Ch. 1 Enviromental Engineering

8/9/2019 Ch. 1 Enviromental Engineering

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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

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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 isdue ro the intentional andurunrennona mpur from If'leagues and students In a ticulgroup a veryopen, mtelligent, and dedicated col-

and Eduardo Saez of'the Upamcuar, wfeAwouldlike to thank Professors Bob ArnoldmvefSltya mona for rhei iticali .

o ve rs ig ht At S ta nf or d U' . D R e rr c nn ca m pu r a nd c ons is re nr . mversIfY r. oyal K p d ided i . .

and assistance with the chapte ' I ' a peru proVI ed invaluahlr- insightsrs on air qua tty and li hpublisher would also like to thank th f II' c rrnare c 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~cohraadAC001 of Mines, David L. Sedlak, UC

M ' , an very Demo d U' , f .os r Im po rt an tl y b ot h a ut ho f f . n c, m ve rs iry 0 Michigan.K h ' rs a er special th k '

aren, w 0 served asghost~write di . an s to our WIves, Mary andrhi ff T hei rs, e itors motlVat d .

IS e art. elf support enco ' Drs an , too often victims ofI . ,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 Fahrenheitas well as the more recommended International System of units. Both are used in the

practice of e nvironmental 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.


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2 Chapter 1 Mass and EnergyTransfer

Libr

MasIn

Mas

, In a similar way, the law of conservation of energy i al 0an nnal a o u n r-

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 ro d "

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 Symbol10 15 femro f10-12 pI CO P10-9

nano n10-6 mICro

J . L10-3 milli m

10-2 centi c10-1 deci d10 deka da102 hecro h103 kilo k10' mega M109 glga G1012 tera T1015 peta p

1018 exa E1021 zetta Z1024 yotta y

IrI~

1

W eiT

6In t he Un it ed St at es e nv iro nm I ' ,

the U,S, Customary System (US~~a q~a~tItles are measured and reported in b o t his important to be familiar with b) ~n I t ~Inter natIo nal System o f U n it s ( IJ so II

although the U.S. system wil! b otd

" n tms book, preference i given 10 I units,, fIe use In some CI -r- b .

sion actors between the 51a d uses rcumstances, Ia le 1.1 II t conver-'II n systems f f

WI b e e nco unt ere d. A mo re e t ' b o r s ome 0 me mo r ba ic units ro a r"U fie x enSIve ta Ie of co ' , , .

se u auversian Factors." nversions 1 g iven In Appendix A1

In the study of environmental en inee ' "extremely large quantities and gl nng, It IScommon ro encounrer b o t hroxi b extreme y small ThOXICsu stance may be meas di 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 usefulf accompa h ' ,

pre ixes are presented in Table 1 2 ny t e urnrs, Some of the most importantOfte " h ..

n, It IS t e < on ce nt .interesr U' h ratIon of some sub "

b d' SIng t e metric s yslem i ' h srance In air Or warer thar is of

ase On mass (II n eit er medimol) h i h u su a y mg Or g ) v el um ( Il Iu m, c onc en tra ri on s ma y b e,W IC can le d ' e usua y1 3 II

t har o ne I a t o so me c on fu si on 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 weighrIqUi 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 ,Se n

Ed ;

SerA Il

CoAr

M,

M.

SeI

A I

',1

re

IA

P

p

1

ile

t

ConCentrations of subStances dis I d'

mass or number er u . So ve InWaterp 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 highthar 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 k rn 3.2808AI CelSius g

ea 'C 2.2046Volurne square rneter rn' 1.8 (OC) + 32E Cubic meter

nergy kilojoule rn3 10.7639POwer k 35.3147VelOCity Watt J 0.9478Flow rate rneterlsec 3.4121D . meter]/sec rnls

enslty kilogrr n J rn3 /s 2.23694S. a meter] 35

AppendIX A fora m kg/m3 .3147Ore COlllplete list. 0.06243

uses unirsfr

IbOF

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), Ifsodium fluoride (NaF) ispurchased 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)


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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/molmass 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 IIIwater 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 genericallyEXAMPLE 1.2 Volume of an Ideal Gas

c=mV (1.4)

where m i s t he m ass o f t he s ub st an ce a nd V i s th e v I . _ a nd th e r esults of the t I I ' 0 ume of the fluid. Using (1.4)

wo ca cu anons above the I f h btreated 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 OC temper-

ature. Repeat the calculation for 1 atm and 25C.

