Fum/amenta!s

1

Engineering Thermodynamic.>MICHAEL

J. MORAN

HOWARD N. SHAPIRO

EXECUTIVE EDITOR EXECUTIVE PUBLISHER ASSOCIATE PUBLISHER EDITORIAL ASSISTANT MARKETING MANAGER SENIOR PRODUCfION EDITOR CREATIVE DIRECfOR SENIOR DESIGNER MEDIA EDITORS PRODUCflON MANAGEMENT SERVICES COVER PHOTOS

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Joseph Hayton Don Fowley Daniel Sayre Sandra Kim Phyllis Cerys Sandra Dumas Harry Nolan Maddy Lesure Steve ChaseylLauren Sapira Ingrao Associates/Suzanne Ingrao (Flame) PixtaVAgeFoloslock (Thermographic profile) Zap ArtfTh:ciJGeuy Images. Inc. (Iceberg) Rod Salmff3JI:iJGetty ImageC thcrmod) n.lml~ 1 . .. Prcparcsstu d en I s .' 1 I ddLOChap,>.8-I~arec{)m licatIOn areas. nc u t: o r t hcap P .. Sound d.c\c1opmcnIS d refriocration I..,de r"H:hromctri tsofro'~t:rJn ~ .prchcn.. l....e dcve Iopmen,. rs C'lIl ChOlISC \aflOU'> k\cb of Co\cra~c and c.ombu..tion from whIch .lIlo;,lructo -dcP' th ,>lUdic'" ntroductlons to LO rangIng rrom sh ' I or

vII

viii

Preface.... Elllphas~s ~n enginc~ring dcsign and ~naly'sis. Specific text material on the de...ign process IS mcluded m Sec. I.S: ElIgmeermg Desig" (//1(1 Allalysi\ and Sec. 7.7: Thermoeconolllic.'i. Each chapter also provides carefully crafted Dnign {I/1(J open ende{~ problellu that allow stu~ents to d.evelop an appreciation of engineering practIce and to enhance a vanety of skills such as creativity. fonnulating and solving problems. making engineering judgments. and communicating ideas. .... Employs an effective de\'elopment of the second law of thcrmod\'nami~The text features the etl/ropy balance (Chap. 6). recognized globally as the most effecti\e way for students to learn how to apply the second law. Also. the presentation of exergy analysis (Chaps. 7 and 13) has become the state-of~the-art model for learn. ing that subject matter. Uses software to enhance problem solVing for deeper learning. In e\'el') chapter. end-of-chapter problems are identified by a special icon to signal the use of computer software. All of these problems can be solved using ImeraClII't' Thermodynamics (IT). the interactiv'e software program a\'ailable with the text. Tutorials on the use of IT are given in several places throughout lhe text. beginning in Sec. 3.7. As in previous editions. the text presentation is carefull~ structured so instructors wanting to omit the use of software can do so seamlessl~. Encourages nexibilit)' in units. The text allo\\s an 51 or mi'\O:ed 51 Engli...h presentation. Careful use of units and systematic application of unit comer"ion factors t) emphasized throughout the te'\O:t. A handy )et of comersion factors is inserted inside the front cover of the book for eas~ access. Provides many text features that are pro.. en to enhance student learning. We recog nize that the first course in thermodynamics covers a lot of material in a short period of time. To help students learn and study the material more effectively. \\e haw includ ed a variety of features to assist them. For easy reference. these features are iru.ened inside the front cover under the heading "How to Use This Bool.: Effecti\l.:I~

....

....

....

New in the Sixth Edtlion.... .... .... .... .... Each chapter begins with a set of Learning Objecti\c~ Each solved example concludes with a list of the Sl.iIIs DClclollcd in ,ohing the example that reinforce specific skills related to the le:lrning ohJecti\'" Each solved example ha... a followup Quick Qui, that gi\c... the . . tudelll an lll1me diate check of undcr:-.tanding of an import,lllt ...kill. A :-.ct of Key Equation'! has been added at the dose of each chapter tor 1,'3'\ rd erence and to as:-.ist in "'ludying. Engaging nc\\ feature,> called 8io...cullncctiolh. Ene'K" So. EI1\ irollmenl, Jnd lIoriLOns ha\e heen introow,:cd throuchout 111C\ tic the ,ubJel:t mailer hl 1:~1l1 temporary apphcatlon... in biolllcdiunc: hlllCn~llH:enn!!. ener!!\ rl."~lUn:c U'l.' en\1 ronmental Cn~IIICCTln!!. and emergin!! technolol!ll.' , t.ndorChaplcr problem... in c.lch nl three IlHxle . cnnceptual. ,l..111 l'oulldlll!!, anJ dcsil!1I h as AS~latc \'ICC !)r":'ld..:nt lor t I1dl,:r~radudtl,: Pr~grams and Gencr~1 Education and Profes or of \kl'halllcal I n!!-lllccnn~haf I J "",un mQrAn 4( du 11,11I aId \ \1J4p""

"II

rnMl'U ..,

D called SI that take.. rna!>, length. and time as primary dimensions and regards force as secondary. SI is the abbreviation for Systeme International d'Un;tcs (International System of Units). which is the legally accepted system in IllOSl countries. TIle COI1\ entions of the SI are published and controlled by an international treaty organization. The 51 base un"ls for mass. length. and time are listed in Table 1.2 and discussed in thc follo\\ ing paragraphs. The SI base unit for tcmperature is the !..e1vin. K -Inc SI base unit of mass i-; the kilogram. kg. II is equal to the mass of a particular cylinder of pl steadily across the turbine blades. electricit)' is generated. The electrical output of the generator is fed to a storage baiter)'. yoh/ems: developl'nJ e11jineen'l1} sktllsExploring System Concepts1.1 l Ill,!! th.. Inkrnet. ohl;lin infornliltmn ilhtlUl tho: opo:r,llloll ,j III lpplu.:.ltilln h,t..:d or ...ho.... n 111 lahl..: 1.1. Obtalll "ulll II: t ml"rmallon to pW\lde a lull d":-"':lIptltm of the appll atum logclh\:r .... Ith rdnant th\:nmitJ~n.. ml( a,pecl rna'~ in Kg

1.17

An a,tronaUI welj!.h, 700 ""J on [arlh where", '1 I Ill' \\'hat IS the il,tron.tut\ wei~ht. In \i. on .tn l)rl'llllll( ...pd......

26

ei"'pl"rIn

Getting Started: Introductory Concepts and Definitions(a) Determine the specific volume. in m1/kg, of the CO, in the cylinder initially. Repeat for the CO 2 in the cylinder after the 15 kg has been added. (b) Plot the amount of CO 2 that has leaked from the cylinder. in kg. versus the specific volume of the CO 2 remaining in the cylinder. Consider fJ ranging up to LOm 1/kg.t.26 Go to http://www.wearher.goY for weather data al three locations of your choice. At each location expre~ the local atmospheric pressure in bar and atmospheres.

2 statIon where the acceleration of gravity is 6 O1/s 1 Expresseach weight lbf.

1.18 If the \anation of the acceleralion of gravity, in m/s 2, with 6 ele... ation;:, in m.abovc sea Ic\'c1 isg"'" 9.81 - (3.3 x 10- )y~tem consists of n.s kmol of ammonia occupying a \olume of 6 m'. Determlnc (a) the wcight of the systcm. In N. ll hcl.... eln the pi,lon ,IIlU th... C\lll1uer .... all J t mlln thl: prc.....ure of Ihe air .... lthll1 Ihe C\!;nJer. In h.lr the Pi lun I In II mIlia) J'KNllO>O Re~al .... hen lhe PI\_ Ion In II luUtI f'l"1>IIl'N\ Thl: 11>0,;,11 /Jut'Jeralhm of j!;r.I\II\ I

",h~'rc

k

_nl'1'l

1m!

Design & Open Ended Problems: Exploring Engineering Practicelh,1I th .... " 0.9"0

29

'p....ClIII.:

\olume of ,>ca\\i.ltcr I'> con"lanl,

x

III

Ill'''~

I'

In c,h.:h C can experience before "eriou\ medical compllcalloll~ result?

1.58 For liqUid-ing)a....., thermomde,", lhe thermomctrit propcrt) i",thc change in length of the thermometer liquid .... llh l..-m 1"'l:ratul"c. 110\\c\''':I". other dlcel!; are prcscnt that can affect the ternperalUic r..:adll1g of ..,uth thcrmometers. Whal arc \Orne 01the~e'!

