol) and has a ga ro s number of molecules in it
L' 'd mass equal to its molecular weighr
IqUi S .
micrograms (!'-g), or moles (mol) of substance per liter (L) of mixture. Ar times, they
may be expressed in grams per cubic merer (g/m ').
Alternatively, concentrationsin liquids are expressed as mass of subsrance per
mass of mixrure, with the most common unirs being parts per million (ppm) or parrs
per billion (ppb). To help put rhese unirs in perspecrive, 1 ppm is abour the same as
1 drop of vermouth added to 15 gallons of gin, whereas 1 ppb is abour rhe same as
one drop of pollutant in a fairly large (70 m3
) back-yard swimming pool. Since mosr
concentrations of pollutants are very small, 1 Iirer of mixture has a mass thar is
essentially 1,000 g, so for all practical purposes, we can write
Ed,
Sen
Ed;
Ser
AIl
Co
Ar
M,
M.
SeI
AI
',1
re
I
A
P
p
1
il
e
t
ConCentrations of sub
Stances dis I d'
mass or number er u . So ve In Water
p nIt volume of mixture M are usually expressed in rerms of
TABLE . OStoften th ' )
1.1 e umrs are milligrams (mg,
~Some Basic Units and C In unusual circumstances, the concentration of liquid wastes may be so high
thar rhe specific gravity of the mixture is affected, in which case a correction to (1.1)
and (1.2) may be required:
mgll = ppm (by weighr) x specific gravity of mixture (1.3)
1 mgIL = 1 glm3
= 1 ppm (by weighr)
1 !,-glL = 1 mglm3
= 1 ppb (by weighr)
Mass meter ConverslOn factor
Ternperature kilograrn krn 3.2808
AI CelSius g
ea 'C 2.2046
Volurne square rneter rn' 1.8 (OC)+ 32
E Cubic meter
nergy kilojoule rn3 10.7639
POwer k 35.3147
VelOCity Watt J 0.9478
Flow rate rneterlsec 3.4121
D . meter]/sec rnls
enslty kilogr rnJ rn3/s 2.2369
4S«. a meter] 35
AppendIX A fora m kg/m3 .3147
Ore COlllplete list. 0.06243
uses unirs
fr
Ib
OF
ft2
ft3
Btu
Btu/hr
milhr
ft3/s
EXAMPLE 1.1 Fluoridation of Water
The fluoride concentration in drinking water may be increased ro help prevent
rooth decay by adding sodium fluoride; however, if roo much fluoride is added,
it can cause discoloring (mottling) of lhe teeth. The optimum dose of fluoride in
drinking warer is abour 0.053 mM (millimolelliter), If sodium fluoride (NaF) is
purchased in 25 kg bags, how many gallons of drinking water would a bag !reat?
(Assume there is no fluoride already in lhe water.)Ib/fr'
(1.1)
(1.2)

4 Chapter 1 Mass and Energy Transfer 1.2 Units of Measurement
L
~ R = ideal gas constant = 0.082056 L· arm- K-1 • mol-1
T = absolute temperature (K)
The mass in (1.6) is expressed as moles of gas. Also note the temperature is
exptessed in kelvins (K), where
Solution Note that the mass in the 25 kg bag is the sum of the rna s of the
sodium and the mass of the fluoride in the compound. The atomic weight of
sodium is 23.0, and fluoride is 19.0 (values are given in Appendix B), so the
molecular weight of NaF is 42.0. The ratio of sodium to fluoride atoms in aF
is 1:1. Therefore, the mass of fluoride in the bag is
19.0 g/mol
mass F = 25 kg X = 11.31 kg
42.0 g/mol
Conve~ting the m~lar .concentration to a mass concentration, the optimum con-
centration of fluonde III water is
There are a number of ways ro express pressure; in (1.6), we have used atmospheres. One
atmosphere of pressure equals 101.325 kPa (Pa is the abbreviation for Pascals). One
atmosphere is also equal to 14.7 pounds per square inch (psi), so 1 psi = 6.89 kPa.
Finally, 100 kPa is called a bar, and 100 Pa is a millibar, which is the unit of pressure
often used in meteorology,
x K = °C + 273.15
F = 0.053 mmollL X 19.0 g/mol X 1,000 mg/g
1,000 mmollmol = 1.01 mgfL
The mass concentration of a substance in a fluid is generically
EXAMPLE 1.2 Volume of an Ideal Gas
c=m
V (1.4)
where m is the mass of the substance and Vis the v I ._
and the results of the t I I ' 0 ume of the fluid. Using (1.4)
wo ca cu anons above the I f h b
treated is ' vo ume 0 water t at can e
Find the volume that 1 mole of an ideal gas would occupy at standard tempera-
ture and pressure (STP) conditions of 1 atmosphere of pressure and O°C temper-
ature. Repeat the calculation for 1 atm and 25°C.
Solution Using (1.6) at a temperature of O°C (273.15 K) gives
V = Imol X 0.082056 L'atm'K-
1
'mol-
J
X 273.15 K = 22.414 L
1 atm
V = 11.31 kg X 106
mg/kg
1.01 mgIL X 3.785 Ugal = 2.97 X 106
gal
The bag would treat a day's supply of d - ki
the United States! rm 109Waterfor about 20,000 people in
and at 25°C (298.15 K)
V = 1 mol X 0.082056 L· atm' K-
1
• mol-
1
X 298.15 K = 22.465 L
1 atm
Gases
For most air pollution work it is custo
I· , mary to ex II
vo umetnc terms. For example th . press po utanr concentrations in
II' (ppm} i , e concentration ofrru Ion ppm) ISthe volume of poll t . _ a gaseous pollutant in parts per
u ant per million I f
1 I vo umes 0 the air mixture:
vo ume of gaseous pollutant
10
6
volumes of air = 1 ppm (by volume) = 1 ppmv (1.5)
To help remind us that this fraction is b d
to the ppm, giving pprnv, as suggested ina;~5)n volume, it is common ro add a "v"
At ~Imes,conc~ntrations are expressed' . .
or mg/m . The relationship between p adsmass per umr volume, such as J-Lg/m3
perature and I I' pmvan mg/rn! d d
'. mo ecu ar weIght of the 011 _ epen s on the pressure, tern-
that relationship' p utant. The Ideal gas la h I bli h. w e ps us esta 15
From Example 1.2, 1 mole of an ideal gas at O°C and 1 atrn occupies a volume
of 22.414 L (22.414 X 10-3 rn'}. Thus we can write
1 m3
pollurant/TU'' m3
air mol wt (g/mo!) 3
mg/rn ' = ppmv X ppmv X 22.414 X 10 3 m3/mol X 10 mg/g
Of, more simply,
3 _ ppm v X mol wt
mg/m - 22.414 (at O°C and 1 atm)
Similarly, at 25°C and 1 atrn, which are the conditions that are assumed when air
quality standards are specified in the United States,
3 _ ppmv X mol wr (250C d 1) (1.9)
mg/m - 24.465 at an armPV=nRT
where
P = absolute pressure (atm)
V = volume (rrr')
11 = mass (mol)
(1.6)
In general, the conversion from ppm ro mg/rn:' is given by
3 ppm v X mol wt 273.15 K
mg/m = 22.414 X T(K) X
P(atm)
1 atm
(1.10)
5
(1.7)
(1.8)

6 Chapter 1 Mass and Energy Transfer 1.3 Materials Balance
Control volume
boundaryEXAMPLE 1.3 Converting ppmv to mg/m'
The U.S. Air Quality Standard for carbon monoxide (based on an 8-hour mea-
surement) is 9.0 ppmv. Express this standard as a percent by volume as well as in
rng/nr' at 1 arm and 25'C.
.......... Accumulationt
Inputs~ --t-+- Outputs
Reactions: Decay +
and generation t
Solution Within a million volumes of this air there are 9.0 volume of CO, no
matter whar the temperature or pressure (this is the advantage of rhe ppmv units).
Hence, the percentage by volume is simply
CO
9.0
percent = 6 X 100 = 0.0009%
1 x 10
To find the concentration in mg/rrr', we need the molecular weight of CO, which
tS28 (the atomic weights of C and 0 are 12 and 16, respectively). Using (1.9) gives
CO = 9.0 x 28 _ 3
24.465 - 10.3 mg/m
Actually, the standard for CO is usually rounded and listed as 10 mg/rrr',
the region, as is suggested in Figure 1.1, we can then begin to quantify the flow of
materials across the boundary as well as the accumulation and reaction of materials
within the region.
A substance that enters the control volume has four possible fates. Some of it
may leave the region unchanged, some of it may accumulate within the boundary,
and some of it may be converted to some other substance (e.g., entering CO may be
oxidized to CO2 within the region). There is also the possibility that more substance
may be produced (e.g., CO may be produced by cigarette smoking within the con-
trol volume of a room). Often, the conversion and production processes that may
occur are lumped into a single category termed reactions. Thus, using Figure 1.1 as
a guide, the following materials balance equation can be written for each substance
of interest:
FIGURE 1.1 A materials balance diagram.
The fact that 1 mole of every ide I ies tha gas Occupies t e same volume (under the
same temperature and pressure co dition) id
I' , n I Ion provi es several other interpretations of
vo umetnc concentrations expressed as ppmv. For ell '1 I f
pollutant per million volumes of air whi h ' . xamp e, ppmv IS vo ume 0
rant per million moles f ' S' '1'1 IC ISequivalenr to saying 1 mole of pollu-
o air, irnuarty smcee hi' f
molecules 1 ppmv also Cd' ac mo e contains the same number 0
, orrespon s 10 1 mol I f IIof air. ecu e 0 po utanr per million molecules
(
ACCUmulation) = (Input) _ (output) + (Reaction)
rate rate rate rate
(1.12)
The reaction rate may be positive if generation of the substance is faster than its
decay, or negative if it is decaying faster than it is being produced. Likewise, the ac-
cumulation rate may be positive or negative. The reaction term in (1.12) does not
imply a violation of the law of conservation of mass. Atoms are conserved, but there
is no similar constraint on the chemical compounds, which may chemically change
from one substance into another. It is also important to notice that each term in
(1.12) quantifies a mass rate of change (e.g., mg/s, lb/hr) and not a mass. Strictly, then,
it is a mass rate balance rather than a mass balance, and (1.12) denotes that the rate
of mass accumulation is equal to the difference between the rate the mass enters and
leaves plus the net rate that the mass reacts within the defined control volume.
Frequently, (1.12) can be simplified. The most common simplification results
when steady state or equilibrium conditions can be assumed. Equilibrium simply
means that there is no accumulation of mass with time; the system has had its inputs
held constant for a long enough time that any transients have had a chance to die out.
Pollutant concentrations are constant. Hence the accumulation rate term in (1.12) is
set equal to zero, and problems can usually be solved using just simple algebra.
A second simplification to (1.12) results when a substance is conserved within
the region in question, meaning there is no reaction occurring-no radioactive
decay, bacterial decomposition, or chemical decay or generation. For such conserva-
tive substances, the reaction rate in (1.12) is O. Examples of substances that are typ-
ically modeled as conservative include total dissolved solids in a body of water,
heavy metals in soils, and carbon dioxide in air. Radioactive radon gas in a home or
1 mol of pollutant = 1 molecule of pollutant
10 mol of air 106 I I .
rna ecu es of air
1ppmv =
(1.11)
1.3 I Materials Balance
Everything has to go somewhere is a sim I
mental engineering principles Mo p e way to express one of the most funda-
h h
' . re preCisely th If'
t at w en chemical reactions tak I ,e .aw 0 conservauo« of mass says
(th h 'I e p ace, matter IS ith d
aug In nuc ear reactions mass b nett er created nor destroye
II ' ' can e convert d
a ows us to do IStrack materials f e to energy). What this concept
irh b J ' or example poll f
WI mass a ance equations Thi , utants, rom one place to another
II' . ISISone of the m id
po .urants 10 the environment and' h basi ost WI ely used tools in analyzing
b d d ' ISt e asis for f
e intro uce 10 later chapters. many a the approaches that will
The first step in a mass b I
h - a ance analys' def
s~ace t at IS to be analyzed. This is often ;~ ~s ~ efine the particular region in
t e ~ontrol volume might include anything ~a e t e COntrol volume. As examples,
nb~mg tank, to an entire coal-fired po rlo
m
a glass of water or simple chemical
asm above a cit hI' wer P ant a lak '
y, Or t e g abe Itself B ' .' e, a stretch of stream an air
. YP1ctunng an . . '
Imagmary boundary around
7