Solution Using (1.6) at a temperature of OC (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 - kit he U ni te d S ta te s! rm 1 09 Wa te r fo r a bo ut 2 0, 00 0 p eo pl e i n

and at 25C (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 custoI , mary to ex IIvo umetnc terms. For example th . press po utanr concentrations inII' (ppm} i , e concentration of

rru Ion ppm) ISthe volume of poll t . _ a gaseous pollutant in parts peru ant per million I f

1 I vo umes 0 the air mixture:vo ume of gaseous pollutant

106

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/m3perature and I I' pmvan mg/rn! d d

'. mo ecu ar weIght of the 0 1 1 _ epen s on the pressure, tern-that relationship' p utant. The Ideal gas la h I bli h

. w e ps usesta 15

From Example 1.2, 1 mole of an ideal gas at OC 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 a ir 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 wtmg/m - 22.414 (at OC and 1 atm)

Similarly, at 25C 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 mo l wt 27 3. 15 K

mg/m = 22.414 X T(K) X

P(atm)

1 atm(1.10)

5

(1.7)

(1.8)


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6 Chapter 1 Mass and Energy Transfer 1.3 Materials Balance

Contro l vo l ume \ 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 i n

rng/nr' at 1 arm and 25'C.

.......... AccumulationtInputs~ --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

CO9.0

percent = 6 X 100 = 0.0009%1 x 10

To find the concentration inmg/rrr', we need the molecular weight of CO, which

tS28 (the atomic weights ofC and 0 are 12 and 16, respectively). Using (1.9) gives

CO =9.0 x 28 _ 324.465 - 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 thesame temperature and pressure co dition) id

I' , n I Ion provi es several other interpretations ofvo umetnc co nce nt ra ti on s e xp res sed as p pmv . F or el l ' 1 I fpollutant per million volumes of air whi h ' . xamp e, ppmv IS vo ume 0rant per million moles f ' S' '1'1 I C IS equivalenr to saying 1 mole of pollu-

o air, irnuarty smcee hi' fmo lec ul es 1 p pmv al so C d' ac m o e co nt ai ns th e s ame n umb er 0

, orrespon s 101 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) isO . 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 Imental engineering principles Mo p e way to express one of the most funda-

h h ' . re preCisely th I f '

t at w en chemical reactions tak I ,e .aw 0 conservauo o f mass says

(th h 'I

e p ace, matter IS ith

daug In nuc ear reactions mass b nett er created nor destroyeII ' ' can e convert d a ows us to do IStrack materials f e to energy). What this conceptirh b J ' or example poll f

WI mass a ance eq uat io ns T hi , ut ant s, r om o ne p lac e t o an ot he r II' . IS ISone of the m id

po . urants 10 the environment and' h basi ost WI ely used tools in analyzingb d d ' ISt e asis for f

e intro uce 10 later chapters. many a the appr oaches that will

The first step in a mass b Ih - a ance analys' def

s~ace t at ISto 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 rlom

a glass of water or simple chemicalasm above a cit hI' wer P ant a lak '

y, Or t e g a be Its el f B ' . ' e, a s tr et ch o f s tr eam a n ai r . Y P1ctunng an . . '

Imagmary boundary around

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8 Chapter 1 Mass and Energy Transfer 1.3 Materials Balance