1.10 llle i"sue ol1:lolllllll'lIrllll/tl; '" rel'Clving. con.,iderabk allentlon thc'oC day,>- Write J technical rcporl indudmg at least lhree rekrem:c" 011 the ..,ubJect of glo!:lal w'l11ll1ng. "plain \\hat is ll1t:ant h) lhe tcrm gloh,11 "'arrnlll!! and discu"s ObjCCII\c1y the loClClllllic c\ldcm.:c that ,., l'lled .." the ha"ls lor the argument th.1l l!lnhal .... :mll1n~ I.. Ol1.:U1 rlllA 1.ao Iluumeln ,l1\d hqUld.l1l-g1a.... thermomele"'" h,IH' (u.. lom","I) u'>Cd lllcTl.:UT\ ..... 111(11 is '1(IW rceo~mIed ,I.. rdc\.lnl t do.i n \l, h II lire the Il1lrtlrl..IIlt nll1tn!:lulllr-. III ("I'll th.ll

nude,11 po..... el. h~d.ullthe ki"rf;c enugy. KE. of the body. Kinetic energy is a scalar quanllt~ lbe challge In kinetic energy. .lKE. of the hody is'.lKE =

kinetic rnergy

KE~ ~

leE l =

~m(v~ - vi)

(2.5)

The integral on the right of Eq. 2.3 is the work of the force F, as the body movcs from ,$1 to .\~ along the path. Work is also a scalar quantity. With Eq. 2.4. Eq. 2.3 becomes

~m(v~ ~

Yi) =

r

F ds

(2.6)

\\here the c;:xprl,lssion for work has been written in terms of the scalar product (dot product) of the force \'ector F and the displacement vector ds. Equation 2.6 states that the work of the resultant force on the body equals the change in its kinetic energy. When the body is accelerated by the resultant force, the work done on the body can he considered a Irlltuft>r of energy to the body. where it is srorl'd as kinctic encrgy. Kinetic energy can be assigned a \alue knowing only the mass of thc body and the mag.nitude of its instantaneous velocity relative to a specified coordinate frame. \\ithout regard for how this velocity was attained. Hence. kille/ic ellergy is a property of the body. Since kinetic energy is associated with the body as a whole. it is an txterl.'jin' property.

Did you ever wonder what happens 10 the kinetic energy when you step on the brakes of your moving car? Automotive engineers have, and the result is the hybrid electric vehicle combining an electric motor with a small conventional engine. When a hybrid is braked, some of its kinetic energy is harvested and stored in batteries. The electric motor calls on the stored energy 10 help the car start up again. A specially designed tranSmission provides the proper split between the engine and the electric motor to minimize fuel use. Because stored energy assists the engine, these cars get beller fuel economy than comparably sized conventional vehicles. Tomorrow's hybrids may get even beller fuel economy if plans develop for a plug.in version whereby the storage batteries are charged from an electrical outlet when the vehicle is idle, thereby allowing drivers of plug in hybrids to get most of the energy theyne~ for transportation from the electricity grid-not by way of the gas pump. Better fuel economy not only stretches increasingly scarce and costly oil supplies but also

EneYj;l I'-'Environment

reduces the release of CO into the atmosphere, which has been linked to global worming. Each2

pl'on of gasohne burned by a vehicle produces about 9 kg (20 Ib) of CO. Annually, a conventional vehicle produces several tons of CO 2 , Fuelthrifty hybrids produce much less.

2,1.2 Potential EnergyE4lJalllln 2.tI ,.. a prinCipal result of the prc\iou!> section, ~C:=fl\ed Irom .... cw!tln "Ce lIkJ law. the cqu,ltlon gi\c!> a relationship !let"ccn t"O tlt'lml'tI concept' 1..111dtC cncrg~ lind "ork In thi 'leCtinn It is used a" a point of departure to c\.tcnd thc conccpt of

_.

34

cJ..pter2

Energy and the First Law of Thermodynamicsenergy. To begin. refer to Fig. 2.2. which shows a body of mass m that move \I~r. tlcally from an elevation =1 to an elevalJon 'urface of the earth 1',:,"'0 forces, are shown acting o~ the system:. a down~ard force due 10 gravity with magmtude mg and a vertical force with magmtude R representing the resultant of all other forces acting on the system. The work of each force acting on the body shown in Fig. 2.2 can be determined by using the definition previously given. The total work is the algebraic sum of these individual values. In accordance with Eq. 2.6. the lOla I work equals the change in kinetic energy. That is

R

...Eanh'~ ~urfacr

2m (V; - V,) ~ I ,

'f"

" Rd, -

_,

f"l,

mgdz

(2.7)

Fig. 2.2 Illustration used to introduce the potential energy concept.

A minus sign is introduced before the second term on the right because the gravitational force is directed downward and z is taken as positive upward. The first integral on the right of Eq. 2.7 represents the work done by the force R on the body as it moves vertically from ZI to l2' The second integral can be evaluated as follows:

f'

mg ell = mg(Z2 - ll)10

(2.8)

where the acceleration of gravity has been assumed incorporating Eq. 2.8 into Eq. 2.7 and rearrangingI -m(V~ -

be constant with elevalion. 8)

gravitational pOlential enugy

(2.9) VD + mg(l2 - ll) = f" R ell 2 " The quantity mgz. is the gravitational poftntial energy, PE. The change in gravita. tional potential energy, ~ PE. is(2.101

POlCntial energy is associated wilh the force of gravity and is therefore an attribute of a system consisling of Ihe body and tbe earth together. Ho\\-e\'er. e\'alu~uJlg the force of gravity as mg enables the gravilational potential energy 10 be detemll ned for a specified value of g knowing only the mass of the body and its elevation. \\~lh this view. potential energy is regarded as an extensh'e properl)' of the body. Through out this book it is assumed that elevation differences are small enough that the gf3\itational force can be considered constant. The concept of gravitational pOlenl.lal energy can be formulated 10 account for the variation of the gravitational force "Ith elevation, however. To assign a value to the kinetic energy or the potential energy of a system. it is nee; e~sa~ to assume a.datum and specify a value for the q~antilY at. the datum. values~ kmetlc and potential energy are then detennined relall\e to thiS arbltraT) chOlet datum and reference value. Howcver, since only cha"ges in kinetic and potenlial enef!.." between two states arc required, these arbitrary reference specifications cancel.

2.1.3 Units for EnergyWork has units of force times distance . The units of kinetic cner\ and JX1(~'nl}; energy arc the samc as for work. In SI, the energy umt IS the nt:\\ton-metl'r. 1\ called the joule. J. In this book it is convenient 10 usc the kilojoule, "J (\11111111111 U"-ed fngli h units for work, kinetic energ\. and potential energ\ are ,he "'1)1 J"'U force. ft . lbe. and the British Ihermal unit, Blu-

~-

,,~

2,1 Reviewing Mechanical Concepts of EnergyWhcn a 'l\'\lcm undergocs a r ,. h tinl cncrg\, ~recial care i~ re ~ ~ess \lob e~e there ~re changes in kinelic and poten. . qUlreu to 0 lalll a conSIstent set of units lcr: :~I~"I~x::.~PSl)~~;, to tlu~lra{e the proper use of units in the cllJculatiol1 of such 10 10 mis wh 1' ~I - _I em la\~ng a mass of I kg whose velocity increases from 15 m/s I C I ~ e cvallon ecrcases by 10 mal a localion where g = 9.7 Ill/5 2 Then

3S

~KE = ~11I(V~= =

Vi}

I~(I kgl [(30!".)' _ (15~)'] s s 0.34 kJ2

I

I kg . m/s 2

1N

II

10' N .

I kJ

III

I

..1PE = mg(l.2 - l.l)

= (I kg)(97~)(-lOmll1 kg1.Nm/s2 s = -O.lOkJ

II

1 kJ loJ N . m

I1B"

Fo~ a ~ystem h~\ling a mass of 1 Ib whose velocity increases from 50 fl/s to 100 fl/s \\hlle liS elevation decreases by 40 ft at a location where g = 32.0 ftJs 2, we have

.lKE = ~(llb)[(IOO!!.)' _ (50!!.)'] I2 = 0.15 Btu

s

s

32.2Ib fl/s 2

Ilbf

II_ I 778 ft Ibf

.lPE = (l,b)(32.ort)(-40ftll 32.2 Ilbffl/s 2 S2 Ib . = -0.05 Blu . .