8 Chapter 1 Mass and Energy Transfer 1.3 Materials Balance
Stream ~ .: --"
,/ Accumulmion e 0 C e,r 1 m'
I :~Mixlure
Reaction = 0 "
Wastes ~Q "
w' w _
Control volume
boundary
C,~20,OmglL ----
,
C =1m '
FIGURE 1.2 A steady-state conservative system. Pollutants enter and leave [he region at
the same rate.
Q= flow rate
C = concentration
of pollutant
decomposing organic wastes in a lake are examples of nonconservarive substances,
Often problems involving nonconservative substances can be simplified when rhe
reaction rate ISsmall enough to be ignored,
FIGURE 1.3 Sketch of system, variables, and quantities for a stream and tributary
mixing example.
Next the mass balance equation (1.12) is written and simplified to match the
problem's conditions
Steady-State Conservative Systems
The simplest systems to analyze are those in which steady state can be assumed (so
the accumulation rate equals 0) and th b ' " ,. ' e su stance in question IS conservative (so the
reaction rate equals 0), In these cases, (1.12) simplifies to the following:
Input rate ~ Output rate (1.13)
Consider the steady-state con' hown i
, d ithi ,servatlve system sown m Figure 1,2, The system
contame Wit m the boundaries mi ht b I k '
O th f ai b ' g e a a e, a section of a free flowing stream,
r e mass 0 air a ave a cay One' h .
for instance) with fl ' input to t e system ISa stream (of water or air,
I a ow rate Q (vol I' ) d
( I I ) Th ' " urne time an pollutant concentration C
mass vo ume . e other mput IS assumed to be a . S
and pollutant concentratio C Th ,waste stream With flow rate Qw
n W' e output ISa rni t ' h fl Q d
pollutant concentration C If h 11 ' x ure Wit ow rate m an
rn- t e po utant IS c ' d ifsteady state conditions then a b I onservanve, an 1 we assume
following: ,mass a ance based on (1.13) allows us to wrire the
~ (Input) (Output) ~
~= rate - rate +~
The simplified (1.12) is then written in terms of the variables in the sketch
o = C,Q, + CwQw - CmQm
The next step is to rearrange the expression to solve for the variable of interest-
in this case, the chloride concentration downstream of the junction, Crn' Note
that since the mixture's flow is the sum of the two stream flows, Q, + Qw can be
substituted for Qm in this expression,
C,Q, + CwQw C,Q, + CwQwC = = --,"""-_--"-"'c:'
m Qrn Q, + Qw
The final step is to substitute the appropriate values for the known quantities into
the expression, which brings us to a question of units, The units given for Care
mg/L, and the units for Q are m3/s, Taking the product of concentrations and
flow rates yields mixed units of mg/L> m3/s, which we could simplify by applying
the conversion factor of 103 L = 1m'. However, if we did so, we should have to
reapply that same conversion factor to get the mixture concentration back into
the desired units of mg/L. In problems of this sort, it is much easier to simply
leave the mixed units in the expression, even though they may look awkward at
first, and let them work themselves out in the calculation, The downstream con-
centration of chloride is thus
(20.0 X 10,0 + 40,0 X 5,0) rng/L: m3
/s
C = '----.:......-------oc.:.-.:::...-- = 26,7 mgIL
m (10,0 + 5,0) m3/s
, C,Q, + CwQw ec CmQm (1.14)
The followmg example illustrates th f hi ,
provides a general algorithm for d e use 0 t ISequation. MOte importantly, it also
omg mass balance problems,
EXAMPLE 1.4 Two Polluted Streams
Astream flowing at 10,0 m3/s has ib
Th' a tn utaryfeed' , , , 3
e stream s concentration of chi id mg inro It WItha flow of 5,0 ra'!».
the, tributary chloride concentrat:~ ~1~.t~e~mof the junction is20,0 mg/L, and
vanve substance and assuming I ' gIL, Treatmg chlonde as a conser-
do, hi ' comp ete mlxmg f h
vnstream conde concentration, 0 t e two streams, find the
SolutIon The first step in s I '
I id ify 0 vmg a mass b I
em, I ent the "region" or control volu a ance problem is to sketch the prob-
the variables as has been done in Figu 1~ef that we Want to analyze, and label
re , or thISproblem.
This stream mixing problem is relatively simple, whatever the approach used,
Drawing the system, labeling rhe variables and parameters, writing and simplify-
ing the mass balance equation, and then solving it for the variable of interest is
the same approach that will be used ro solve much more complex mass balance
problems later in this chapter and later in the book,
9

- - --~---~---- -. -- --~~~-'--~-- --====-- ---- ----
.-
10 Chapter 1 Mass and EnergyTransfer 1.3 Materials Balance
1
I Batch Systems with Nonconservative Pollutants
The simplest system with a nonconservative pollutant is a batch system, By defini-
tion, there is no contaminant flow into or out of a batch system, yet the ontami-
nants in the system undergo chemical, biological, or nuclear reaction fa r enough
that they must be treated as nonconservative substances. A batch system (reactor)
assumes that its contents are homogeneously distributed and is often referred ro as a
completely mixed batch reactor (CMBR). The bacterial concentration in a closed
water storage tank may be considered a nonconservative pollutant in a batch reac-
tor because it will change with time even though no water is fed into or withdrawn
from the tank. Similarly, the concentration of carbon dioxide in a poorly ventilated
room can be modeled as a nonconservative batch system because the concentration
of carbon dioxide increases as people in the room breathe. For a batch reactor (1.12)
simplifies to '
200
c 150
0
'"g
100e
~u
c
0
u
50
0
0
Decay
Production -k
20 40 60 80 100
Time
FIGURE 1.4
kinetics.
Concentration of a substance reacting in a batch system with zero-order
Solving for concentration gives usAccumulation rate ~ Reaction rate (1.15)
As discussed before, the reaction rate is the sum of the rates of decay, which
are negative, and the rates of generation, which are positive. Although the rates of
reacnon can exhibit many dependencies and complex relationships most nuclear,
chemical and biocherni I eacti b . , ' ., . rca r Ionrates can e approximated as either zero-, [irst-,
or second-order reactton rates. In a zero-order reaction, the rate of reaction r(C) of
the substance IS not dep dent h f h ' ,
en On t e amount ate substance present and can be
expressed as
C ~ Co - kt (1.18)
where Co is the initial concentration. Using (1.18) and its analog for a zero-order
generation reaction, Figure 1.4 shows how the concentration of a substance will
change with time, if it is reacting (either being generated or destroyed) with zero-
order kinetics.
For all nonconservative pollutants undergoing a reaction other than zero-
order, the rate of the reaction is dependent on the concentration of the pollutant
present. Although decay and generation rates may be any order, the most commonly
encountered reaction rate for generation is zero-order, whereas for decay it is first-
order. The first-order reaction rate is
r(C) ~ k (generation) or r(C) ~ -k (decay) (1.16)
where k is a reaction rate coefficie hi h h h . I I
( .L-I -I nt, W ic as t e uruts of mass' volume- . time-
e.g., mg . s ). The rate of ev ti f f
reaction beca th f I apoea Ion a water rom a bucket is a zero-order
use e rate a ass of the ' d f
water in the bucket bu- : I d water IS not ependent on the amount 0
water exposed to the Ut ISon y ependenr on the nearly constant surface area of the
air,
Using (1.15) and (1.16), the mass b I '
stance in a batch reactor is a ance for the zero-order reacnon of a sub-
riC) = kC (generation) or riC) ~ -kC (decay) (1.19)
where k is still a reaction rate constant, but now has the units of reciprocal time
(time-I). Radioactive decay of radon gas follows first-order decay-the mass that
decays per given time is directly proportional ro the mass that is originally present.
Using (1.15) and (1.19), rhe mass balance for a pollutant undergoing first-order
decay in a batch reacror is
dC
V- ~ -Vk
dt
The equation is written as a zero-orde d
fight-hand side of the' r ecay, denoted by the negative sign on the
, equanon, So that each t in th
recr umts of massltirne both th I ' errn m t e mass balance has the cor-
hi' e accumu anon d '
t e vo ume of the batch reactor. Al h h i an reaction terms are multiplied by
dl. appears by dividing both side ~ ~g 10 a batch sYStem, the volume coefficient
initial balance equation because s y h ' It IS worth remembering its presence in the
diff . I' 10 ot er system .
I erenna equanon, the variabl s It may not cancel out. To solve the
es are separated and inr degrate as
C t
JdC~ -kJdt
Co 0
dC
V- = -VkC
dt
This equation can be integrated by separation of variables and solved similarly ro
(1.17). When solved for concencration, it yields
(1.20)
That is, assuming a first-order reaction, the concentration of the substance in
question decays exponentially. This exponential function will appear so often in
this text that it will be reintroduced and explored more fully in Chapter 3,
"Mathematics of Growth." The first-order rime dependence of a nonconservative
pollutanc's concentration in a batch system can be seen in Figure 1.5.
Although not nearly as common as first-order processes, sometimes a sub-
stance will decay or be generated by a second-order process. For instance, hydroxyl
(1.171
which yields
C - Co ~ -kt
11