Stream ~ .: --"\

,/ Accumulmion e0 \ 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 rateC=concentration

of pollutant

decomposing organic wastes in a lake are examples of nonconservarive substances,

Often problems involving nonconservative substances can be simplified when rhereaction rate ISsmall enoughto 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 (sothe accumulation rate equals 0) and th b ' " ,

. ' e su stance in question IS conservative (so thereaction rate equals0) , In these cases, (1.12) simplifiesto the following:

Input rate ~ Output rate (1.13)

Consider the steady-state con' hown i, d ithi ,servatlve system sown mFigure 1,2, The systemcontame Wit m the boundaries mi ht b I k 'O th f ai b ' g e a a e, asection of a free flowing stream,r e ma ss 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 Cmass vo ume . e other mput IS assumed to be a . Sa nd p ollu ta nt co nc en tr atio C T h , wa ste str ea m With f lo w r ate Q w

n W' e output ISa rni t ' h fl Q dpollutant concentration C If h 1 1 ' x ure Wi t ow rate m a n

rn- t e po utant IS c ' d ifs te ad y s ta te c on di ti on s t he n a b I o ns er va nv e, a n 1 we assumefollowing: ,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 besubstituted 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 mgILm (10,0 + 5,0) m3/s

, C,Q, + CwQw ec CmQm (1.14)

T he f ollo wmg e xa mp le i llustr ates th f h i ,provides a general algorithm for d e use0 t ISequation. MOte importantly, it also

omg mass balance problems,

EXAMPLE 1.4 Two Polluted Streams

Astream flowing at 10,0 m 3/s has ibTh' a tn utaryfeed' , , , 3e stream s concentration of chi id mginro ItWItha flow of5,0 ra'!.

the,tributary chloride concentrat:~ ~1~.t~e~mof the junction is 20,0 mg/L, and

v an ve s ub st an ce a nd a ss um in g 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 0vmg a mass b Iem, I ent the "region" or control volu a ance problem is to sketch the prob-

the variables as has been done in Figu 1~efthat we Want to analyze, and labelre , 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


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~ ~ ~ -----


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----~

-----~. - -~ ------ ------ -~ -----

12 Chapter 1 Mass and Energy Transfer 1.3 Materials Balance 13

200

c 150

0

.~

b100s

oc0

u50 I \-

I~ Y Production

./aa 2 0 40 60 80 100

Time

200 ~- - - - - - - - - - - - - - - - ,

Decaye 150

0

.~

100c

"cc0

u50

a

0

Production

20 40 60 80 100

Time

FIGURE 1.5 Concentration of a substance reacting in a batch system with first-orderkinetics.

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-orderreaction rateIS

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 =VdC/dt =Vr(C) , So summarizing (1.16), (1.19), and (1.21), we

can write the reaction rate expressions for anonconservative substance:

r(C) =kC2

(generation) or r(C) = -kC2 (decay) (1.21)

where k is n OW a re ac ti on ra te c ons ta nt w it h u ni ts o f ( 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 thesecon -or er ecay 0 a nonconservative substance in a batch reactor

dCVdt = -VkC

2

wh ic h c an b e i nt eg ra te d a nd t he n s ol ve d f h 'or t e COncentratiOnto yield

C = _-,Co, - -

1+ Cokt

Figure 1.6 shows how the cOncentration ofdecays 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 tia ive Pollutants

Ifwe assume that steady-state conditions "Iservative, then (1.12) becomes prevai and treat the pollutants as noncon-

o = Input rate - Outpur rate + Reaction rate (1.23)

Thebatch reactor, which has just b disystem 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-IXmgIIIthe 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 + kC2V (1.30)


9/25

1.3 Materials BalanceEXAMPLE 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 m3/s and pollutant concentration equalto 10.0 mgiL (Figure 1.7). There is

also a sewage outfall that discharges 0.5m 3/s of wastewater having a pollutant

concentration of 100 mg/L. The stream and sewagewastes 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- tarepollutant 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 aCSTR-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 reactioncoe 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, about1.4 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 =Q m C m =(Q , + Q w ) C(1.31) then becomes