II

778 ft . Ibf

1B"

I

2.1.4- Conservation of Energy in MechanicsEquation 2.9 states Ihat the lotal work of all forces acting on the body from the surroundings, with the exception of the gravitational force, equals the sum of the changes in the kinetic and potential energies of the body. When the resultant force causes the elevation to be increased, the body 10 be accelerated, or both, the work done by the force can be considered a transfer of energy to the body, where it is stored as gravitational potential energy :;llld/or kinetic energy. lhe notion that {'flergv i\ cumerl'cd underlie'i this interpretation. lne interpretation of Eq. 2.Y as an expression of the conservation of energy principle can be reinforced by considering the special case of a body un which the only force acting is lhat due to gravity. for then Ihe right side of Ihe equation vanl~he~ ,lnd the equal ion reduces to

0'

(2.11 )

2 ",V~

I

,

+ mg:.~

1 , mVj + 2

mg:.:,

n

l nder lhe"C condlliom Ihe 111m of the kinetic and grantational potential encrt:lc.. rt'nIlJim (Ofl\ltlllf. [quatlun 2.11 abo illustrates thai encri::t can ~ (mu aft'll Inlm one furm 10 anolher; For an ohJcct falhng under (he ll1fluencc (ll grfer

Q > 0: heat transfer ro the system Q < 0: heat transfer from the systemThis sign convention is used throughout the book. However. as was indicated for work, it is sometimes convenient to show the direction of energy transfer by an arroW on a sketch of the system. Then the heat transfer is regarded as positive in the direction of the arrow. The sign convention for heat transfer is just the reverse of the one adopted for work, where a positive value for W signifies an energy transfer from the system to the surroundings. These signs for heat and work are a legacy from engineers and scientists who were concerned mainly with steam engines and other devices thai dL'\ clop a work output from an energy input by heat transfer. For such applications.. it \\a~ convenient to regard bolh the work developed and Ihe energy input by heal lran~' fer as positive quantities. The value of a heat transfer depends on the details of a process and nut just the end !>tates. Thus. like work, hem is not II property, and its differential i... \\riltcn a~ SQ. The amount of energy transfer by heat for a process is gi\cn h~ thl' intl'~ralQ

heat i!> not a property

~

r

OQ

2.4 Energy Transfer by Heat",here the

49

at tho~c slalc". A... for \\Ork.lhe nohon of "heal'" al a state has no meaning, anllthe mte.:gral 'lhould tina be evaluated as Q2 - Qt. 111e net ral~ of h~al transfer is d~noted by Q. In principle. the amount of energy (ransfer b~ heal (junng a penod of time can be found b) inlcgrating from time II totime

11l111h

mean 'trom Siale I to state

r

and do not refer 10 the value!> of heal

rale of heat Iransju

'2

(2.29)

To perform the integration. II is necessary to know hO\~ the ralc of heal transfer varies\\ilh lime.

In Mlrne cases it is comenient to use the heat flux, which is the heal transfer rate per unit of system surface area. The net rate of heat transfer. Q. is related to the heat flux q by the integral

i,.

Q~Li/dA

(2.30)

",here A represents the area on the boundary of the system "here heal Iransfer occurs. The units for heat transfer Q and heal transfer rate Q are Ihe same as Ihose inlroduced previously for W and ~V. respectively. The units for Ihe heal nux are those of the heat transfer rale per unil area: kW/m 2 or Blulh . (t 2. The word adiabatic means I\'irhour IIeat mlnsfer. ThUs. if a system undergoes a process in . . olving no heat Iransfer with ils surroundings. Ihat process is called an (ulillhatic procen.

adiabatic

Medical researchers have found that by gradually increasing the lemperature of cancerous tis.sue to 41-4SoC the effectiveness of chemotherapy and radiation therapy is enhanced for some patients. Different approaches can be used, including raising Ihe temperature of the entire body with heating devices and, more selectively, by beaming microwaves or ultrasound onto the tumor or affected organ. Speculation aboul why a temperature increase may be benefi cial varies. Some say it helps chemotherapy penetrate certain tumors more readily by dilating blood vessels. Others think it helps radial ion therapy by increasing the amount of oxygen in tumor cells, making them more receptive to radiation. Researchers report that further study is needed before the efficacy of this approach is established and the mechanisms whereby it achieves pas iti~ results are known.

B,u...&QHJ1&cbUJ15

2.4.2 Heal Transfer ModesMethod!> based on experiment are available for evaluating energy tran.,(er h) heat. TlIc'>e methods recogni7e two basic transfer mech'll1i~.ms: COI/lIII('/IQII and f/~l'r:I/lt1 radiallOIl. In addilion. empirical relationships are a\31lable for e"alu.llmg cnt:rg)' Iran fer invol"ing ,I wmhil1e(/ mode called COl~I'('CljOIl. A bnef dc'\Cn.pl:on, ~)I each ,of lhc~ I~ gi"'cn next. A detailed consideration IS left 10 a cour:.c III cn~llleenny: heal (r > T f In this casco energy is transferred m the direction illdicated by the arrow due to the combined effects of conduction within the air and the bulk motion of the air. The rate of energy transfer from the surface to the air can be quantified by the following empirical expression:(2.34)

known as Newton's Jaw oj cooling. In Eq. 2.34, A is Ihe surface area and the pro-ponionalily factor h is called the heat transfer coefficiem. In subsequent applications of Eq. 2..34. a minus sign may be introduced on the right side 10 conform to the sign com'ention for heat transfer inlroduced in Sec. 2.4.1. The heat transfer coefficient is not a thermodynamic property. It is an empirical parameter that incorporates into the heat transfer relationship the nature of the flow pattern near the surface, the nuid properties. and the geometry. When fans or pumps cau'le the fluid to move. the value of the heat transfer coefficient is generally greater than \\-hen relatively slow buoyancy~induced motions occur. These IwO general calegories are called forced and free (or nalural) convcction. respectively. Table 2.1 provides typical "alues of the cOlweclion heat transfer coefficient for forced and free conveclion.

Newton's law oj cooling

2.4.3 Closing CommentsThe first step in a thermodynamic analysis is to defme lhe system. It is only after Ihe system boundary has been specified that possible heat interactions with the surroundings are considered. for these are always evaluated at the system boundary. In ordinary conversation. the term heal is often used when the word energy would be more correct thermodynamically. FOT example. one might hear, "Please close the door or 'hear will be lost.... In thermodynamics, heat refers only to a particular means \\hereby energy is transferred. It does nOI refer 10 whal is being Iransferred belween systems or to what is stored within systems. Energy is transferred and stored, not heat.

Tabl,

2.1

Typi(,;ll ValuM of the Convection Heat Transfer Coefficient

Applications Free nJn\ttlion

h (W/m!' K)

h (Btu/h . fl' R)

G,L.....

2-2550---1000

Furccd ~lX"tl(1Il

0.35-4.4 8.8---180 44-+1fU~-3500

e-.

L.&qllld,

25-25050--20,000

S2

cJr.,.trr2

Energy and the First law of ThermodynamicsSometimes the heat transfer o{ energy to. or from. a S)Slem can be neglt:cted llus might occur {or se"'eral reasons related to the mechanisms for heat transfer dl~u d aho\e. One might be that the materials surrounding the sy~tem are good in:perimenlal uncerl,lintieurements Ihtl! the nd work i~ ('_wull the same lor /If! ,tdi.lbatic pmce..se., bl:lween the ~a1Tle end stales. Howe)'cr. thc prepon dcrann: \11 experimcnt'tl findings supports this conclusion. so il is (uJopled as a lundamenlal principle Il1nt thl; \\ork actually is Ihe same. nlis prinCIple 1\ an ;ll1l'rn,lli".... lormulation of lhe firJI /(/11', and has been used hy subsequenl ..ci ell! i.. l.. ;lnd .... ng.ineers as a ..pri ngboard for dev(doplng Ihe rOIl \'l'fI'flliO/l (If t'Ilel"gy concept and Ihe ('/wI"gy balllll('(' as we know them loday.

ilrpr:,us helore H' bccau.,e energy transfcr by work [mm thl' ..yslem 10 lhe surroundmg... is takcn 10 be posili\'c. A plu~ sign appears before Q because 11 1\ rcg;lrdcd In he pO..ili\c "hl.'n the heat lrilllsfcr of energy is 111/0 l11c sy..l..::m from Ihc ..ur fll unJlI1\!:.'

2.5./ Important Aspects of the Energy Balance\ ""'1t

~

";_

fl':1.1,tllllrm the heat transfer. in kJ. leeringr

Ans. 20....9 UIn the ne,,( e,ample. \\e follo\\ up the discussion of Fig. 1.15 by considering t\\O ahematl'\; sYlrllems. ThIs example highlighb the need 10 account correct I\' for Ihe heat and .... ork interactions occurring on the boundary as \'ell as the energ)- change_

il

Examp e 2.3

ONSIDERING ALTERNATIVE SYSTEMS

Air is conlain,;-d in a \crtical piston- n'_lit BrWIb

,',

Ag. E'.3

58

Ch"flrr2 Energy and the First Law of Thermodynamics

Engineering Model: I. T,,-o c1o~cd s~slems are under consideration. as shown in the schematic. 1. The onl) significant heat transfer is from the resistor to the air. dunng which the air expands slowly and ih pre UTe remams constant. 3. There is no net change in kinetic energy: the change in potential energy of the air is negligible: and since the piston matcrial is a good insulator. the internal energy of the piston is not affected by the heal transfer. 4. Friction bet,,"cen the piston and cylinder wall is negligible. 5. The acceleration of gravity is constant; g = 32.0 fr/s~.Analysis: Tal-ing the ajr as thesy~tem.

the energy balance. Eq. 2.35. reduces with assumption 3

10

(~D + .weD + ~U)w ~ Q _ 1\'Or. soh-jog for Q

Q = lV +

~U..