- -- -~ ----- ~. - -~ -- ----~------ -~ -----
12 Chapter 1 Mass and Energy Transfer 1.3 Materials Balance 13
200
c 150
0
.~
b
100so
c
0
u
50 I-
I~Y
Production
./a
a 20 40 60 80 100
Time
200~----------------,
Decaye 150
0
.~
100c
"c
c
0
u
50
a
0
Production
20 40 60 80 100
Time
FIGURE 1.5 Concentration of a substance reacting in a batch system with first-order
kinetics.
FIGURE 1.6 Concentration of a substance reacting in a batch system with second-order
kinetics.
radical reactions with volatile organic pollutants is a key step in smog generation,
However, If two hydroxyl radicals collide and react, they form a much less potent
hydrogen peroxide molecule, This is a second-order reaction, since two hydroxyl
radicals are consumed for each hydrogen peroxide produced, The second-order
reaction rate IS
two orher ideal reactors, In the first, the substances within the system are still
homogeneously mixed as in a batch reactor, Such a system is variously termed a
continuously stirred tank reactor (CSTR), a perfectly mixed flow reactor, and a
complete mix box modeL The water in a shallow pond with an inlet and outlet is
typically modeled as a CSTR as is the air within a well-ventilated room, The key
concept is that the concentration, C, within the CSTR container is uniform through-
out, We'll give examples of CSTR behavior first and later discuss the other ideal
reactor model that is often used, the plug flow reactor (PFR),
The reaction rate term in the right-hand side of (1.23) can denote either a
substance's decay or generation (by making it positive or negative) and, for most en-
vironmental purposes, its rate can be approximated as either zero-, first-, or second-
order, Just as for the batch reactor, for a CSTR, we assume the substance is
uniformly distributed throughout a volume V, so the total amount of substance
is CV, The total rate of reaction of the amount of a nonconservative substance is
thus d(CV)/dt = V dC/dt = Vr(C) , So summarizing (1.16), (1.19), and (1.21), we
can write the reaction rate expressions for a nonconservative substance:
r(C) = kC
2
(generation) or r(C) = -kC2 (decay) (1.21)
where k is nOWa reaction rate constant with units of (I . -I." -I)
Agam substit ti (121)" vo ume mass time '
d d I ~ mg f' into (1.15), we have the differential equation for the
secon -or er ecay 0 a nonconservative substance in a batch reactor
dC
V
dt = -VkC2
which can be integrated and then solved f h '
or t e COncentratiOnto yield
C = _-,Co,--
1 + Cokt
Figure 1.6 shows how the cOncentration of
decays or is produced by a second- d ,a substance changes with time if it
or er reacnon 10 a batch reactor.
(1.22)
Zero-order, decay rate = - Vk (1.24)
Zero-order, generation rate = Vk (1.25)
First-order, decay rate = -VkC (1.26)
First-order, generation rate = VkC (1.27)
Second-order, decay rate = -VkC2 (1.28)
Second-order, generation rate = VkC2 (1.29)
Steady-State Systems with Nonconserv ti
a ive Pollutants
If we assume that steady-state conditions "I
servative, then (1.12) becomes prevai and treat the pollutants as noncon-
o = Input rate - Ou
tpur rate + Reaction rate (1.23)
Thebatch reactor, which has just b di
system With a nonconservative substan ee~ Iscussed, can't describe a steady-state
Although there are an infinite number ofc~th~~ausenow there is input and output,
ploymg two other types of ideal reactors II pOSSIbletypes of reactors simply em-
ronmental processes, The type of m' , a ows us to model a great number of envi-
IXmgIII the s ' ,
ystem dlsltnguishes between the
Thus, for example, to model a CSTR containing a substance that is decaying with a
second-order rate, we combine (1.23) with (1.28) to get a final, simple, and useful
expression for the mass balance involving a nonconservative pollutant in a steady-
state, CSTR system:
Input rate = Output rate + kC2
V (1.30)

EXAMPLE 1.5 A Polluted Lake
l
~ Consider a 10 X 106
m3 lake fed by a polluted stream having a flow rate of
5.0 m
3
/s and pollutant concentration equal to 10.0 mgiL (Figure 1.7). There is
also a sewage outfall that discharges 0.5 m3/s of wastewater having a pollutant
concentration of 100 mg/L. The stream and sewage wastes have a de ay rate coef-
ficienr of 0.20/day. Assuming the pollutant is completely mixed in the lake and
assuming no evaporation or other water losses or gains, find the steady- tare
pollutant concentration in the lake.
Idealized models involving nonconservative pollutants in completely mixed,
steady-state systems are used to analyze a variety of commonly encountered water
pollution problems such as the one shown in the previous example. The same simple
models can be applied to certain problems involving air quality, as the following
example demonstrates.
EXAMPLE 1.6 A Smoky Bar
Solution We can conveniently use the lake in Figure 1.7 as our control volume.
Assuming that complete and instantaneous mixing occurs in the lake-it acts as a
CSTR-lmphes that the concentration in the lake, C, is the same as the concen-
rranon of the mix leaving the lake C The units (day" I) f h d .
ffi '. ...' m- 0 t e ecay reaction
coe ficient indicates this IS a first-order reaction. Using (1.23) and (1.26):
Input rate = Output rate + kCY
We can find each term as follows:
There are two input sources, so the total input rate is
Input rate = Q C + Q Cssw w
(1.31)
A bar with volume 500 m3 has 50 smokers in it, each smoking 2 cigarettes per
hour (see Figure 1.8). An individual cigarette emits, among other things, about
1.4 mg of formaldehyde (HCHO). Formaldehyde converts to carbon dioxide
with a reaction rate coefficient k = 0.40/hr. Fresh air enters the bar at the rate
of 1,000 m3/hr, and stale air leaves at the same rate. Assuming complete mixing,
estimate the steady-state concentration of formaldehyde in the air. At 25"C and
1 arm of pressure, how does the result compare with the threshold for eye irrita-
tion of 0.05 ppm?
Solution The bar's building acts as a CSTR reactor, and the complete mixing
inside means the concentration of formaldehyde C in the bar is the same as the
concentration in the air leaving the bar. Since the formaldehyde concentration in
fresh air can be considered 0, the input rate in (1.23) is also O. Our mass balance
equation is then
The output rate is
Output rate = QmCm = (Q, + Qw)C
(1.31) then becomes
Q,C, + o,c, = (Q, + Qw)C + kCY
And rearranging to solve for C,
C = Q,C, + QwCw
Q, + Qw + kY
_ 5.0 m
3
/s X 10.0 mgIL + 05 31
- . m s X 100.0 mg/L
(5.0 + 0.5) m3/s + 0.20/d X 10.0 X 106 m3
S 24 hrld X 3600 slhr0,
Output rate = Reaction rate (1.32)
However, both a generation term (the cigarette smoking) and a decay term (the
conversion of formaldehyde to carbon dioxide) are contributing to the reaction
rate. If we call the generation rate, G, we can write
G = 50 smokers X 2 cigslhr X 1.4 mg/cig = 140 mg/hr
We can then express (1.32) in terms of the problem's variables and (1.26) as
QC = G - kCY
so
C =..!QQ.._
28.65 - 3.5 mgIL C = G = __ ---;;...,-_14-'0~m-"'.g/_:hr---___;_
Q + k V 1,000 m3/hr + (0.40/hr) X 500 m3
= 0.1l7mg/m3Outfall
Qw=0.5 m3/s
Cw = 100.0mg/L
---.;:
Lake
/ Oasis Incoming
_sl_"_Ufl_' .-/ (r Indoor concentration C
V:::: 500 m3
- ,-,J
~ V=10.OxlO'm3 ~
'..~ k= 0.20/day
C=? ff----
Cm=?
Qm=?
A lakewitha
nonconservative pollutant,
FIGURE 1.8 Tobacco smoke in a bar.
Q = 1,000 m3/hr Q = 1,000 m3/hr
-t--
C=?
Outgoing
Qt= 5.0 m3/s
C, = 10.0 mglL
Fresh air
140 mglhr
I !l
k·O.40/hr
14
FIGURE 1,7
15

16 Chapter 1 Mass and Energy Transfer
We will use (1.9) to convert mg/rrr' to ppm. The molecular weight of formaldehyde
is 30, so
C (mg/rrr') X 24.465 0.117 X 24.465 _ 0095
HCHO = iff = -. ppm
mol wt 30
This is nearly double the 0.05 ppm threshold for eye irritation.
Lake
Q = 700 m3
/rnin
.....-- Co = 7 fishlm3
Q = 7(J() m
3
/min 75 <In :::=---I.--; L=4.
c=? r--- A=20m'
Besides a CSTR, the other type of ideal reactor that is often useful to model
pollutants as they flow through a system is a plug flow reactor (PFR). A PFR can be
visualized as a long pipe or channel in which there is no mixing of the pollutant
along its length between the inlet and outlet. A PFR could also be seen as a conveyor
belt carrying a single-file line of bottles in which reactions can happen within each
bottle, but there is no mixing of the contents of oue bottle with another. The behav-
ior of a pollutant being carried in a river or in the jet stream in the Earth's upper
atmosphere could be usefully represented as a PFR system. The key difference be-
tween a PFR and CSTR is that in a PFR, there is no mixing of one parcel of fluid
with other parcels in front or in back of it in the control volume, whereas in a
CSTR, all of the fluid in the container is continuously and completely mixed. (1.23)
applies to both a CSTR and PFR at steady state, but for a PFR, we cannot make the
simplification that the concentration everywhere in the control volume and in the
fluid leaving the region is the same, as we did in the CSTR examples. The pollutant
concentration in a parcel of fluid changes as the parcel progresses through the PFR.
Intuitively, it can then be seen that a PFR acts like a conveyor belt of differentially
thin batch reactors translating through the control volume. When the pollutant en-
ters the PFR at concenrranon, Co, then it will take a given time t to move the length
of the control volume and will leave the PFR with a concentra;i~n just as it would if
It had been 1B a batch reactor for that period of time. Thus, a substance decaying
With a zero-, firsr-, or second-order rate will leave a PFR with a concentration given
by (1.18)~ (1.20), and (1.22), respectively, with the understanding that t in each
equation IS the residence time of the fluid in the co t I I d i b
n ro vo ume an IS gIven y
t = tl» = V/Q (1.33)
whderQelis thhefllendgtfhlof the PFR, v is the fluid velocity, V is the PFR control volume,
an IS t e ill owrate,
Ocean
FIGURE 1.9 Birds fishing along a salmon srream.
Solution First we draw a figure depicting the stream as the control volume and
the variables (Figure 1.9).
Since the birds eat the salmon at a steady rate that is not dependent on the
concentration of salmon in the stream, the rate of salmon consumption is zero-
order. So,
k = 10,000 fish· km~l. hr~l = 0.50 fish. m~3. hr~l
20 m2
X 1,000 m/km
For a steady-state PFR, (1.23) becomes (1.18),
C = Co - kt
The residence time, t, of the stream can be calculared using (1.33) as
t = ~ = 4.75 km X 20 m
2
X 1,000 m/km = 2.26 hr
Q 700 m3
/min X 60 minlhr
and the concentration of fish reaching the ocean is
C = 7 fish/m3 - 0.50 fish- m~3. ht~l X 2.26 hr = 5.9 fish/m3
Step Function Response
So far, we have computed steady-state concentrations in environmental systems that
are contaminated with either conservative or nonconservative pollutants. Let us
now extend the analysis to include conditions that are not steady state. Quite often,
we will be interested in how the concentration will change with time when there is a
sudden change in the amount of pollution entering the system. This is known as the
step function response of the system.
In Figure 1.10, the environmental system to be modeled has been drawn as if it
were a box of volume V that has flow rate Q in and out of the box.
Let us assume the contents of the box are at all times completely mixed
(a CSTR model) so that the pollutant concentration C in the box is the same as the
concentration leaving the box. The total mass of pollutant in the box is therefore
EXAMPLE 1.7 Young Salmon Migration
~~::~e:r~~~~oe~:i~geaagUlllkS,eaglehs,and Other birds mass along a 4.75 km stretch
aetoteoceanto h the f h
migrate downstream to the e The hi catc t e mgerhng salmon as t ey
surne 10,000 fingerlings per ~il:met~r ~;ds are efficient fishermen and will con-
ber of the salmon in the stream I h stream each hour regardless of the num-
birds are only limited by how f' nhot er words, there are enough salmon; the
ast t ey can cat h d' ,
average Ctoss-sectional area is 20 2 d h c an eat the fish. The stream s
stream's tlow rate of 700 m3/m'I mlf'han t e salmon move downstream with the
. n. t ere are 7 f I' 3
termg the stream what is the co ' inger lOgs per m in the water en-
h bi ' ncentratlon of I h h
t e irds are feeding? sa man t at reach the ocean w en
17