Q,C, + o,c, = (Q, + Q w ) C + kCYAnd rearranging to solve forC

,

C = Q,C, + Q w CwQ, + Q w + k Y

_ 5.0 m3/s X 10.0 mgIL+ 05 31

- . m s X 100.0 mg/L

(5.0 + 0.5) m 3/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. Ifwe 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

Q C =G - kCY

so

C = ..! Q Q .._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

---.;:

L ak e \ / Oasis \In c o min g

_ s l _ " _ U f l _ ' . - / (rI n d o or c o n c en t r at i o n C

V : : : : 500 m 3

- ,-,J~ V =1 0. Ox lO '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=1,000m3/hr

-t--C=?

Ou tg o in gQt=5.0 m 3/s

C, = 10.0 mglLFresh air

1 4 0 mg lh r

I !lkO.40/hr

14 FIGURE 1,7

15


10/25

16 Chapter 1 Mass and Energy Transfer1.3 Materials Balance

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 _ 0095HCHO = iff = -. ppm

mol wt 30

This is nearly double the 0.05 ppm threshold for eye irritation.

Lake

Q =700m 3/rnin

. .. .. -- C o = 7 fishlm3

Q = 7(J () m3/min 75 \


11/25

18 Chapter 1 Mass and Energy Transfer1.3 Materials Balance

then

dy dC

dt dt

Flow OUtI

Q.C

Flow in,

(1.39)

Control vo lume VConcentration C

Q,C,

so (1.37) becomes

Decay coef f ic ient kd

Generation coefficient k g

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 togive

VC, and the rate of accumulation of pollutant in the box is VdC/dt. Ler u designaterhe concentration of pollutant entering the box as C, . We'll also a ume there areboth 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 ofy at t =O . Integrating gives

y =yoe - (kd+~) t (1.42)

IfCo is the concentration in the box at time t=0, then from (1.38) we get

(1.34)Yo =C o - C oo (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 mi g t e

encountered; any consistent set will do.

An easy way to find the steady-state solution to (1.34) is simply to setdC/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 + (C o - C oo )e xp [ - ( k d + ~)] (1.45)

Equation (1.45) should make some sense. At time t =0, the exponential function

e qu al s 1 an d C =Co. At t =00, the exponential term equals 0, and C =Coo.

Equation (1.45) is plotted in Figure 1.11.C oo =Q C i + kgV

Q + kd Vwhere C o o is the concentration in the b ' hi s w it h t h e co nc en tr at io n b f ' o x a t t im e t =DC. Our concern now, thoug ,

Rearranging (1.34) gives e ore It reaches steady state, so we must solve (1.34),

~ ~= -( ~+ k} (C _ Q C' +k gV ) (1 .361

hi h . Q + kdVw IC , usmg (1.35) can be .) rew ntten as

~~ = -( ~ + kd)- (C - Coo) (1.371

One way to solve this differential ' .equation 1S to m ak e a h f vari ble If we letc ange 0 vana .

Y = C - C oo (1 .3 8 1

(1.35)

c " " - - - - - - - - - - - - - - - - - - - - - . - - - - - - - - - - - - -

C = . : : Q ; ; - C - , : - i+-ck-;,,-v~ Q +kdV

o

Time, 1

F I G U R E 1 . 11 Step function response for the CSTR box model.

19


12/25

20 Chapter 1 Mass and Energy Transfer

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 m3

/hr. Suppose when the bar opens at 5 P.M., rhe air i lean. Ifformalde-

hyde, with decay rate kd = OAO/hr,is emitted from cigarette smoke ar the constant

rate of 140 mg/hr starting at 5P.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 + k dV Q + k dV 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 atany nrne after 5 P.M., we can apply ( 1045) with Co =O.