For this system. work is done by the force of the pres.sure p acting on the bOllom of tbe piston as the air expands. With Eq. 2.17 and the assumption of constant pressure

To determine the pressure p, we use a force balance on the slowly moving, frictionless piston. The upward force exerted by the air on the bOllom of the piston equals the weight of the piston plus the downward force of the atmosphere acting on the top of the piston. In symbols

Solving for p and inserting valuesp~

-A-pi\\on

f1I P"lon

g

+ Palm2

~ (100 Ib)(32.0 fti,')1 ft

I

Ibf Ibf I Ibf II~I + 14.7:---; = 15.4-.-, 2 2 10. 10.32.2Ib ft/s 144 in.

Thus. the work is\V = p(V2-

Vd

~ (15.4 .lb~)(1.6 fll) IU4 i?21110.

I ft-

778 ft . Ibf

1 Blu

I

= 4.56 Blu

With J.U. 1f

m.,,(j,u.l,r)' the heat transfer is

Q = W + m ..,,(J.uu )=

4.56 Blu + (O.6Ib{ 18 ~~u) = J5A Btuj,

fb) Con ider nC:>..t a system consisting of the air and Ihe piston. The energy change of the o\erall ..~ stem ()f Ihe cnerg) changes of the air and the piston. Thus, the energ) balance, Eq. ~.35. reads(~II

the

'Uffi

+

~

IJ

+ J.V)., +Q~

(~

O

+ J,PE +

JJ j-(J ) . _

Q

U

v. hen.: the indICated term.. drop out h) as..umption 3. Solnng for Q

II +

(~pr>,....

PI )w

2.5 Energy Accounting: Energy Balance for Closed Systems

59

For thIs 'i' h:m. "wrk t'i LInne: at tht.' lop of the pISton as it pusht.'s a!>idc the surrounding atmSion .... ark related to fluid pressure. See Fig. 2.4.

1\'

f

pdV

Thermod~

\\ ,,ilge, 111e m~entor al-,l) mamtams thaT this approach not only give~ a coeffiCient of performance superior to those of o/r5Quru heat pumps but also a"oids the installallon costs associated with grounllSOl/reI' heal pumps. In all. significanl COSI saving~ result. the inventor sayoo.. Critically c"a]uate the inventor's claim... Write a report including at least Ihree references.

H~bod

IlltenOf of dwell'llg

Q~

~rw mains

~=:::::t

'V'crJ+

Q.'". .

I. .i

=~

Heat pump

1.

Cold body:-.J

~menl

waler ,wrage tank

Fig. P2.1OD

Evaluatt'nj Pyoferties~ ....~ tern and how the properties are related. The objectives of this chapter are to introduce property ~ relations relevant to engineering thermodynamics and provide several examples illustrating the use~\j

~

To apply the energy balance to a system of interest requires knowledge of the properties of the sys

of the closed system energy balance together with the property relations considered in this chapter.

.~

I.(j

~When you complete your study of this chapter you will be able to ...

pure substance, state princi ~ pIe for slmp.le ~ompre5sl~le systems,. p-v- T surface. saturation temperature and saturation pressure, ~ two-phase hquldvapor mixture, quality, enthalpy, and specific heats. .........

!tl,/ dem~nstrale underst.anding of key concepts... including phase and ~

~ tI' apply the closed system energy balance with property data.

~ ~

~

tl'sketch Tv, pv, and phase diagrams, and locate states on these diagrams.

tI' retrieve property data from Tables A'l through A'23. tI' apply warranted.gas model for thermodynamic ana IySIS, Including determining when use of the the ideal " . model is

80

3.1 Getting Started

81

111

In thl~ 'tC(tlon. \\~ intruJuce concepts thai support d f cluJ lilt! pha,c. pure ~uhstance and th . . our !>IU y 0 property relations. e slate pnnclple for ~imple ~ystems.

3,1.1 Phase and Pure SubstanceThe term phas~ refers ~o a quantit~ of matter that is homogeneous throu hout in bolh chemical composition and .phvslcal structure Hom .. h 1g h' . , . . ogenclty III p YSlca 'itructure means that t c maucr IS all mild. or all liquid. or all ~'apor (or equivalent I all liS). A sy~tem can COnlalll one or more phases.. y g_fOR EXAMPLE a sy:slem of liqui~ water and water '-apor (l>Iearn) contains t\\-O Phases.. A system of hqUld wate.r and Ice. including the ca'" 0 r ,IWiII. a 1 contains _ -so (it'O phases. Gases. oxygen and n".rogen for example. can be mixed in am proportion to form a ~""g/e ~as .rh:l se . Certam liquids.. such as alcohol and waler. c~n be mixed 10 form a s/1lgl~ h~uld phase. But liquids such as oil and "-ater. which are not misci/:lie. form (WQ ItqUld phasc!>. . -

phau

sltb/mlQ/lQIl.

T\\o ~hasc!> coexist during Ihe changes in phase called ~'llpon::'Ofioll, meltillg. andpur~

A purr subslanu is o~e t~at is unifonn and im.ariable in chemical compo..ition. A pure substance can eXI!>t ill more than one phase. but its chemical composition must be the same in each phase. - fOR EXAMPLE... if liquid water and water vapor fonn a system with h\O phases.. the syslem can be regarded as a pure substance because each phase has the same composilion. ':' uniform mixlUre of gases can be regarded as a pure substance pro\ldcd It rcmalllS a ga~ ~nd do~sn't react chemically. Air can be regarded as a pure substance as long as 11 13 a lllixture of gases; but if a liquid phase should form on cooling. lhe liquid would have a diffcrenl composition than the gailion due 10 chemical reaction are comidercd in Chap. 13.

subslanCt!

3.1.2 Fixing the StateThe stale of a closed system at equilibrium is its condition as described by Iht.: \31ues of its thermodynamic properties. From observation of many thermodynamic syslems. it is known that nOI all of these properties arc inuependent of one another. and the Slate can be uniquely determined by giving the valucs of the il1llepelUlcllf propertics. Values for all olher thermodynamic properties arc determined once thi~ independent subset is specified. A general rule known as the Siali! principii! ha~ l~ .' ..', t Ie ,>;tlm ated 'apl)" \t.ltl'

r'

3.3 Studying Phase Change

87

n

n...,.... "ll:r \.opor

... n ......."" J:]

\\"lo,:r

\'*1"'"

L"IuiJ \later

'0'

(bl

1 llf 0.1\1 CUean bl ,r'a n. Tahle\ A-7 throu~h .VH.......

JUDm fDbfa

3.5 Evaluating Pressure, Specific Volume, and TemperatureRdng..:rant ~:!. Tables A-IO throu!?,h A-12 for Refrigerant l>la. Tables A-13 throu h \-1." lor ammonia, and Tables f A-16 through A-I8 fo r propane are used similarly g f Ih bs a'. an: tahks or 0 cr sou lances ouod in the engineering literature. Tables are rovided 10 the Append.\: 10 $1 and English UOItS.. Tables in Eng.lish units are designat~d with a ktl a E. For e\.umpk the steam tablc~ In Ennl;sh units are T13 hi es A -2E I h rough A-6E " 5U The wbstan~es ror \\ h .Ich tabulat~d data are provided in this book have bee~ ..elected because of thelT Importance 10 current practice . srll. th ey are merely repref d f . d . I ,colao\,c 0 a WI e range 0 I~ ustnally-important substances. To meet changingreqUIrements

89

and address specIal needs' new substances a re rrcqucnt I 'mired uce d y \\hile olhers become obsolete.

Naturally-occurring refrigerants like ammonia, carbon dioxide, and certain hydrocarbons ~re used in refrig.eration s~s~ems a century ago. These refrigerants were displaced by safer-to-use chlOrine-containing refrigerants, known as chlorofluorocarbons (CFCs), paving the way for the refrigerators and air conditioners we enjoy today. Recently, beciluse of concerns over the effects of chlorine on the earth's protective ozone layer, international agreements have been ~mplemented to phase out use of Cf(s. CFCs have been replaced by refrigerants such as Refngerant 22 (CHClFJ and Refrigerant 134a (CFJCH.F), having less potential to deplete the ozone layer. But studies now suggest that natural refrigerants may be preferable beciluse of even lower overall impact on global warming. Although decades of research and development have gone into today's refrigerators and air conditioners, and the use of natural refrigerants pose significant engineering challenges, naturals are again being actively considered. Ammonia, once widely used but dropped for domestic use owing to tox.icity, is receiving renewed interest because it contains no chlorine. New energy-efficient refrigerators using propane are available on the global market, but are still of concern in the U.s. because of propane's flammability. Carbon-dioxide is considered promising for small, lightweighl systems, such as automotive and portable air conditioning units. Although CO. is a global warming gas, only a tiny amount would be required in a typical unit, and even this would be contained under proper maintenance and refrigeration unit disposal protocols, experts say.

En lhe corresponding ,aturation temperaturc.'" FO" EXAMPLE.. In Tables AA and A5. at a pressure 01 10.0 MPa. the t fallon h:mpe:raturc I~ liMed ali 311.06 C. In Table~ A--tE and A-~E. at a pre.... urc fSl'JJlhfm the.: aturation temperature is lisled a.. 467.1 F. . .