then
dy dC
dt dt
Flow OUt
I
Q.C
Flow in
, (1.39)
Control volume V
Concentration C
Q,C,
so (1.37) becomes
Decay coefficient kd
Generation coefficient kg
FIGURE 1.10 A box model for a transient analysis.
dy
dt
(1.40)
This is a differential equation, which we can solve by separating the variables to give
VC, and the rate of accumulation of pollutant in the box is VdC/dt. Ler u designate
rhe concentration of pollutant entering the box as C,. We'll also a ume there are
both production and decay components of the reaction rare and de ignare the decay
coefficient kd and the generation coefficient kg. However, as is mo t common, the
decay will be first-order, so kb units are time-I, whereas the generation is zero-
order, so kg's units are mass' volume -1. time-I From (1.12), we can write
(
ACCUmulation) = (Input) _ (Output) + (Reaction)
rate rate rate rate
(1.41)
where Yo is the value of y at t = O. Integrating gives
y = yoe-(kd+~)t (1.42)
If Co is the concentration in the box at time t = 0, then from (1.38) we get
(1.34)
Yo = Co - Coo (1.43)
Substituting (1.38) and (1.43) into (1.42) yields
C - Coo = (Co - Coo)e-(kd+~)t
where
v = box volume (rrr')
C = concentration in the box and exiting waste stream (g/rrr')
C, : concentranon of pollutants entering the box (g/rrr')
Q - the total flow rate in and out of the box (m3/hr)
kd = decay rate coefficienr (hr-I)
kg = production rate coefficient (g. m . hr -1)
The units given in the preceding list ate rep' h b, resentatrve of those that mig t e
encountered; any consistent set will do.
An easy way to find the steady-state solution to (1.34) is simply to set
dC/dt = 0, which Yields
(1.44 )
Solving for the concentration in the box, writing it as a function of time Cit), and
expressing the exponential as exp( ) gives
C(t) = Coo + (Co - Coo)exp[ -( kd + ~)] (1.45)
Equation (1.45) should make some sense. At time t = 0, the exponential function
equals 1 and C = Co. At t = 00, the exponential term equals 0, and C = Coo.
Equation (1.45) is plotted in Figure 1.11.
Coo = QCi + kgV
Q + kdV
where Coo is the concentration in the b ' h
is with the concentration b f ' ox at time t = DC. Our concern now, thoug ,
Rearranging (1.34) gives e ore It reaches steady state, so we must solve (1.34),
~~=-(~+k}(C _ QC'+kgV) (1.361
hi h . Q + kdVw IC , usmg (1.35) can be .) rewntten as
~~ = -( ~ + kd)- (C - Coo) (1.371
One way to solve this differential '.
equation 1S to make a h f vari ble If we letc ange 0 vana .
Y = C - Coo (1.381
(1.35)
c"" - -- - - -- --- - -- - -- - ----.- -- --- - -- - ---
C = .::Q;;-C-,:-i+-ck-;,,-v
~ Q+kdV
o
Time, 1
FIGURE 1.11 Step function response for the CSTR box model.
19

EXAMPLE 1.8 The Smoky Bar Revisited
The bar in Example 1.6 had volume 500 m3 with fre hair enrermg at rhe rate of
1,000 m
3
/hr. Suppose when the bar opens at 5 P.M., rhe air i lean. If formalde-
hyde, with decay rate kd = OAO/hr, is emitted from cigarette smoke ar the constant
rate of 140 mg/hr starting at 5 P.M., what would the concentration be at 6 P.I.?
Solution In this case, Q = 1,000 m%r, V = 500 m3, G = kg, = 140 mglhr,
C; = 0, and kd = OAO/hr. The steady-state concentration i found u ing (1.35):
Coo = QC; + kgV = G = 140 mg/hr
Q + kdV Q + kdV 1,000 m3/hr + OAO/hr X 500 m3
= 0.117mg/m3
This agrees with the result obtained in Example 1.6. To find the concentration at
any nrne after 5 P.M., we can apply (1045) with Co = O.
C(t) = Coo{l- eXP[-(kd + ~}]}
= 0.117!1 - exp[-(OAO + 1,000/500)t]}
at 6 P.M., t = I hr, so
C(lhr) = 0.117[1 - exp(-2A X l)J 0.106mglm3
To further demonstrate the use f (14 .
Example 1 5 This tim 'II o. 5), let us reconsider the lake analyzed m
. . I e we WI assume th h fl"
the lake, so its contribution t h I ke' at t e out a I suddenly stops draining mID
ate a e s pollutIOn stops.
EXAMPLE 1.9 A Sudden Deer .
ease in Pollutants Discharged inro a Lake
Consider the 10 X 106 rrr' I k I .
tions given, was found to h a e ana yzed in Example 1.5 which under a condi-
, ave a steady- t II' '
The POllutIOn is nonconse . . s ate po unon concentration of 3.5 mg/L.
Suppose the condition of thelVlatklveWdlthreaction-rate constant kd
= 0.20/day.
, . d id a e IS eemed I
It IS eCI ed to completel di h unacceptable. To solve the prob em,
y Ivert t e sewa f II ' .
as a SOurce of pollution Tbe : . ge out a from the lake e1iminatmg It
. . e mcommg t 'I'd
concenrrallon C = 100 gIL's ream sn I has flow Q = 5 0 m3/s an
fl s. m With h ' .
ow Q is also 5.0 m3/s Ass', t e sewage outfall removed the outgoing
. f . ummg co I' ,
tlon 0 POliutanr in the lake kmp ete-mlx conditions find the concenrra-
stead one wee afte h d" , Iy-state concentration. r t e Iversion, and find the new flna
Solution For this siruation
,
Co = 3.5 mg/L
V = 10 X 106 m3
Q=Q -503
,- . m /s X 3 600 sib
' r X 24 hr/day = 43.2 X 104 m3/day
1.4 Energy Fundamentals 21
C, = 10.0 mgIL = 10.0 X 103 mg/m '
kd = 0.20/day
The total rate at which pollution is entering the lake from the incoming stream is
Q,C, = 43.2 X 104m3/day X 10.0 X 103mglm3 = 43.2 X 108mglday
The steady-state concentration can be obtained from (1.35)
QC, 43.2 X 108 mg/day
C - --:::---=--:"-:-::
00 - Q + kdV 43.2 X 104 m3/day + 0.20/day X 107 rrr'
= 1.8 X 103mglm3 = 1.8mglL
Using (lAS), we can find the concentration in the lake one week after the drop in
pollution from the outfall:
C(t) = Coo + (Co - Coo)exp[ -( kd + ~)t]
C(7days) = 1.8 + (3.5 - 1.8)exp[-(0.2/day + 43.2 X 10
4
;da
y
) X 7days]
10 X 10 m
C(7days) ~ 2.1mg/L
Figure 1.12 shows the decrease in contaminant concentration for this example.
33,51'---'"S
§ 2.1 -~-----------~_~~~~~ ~
~ 1.8 -------- ~
s
~ o '-------=-~--------
o 7 days
Time,t
FIGURE 1.12 The contaminant concentration profile for Example 1.9.
1.4 I Energy Fundamentals
Just as we are able to use tbe law of conservation of mass to write mass balance
equations that are fundamental to understanding and analyzing the flow of materi-
als, we can use the first law of thermodynamics to write energy balance equations
that will help us analyze energy flows.
One definition of energy is that it is the capacity for doing work, where work
can be described by the product of force and the displacement of an object caused by
that force. A simple interpretation of the second law of thermodynamics suggests
that when work is done, there will always be some inefficiency, that is, some portion
of the energy put into the process will end up as waste heat. How that waste heat

affects the environment is an important consideration in the study of environmental
engineering and science. .
Another important term to be familiar with is power. Power I the rate of
doing work. It has units of energy per unit of time. In Sl units, power i given in
joules per second (Jisl or kilojoules per second (kj/s}, To honor the Scottish engineer
James Watt, who developed the reciprocating steam engine, the joule per second has
been named the wart (1 Jls = 1 W = 3,412 Btu/hr).
. In many applications of (1.46), the net energy added to a system will cause an
increase m temperature. Waste heat from a power plant, for example, will raise the
temperature of cooling water drawn into its condenser. The amount of energy needed
to raise the temperature of a unit mass of a substance by 1 degree is called the specific
heat. The specific heat of water is the basis for two important units of energy, namely
the British thermal unit (Btu), which is defined to be the energy required to raise 1 Ib
of water b~ 1°F, and the kilocalorie, which is the energy required to raise 1 kg of
water by 1 C. In the definitions Just gIven, the assumed temperature of the water is
15°C (59°F). Since kilocalories are no longer a preferred energy unit, values of spe-
cific heat in the SI system are given in kJlkgOC, where 1 kcal/kg'tC = 1 Btu/lb'T =
4.184 kJ/kg°C.
For most applications, the specific heat of a liquid or solid can be treated as a
simple quantity that varies slightly with temperature. For gases, on the other hand
the concept of specific heat is complicated by the fact that some of the heat energy
absorbed by a gas may cause an increase in temperature, and some may cause the
gas to expand, doing work on its environment. That means it takes more energy to
raise the temperature of a gas that is allowed to expand than the amount needed if
the gas is kept at constant volume. The specific heat at constant volume c, is used
when a gas does not change volume as it is heated or cooled, or if the volume is al-
lowed to vary but is brought back to its starting value at the end of the process.
Similarly, the specific heat at constant pressure cp applies for systems that do not
change pressure. For incompressible substances, that is, liquids and solids under the
usual circumstances, c, ~nd.cp are identical. For gases, cp is greater than Cv'
The added complications associated with systems that change pressure and
volume are most easily handled by introducing another thermodynamic property of
a substance called enthalpy. The enthalpy H of a substance is defined as
H = U + PV (1,48)
where U is its internal energy, P is its pressure, and V is its volume. The enthalpy of
a unit mass of a substance depends only on its temperature. It has energy units (kJ or
Btu) and historically was referred to as a system's "heat content." Since heat is cor-
rectly defined only in terms of energy transfer across a system's boundaries heat
content is a somewhat misleading descriptor and is not used much anymore. '
. When a process occurs without a change of volume, the relationship between
Internal energy and temperature change is given by
!!.U=mc,!!.T (1.49)
The analogous equation for changes that occur under constant pressure involves
enthalpy
The First Law of Thermodynamics
The first law of thermodynamics says, simply,that energy can neither be created nor
destroyed. Energy may change forms in any givenprocess, as when chemical energy in
a fuel isconverted to heat and electricityin a power plant or when the potential energy
of water behind a dam is converted to mechanical energy as it spins a turbine in a hy-
droelectric plant. No matter what is happening, the first law says we should be able to
account for every bit of energy as it takes part in the process under study, so that in the
end, we have just as much as we had in the beginning. With proper accounting, even
nuclear reactions involving conversion of mass to energy can be treated.
To apply the first law, it is necessary to define the system being studied, much
aswas done in the analysis of mass flows. The system (control volume) can be any-
thing thar we want to draw an imaginary boundary around; it can be an automobile
engme, or a nuclear power plant, or a volume of gas emitted from a smokestack.
Later when we explore the topic of global temperature equilibrium, the system will
be the Earth Itself. After a boundary has been defined the rest of the universe be-
comes the surroundings. Just because a boundary has 'been defined, however, does
not mean that energy andlor matenals cannot flow across that boundary. Systems in
which both energy and matter can flow across the boundary are referred to as opell
systems, whereas those III which energy is allowed to f1 h b d y but. ow across t e Dun ar ,
matter ISnot, are called closed systems.
h d
Sinfcedenergyis conserved, we can write the following for whatever system we
ave e me :
(
:~::i~;e~::ndarY) + (~~:~:;ergy ) _ (T~tal energy) (Net Change) (1.46)
h
0 mass = of energy In
as eat and work entering system leavi
eavmg system the system
For closed systems there is no move f
and third term drop out of th ment a mass across the boundary, so the second
e equation The ac I' f d b
the right side of (1 46) m h . , cumu anon a energy represente Y
. ay cause c anges 10 the b bl f f
energy, such as kinetic and pot 'I ' a serva e, macroscopic arms a
. entia energies or mi . f I d the
atorruc and molecular srruct f h ' ICroscoplc arms re ate ro
. ure ate system Th . , f
include rhe kinetic energies of I I . ose microscopic forms a energy
. b rna ecu es and the . , h f S
acting erween molecules b tw ,energIes associated with t e orce
f
' e een atoms With' I I Th
sum 0 those microscopic for f ,Ill rna ecu es, and within atoms- e
is represented by the symbol ~Oh energy IScalled the system's internal energy and
d ib d h . e total energy E th b beescn e t en as the sum of' . at a su stance possesses can
, I ItSIOtemal ene U' k d .potentia energy PE: rgy ,1lS inetic energy KE, an ItS
!!. H = m cp!!. T (1.50)
For many environmental systems, the substances being heated are solids or
liquids for which c, = cp = c and AU = !!.H. We can write the following equation
for the energy needed to raise the temperarure of mass m by an amount !!.T:
Change in stored energy = m cS'T (1.51)
Table 1.3 provides some examples of specific heat for several selected sub-
stances. It is worth noting that water has by far the highest specific heat of the
substances listed; in fact, it is higher than almost all common substances. As will be
E = U + KE + PE (1,471
23