C(t) = Coo{l- e X P [ - ( 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 t im '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' a t t e o ut a I suddenly stops draining mID

ate a e s pollutIOn stops.

EXAMPLE 1.9 A S ud de n D ee r .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' 'T he P Ol lu tI On i s n on co ns e . . s a te p o unonconcentration of 3.5 mg/L.

Suppose the condition of thelVlatklveWdlthreaction-rate constant kd

= 0.20/day., . d id a e IS eemed IIt IS eCI ed to completel di h unacceptable. To solve the prob em,

y I vert 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 A ss ', t e s ew ag e o ut fa ll r em ov ed t he o ut go in g. f . ummg co I' ,tlon 0 P Ol iu tan r i n th e l ake k mp ete-mlx conditions find the concenrra-stead one wee afte h d" , I

y -s tat e co nc en tr at ion . r t e I ver si on, a nd fin d the n ew f lna

Solution For this siruation ,Co = 3.5 mg/L

V= 1 0 X 10 6 m3

Q=Q -503,- . m /s X 3 600 s ib

' 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/dayC - --: : : ---= --:" -: -: :

00 - Q + k dV 43.2 X 104 m3/day + 0.20/day X 107 r r r '=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 104

;\day

) X 7days]10 X 10 m

C(7days) ~ 2.1mg/L

Figure 1.12 shows the decrease in contaminant concentration for this example.

3

3,5 1 ' - - -' "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


13/25

22 Chapter 1 Mass and Energy Transfer1.4 Energy Fundamentals

affects the environment is an important consideration in the study of environmental

engineering and science. .Another important term to be familiar with ispower. 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 (1Jl s =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~ 1F, 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

15C (59F). 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/kgC.

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. Ithas 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 ThermodynamicsThe first law of thermodynamics says,simply,that energy can neither be created nor

destroyed. Energy maychange 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 isconverted to mechanical energy as it spins a turbine in a hy-

droelectric plant. No matter what ishappening, the first law says we should be able to

account for everybit of energyas 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 toas opell

systems, whereas those IIIwhich energy is allowed to f 1 h b d y but. ow across t e Dun ar ,matter ISnot, are calledclosed systems.

h dSinfcedenergyis conserved, we can write the following for whatever system we

a ve e me :

(:~::i~;e~::ndarY) + (~~:~:;ergy ) _(T~tal energy) (Net Change) (1.46)

h 0 mass = of energy In

a s e a t a nd wo rk e nt er in g s ys te m leavieavmg system the syste m

For closed systems there is no move fand third termdrop out of th ment a mass across the boundary, so the second

e equation The ac I' f d bt he r ig ht s id e o f ( 1 46 ) m h . , c um u a no n a e ne rg y re pr es en te Y. a y c au se c an ge s10the b bl f f e ne rg y, suc h a s k in etic a nd p ot ' I ' a se rv a e , mac ro sc op ic a rms a

. entia energies or mi . f I d thea to rr uc a nd m ol ec ul ar s rr uc t f h ' IC ro sc op lc a rm s re a te ro. ure ate system Th . , f inc lu de r he k in etic e ne rg ie s o f I I . o se mic ro sc op ic f or ms a e ne rg y

. b rna ecu es and the . , h f Sacting 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 energyE th b beescn e t en as the sum of' . at a su stance possesses can, I ItS IOtemal ene U' k d .

potentia energyP E: rgy ,1lS inetic energy KE , a n I tS

!! .H=mcp!! . 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 mby an amount !! . T:

Change in stored energy =mc S'T (1.51)

Table 1.3 provides some examples of specific heat for several selected sub-

stances. Itis 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 beE = U + K E + PE (1,471

23


14/25

24 Chapter 1 Mass and Energy Transfer1.4 Energy Fundamentals

TABLE 1.3There 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,B tu /l b' F) ( kj /k g' C)

Water (15C)