90

ch'V't'Y J Evaluating Propertiescom~stem under conliideration

91

xamp e 3

HEATING AMMONIA AT CONSTANT PRESSURE

A \alical piston-cylinder assembly containing 0.1 Ib of ammonia. initiall) a saturated vapor. is placed on a hot plate. ~~e 10 the \\eight of the piston and the s~rrounding atmO!ipheric pre~ure. the pressure of the ammonia is 2(llof/m: Hcatlllg occurs slowly. and the ammOllla expands at constant pressure until the fmal temperature IS 77 F. Sho\\ the initial and final states on T-v and p-u diagrams. and determine (_) the volume occupied by the ammonia at each end statc. in ft l .(b)

the work for the process. in Btu.

SolutionKnown: Ammonia is heated at constant pressure in a vertical pislon--cylinder assembly from the saturated vapor stale to a known final temperaturc. Find: Sho\\- the initial and final 3turation table, Ta!:lk .\-2 dnd \.1, Ii t rh)rert~ \,tlue lor thl' .Ilurill.,;d II"!. uid and \apor stat!:'!. The rr('Jrert~ \3Iu..:.. at th..:.,; latl" arc denokd f,\ th.,; !>u~nrt f and g. re"pecli\el~. Tat'!k A':~ i~ calkd the IUtrp rtltllrt tdhh. ~cau..1.: Il.:mp..:rpectlon of Table A-t .... e ge T = 120~C .l.' _ m g.andh -="I7166kJ'k A I ' loJ b) Ihe defmition of h -. g. tematfwly. " and /I are re al

T.

h=/I+ln'

3.6 Evaluating Specific Internal Energy and Enthalpy\' amlthcr illuu

v3=V1 t Px( "Water/Steam "P3'} , , VJ '" vsa determine the heat transfer I f Energy balance to m. (U2 -Ul) =Q_W

o

W=o

f = 0.8475 m'/kg and . ' _ b The program returns values 0 VI. Click the Sohe button (0 OhlaLO a solution for Ih.- 1.5 ar~ 1l1ese values agree with the values deter. 111 '= 0.59 kg. Also. at P.l = 1.5 bar. the program gives m~ - 0.4311 kg. mined in Example 3.2. . the Ex lorc button to \'aT)' pressure from 1 to 2 bar in Now that the computer program has been vcnfied. use . p. The results are: steps of 0.1 bar. Then, use the Graph button to construct the required plots.

6OO~--------,500

0.6~--------0.5

"

~300200100

""

04

1.'

,

.. 0.3

0.20.1

0.

1.I 12 U

1.4 15 1.6 1.1 1,8 1.9

0\

Pressure. bar

1,1 12 \.3 IA 15 I~ 1.1 I~ 1,9 Pres,ure. bar

Fig. E3.5

We conclude from the first of these graphs that the heal transfer to the water varies directly with the pressure. The plot of "'~ shows that the mass of saturated vapor present also increases as the pressure increases. Both of these results are in accord with expectations for the process.

o

Using the Browse button, the computer solution indicates that the pressure for which the quality becomes unity is 2.096 bar. ThUs. for pressures ranging from 1 ;;' to 2 bar. all of the states are in the two-phase liquid-vapor region.

QuiCK QuizIf heating continues at constant speci~c volume to a state where the pressure is 3 bar. modif m to gJ\;e the temperature at that state, In 0c. y the IT prograAns. V4 - VI P4 '" 3/1barV4 v]1 ("Water/Steam", P4, T4) 14282.4 C

J.1 Introducing Specific Heats c, and cp

105

;. '1 IntYot!udn!J Sfeclfic Heats Cu ant! COnein SeC. _.

I'

s.;,c(;ll propCrllC:i related to internal energ" arc impOriant in th d lllthcse is the property enthalpy introd~eed in Sec 361 "r. . crmho ynkamlcs. J 'd" . . . . ,wo at ers, nown a' ,ptYI{it' leat5, are, con~1 ered m .thls section. The specific heats are particularly u,t"tul torl'lIhermod}nallllC calculatIons involving 'he i(/ea I gas mo d eI . mtro d uce d~'.

The intensi\c 1properties c, and cp arc defined for pure . s,mplc compresslhl e su b~ . d . . tances as partll.\nl,;e, wltm thc' (,"h 111[11111 (I/lM tll"Ide oj Ihi... cxpres..ion is nghl idl'l!.i\e"('1'

c,. b}c,

definilion (Eq. 3,9). "0 u...inv. [4 J, l:'i on the (incompre.....iNe)(.lP)

. . ""r\ Thus, lor an Incompre'\iilk substance il 1lanll:.... p..te kd con 1.lnl \\llhoul . I '>(.nuu It sol acl:UnlL....

108

cJ,''ffe,-3 Evaluating Properties

. efic internal energy anrJ ~r( Using Eqs. 3.15 and 3.16, Ihe changes 111 ~pe I b enthalpy belween IWO slates are given, resp~cllvely, Y112- 11 1=

lfll;

h2

-

hi ==

1/2 -

f'-

f"'C(T)dT111

(incompressihle)

( l>lj

+ U(P2 - PI)

e(T) dT

+ v(I', -

1',)

(incompressible)

('

I~I

T,

If the specific heatIII III -

C IS

.

E ' 318 and 3.19 become, respecti\t:I' taken as constant. qs. . .Tl ) Tl)

c(T2

------PI)

(incompressible. constant c)

h! - hI = c(T2

+

U(P2

(3.20aj (3.200)

In Eq. 3.20b. the underlined term is oflen small relative to the filSt term on the right side and then may be dropped. . The next example illustrates use of the incompress~ble sUb~tance m~el In an application involving the constant-\'olumc calorimeter discussed In Sec. 2.J.1.

xample 3.6

MEASURING THE CALORIE VALUE OF COOKING OIL

One-Ie nth milliliter of cooking oil i; pi"aced in the chamber of a constant\,olume calorimeter filled with sufficient oxygen for the oil 10 be complelely burned. The chamber is immersed in a water bath. The mass of the water bath is 2.15 kg. For the purpose of this analysis.. the mctal parts of Ihe apparatus are modeled as equivalent to an addi tional 0.5 kg of water. lbe calorimeter is well-insulated. and initially the temperature throughout is 25C. The oil is ignited by a spark. When equilibrium is again attained, the temperature throughout is 2S.3C. Determine the change in internal energy of the chamber contents, in kcal per mL of cooking oil and in kcal per tablespoon of cooking oil. Known: Find: Data are provided for a constant-volume calorimeter tc~ting cooking oil for caloric value. Dctermine the change in ir1!ernal energy of the contents of the calorimeter chamber.

Schematic and Given Data:Thermometer

Engineering Model:1. lhe closed system is shown by the dashed

C ,r, ----------, , , , r-=' ,0,

-

,

, , , ,,

, ,

...Igmter

--

r1n,ulation

line in the accompanying figure. 2. The total volume remains constant including the chamber, water bath, and the a~ount i waler modeling the metal parts. 3. Water is modeled as incompressible with constant specific beat c. 4. ~eat transfer with the surroundings is negli glble. and there is no change in kinetic or potentia) energy.

--

--,,

/~

~

~-

, , , - ------------------ -- I

"""

-Coolem)

, , , ,

, , , , , , , , , ,

BounoJ3l)

Fig. E3.6

3.11 Generalized Compressibility Chart

109

Analysis:

\\ Ith 'he a !l.umptiono;, Ii\h~d th I . 'd. e C OSt:

~

. km energy balam':l" fl"ad

,rthu"

-~

,,

( ..HI)("."~",, (.i U)'""n (a) The change in inlCrnal energy of the COnt t I en S IS equa and opposite to the change in internal energy of the "alcr ~ Sinc~ water i.. modeled as incompressibl E 320 . , - e. q. . a IS used to evaluate the right side of Eq. (a). giying

08

.iL")

m .. c.. lT1"=

-

r3.32 kJ

(hi

\\lth m.

::0

~.15 kg + 0.5 kg

'=

2.65 kg. (T1'=

-

TI )

0.3 K. and c..

=

4.11'1 kJ kg. K from Table A-19, Eq. lb) gi\e...

(.it. _ _\\~

~(2.65 kg)l4.IX kJ kg K~n.3 K

Ct.1merling to keal. and expressing the result on a per milliliter of oil ba..is using the oil \olume. 0.1 mL,gt:l

(~U)._ ~V,,,I'=

3.32 kJ 1 kcal 0.1 mL 4.1868 kJ

I

I

-7.t.) kcal/mL

The calorie \-alue of the cooking oil \os the magnitude-that is.. 7.9 kealJmL. Labels on cooking oil ~ontainers usually giVt~ calorie \alue for a \eT\ing size of I tablespoon li5 mL). Usmg the calculated YaIUt~, \\C get 119 keal per tablespoon_

o The change in inlt.,:mal energ~

for \\ater can ~ found alternati\cl\ u me Eq. 3.1::! together \\ ith saturated liquid internal en~rg~ data from Table A-2.

f) The change in inlernal energy of lhc chamber contents cannot be e\aluated

using a specific heat because specific heats are defined (Sec. 3.9) only for pitre substances-that is.. substances that are unchanging in compollll dlo"o.\e I Water A\C,~~e curve b;A,;ai ('II data llll hydrocarbo.lIl'

_

02

01 oL---'o-,__,~.o-~I~,-~,~.o-~,~,-~,~o-~,~,--.,7..,,--',',-',"ii.o~""~'6;:;O)'6~'\--;7_URCUlIlCUr-rc'~lI"""1I.