TABLE 1.3
There are two key assumptions implicit in (1.51). First, the specific heat is
assumed to be constant over the temperature range in question, although in actuality
it does vary slightly. Second, (1.51) assumes that there is no change of phase as
would occur if the substance were to freeze or melt (liquid-solid phase change) or
evaporate or condense (liquid-gas phase change).
When a substance changes phase, energy is absorbed or released without a
change in temperature. The energy required to cause a phase change of a unit mass
from solid to liquid (melting) at the same pressure is called the latent heat of fusion
or, more correctly, the enthalpy of fusion. Similarly, the energy required to change
phase from liquid to vapor at constant pressure is called the latent heat of vapor-
ization or the enthalpy of vaporization. For example, 333 k] will melt 1 kg of ice
(144 Bru/lb}, whereas 2,257 k] are required to convert 1 kg of water at 100'C to
steam (970 Btu/lb}, When steam condenses or when water freezes, those same
amounts of energy are released. When a substance changes temperature as heat is
added, the process is referred to as sensible heating. When the addition of heat
causes a phase change, as is the case when ice is melting or water is boiling, the
addition is called latent heat. To account for the latent heat stored in a substance, we
can include the following in our energy balance:
Specific Heat Capacity c of Selected Substances
(kcaVkgOC,Btu/lb'F) (kj/kg'C)
Water (15°C)
Air
Aluminum
Copper
Dry soil
Ice
Steam (100°C)'
Water vapor (20'C)'
1.00 4.18
0.24 1.01
0.22 0.92
0~9 039
0,20 0.84
0.50 2.09
0.48 2.01
0.45 1.88
"Constant pressure values.
noted in Chapter 5, this is one of water's very unusual properties and is in large part
responsible for the major effect the oceans have on moderating temperature varia-
tions of coastal areas.
Energy released or absorbed in phase change ~ m L (1.52)
EXAMPLE 1.10 A Water Heater
where m is the mass and L is the latent heat of fusion or vaporization.
Figure 1.13 illustrates the concepts of latent heat and specific heat for water as
it passes through its three phases from ice, to water, to steam.
Values of specific heat, heats of vaporization and fusion, and density for water
are given in Table 1.4 for both SI and USCS units. An additional entry has been
included in the table that shows the heat of vaporization for water at 15'C. This is
a useful number that can be used to estimate the amount of energy required to cause
How long would it take to heat the water in a 40-gallon electric water heater
from 50'F to 140'F if the heating element delivers 5 kW? Assume all of the elec-
trical energy is converted to heat in the water, neglect the energy required to raise
the temperature of the tank itself, and neglect any heat losses from the tank to the
environment.
Solution. The first thing to note is that the electric input is expressed in kilo-
watts, which IS a measure of the rate of energy input (i.e., power). To get total
energy delivered to the water,. we must multiply rate by time. Letting t.t be the
number of hours that the heating element is on gives
Energy input ~ 5 kW X At hrs ~ SAt kWhr
Assuming no losses from the tank and no water withdrawn from the tank during
the heating penod, there is no energy output:
Energy output ~ 0
i~~,~h~ge i~1e~;~g~ stored corresponds to the water warming from 50'F to
. smg . a ong With the fact that water weighs 8.34 lb/gal gives
Change in stored energy ~ rncAT
~ 40 gal X 8.34lb/gal X 1Btu/lb'F X (140 - 50)'F
~ 30 X 103Btu
Setting the energy input equal to th h '. .
units using Table 1.1 yields e c ange 10 Internal energy and converting
5At kWhr X 3412 B u!k, t Whr ~ 30 X 103 Btu
~~--~~- ~A~t..:~,-I~.7~6~h~r~ ___
I--------Latent heatofvaporization,2,257kJ------1111/
Boiling water
Steam 2.0 kJjOC
100 --------------
-,
Water 4. 18 kJreMelting ice
latent
heat of
fusion
333 kJ
1 ~
o I,
'2.1 kWC Ice
Heat added to I kg of ice (kJ) -
FIGURE 1.13 Heat needed to convert 1 kg of ice to steam. To change the temperature of 1 kg of ice,
2.1 kJ/'C are needed. To completely melt that ice requires another 333 k] (heat of fusion). Raising the
temperature of that water requires 4.184 k]rC, and converting it to steam requires another 2,257 k]
(latent heat of vaporization). Raising the temperature of 1 kg of steam (at atmospheric pressure)
requires another 2.0 kJ/'C.
25

26 Chapter 1 Mass and Energy Transfer 1.4 Energy Fundamentals
TABLE1.4 Many practical environmental engineering problems involve the flow of borh
matter and energy across system boundaries (open sysrems). For example, it is com-
mon for a hot liquid, usually water, to be used to deliver hear to a pollution conrrol
process or, the opposite, for warer to be used as a coolant to remove heat from a
process. In such cases, there are energy flow rates and fluid flow rates, and (1.51)
needs to be modified as follows:
Important Physical Properties of Water
51 Units nilSProperty
1.00 BruJlb"F
9 2 BruJlb
1,060 Brullb
1-14 BruJlb
62.4 Iblft-' ( .341b1g>!
4.184 kJlkg'C
2,257 kJlkg
2,465 kj/kg
333 kJlkg
1,000 kg/m3
Specific heat (15'C)
Heat of vaporization (100'C)
Heat of vaporization (15'C)
Heat of fusion
Density (at 4'C)
Rate of change of stored energy = m c ~ T (1.53)
where m is the mass flow rate across the system boundary, given by the product of
fluid flow rate and density, and ~ T is the change in temperature of the fluid that is
carrying the heat to, or away from, the process. For example, if water is being used
to cool a steam power planr, then m would be the mass flow rate of coolanr, and ~ T
would be the increase in temperature of the cooling water as it passes through the
steam plant's condenser. Typical units for energy rates include watts, Btu/hr, or kj/s,
whereas mass flow rates might typically be in kg/s or lb/hr.
The use of a local river for power plant cooling is common, and the following
example illustrates the approach that can be taken to compute the increase in river
temperature that results. Some of the environmenral impacts of this thermal pollu-
tion will be explored in Chapter 5.
surface water on the Earth to evaporate. The value of 15· ha been picked a the
starting temperature sincethat is approximately the currenr average urface temper-
ature of the globe.
One way to demonstratethe concept of the heat of vaporization hile at the
same time Introducingan important component of the global energy balance that
WIll be encountered in Chapter 8, is to estimate the energy required ro power the
global hydrologiccycle.
EXAMPLE 1.12 Thermal Pollution of a River
A coal-fired power plant converts one-third of the coal's energy inro electrical
energy. The electrical power output of the planr is 1,000 MW. The other two-
thirds of the energy content of the fuel is rejected to the environment as waste
heat. About 15 percenr of the waste heat goes up the smokestack, and the other
85 percenr is taken away by cooling water that is drawn from a nearby river. The
river has an upstream flow of 100.0 m3
/s and a temperature of 20.0·C.
a. If the cooling water is only allowed to rise in temperature by 10.0·C, what
flow rate from the stream would be required?
b. What would be the river temperature just after it receives the heated cool-
ing water?
ay yr X 24hr/day X 3600s/h = 78.0 W/m2
which is equivalent t I ' r X 5.10 X 10
14
m2
the Ea h' f a a most half of th 168 2 'k'1l8
. rt s Sur ace (seeFig 8 e W/m of incoming sunlight srn I
quired to raise the watet va~:~hi·
14
).It might also be noted that the energy~'
~gh~ble compared to the heat ~; into the atmosphere after it has evaporated Iit IS C apter). vapomation (see Problem 1.27 at the end0
------
Solution Since 1,000 MW represents one-third of the power delivered to the
planr by fuel, the total rate at which energy enrers the power planr is
Output power 1,000MW,
Input power = Effici 1/3 = 3,000 MW,lCleney
Notice the subscript on the input and output power in the preceding equation. To
help keep track of the various forms of energy, it is common to use MW, for
thermal power and MW, for electrical power.
Total losses to the cooling water and stack are therefore 3,000 MW - 1,000 MW =
2,000 MW. Of that 2,000 MW,
Stack losses = 0.15 X 2,000 MW, = 300 MW,
27

28 Cha prer 1 Mass and Energy Transfer
Stack heat
JOOMW,
Hot reservoir
Th
Qc Waste heat
Cold reservoir
T,
Qs = 100.0 m3/s
T~= 20.0 -c
Q = IOO.Om"
T =2~.1 C
FIGURE 1.15 Definition of terms for a Carnot engine.Stream ........
FIGURE1.14 Coolingwaterenergybalanceforthe33.3 percent efficient, 1,000 ~lW,
powerplantinExample1.12.
to work with 100 percent efficiency. There will always be "losses" (although, by the
first law, the energy is not lost, it is merely converted into the lower quality, less use-
ful form of low-temperature heat).
The steam-electric plant just described is an example of a heat engine, a device
studied at some length in thermodynamics. One way to view the steam plant is that
it is a machine that takes heat from a high-temperature source (the burning fuel),
converts some of it into work (the electrical output), and rejects the remainder into
a low-temperature reservoir (the river and the atmosphere). It turns out that the
maximum efficiency that our steam plant can possibly have depends on how high
the source temperature is and how low the temperature is of the reservoir accepting
the rejected heat. It is analogous to trying to run a turbine using water that £lows
from a higher elevation to a lower one. The greater the difference in elevation, the
more power can be extracted.
Figure 1.15 shows a theoretical heat engine operating between two heat reser-
voirs, one at temperature Th and one at TO" An amount of heat energy Qh is trans-
ferred from the hot reservoir to the heat engine. The engine does work Wand rejects
an amount of waste heat Qc to the cold reservoir.
The efficiency of this engine is the ratio of the work delivered by the engine to
the amount of heat energy taken from the hot reservoir:
ff . WE lClency1) = Qh (1.54)
The most efficient heat engine that could possibly operate between the two
heat reservoirs is called a Carnot engine after the French engineer Sadi Carnor, who
first developed the explanation in the 1820s. Analysis of Carnot engines shows that
the most efficient engine possible, operating between two temperatures, Th and To
has an efficiency of
and
Coolant losses= 0.85 X 2,000 MW, = 1,700 MW,
a. Finding the cooling water needed to remove 1 700 MW with a tempera-
ture .increase sr of 1O.0'C will require the u~e of (1.53) along with the
specificheat of water,4,184 Jlkg'C,given in Table 1.4:
Rate of change in internalenergy = til c !l.T
1 700 MW = . k"1
, ,m "'s X 4,184J/kg'C X 10.0'C X 1 MW/(106J/s)
til = 1,700
4,184 X 10.0 X 10-6 = 40.6 X 10
3
kgls
or, since 1,000 kg equals 1 3 f
b T f d h m 0 water, the £low rate is 40.6 m3
/s.
. 0 in t e new temperatureof th . MW
being released into the ri hi e fiver,we can use (1.53) with 1,700 ,
e fiver,w ich again has a flow rate of 100.0 m3/s.
Rate of change in internalen _.
ergy - mc!l.T
1,700MW X (1 X 106J/S)
!l.T= MW
100.00 m3/s X 103k'{/ 3 = 4.1 'C
so the temperature of the ri . m X 4,184 ]lkg'C
The results of the cal la fiverwill be elevated by 4.1 'C making it 24.1'C.
------..:.=.:::.::::...'~~~cu~at~lo~n~s~I~US~t~pe~rf~o~rm~ed~a~re~sh~oiw~n~i~n~F~,~g~u~r~ei1~.l~~
The Second Law of Ther d
rno ynamics
In Example 1.12, you will notice th . .
contained m the coal actually w at a relativelymodest fraction of the fuel enetg)
ahnda rather large amOUntof t~: 7nvlerted to the desired output elecreical power,
t e enVlfonment Th ue energy e d d ' . edto
b . e second law of h n e up as waste heat reJecr
e some waste heat; that is, it is imp tbelrmodynamics says that there will alwaY'
~eto& ~
'iiiiiiiiii lllliiiii _iiiiii_==-__ Visea machine that can convert
TJmax = 1 (1.55)
where these are absolute remperatures measured using either rhe Kelvin scale or
Rankine scale. Conversions from Celsius to Kelvin, and Fahrenheit to Rankine are
K = 'C + 273.15
R = 'F + 459.67
(1.56)
(1.57)
29