Air

Aluminum

Copper

Dry soil

IceSteam (100C)'

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 2.090.48 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 ~ mL (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 toheat 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 thenumber 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 ~ rn cAT

~ 40 gal X 8.34lb/gal X 1 Btu/lb'F X (140 - 50)'F

~ 3 0 X 103Btu

Setting the energy input equal to th h '. .u ni ts us in g T ab le 1. 1 y ie ld s e c an ge 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. 18kJreMelting ice

latent

heat of

fusion

333 kJ1 ~

o I,'2.1 kWCIce

Heat added to I kg of ice (kJ) -

FIGURE 1.13 Heat needed toconvert 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


15/25

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

5 1Units nilSProperty

1 . 0 0 B r u J l b " F

9 2 B ru Jlb

1,060 B ru l lb

1 -1 4 B ru Jlb

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 iscarrying 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 Earthto evaporate. The value of 15 ha been pickeda the

starting temperaturesincethat isapproximately 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 balancet h a t

WIll be encountered in Chapter 8, isto estimate the energy required ro powert h eglobal hydrologiccycle.

EXAMPLE 1.12 Thermal Pollution of a River

A coal-fired power plant converts one-third of the coal's energy inro electricalenergy. 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 20.0C.

a. If the cooling water is only allowed to rise in temperature by 10.0C, 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/m 2

whic h is e quivale nt t I ' r X 5.10 X 1014

m2

the Ea h ' f a a most h alf o f th 1 68 2 'k'1l8. rt s S ur a ce (see Fig 8 e W/m of incoming sunlight srn I

quired to raisethe watet va~:~hi14

).It might also be noted that the energy~'

~gh~ble compared to the heat ~; into the atmosphere after it has evaporatedI it IS C apte r). vapoma tion (see P roblem 1.27 a t the end 0

- - - - - -

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


16/25

28 Cha prer 1 Mass and Energy Transfer 1.4 Energy Fundamentals

Stack heat

JO O M W ,

Hot reservoir

Th

Qc Wasteheat

Cold reservoirT,

Qs = 100.0 m3/s

T~=20.0 - cQ = IO O .O m "

T =2~.1 C

FIGURE 1.15 Definition of terms for a Carnot engine.S tr eam . . . . . . . .

FIGURE1.14 Coolin gwateren ergybalanc efor the33.3 percent efficient, 1,000 ~lW,powerplant inExample1.12.

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 aheat 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). Itturns 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. Itis 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 T O " An amount of heat energy Q h 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) =Q h (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, whofirst developed the explanation in the 1820s. Analysis of Carnot engines shows that

the most efficient engine possible, operating between two temperatures, Th a nd To

has an efficiency of

and

Coolant losses= 0.85 X 2,000 MW, = 1,700 MW,

a. Finding the coolingwater neededtoremove 1 700 MW with a tempera-

ture .increase srof 1O.0'C will require the u~e of (1.53) along with thespecificheat of water,4,184 Jlkg'C,given in Table 1.4:

Rate of changein 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 103 kgls

or, since 1,000 kgequals 1 3 fb T f d h m 0 water, the lowrate is 40.6 m3/s.

. 0 in t e new temperatureofth . M Wbeing released intothe ri hi efiver,we can use (1.53) with 1,700 ,

e fiver,w ich again has a flow rate of 100.0 m 3/s .Rate of changein internalen _.

ergy - mc!l.T

1,700MW X (1 X 106J/S)!l.T= MW

100.00 m 3/s X 103 k'{/ 3 = 4.1 'Cso the temperature ofth e 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 drno ynamics

In Example 1.12, you willnoticeth . .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 ' . ed t ob . e second law of h n e up as waste heat reJecr

e some waste heat; that is, it isimp tbelrmodynamics says that there will alwaY'~eto& ~