Fig. 3.12

Generalized compressibility chart for various gases.

pseudoreduced specific volume

shows experimental data for 10 different gases on a chart of this type. TIle solid hnes denoting reduced isotherms represent the best curves f~lted to the data. . A generalized chari more suitable for problem soh-eng than Fig. 3.12 IS gi\cn in the Appendix as Figs. A-I, A2. and A3. In Fig. A-I. PR ranges from 0 to 1.0: in Fig. A2. PR ranges from 0 to 10.0: and in Fig. A~3. PR ranges from 10.0 to 40.0. At anyone temperature. the deviation of obsencd values from those of the generalized chari increases with pressure. Ho\\c\'er. for the 30 gases used in developing the chart. the deviation is at most on the order of 5% and for most ranges is much less. 1 From Figs. A-I and A-2 it ean be seen that the value of Z tt=nds to unity for all tempera tures as pressure tends to zero. in accord with Eq. 3.26. Figure A3 shows Ihat Z also approaches unity for all pressures at very high temperatures. Values of specific volume are included on the generalized chart through the vari able uR. called the p~'eudoreduced specific tlolume. defined by, Ii =---

UR

RTJp,

(J.291

For correlation purposes, the pseudoreduced specific volume has been found to Ix preferable ~o the r(!(/~ct!d specific volume UR = v!V the temperature range 300 to 1000 K (540 to 1800'R). _FOR EXAMPLL. to Illustrate ,the use of Eq. 3.48, lei us evaluate thc change in ~JX"C1fic enlhalp). to kJ/kg. of air modeled as an ideal gas (rom a Mate where T, '" .woK to a statc wher~ T2 = ?OO K, Inserting the cxpression for cp (7) givcn by Eq, 3A8 mto Eq. 3.43 and mtegratmg with respect to temperaturehe - hi ==

~

If

2 (a + (3T + yT + ST 1 + Er)dT2 -

==

_~ [a(T

"

TI ) +

~(T~ -

Ti) +

~(T~ -

Tn +

~(7i - n) + ; (T~ -

Ti)]

"here Ihe molecular weight .\.f has been introduced to obtain the result on a unit mass basis. With values for the constallts from Table A-21~ 289-7 3.653(900 - 400) - --1(900)' - (400)'

8.314{ . 3.294 + 3(LO),I(900)' - (400)'] 0.2763

1.337 2(10)' 1913 4(10),1(900)' - (400)'1

+ 5(10)",(900)' - (400)']

}

~ 531.69 kJ/kg . .

Specific heat functions cu(71 and cl,(T) are also available in IT Imeractive Thermodynamics in the PROPERTIES menu. These functions can be integrated usingthe integral function of the program to calculate all and tJ.h, respectively.... FOR EXAMPLE... let us repeal the immediately preceding example using IT. For air. the IT code iscp'" cp_T ("Air",T)

delh '" Integral(cp, T) Pushing SOLVE and sweeping T from 400 K to 900 K, the change in specific enthalpy is delh =; 531,7 kJ/kg. which agrees closely with the value obtained by integrating the specific heal function from Table A~2L as illustrated above. ..The source of ideal gas specific heat dala is experiment. Specific heats can be dt:termmcd macroscopically from painstaking property measurements. In the limit as pressure tends 10 zero. the properties of a gas tend to merge into those of its ideal ga!l model. 50 macroscopically determined specific heats of a gas extrapolal~d 10 \er) low pressures may be called either zeropressure specific heats or 'Ileal RO) ~recific heats. Although zero-pressure specific heats can be obtained b) utrsures of 16(X) lbflin.' and 160lbfrin. (b) for c;Jrbon dioxide at S65 K and pressure!> of 75 bar and J b;Jr. J.I0) Dt:termine lhe h;;mperature. in K. of oxygen (O~) at ~50 bar and a ~pccific \'olume of 0.003 m'tkg using generalind compressibilily dlila and compare with the "alue obtained using lhe ideal ga!> modeL J.I04 A clo~cd. rigid lank is filled with a gas modeled as an lueal gas, lnltlally al 27 C and a gage pressure of 300 kPa. If Ihe ga~ IS heated 10 77 C. delermine the final pre..sure. expressed a~ a gage pressure. in kPa.111e local atmospheric pressure IS I aIm.

ICopper ba\c. '" '" 0.25 kg

Fig. P3.90

Using Generalized Compressibility Data3.91 DClerminc the eompressibilily factor for water vapor al 200 bar and 470 C. u~ing (a) dala from lhe coml>re~sibility chart. (b) dala from the Sleam tables. J92 Determine the volume. III gen (!...q at 17 1'>Wa, 180 K.

3. 1 05 A l~nk conlains 10 lb of ;Iir at 70 F with a pressure of :10 lhflm.' Dctcrmfflc the vOlulTle of the air. in ft 1. Verih lhalIdeal ga.. h-dlil\ ior can he a,sumeu for air under 'these condllion..3106 ('omp,lre Ihe deO'>llie\(lre~ no energy.

A,r. mill.lliv at 1~ C 01 \lPoI

Fig. P3."?),118 A piston.cylinder a~sembly contains nitrogt:t1 (~-,). initially at 20 lllflin.:. KO F and occupyinj! a \(Ilume ol.! II 'Illc

Inll;31I). PI = .lO Ibf/in,~. T ,

=~'F. V, =~ It'

Finall), p~ =:!() Ibllin.:. V~ =~.5 fl'

Fig. P3.l13

nnrogen is oompressed to a final state "'here the prc'Surc: is 160 Ibl in.: and thc temperature is ~oci F. Duringromprcs"lon heat uansfer 01 magmtude 1.6 Btu occur,n in Fig P1C(l, a t;lllk htkJ 1\lth an \l.:.-tn,'JI rck ma"" holds ~ l!! ~llllllT\I~Cll" 1I1I.&lh at 17 (" 0.1 ~IP.I O'cr 3 period nl III nllnut.: dcdn",t\ I~

142

cJ,itpltr 3

Evaluating Propertiesf ned to onc side of a rigid contamer dlVldcd 3,123 Air i~ .con as ~ Ilawn h, ~'" P3.123. The other "Ide I" tn, I _ rt~. " by a part~~:~~d !'he air is inttially at Pt '" 5 bar. T; ~f'J k. twllyeva '\ Wh the partlllOn IS removed. the an and VI =: f;l~he ent~r: chamber. Measurements show that expands to I _ 14 A~wmlOg the aIT behaves a~ an V "'2V, and PZ-PI . . K 2 . , (.) the finaltcmperature. In ,and (ttl the ideal gas. deternune heat trander, kJ.

J-'wnJcJ to the reslSlor at II rale of 0 12 kW During this same period, a heat transfer of magJ\ltude 12.59 kJ occurs !romlhe nitrogen to Its surroundings. Assuming Ide~J gas behav.lOr, determme the nitrogen's finlll lemperature, me. and hnat pressure, in M Pa

~1lJ\:lilen.l\.

'",,::! 4: Tt "lrc ptsO,1 MPa

Fig. P3120

r",5 bar

Remo~3ble

partltlon

3,121 A closed, rigid tank fined with a paddle .....heel contains 0.1 kg of air. mitially at 300 K. 0.1 MPa. The paddle wheel stirs the air for 20 minutes. with the power input varying with time according to IV :: -lOt, ..... here W is in .....atts and r is time. in mmutes. The final temperature of the air is 1060 K. Assuming ideal gas behavior and no change in kinetic or potential energy. determine for the air (a) the final preMure, in MPa, (b) the work, in kJ. and (c) the heat transfer, in kJ. 3.U:a As shown m Fig. P3.122, one side of a rigid. insulated container 100tially holds 2 mJ of air at 2rC, 0.3 MPa. The air is separated by a thin membrane from an evacuated \'olume of 3 m]. Owing to the pressure of the air, the membrane stretches and evenlUally bursts, allowing the air to occupy the full \'olume. Assuming the ideal gas model for the air, determine (a) the mass of the air, in kg. (b) the final temperature of the air, in K, and (c) the final pressure of the air. in MPa

'" Pt

\:1 ",0.2m'

Vacuum

T.",500K

IV2 == 2V t

I

1'1'"

~Pl

Fig. P3.'233.124 Two kilograms of air, initially at 5 bar. 350 K and 4 kg of carbon monoxide (CO) initially at 2 bar,450 K are con fined to opposite sides of a rigid, well-insulated container b) a partition, as shown in Fig. P3.!24. The partition is free to move and allows conduction from one gas to the other without energy storage in the partition itself. The air and CO each behave as ideal gases with constant specific heat ratio. k '" 1.395. Determine at equilibrium (a) the temperature, in K, (b) the pressure, in bar, and (c) the volume occupied by each gas, in m'.Insulation

Air 31

17'C. 0.3 MPa

Th,n~membrane

I

lntllally evacuated volume.