1.4 Energy Fundamenrals
requiring less water, involves the use of cooling towers that transfer the heat directly
into the atmosphere rather than into a receiving body of water. In either case, the re-
jected heat is released into the environment. In terms of the heat engine concept
shown in Figure 1.15, the cold reservoir temperature is thus determined by the tem-
perature of the environment.
Let us estimate the maximum possible efficiency that a thermal power plant
such as that diagrammed in Figure 1.16 can have. A reasonable estimate of Th might
be the temperature of the steam from the boiler, which is typically around 600°C.
For To we might use an ambient temperature of about 20°C. Using these values in
(1.55) and remembering to convert temperatures to the absolute scale, gives
Steam-"
.....-- Cool water In
(20 + 273)
TJmox = 1 - (600 + 273) = 0.66 = 66 percent
New fossil fuel-fired power plants have efficiencies around 40 percent. Nuclear
plants have materials constraints that force them to operate at somewhat lower tem-
peratures than fossil plants, which results in efficiencies of around 33 percent, The
average efficiency of all thermal plants actually in use in the United States, including
new and old (less efficient) plants, fossil and nuclear, is close to 33 percent. That
suggests the following convenient rule of thumb:
For every 3 units of energy entering the average thermal power plant,
approximately 1 unit is converted to electricity and 2 units are rejected
to the environment as waste heat.
Fuel__ L----.J
t
Cooling water
Boiler
feed pump
U:~~;:::=::::,;::;w~annYO aterQUI
Condenser
Air
FIGURE 1.16 A fuel-fired,steam-electric power plant,
, One immediate observation that can be made from (1.55) is that the maximum
possible heat engine efficiency increases as rh ', increases as t e temperature of the hot reserves
increases or the temperature f th Id ' , '
f , I hoe co reservoir decreases. In fact since neirher in-
mire y at temperatures no b I '
I d h
,r a so ute zero temperatures are possible we must coo-
cue t at no real engine ha 100 ff' , 'f
h d I
s percent e iciency which is )'ust a resrarement 0
t e secon aw. '
Equation (1.55) can help us d d he seemi , ' f h
1 I
un erstan t e seemingly low efficiency a t er-
rna power p ants such as the a di " '
burned in a firing ch b ne iagramma] 10 Figure 1.16. In this plant, fuel IS
am er surrounded bib' " hr h
this boiler tubing is c d hi y meta tu mg. Water Circulating t oug
onverte to igh p hi h ' h
conversion of chemical t h I - ressure, Ig -temperarure steam. Dunng t IS
a t erma ener I h
due to incomplete c b ' gy, asses on t e order of 10 percent occur
am usnon and 10 f h h II
consider local and regional' II' ss a eat up the smokestack. Later, we s a
their possible role in glob Iarr po, unon effects caused by these emissions as well as
a warmmg.
The steam produced in th b 'I
ways similar to a child' , h aller then enters a steam turbine which is in some
s pmw eel. Th hi h "through the turbine blad ' e Ig -pressure steam expands as It passes
, es, causmg a sh ft h ' to
spm. Although therurb' 'P' a t at IS connected to the generator
bi me 10 igure 1 16' h '
mes have many stages with .., ISS own as a single unit in actualtty, tur-
d
, I steam eXIlmg 'd lIy
expan 109 and COolingas 't T one stage and entering another, gra ua
" , I goes. he g f
Spmnmg shaft IOta electrical enerator converts the rotational energy a a
b ti A II' power that g ", d'"u ion, we -deSIgned t b' oes out onto transmission lines fat IStrt
h ur me may h
w ereas the generator may h ave an efficiency that approaches 90 percent,
Th ave a convers If
e spent steam from th ' Ion e IClency even higher than that.
state as it is cooled in the conde turbme undergoes a phase change back to the liquid
helps pull h enser. ThIS ph h hal
steam t rough the t b' ase c ange creates a partial vacuum t
condensed steam is then pump~~ ~n thereby increasing the turbine efficiencY. The
The heat released when the ac to the boiler to be reheated.
that clhrculates through the conden steaurncondenses is transferred to cooling wale!
rtver, eated in th ser. sually I' I ke al
e condenser ad' coo mg water is drawn from a a
once-through cooling. A rna' n teturned to that body of water which is called
re expensive a 'of
pproach, which has the advantage
The following example uses this rule of thumb for power plant efficiency com-
bined with other emission factors to develop a mass and energy balance for a typical
coal-fired power plant.
EXAMPLE 1.13 Mass and Energy Balance for a Coal-Fired Power Plant
Typical coal burned in power plants in the United States has an energy content of
approximately 24 kJ/g and an average carbon content of about 62 percent, For
almost all new coal plants, Clean Air Act emission standards limit sulfur emis-
sions to 260 g of sulfur dioxide (S02) per million k] of heat input to the plant
(130 g of elemental sulfur per 106 kJ). They also restrict particulate emissions to
13 g/106 k]. Suppose the average plant burns fuel with 2 percent sulfur content
and 10 percent unburnable minerals called ash, About 70 percent of the ash is
released as fly ash, and about 30 percent settles out of the firing chamber and is
collected as bottom ash. Assume this is a typical coal plant with 3 units of heat
energy required to deliver 1 unit of electrical energy.
a. Per kilowatt-hour of electrical energy produced, find the emissions of 502,
particulates, and carbon (assume all of the carbon in the coal is released to
the atmosphere).
b. How efficient must the sulfur emission control system be to meet the sulfur
emission limitations?
c. How efficient must the particulate control system be to meet the particulate
emission limits?
31

Solution
a. We first need the heat input to the plant. Becau e 3 kWhr of hear are
required for each 1 kWhr of electricity delivered,
Heat input = 3 kWhr heat X 1 k]/s X 3 600 slhr = 10 00 k]
kWhr electricity kW'
The sulfur emissions are thus restricted to
To atmosphere
1.4 g S (2.8 g S02)
yash
C
,080 kJ
I kWhr electricity 0.14gfl
(3,600 kJ) 280 g
)))1
~
3 kWhr 33.3% 85% S,
(10,800 kJ) efficient 99.5%
450 g coal power 280 g C particulate
(including: 280 g C plant 31.5 g fly ash removal
45 gash, 9 g S)
l 9 g S
L-
13.5 g bottom ash
6 120 kJ to coolin water
31.36 gash
0
.. 130 gS
S emissions = -6- X10,800 k]lkWhr = 1.40 g IkWhr
10 k]
The molecular weight of S02 is 32 + 2 X 16 = 64, half of which is sulfur.
Thus, 1.4 g of S corresponds to 2.8 g of S02, so 2.8 g SO/kWhr would be
emitred. Particulate emissions need to be limited to:
g 7.6 I:> S
to disposal
FIGURE 1.17 Energy and mass balance for a coal-fired power plant generating 1 kWhr
of electricity (see Example 1.13).
Particulate emissions = 1~ g X 10,800 k]lkWhr = 0.14 glkWhr
10 k]
To find carbon emissions, first find the amount of coal burned per kWhr
. 10,800 k]lkWhr
Coal Input = = 450 g coallk Whr
24 k]lg coal
Therefore, since the coal is 62 percent carbon
C b
" 0.62 gC 450 gcoal
ar on emissions = X = 280 g C/kWhr
g coal kWhr
b. Burning 450g coal containing 2 percent sulfur will release 0.02 X 450 =
9.0 g of S. Smce the allowable emissions are 1.4 g, the removal efficiency
must be
The Carnot efficiency limitation provides insight into the likely performance
of other types of thermal power plants in addition to the steam plants just described.
For example, there have been many proposals to build power plants that would take
advantage of the temperature difference between the relatively warm surface waters
of the ocean and the rather frigid waters found below. In some locations, the sun
heats the ocean's top layer to as much as 30°C, whereas several hundred meters
down, the temperature is a constant 4 or 5°C. Power plants, called ocean thermal
energy conversion (OTEC) systems, could be designed to operate on these small
temperature differences in the ocean; however, as the following example shows, they
would be inefficient.
EXAMPLE 1.14 OTEC System Efficiency
S removal efficiency = 1 - ~:~ = 0.85 = 85 percent
c. Sflince
h
10percent of the coal is ash, and 70 percent of that is fly ash the total
y as generated Will be '
Fly ash generated = 0.70 X 0.10 X 450 g coallkWhr
= 31.5 g fly ash/kWhr
The allowable particular . .
must be installed th the ~ttferlls restricted to 0.14 glkWhr, so controls
a ave teo lOWingremoval efficiency:
Particulate removal efficienc - 1 0.14
y - - 31.5 = 0.995 = 99.5 percent
In Chapter 7, we will see how th '.
ese emiSSIOncontrol systems work.
Consider an OTEC system operating between 30°C and 5°C. What would be the
maximum possible efficiency for an electric generating station operating with
these temperatures?
SolutIon Using (1.55), we find
(5 + 273)
'1n", = 1 - (30 + 273) = 0.08 = 8 percent
An even lower efficiency, estimated at 2 to 3 percent for a real plant, would be
expected.
Conductive and Convective Heat Transfer
" The complete mass and ener b I. .
Figure 1.17. In this diagram it h r b a ance for this coal plant is diagrammed In
. , I as een assum d h 8 h t
IS removed by cooling water d h . e t at 5 percent of the waste ea
(corresponding to the conditi~:; .t e remaining 15 percent is lost in stack gases
given In Example 1.12).
When two objects are at different temperatures, heat will be transferred from the
hotter object to the colder one. That heat transfer can be by conduction, by
convection, or by radiation, which can take place even in the absence of any physi-
cal medium between the objects.
33