' i i i i i i i i i i lllliiiii _iiiiii_==-__ Visea machine that can convert

TJma x = 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


17/25

30 Chapter 1 Mass and Energy Transfer 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 600C.

For To we might use an ambient temperature of about 20C. Using these values in

(1.55) and remembering to convert temperatures to the absolute scale, gives

Steam-"

. .. .. -- C o o l 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

tCooling water

Boiler

fe ed pum p

U : ~ ~ ; : : : = : : : : , ; : : ; w ~ a n nYOa t e rQU ICon den s e r

Air

F IG URE 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 e ffic iency increases as rh ', increases as t e temperature of the hot reservesincreases or the temperature f th Id ' , 'f , I hoe co reservoir decreases. In fact since n e i rh e r i n -

mire y at temperatures no b I 'Id h

,r a so ute zero temperatures are possible we must coo-c ue t a t no re al engine ha 100 ff' , 'f

h dIs percent e iciency which is )'ust a resrarement 0

t e secon aw. '

Equation (1.55) can help us d d he seemi , ' f h1 I u n e rs ta n t e seemingly low efficiency a t er-rna power p ants such as the a d i " 'burned in a firing ch b ne iagramma] 10Figure 1.16. In this plant, fuel IS

am er surrounded bib' " hr ht hi s b oi le r t ub in g i s c d h i y meta tu mg. Water Circulating t oug

onverte to ig h p h i h ' hconversion of chemical t h I - ressure, Ig -temperarure steam. Dunng t IS

a t erma ener I hd ue t o i nco mp let e c b ' g y, a ss es on t e or de r of 10 per cen t o ccu r

am usnon and1 0 f h h I Iconsider local and regional' II' ss a eat up the smokestack. Later, we s atheir possible role in glob Iarr po, unon effects caused by these emissions as wellas

a warm m g.The steam produced inth b 'I

ways similar to a c hi ld ' , h aller then enters a steam turbine which is in somes pmw eel. Th h i h "

through the turbine blad ' e Ig -pressure steam expands as It passes, es, cau smg a sh f t h ' to

spm. Although therurb' 'P ' a t at IS connected to the generatorb i 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 109a nd C Oo li ng a s ' t T o ne s ta ge a nd en te ri ng a no th er , g ra u a" , 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 i on , w e - de SI gn ed t b ' o es o ut o nt o t ra ns mi ss io n l in es f at I St rth ur me may hw 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 liquidhelps pull h enser. ThIS ph h hal

s te am t ro ug h t he t b ' a se c an ge cr ea te s a p ar ti al v ac uu m 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 c on den ser ad ' c oo mg wat er i s dr awn f rom a aonce-through cooling.A rna' n teturned to that body of water which is called

r e exp en s ive a 'o f 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


18/25



Solution

a. We first need the heat input to the plant. Becau e 3 k Wh r o f h e ar a re

required for each 1 kWhr of electricity delivered,

Heat input =3 kWhr heat X 1 k]/s X 3 600 slhr = 1 0 0 0 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 gfl y ash removal

45 gash, 9 g S) l 9 g S L-13.5 g bottom ash

6 120kJ tocoolin water31.36 gash

0

.. 130 gS

S emissions =-6- X10,800 k]lkWhr = 1.40 g IkWhr10 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

F IGURE 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 glkWhr10 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/kWhrg 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 30C, whereas several hundred meters

down, the temperature is a constant 4 or 5C. 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. Sflinceh

10percent of the coal is ash, and 70 percent of that is fly ash the totaly 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 0.14y - - 3 1.5 = 0.995 = 99.5 percent

I n C ha pt er 7 , w e w il l s ee h ow t h ' .ese emiSSIOncontrol systems work.

Consider an OTEC system operating between 30C and 5C. 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 tIS removed b y cooling water d h . e t a t 5 p er ce nt o f th e w as te e a

(corresponding to the conditi~:; .t e remaining 15 percent is lost in stack gasesgiven In Example 1.12).