Vz2m\

V::3m]

,

r

Movable panltion

Air

"'::2kgp=5bar To:: 350 K

Fig. P3.'243.~2~oi;0Iutninsulated. rigid tanks contain air 100tiall), tanL900' R ~ b of atr at !.t4(), R. and lank B has ::! Ib of air ,t e milia! preSSure 10 each tank is 50 lbf in - A \al"in the hne connectmg e two tanks IS openeJ anJ the 1:...... I eots arc allo'Aeu 10 E ta-k' C .. m.'(. \cntualh. the contcn'" ~I( th~ "" orne 10 cqulilb . . rOunulOg 'i20 RAnum at thc tcmpcraturl.' 01" tho: 11' . . s,umlOg the tJeal ga nlllJd. J a constanl current of 10 amp!; at a voltage 01 12 V for 10 mlllutcs. Measurements indiute that \\h5(:mbly contains carbon monolCide modeled as an ideal gas wilh constant specific heat ratio. k"'" 1.4. The carbon monoxide undergoes a polytropic expansion wrth n = k from an initial stale_ where T I = 200"F and p, = 4OIbfin,'. to a final state......here thc \oolume is twice the initial \olume. Determine (a) the fmal temperature, III QF. and final pressure. in IbfJin.~, and (b) the .... or); and heal transfer. each In Btu/lb 3.133 Two-tenths kmol of nitrogen (N z) in a piston--cylinder assembly undergoes two processes in series as follo.....s: Process I:Z: Constant pressure al 5 bar from VI = 1.33 m' toV 2 ", I ml .

Process 23: Constant volume to p, = 4 bar. Assuming ideal gas behavior and neglecting kinetic and potential energy effeCls, determine the .... ork and heat transfer for each process, in kJ. 3.134 One k.ilogram of air in a piston-cylinder assembly undergoes t,>,o processes in series from an initial srate where PI = 0.5 MPa. T, = 227'C: Process 12: Constant-temperature expansion until the volume is twice the iniiial volume. Process 2-3: Constant-volume heating until the pressure is again 0.5 MPa. Sketch the two processes in series on a p-u diagram. Assuming ideal gas behavior. determine (a) the pressure at state 2. in MPa. (b) the temperature al state 3, in C. and for each of the processes (c) the work. and heat transfer. each in kJ. ]-1]5 A piston-cylinder assembly conlains air modeled as an ideal gas WIth a constant specific heat ratio. k = lot The air undergoes a power cyele consisting of four processes in series: Process 1.2: Constant-temperature expan,ion at 600 K from PI = 0.5 MPa 10 p, = 0,4 ~'IPa. Process:z.]: Polylropic e:,,_,

1

Design & Open Ended Problemsrr':"Un-nWnll'lrm!! '''l\:m, Illdudmg Ihe merils of balll'r) ~,'l.:nl '" nlc ,I lcrnrt meludmg at least three refcrences. ).100 Cart'l.lndlll\ldc capture and ..torage I~ under con~ider ,llInn t,lda, !In k"enlll~ Ihe dlect, on global wannmg and (hm3tc chan!!c traccallic hi burmng. fmsil fuels.. One In::quenth-menU()llCd approach. sho"n in Fig_ 3.10D. is to r.:molc CO from the nhaust gas of a po"cr plant and 101'-. C'onsida thl.: leadmg ml.:lh at the inlets.

Quitk QuizEvaluate the volumetric flow rate. in m 3/s. at each inlet. Ans. (AV)1 = 12 ml/s_ (AVh = 0.01 m 3/s

4.3.2 Time-Dependent (transient) ApplicationMan) de\'ices undergo periods of operation durin!!. \\hich the state chanee, \\ltb hmc--:-for exa~ple. the startup and shutdown of mot~rs.. Additional c\.ampk~ induJ( contamers bem~ filled or emptied. The steady-state model is nOI appropriale \\h.:n anal)Llng such time-dependent (transient) cases. Example 4.2 illustrates a tim"edependent or Iran'l" t: apr \ " . 'nl I -> Icatlon 0 r t he rna,:. fa.: "alance. In thiS case, a barrel IS filled with Water.

4.3 Applications of the Mass Rate Balance

153

xample

TllllNG.A BARREL WITH WATER

\\alcT 111)\\' illlo th; the rate of now out of the barrel. From tht.: graph, the limiting value of L is 3.33 ft. which also can be verified by taking the limit of the analytical solution as t --+ 00.

o

Alterna.. vely. this differential equation can be solved using Imeracli\'t' Tllermmll'lwmiCf. IT. The differential equation can be expressed a~der(l, I) + (9 L)/(rho A) - 30/(,ho A)

rho" 62.4 Illb/h} A-Jllfe "'here der(l.t) i~ dl-/dl. rho IS den"ity p. and A is area. U~lI1g the Explort' hutton, . and SOlving for

We- ""

O~~5[ (h

h) + (Vf; Vi) + g(Zl ~ Z2)] l

(a)

Since the lemperature of the waler does nOI change . 'f I Jrop~ out. Evaluating the kinetic energy term sigm leanl y and 11 "" hf(T) (assumption 4). the enthalpy term

vi-vl=2

(122)'-(1956)'(~)'12

IN

1 kg. m/s 2

II

IkJ

to' N . m

I=:

-0.191 kJ/kg

The potential energy term isg(z, - z,) = (9.81

mJ"XO -

lO)ml Ikgm/s2 IN

II

I kJ to) 'm

I

= -0.098 kJ;kg

In:.erling \-'alues into Eq. (a)We- ""

.

(13.8 kg/,) 0.95 :0 - 0.191

O.098/kJ)\1 kW ;\kg 1 kJ/s

"" -4.2 kW .... hcre the minus sign indicates that power is provided to the pump.

o Alternatively. VS

I can be evaluated from the volumetric now rate at 1. This is left as an exercise. Since power is required 10 operale the pump, W is negative in accord with our cv sign convenlion. The energy transfer by heal is from Ihe conlrol volume 10 the surroun~ings., and I~US QCY is negative as well. Using the value of ~'vC" found below, Q" = (0.05)WN ~ -0.21 kW.

Quitk Quiz

If Ihe nozzle were removed and water exiled directly from the hose. whose diameter i!> 5 em. determine the ,docil~ at the exit. in mJs., and the power required. in kW. keeping all other d&'_a_l'n~'_h~a;n~g~'~d_. MS. 7.1)4 m/... 1.77 kW.

-

4.1 Heat Exchal1JeysHtlll txChangt!TS have innumerable domeslic and industrial applicalions. including1J5c: in home heating and cooling systems, automotive systems. electr.ical. pow.er gen-

htal txchangtr

eratum, and chemical processing. Indeed. nearly every area of application listed 10 lihle I.l Cp. 3) involves heat exchangers.

172

dt4plu4 Control Volume Analysis Using Energy

~-l~~~~-~;~'~~:-i:~7.-"--'t::i"ii-~- ;;:1_ _~ - -

-

,.,,,_.,,~--------

'"

-

"~I

'JI

fig. 4.1) (ommon heal exchanger types. (0) Direct contact ~e~t exchanger. (b) Tubewithinatube counterflow heat exchanger. (d Tube,wlthln'a-tube parallel flow heat exchanger. (d) (rossflow heal exchanger. One common Iype of heat exchanger h a mixing chamber in which hot and cold streams are mixed directly as shown in Fig. 4.130. Thc open feedwaler healer. \\hKh is a componenl of the vapor power systcm.. considered in Chap. 8. is an example of Ihis type of device. Another common type of exchanger is one in which a gas or liquid is sepomretl from anolher gas or liquid by a wall through \\ hich energy is conducted. These heat exchangers, known as recuperators. takc many diffcTI::nl forms. Counterflow and parallel tuhcwithin-a-Iubc configurations are shown in Figs. 4.l3b and 4.13(", respectively. Other con figurations include crosstlow. as in automobile r"dialOrs. and multiple-pass shell-and-tuhc condensers and evaporators. Figure 4.13d iIlu!>trates a cross-flow heat exchanger.

8/0... coY/ned/ons

Inflatable blankets such as shown in Fig. 4.14 are used to prevent subnormal body tern peratures (hypothermia) during surgery. Typically, a heater and blower direct a stream of warm air into the blanket. Air exits the blanket through perforations in its surface. Such thermal blankets have been used safely and without incident in millions of surgical procedures. Stilt, there are obvious risks to patients if temperature controls fail and overheating occurs. Pro' fessional groups have called for reducing such hazards through improvements in the design of thermal blankets and better training of operating room personnel. This biomedical application is a good example of how engineers must bring into the design process their knowledge of heat exchanger principles, temperature sensing and contro\, and safety and reliability requirements.