Radiation
T,
FIGURE 1.18 Heattransfer througha simplewall.
Conductive heat transfer is usually associated with solids, as one molecule
vibrates the next in the lattice. The rate of heat transfer in a solid is proportional
to the thermal conductivity of the material. Metals tend to be good thermal conduc-
tors, which makes them very useful when high heat-transfer rates are desired. Other
materials are much less so, with some being particularly poor thermal conductors,
which makes them potentially useful as thermal insulation.
Convective heat transfer occurs when a fluid at one temperature comes in con-
tact ~ith a substance a~ another temperature. For example, warm air in a house in
the wmter that comes m contact with a cool wall surface will transfer heat to the
,:,all. As that warm air losessome of its heat, it becomes cooler and denser, and it will
sink and be replaced by more warm air from the interior of the room. Thus there is
a continuous movement of air around the room and with it a transference of heat
from the warm room air to the cool wall. The cool wall, in turn, conducts heat to the
cold exrenor surface of the house where outside air removes the heat by convection.
Figure 1.18 Illustrates the two processes of convection and conduction
through a hypothetical wall. In addition, there is radiative heat transfer from objects
m the room to the wall, and from the wall to the ambient Outside. lt is conventional
practice to combine all three processes into a sin I II h f cess
thar is characterized by the following simple g e, overa eat-trans er pro
equanon.
A(T - T)
q = 'R 0 (1.58)
where
q = heat transfer rate through the wall (W) (B /hr)
A = wall area (m2)or (ft2) or tu r
Ti = air temperature on one side of the II ('C
T - bi . wa ) or ('F)
o ::: am lent air temperature ('C) or ('F)
R - overall thermal resistance 1m2_'crw) (h 2
Or r-ft -'F/Btu}
The overall thermal resistance R is called the _ .,
hardware store, It will be designated as h R value. If you buy insulation at the
system (hr-fr-'FlBtu). For example 3~-' a~mg an R-value in the American unIt
marked R-ll, whereas 6 inches of h me -thick fiberglass insulation is usually
t e same material is R-19.
As the following example illustrates, improving the efficiency with which we
use energy can save money as well as reduce emissions of pollutants associated with
energy consumption. This important connection between energy efficiency and
pollution control has in the past been overlooked and underappreciated. However,
as will be described in Chapter 7, that situation has been changing. The 1990
Amendments to the Clean Air Act, for example, provide S02 emission credits for
energy efficiency projects.
EXAMPLE 1.15 Reducing Pollution by Adding Ceiling Insulation
A home with 1,500 ftl
of poorly insulated ceiling is located in an area with an
8-month heating season during which time the outdoor temperature averages
40'F while the inside temperature is kept at 70'F (this could be Chicago, for
example). lt has been proposed to the owner that $1,000 be spent to add more in-
sulation to the ceiling, raising its total R-value from 11 to 40 (fc2-'F-hriBtu). The
house is heated with electricity that costs 8 cents/kWhr.
a. How much money would the owner expect to save each year, and how long
would it take for the energy savings to pay for the cost of insulation?
b. Suppose 1 million homes served by coal plants like the one analyzed in
Example 1.13 could achieve similar energy savings. Estimate the annual re-
duction in S02> particulate, and carbon emissions that would be realized.
Solution
a. Using (1.58) to find the heat loss rate with the existing insulation gives
= A(Ti - To} = 1,500 ft2 X (70 - 40)'F = 4,090 Btu/hr
q R 11 (ftl-'F-hriBtu)
After adding the insulation, the new heat loss rate will be
= A(Ti - To) = 1,500 ft
l
X (70 - 40)'F = 1,125 Btu/hr
q R 40 (ftl-'F-hr/Btu)
The annual energy savings can be found by multiplying the rate at which en-
ergy is being saved by the number of hours in the heating season. Ifwe assume
the electric heating system in the house is 100 percent efficient at converting
electricity to heat (reasonable) and that it delivers all of that heat to the spaces
that need heat (less reasonable, especially if there are heating ducts, which
tend to leak), then we can use the conversion 3,412 Btu = 1 kWhr.
(4,090 - 1,125) Btu/hr
Energy saved = X 24 hr/day X 30 day/mo X 8 mo/yr
3,412 Btu/kWhr
= 5,005 kWhr/yr
The annual savings in dollars would be
Dollar savings = 5,005 kWhr/yr X $0.08/kWhr = $400/yr
Since the estimated cost of adding extra insulation is $1,000, the reduction
in electricity bills would pay for this investment in about 2" heating seasons.
35

b. One million such houses would save a total of 5 billion kWhr/yr (nearly
the entire annual output of a typical 1,000 MW, power plant). U ing the
emission factors derived in Example 1.13, the reduction in air emi ions
would be
Carbon reduction = 280gC/kWhr X 5 X 109kWhrlyr X lO-Jkglg
= 1,400 X 106 kg/yr
SOz reduction = 2.8gS0z/kWhr X 5 X 109kWhrlyr X lO-Jkg/g
= 14 X 106kg/yr
Particulate reduction = 0.14g/kWhr X 5 X 109kWhr/yr X lO-Jkglg
= 0.7 X 106 kg/yr
Radiant Heat Transfer
Heat transfer by thermal radiation is the third way that one object can warm another.
Unlike conduction and convection di ', ra iant energy IStransported by electromagnetIC
waves and does not require a m di h ', e rum to carry t e energy. As IS the case for other
forms of electromagnetic phenomen h di
h
. . a, sue as fa 10 waves, x-rays and gamma rays,
t ermal radiation can be described ith ' ', I f eit er 111 terms of wavelengths or using the par-
ttc e nature 0 electromagneti di . 'I ' c ra ration, 111 terms of discrete photons of energy, All
::~~~~:t~n~;I~hwa~es travel at tdhehspeedof light. They can be described by their
elf requency, an t e two are related as follows:
c = Av (1.59)
where
c = speed of light (3 X 108 mls)
A = wavelength (rn)
v = frequency (hertz ' I, r.e, eye es per second)
When radiant energy is described'
frequency and energy is given by 111 terms of photons, the relationship between
E = h u (1.60)where
E = energy of a photon (J)
h = Planck's constant 16.6 X 10-34 [-s)
Equation (1.60) points Out that hi h f
hi h Ig er- reque h
Ig er energy Content which k h ncy, s arter-wavelength photons have
I
, , maestemp'l
t ungs are exposed to them. otentla ly more dangerous when living
In this book the most I'
. Ch ' mportant ap licari ,
111 apter 8 when the effects of ' p catIon of radiant heat transfer will come
the stratospheric Ozone layer will V~[)~uS gases on global climate and depletion of
tha.:
6
context are roughly in the rang: o/:~~sed. The wavelengths of importance in
10 m, also called 1 micton) F ut O.llLm up to about 100 JLm (llLm IS
electromagneti~ spectrum. . Or perspective, Figure 1.19 shows a portion of the
. Every object emits thermal di .
radiation I bi ra ianon Th I h
a rea a JeCt emits as well . e usua way to describe how muc
, as Other ch '. h
aractenstles of the wavelengt s
1000 urn
1Microwave
100 urn
l~m
[/
0.7 Red
0,6 Orange
Visible Yellow
0,5 Green
Ultraviolet
Blue
0.4 Violet
X-rays
Wavelength
om urn
0.001 urn
0.000] urn
FIGURE 1.19 A portion of the electromagnetic spectrum. The wavelengths of greatest
interest for this text are in the range of about 0.1 JLm to 100 ,urn.
emitted, is to compare it to a theoretical abstraction called a blackbody. A black-
body is defined as a perfect emitter as well as a perfect absorber. As a perfect emit-
ter, it radiates more energy per unit of surface area than any real object at the same
temperature. As a perfect absorber, it absorbs all radiation that impinges upon it;
that is, none is reflected, and none is transmitted through it. Actual objects do not
emit as much radiation as this hypothetical blackbody, but most are close to this the-
oretical limit. The ratio of the amount of radiation an actual object would emit to
the amount that a blackbody would emit at the same temperature is known as the
emissivity, e. The emissivity of desert sand, dry ground, and most woodlands is esti-
mated to be approximately 0.90, whereas water, wet sand, and ice all have estimated
emissivities of roughly 0.95. A human body, no matter what pigmentation, has an
emissivity of around 0.96.
The wavelengths radiated by a blackbody depend on its temperature, as
described by Planck's law:
(1.61)
where
EA
= emissive power of a blackbody (W/mz-JLm)
T = absolute temperature of the body (K)
A = wavelength (JLm)
C, = 3.74 X 108 W-f.Lm4/mz
Cz = 1.44 X 104
JLm-K
Figure 1.20 is a plot of the emissive power of radiation emitted from black-
bodies at various temperatures. Curves such as these, which show the spectrum of
37

2400
70
60
E
7- 50N
f 40
.~ 30
c
~ 20
10
0
0
350 K
E
7-
} 1200
?c
-z
c
~
2000
1600
250 K
3010 20
Wavelength A(urn)
FIGURE 1.20 The spectral emissivepower of a blackbody with various remperarures.
800
where
400
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
Wavelength A (urn)
FIGURE 1.21 The extraterrestrial solar spectrum (solid line) compared with the spectrum
of a 5,800-K blackbody (dashed).
'. I
, I I I
Ultravio et Visible Tnfrared
7% 47% 46%
,"iI-
I'.
I
I

,
I
1I
,
,
'~ - Extraterrestrial solar flux
I
I
~I
I 5,800-K
~I Blackbody
~
/j I ' - --0
1.8 2.0 2.2 2.
wavelengths emitted, have as their vertical axis an amount of power per unit-
wavelength, The way to interpret such spectral diagrams is to realize that the area
under the curve between any two wavelengths is equal to the power radiated by the
object within that band of wavelengths. Hence the total area under the curve is
equal to the total power radiated. Objects at higher temperatures emir more powet
and, as the figure suggests, their peak intensity occurs at shorter wavelengths.
Extraterrestrial solar radiation (just outside the Earth's atmosphere) shows
spectral characteristics that are well approximated by blackbody radiation. Although
the temperature deep within the sun is many millions 01 degrees, its effecrive surface
temperature ISabout 5,800 K. Figure 1.21 shows the close match between rhe actual
solar spectrum and that of a 5,800-K blackbody.
As has already been mentioned, the area under a spectral curve between any
two wavelengths IS the total power radiated by those wavelengrhs. For the solar
spectrum of Figure 1.21, the area under the curve between 0.38 and 0.78 jLm (the
wavelengths visible to the human eye) is 47 p I h I Thar i 47 pet·
f h I
ercenr 0 t e tora area. at IS,
cent 0 t e so ar energy striking th id f ..
. f h e OutSIe a Our atmosphere is in the visible par-
non 0 t e spectrum The ultr . I· d h
inf d I h: . avio et range conrains 7 percent of the energy an t e
I rare wave engt s delive th . . , h
I
. r e remammg 46 percent. The total area under t e
so ar spectrum curve IScalled th I
2 A '11 b h . e so ar constant, and it is estimated to be 1 372 wattS
per m. s WI e sown m Chapt 8 hi' I'
d .. h I er , t e so ar constant plays an important ro e 10
etermmmg t e sur ace temperature of the Earth
The equation describing Plan k' I ( .. .
especially for calculations involvinc h aw 1.61) IS somewhat tricky to maDlpulate,
arion equations howeve g h e area under a spectral curve. Two other radl'
, r, are sttalg tfor d did The
first, known as the Stelan-B It I war an are 0 ten all that is neede .
I' a zmann awof diati . Y
emitted by a blackbody . h f ra tatuvn, gives the total radiant energ
Wit sur ace are A d b Ia an a so ute temperature T:
E ~ O'AT4 (1.621
2,898
Amax = T(K)
Wavelength A. (urn)
FIGURE 1.22 Wien's rule for finding the wavelength at which the spectral emissive power
of a blackbody reaches its maximum value.
39
4
The second is Wien's displacement rule, which tells us the wavelength at which
the spectrum reaches its maximum point:
2,898
Am" (jLm) ~ T(K)E ~ total blackbody emission rate (W)
0' ~ the Stefan.Bolt
_ zmann constant ~ 5 67 X 10-8 2 4
T:= absolute temperature (K) . Wlm -K
A - surlace area of the object (m2)
(1.63 )
where wavelength is specified in micrometers and temperature is in kelvins. Fig-
ure 1.22 illustrates this concept, and Example 1.16 shows how it can be used.