When two objects are at different temperatures, heat will be transferred from the

h ot te r o bj ec t t o t he c ol de r o ne . T ha t h ea t t ra ns fe r c an b e b y conduction, by

convection, or by radiation, which can take place even in the absence of any physi-

cal medium between the objects.

33


19/25


Radiation

T,

FIGURE 1.18 Heat transfer 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 comesm 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

sinkand 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 conventionalpractice to combine all three processes into a sin I II h f cessthar 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 sideof the II ('CT - bi . wa ) or ('F)

o ::: am lent air temperature ('C) or ('F)

R - overall thermal resistance 1m2_'crw ) (h 2Or r-ft -'F/Btu}

T he o ve ra ll t he rma l re si st an ce R i s c al le d t he _ . ,hardware store, It will be designated as h Rvalue. 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 usuallyt 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 11to 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/hrq R 11(ftl-'F-hriBtu)

After adding the insulation, the new heat loss rate will be

=A(Ti - To) =1,500 ftl

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/hrEnergy 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


20/25


21/25


240070

60E7 - 50N

f 40.~30c

~20

10

00

350K

E7 -

} 1200

?c-zc

~

2000

1600

250K

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 IUltravio et Visible Tnfrared

7% 47% 46%

,"iI-

I '.

I \I\

, \

I 1\I,,'~- Extraterrestrial solar fluxI

I

~II 5,800-K

~I Blackbody~/j

I ' - --01.8 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 wellapproximated 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 actualsolar 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 (thewavelengths visible to the human eye)is 47 p Ih I Thar i 47 pet

f h I ercenr0 t e tor a a r ea . a t I S,

cent 0 t e so ar energy striking th id f ... f h e OutSI e a Our atmosphere is in the visible par-

non 0 t e spectrum The ultr . I d hinf d I h: . avio et range conrains 7 percent of the energy an t eI rare wave engt s delive th . . , hI

. r e remammg 46 percent. The total area under t eso ar spectrum curve IScalledth I

2 A ' 1 1 b h . e soar constant,and it is estimated to be 1 372 wattSper m. s WI e sown m Chapt 8 hi' I'd .. h I er , t e so ar constant plays an important ro e10etermmmg t e sur ace temperature of the Earth

The equation describingPlan k' I ( .. .especially for calculations involvinc h aw 1.61)I S somewhat tricky to maDlpulate,

arion equations howeve g h e area under a spectral curve. Two other radl'

, r, are sttalg tfor d did Thefirst, known as the Ste lan-B It Iw ar a n a re 0 ten all that is neede .I' a zmann awof diati . Y

emitted by a blackbody . h f ra tatuvn, gives the total radiant energWit sur a ce a re A d b I

a an a so ute temperature T:

E ~ O ' A T4 (1 .621

2,898Amax =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:

2,898Am" (j Lm ) ~ T(K)E ~ total blackbody emission rate(W )

0' ~ the Stefan.Bolt_ zmann constant ~ 5 67 X 10-8 2 4

T:= a bsolute tempe ra tu re ( 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.


22/25

40 Chapter 1 Mass and Energy Transfer Problems

1.5 Ifwe 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 m3/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 Fof 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 /m 2_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 isreached 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 J L ) 5,800 - 0,481-'m (sun) 25.0 m 3/s

400 ppm

1+ __ FQm3/sOppm

hi hT~istrem~ndous rate of energyemissionby the Earth is balanced by the rate at

wh

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 a r. ISw aveengrh shift plays a crucial role in theg re en o u se e e c t. A s d escr ib ed in Ch 8 b 0I

0 I apter ,car on dioxide and other greenhousegases are reanve y transparent t the i . hthey tend to absorb the out Din0 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, andraises the Earth's temperature.

PROBLEMS

50m3/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) Ifthe farmer irrigates continuously, what will be the steady-state concentration ofselenium 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

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Ch. 1 Enviromental Engineering