40 Chapter 1 Mass and Energy Transfer Problems
1.5 If we approximate the atmosphere to be 79 percent nitrogen (N2) by volume and 21 per-
cent oxygen (02), estimate the density of air (kg/m ') at STP conditions (O'C, 1 atm).
1.6 Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of
a conservative pollutant, is released into a stream having an upstream flow of 10 MGD
and pollutant concentration of 3.0 mg/L.
(a) What is the concentration in ppm just downstream?
(b) How many pounds of substance per day pass a given spot downstream? (You may
want the conversions 3.785 Llgal and 2.2 kg/Ibm from Appendix A.)
1.7 A river with 400 ppm of salts (a conservative substance) and an upstream flow of 25.0 m
3
/s
receives an agricultural discharge of 5.0 m3/s carrying 2,000 mg/L of salts (see Figure P1.7).
The salts quickly become uniformly distributed in the river. A municipality just down-
stream withdraws water and mixes it with enough pure water (no salt) from another
source to deliver water having no more than 500 ppm salts to its customers.
What should be the mixture ratio F of pure water to river water?
EXAMPLE 1.16 The Earth's Spectrum
Consider the Earth to be a blackbody with average temperature 15'C and surface
area equal to 5.1 X 1014 m2• Find the rate at which energy is radiated by the
Earth and the wavelength at which maximum power is radiated. Compare this
peak wavelength with that for a 5,800-K blackbody (the sun).
Solution Using (1.62), the Earth radiates
E = aAT4
= 5.67 X 10-8 W/m2_K4 X 5.1 X 1014 m2 X (15.0 + 273.15 K)'
= 2.0 X 1017 Watts
The wavelength at which the maximum point is reached in the Earth's spectrum is
A (m) = 2,898 = 2,898 _
rnax I-' T(K) 288.15 - 10.11-'m (Earth)
For the 5,800-K sun, 500 ppm
A (m = 2,898 _
max JL) 5,800 - 0,481-'m (sun) 25.0 m3/s
400 ppm
1+__ FQm3/s
Oppm
hi hT~istrem~ndous rate of energyemission by the Earth is balanced by the rate at
w
h
IC It e Eart absorbs energy from the sun. As shown in Example 1.16 however,
t e so ar energy striking the Ea th h h h 'di d b k r as muc Sorter wavelengrhs than energy
ra late ac to space by the E th Thi I .
h ff
ar. ISwave engrh shift plays a crucial role in the
green ouse e ect. As described in Ch 8 b 0
I
0 I apter ,car on dioxide and other greenhouse
gases are re anve y transparent t the i . h
they tend to absorb the out Din 0 e incommg s orr wavelengths from the sun, but
h
g g, longer wavelengths radiated by the Earth As those
green ouse gases accumulate in 0 h .
velops the planet, upsets the radiat:::n~mosp ere, they act like a blanket that en·
alance, and raises the Earth's temperature.
PROBLEMS
50m
3
/s 7//'2,000 ppm
FIGUREPl.7
1.8 A home washing machine removes grease and dirt from clothes in a nearly first-order
process in which 12 percent of the grease on the clothes is removed per minute. The wash-
ing machine holds 50.0 L of water and has a wash cycle of 5.00 minutes before discharg-
ing the wash water. What will be the grease concentration (in mg/L) in the discharge
water if the clothes initially contain 0.500 g of grease?
1.9 Plateau Creek carries 5.0 m3/s of water with a selenium (Se) concentration of 0.0015 mg/L.
A farmer starts withdrawing 1.0 m3/s of the creek water to irrigate the land. During irri-
gation, the water picks up selenium from the salts in the soil. One-half of the irrigation
warer is lost to the ground and plants, and the other half is returned to Plateau Creek. The
irrigation run-off to the creek contains 1.00 mgIL of selenium. Selenium is a conservative,
nonreactive substance (it does not degrade in the stream), and the stream does not pick up
more selenium from any other source.
(a) If the farmer irrigates continuously, what will be the steady-state concentration of
selenium in the stream downstream from the farm (after the irrigation run-off returns
to the stream)?
(b) Fish are sensitive to selenium levels over 0.04 mg/L. The farmer agrees not to use
more warer than will keep the stream selenium level below this critical concentration.
How much water can the farmer withdraw from the stream to use for irrigation?
1.10 When methanol is used to generate hydrogen, it reacts with the following reaction:
2CH20H ---> 2CO + 3H2
The reaction is second order in methanol (CH20H), and it is observed that 100 g of
carbon monoxide (CO) can be produced in one day in a batch reactor, if you start with
200 g of methanol. What is the rate constant for this reaction?
1.1 The proposed air quality standard for Own (0)' 0
e 3 IS .08 ppm
(a) Express that standard in l-'g/m3 t 1 f '
b
a atrn 0 pressure and 25'C
( ) At the elevation of Denver the 0 •
d d
' pressure ISabo t 0 82
ar at that pressure and at a t u , atm. Express the ozone stan-
emperature of 15'C
Suppose the exhaust gas from an aut bOI .
id omo I e co t 0 1 0
rnonoxi e. Express this concentratio 0 g/ 3 n ams . percent by vol ume of carbon
n III m m at 25'C d
Suppose the average concentration of SO . an 1 atm.
1 atm. Does this exceed the (24 h) 0 2 IS measured to be 400 JLg/m3 at 25'C and
for atomic weights.) - r air quality standard of 0,14 ppm? (See AppendiX B
A typical motorcycle emits about 20 f CO 0
(a) What volume of CO would a 5- glo 0 per mile,
(b) rme tnp produce f h l
Per meter of distance traveled h a ter j e gas cools to 25'C (at 1 atm)·
d d f
,w at volum f 0
stan ar 0 9 ppm' eo air could be polluted to the air quality
1.2
1.3
1.4
41

42
1.11
1.12
Chapter 1 Mass and Energy Transfer
k
irh I 10 X 106 mJ is fed by a pollution-free stream with flow rate
A la e Wit constant vo ume .'
50
31 A f tory dumps 5 m3/s of a nonconservative waste with concentranon
m s. ac ff K fOnd
100 mgiL into the lake. The pollutant has a reaction rate coe rcienr 0 .i» ay.
Assuming the pollutant is well mixed in the lake, find the steady-state concenuanon of
pollutant in the lake,
The two-pond system shown in Figure P1.12 is fed by a stream with flow rate 1.0 MGD
(millions gallons per day) and BOD (a nonconservativepollutant) concentratton 20.0 mgIL,
The rate of decay of BOD is 0.30/day. The volume of the firsr pond IS5,0 million gallons,
and the second is 3,0 million,
Assuming complete mixing within each pond, find the BOD concentrarion leaving
each pond,
20 mglL
1 MOD
-
FIGURE Pl.12
1.13 A lagoon is to be designed 10 accomodate an input flow of 0.10 m3/s of nonconservative
pollutant with concentration 30.0 mgiL and reaction rate 0.20/day. The effluent from the
lagoon must have pollutant concentration of less than 10.0 mgiL. Assuming complete
mixing, how large must the lagoon be?
Ozone is sometimes used as a disinfectant for drinking water, Ozone is highly reactive,
It will react With many benign species found in water, as well as the pathogens it is in·
tended to kill, It ISfound that ozone reaction in water nearly follows first-order kinetics
m ozone concentration, such that its concentration decreases by 50 percent in 12 minutes
(t1/2 = 12 min), A water supplier wants to inject Ozoneinto a pipe bringing water 10 the
water treatment plant to predisinfecr the influent The 3 0 f di " 3 400 feet
I
' . . - t. iarneter pIpe IS ,
ong with a steady flow rate of 10000 galimin Wh t ion of (i mgILl. . ,. a concentration 0 ozone III
should be injected at the head of the pipe so that the 'II b rion of
1 0
he ni , " re WI e an ozone concentra I
. mgIL ar t e pipe s exit into the plant) The o: b " s
an ideal PFR. . e pipe may e realistically assumed 10 act a
You~stomach is a chemic~lkreactor.When you devour a fast food 99¢ special hamburger
lI1 ha lout a mll1Ule'h,tacts , e an IOstantaneousinput of 325 g of food entering the srom
ac . n response, t e stomach starts p duci " '
ously excreted into the st h ro ucmg gastric liquids (acids), which are connnu'
The fluid also leaves the s~:~~cha:oat~ate of 12.0 mLimin as the hamburger is digested,
the volume of liquid in the st h e small IOtestmeat a flow rate of 12.0 mLlll1ln, SO
omac stays co r 1 ' inn
rate constant is 1.33 hr-', ns ant at .15 L. The hamburger dlgesno
(a) What kind of ideal reactor would yo d I
(b
' u mo e your stomach as)
) What fractton of the hamburger' , " h
one hour after you eat the hamb s m)assWill remam undigested in your stomar
. urger.
A simple way to model air pOllutionov " ,
plete mixing and limited capabill't f hera CItyIS WItha box model that assumes com-
" y or t e pollur' d' ' IlYexcept 10 the dtrection of the pr 'I' , Ion to Isperse horizot1lally or vertICa
eValmg wmds (f IIyOr example, a town located in a va e
1.14
1.15
1.16
1.17
1.18
1.19
Problems
with an inversion layer above it). Consider a town having an inversion at 250 m, a 20-km
horizontal distance perpendicular 10 the wind, a winds peed of 2 mis, and a carbon
monoxide (CO) emission rate of 60 kgis (see Figure P1.16). Assume the CO is conserva-
tive and completely mixed in the box.
What would be the CO concentration in the box?
Wind ---f-.
2 m1s
c=?
-- ..2 m/s
43
------_.,-
• 60 kgls
I or co
250 m
20km
FIGURE Pl.16
1.20
1.21
Consider the air over a city 10 be a box 100 km on a side that reaches up to an altitude of
1.0 km. Clean air is blowing into the box along one of its sides with a speed of 4 mls. Sup-
pose an air pollutant with reaction rate k = 0.20/hr is emitted inro the box at a total rate of
10.0 kgis. Find the steady-state concentration if the air is assumed to be completely mixed,
If the windspeed in Problem 1.17 suddenly drops to 1 mis, estimate the concentration of
pollutants two hours later.
A lagoon with volume 1,200 m3 has been receiving a steady flow of a conservative waste
at a rate of 100 m3/day for a long enough time to assume that steady-state conditions
apply. The waste entering the lagoon has a concentration of 10 mgiL. Assuming com-
pletely mixed conditions,
(a) What would be the concentration of pollutant in the effluent leaving the lagoon?
(b) If the input waste concentration suddenly increased to 100 mgiL, what would the
concentration in the effluent be 7 days later?
Repeat Problem 1.19 for a nonconservative pollutant with reaction rate k = 0.20/d.
A nuclear power station is situated in Coal Valley, which is a roughly rectangular valley
that is 5 km long, 2 km wide, and 200 m deep. You have been asked to evaluate the effects
of a worst-case scenario where the reactor housing fails, and radiation is released to the
atmosphere. In your evaluation, you determine that 120 kg of Iodine-131 (a radioisotope
that causes thyroid gland and liver damage) could be released into the atmosphere,
(a) Assuming the release of Iodine-131 was very rapid and all of it was uniformly dis-
tributed through the valley's atmosphere with none escaping the valley, what would
the concentration of lodine-131 be in the valley'S air? Your answer should be ex-
pressed in units of ppmv, and you may assume an atmopheric pressure of 1.0 atm and
a temperature of 20"C.
(b) Assuming the Iodine-131 concentration you calculated in part (a) is the initial con-
centration in the valley, you now want to determine the time it will take for tbe con-
centration to decrease to the safe limit of 1.0 X 10-5
ppmv. The average windspeed
through the valley (entering at one end and exiting at the other) is only 1.5 mlmin.
However, Iodine-131 also is removed by two other processes: 1) radioactive decay

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