4.1.1 Heat Exchanger Modeling ConsiderationsAs sh,o"" n b\ Fig. 4.13. heal c.. chanoers can in\oh.' n,ull'pl e III 1 an d e"ts. C a ron . l' . .. Cb rOr Irol \olume encloslllg a heat c"chan~er the onh \\,),' n' " ~. -. . _ ...... 0\\ \\or... . the untlerlincd term f [ 'I' . . ... ~l q .. ~ (repclIed tlt'lo"") In'l' OUI. Iea\ IIlg the enthalp~ anti heal tr3n..ler lerm' h .... " o""n h\ E4. (.1). nut 1

4.1 Heat hchangersil II

173

Q,~ +

L ni,h, L /"',f'r ,

(ojHeJIN.Ill",,~.UIl11

Although high rate... of entlrgy transfer lI'il11i" the heat e)[chan~cr l/CCur. heat transfer \\ ltll.the surrounding!; is often small enough~ to re neglected. Thu'>. the Q" term of Eq. (a) would drop out. leaving jU~1 the enthalpy term~ The final form of the energy ralc balance must be sohed together wllh an appropriate expression of the ma"" rate balance. recognizmg both the number and I~ pc of inlet... and C'l:iIS for the case at hand.

4. 'f.2 Applications to a Power Plant Condenser and Computer CoolingThe next example illustrates ho\\ the mass and energy rate balances are applied to a condenser at steady stale. Condenser, are commonly found in po"'er plants and refrigeration systems..Fig. -,.14 Inflatable thermal blanket.

xamp e 4.7

-!'OWER. PLANT CONDENSER

Steam enters the condenser of a vapor power plant at 0.1 bar with a quality of 0.95 and condensate exits at 0.1 bar and 45 C. Cooling waler enters the condenser in a separate stream as a liquid at 20 C and exits as a liquid al 35 C y,;th no change in pressure. Heat transfer from the outside of the condenser and changes in the kinetic and potential energies of the flowing streams can be ignored. For steady-stale operation. determine (a) the ratio of the mass flow rale of the cooling water 10 (he mass flow rate of the condensing steam. (b) the ralC of energy transfer from the condensing steam to the cooling water. in kJ per kg of steam passingthrough the condenser.

SolutionKnown: Steam is condensed at steady state by interacting with a sepamte liquid water stream.

Find: Determine the ratio of the mass flow rate of the cooling water to the mass now rate of the steam and therate of energy transfer from the steam to the cooling waler.

Schematic and Given Data:Cooden,ale Olbolr 21-----------1 ~5'C 1 1T

Engineering Model:I. Each of the IwO control volumes sho\\1\ on the accompanying sketch is at stcadystate. 2. There is no significant heat transfer bdwecn the u\erall condenser and ib surrounding~ \"\1 of compo ! . . nents se ected fro I . those com mo.n) sec." In practice. These co m \..' mponents are usually encountcrc.d In coml'lnatlon. rather than individuall E . \.. y. ngmeers often I musl creatiVe} comllme components to achieve some 0" lib' . . . ~era 0 JCCll,'e. subject to constralnl" such as minimum total t Th' , . ... _ cos. IS Important w, enginecnng aC~lVlI) IS called System Integration. In engmeenng praC!lCC and everyday life . integ'ated systems arc 'I ,L re,,,,2

'I

1','" I all11:.l00'F

,

r-

1(1' II'/min

----

_

("mIltJ,li,'" p"-,,Iud' "ut I

.Powerco,

T,,,,2hOF' I':'" L,tm I,l

, ,_

,------,

,

WiI\l'r 10

----------->

, , ,

1. The control volume shown on the accompanying figure is at steady state. 2. Heat transfer is negligible, and changes in kinetic and potential energy can he ignored. 3. There is no pressure drop for water flowing through the steam generator. 4. The combustion products are modeled as air as an ideal gas.

wa(l'r om

1'.1",.l0Ibt/in: T, '" ]02'F/;1)'"

2751blmin

Ps'" I Ibflin 1 .I~ '" 93%

Fig. 4.10Analysis:(a) The power developed by the turbine is determined from a control volume enclosing both the steam generator and the turhine. Since the gas and water streams do not mix. mass rate balances for each of Ihe streams reduce. respectively. to give

For this control volume. the appropriate form of the steady-state energy rate balance is Eq. 4.18, which reads

0= Qcv -

w" + 1il{h l +- liI2(11 2 +

1-

+ gZI) + lil1 3 +

(h

~ + gZ3)

+ gZ2) - lil{h S +

'if

+

g:.s)

The underlined terms drop out by assumption 2. With these simplifications. together with the above mass now rate relations. the energy rate balance becomes

W"

= lill{h l

h2) + lil,(11 3 - 11 5)

The mass flow Tfltc 'ill can be evaluated with given data at inlet 1 and the ideal gas equation of state

';'1 =---~

(A V),

u,

(A V)"" (R/M)T,

2 l (2 X 1~5 ft /min)(14.7Ibf/in. ) 1 144 i~.21 1>45 ft lbl )(8600 R) ( 28.97 lb OR= 9230.6 lb/min

I ft

ll1 e specific enthalpies hi and h~ can be found from Table A22E: At 860 R. It l ~ 2~6.46 Btullb. and at 720' R. II I72J9 Btu/lb. At state 3. water is a liquid. Using Eq. 3.14 and saturated hqUld data from Tabk A-lE. h ht(T,) 70 Btu/lb. State 5 is a two-phase liquid-vapor mixture. With data from Table A-3E and the gi\Cll

o

l/ualityhs

hf~

~ 69.74

+ xs{llg5 - 11 15) + IJ.9J(IIJ36.0)

~ 1033.2 Blullb

182

cJ,.,t"+

(ontrol Volume Analysis Using Energy

su"~titutmg \alues inlo the expression for \\1 //(/). Thus me-(l) == Vc.(l)/v(l)(4.251

UC\(t) == 11lC\(t)lI(t) . . d f different phases. the state of each phase When the control volume tS compnse 0 would be assumed uniform throughoUI.

4.123 Transient Analysis Applications Th e f 0 II owmg examp Ies provlde illustralions of the transient analysis of control volh. . urnes usmg t he consen'a ,.Ion of mass and energy principles. In eac case considered. we begin with the general forms of the mass and ~ner~y ~alana:s and re~ucc. them to forms suited for the case at hand, invoking the Ideahzatlons discussed m thiS section when warranted. The first example considers a vessel that is partially emptied as mass exits through a valve.

WITHDRAWING STEAM FROM A TANK AT CONSTANT PRESSUREA tank having a volume of 0.85 m~itially contains water as a two-phase liquid-vapor mixture at 260C and a quality of 0.7. Saturated water vapor at 260"C is slowly withdrawn through a pressure-regulating valve at the top of the tank as energy is transferred by heat to maintain the pressure conSlant in the tank. This continues until the tank is filled with saturated vapor at 260C. Determine the amount of heat transfer, in kJ. Neglect all kinetic and potential energy effects.

SolutionKnown: A t.'lnk in!tiatly holding a two-phase liquid-vapor mixture is heated while saturated water vapor is slowly removed. ThIs contmues at constant pressure until the tank is filled only with saturated vapor. Find: Determine the amount of heat transfer. Schematic and Given Data:Pn:~~ure

,TJ

y -........... regulating valve ~I~

,

J

::

ff:j:(::lo -, , , ,'""3ler_ remo\ed. I "hllelhebnl. : l~ heated: \'apor

I Salurutell

-t----+--'y-'----- 26O"C

t

:

, , , ,I

l"'>-",,-o!'!'" '

"

lmllal. t"lrph;!'oC It'luld-\apol'" ml\llIre

Fig. 4.11

4.12 Transient Analysis

185

Engineering Model:

I. l11c control \{)IUlllC i... udined b~ thl' dashed line on the accompanying diagram. Z. hlr the control \olume. H .... - (l and kinetic and pot~ntial en~rg.~ effects can be neglected.\, \1

the

C\.II

the

~tate

rt:=mams constant.

o ~ Thc initial and fmal sta{e~ of the mass \\ lthin the H~ssel are equilibrium slates..,Analysis: Since Iher~ is a single e,it and no inlet. the mass rate halance Eq. 4.2 takes the formdm", dt

-m,

"ith as~umption 2., the energy rale halance Eq. U5 n:duces toltV...d~-- =

. . Q" - m,h~

Contbllling the mass and energy rate balances results in

B~ assumption 3. the specific enthalp) at the exit is constant. Accordingly. integration of the last equationg.1\es

Sohing for the heat transfer Q",

or"here lilt and III! denote. respectively. the initial and final amounts of mass within the tank. The terms III and nil of the foregoing equation can be e\aluated with property \alues from T5T.

fOR EMERGENCY POWER GENERATION

Steam at a pressure of 15 bar ,md a tcn~turc of 320C is contained in a large vessel. Connected 10 the vessel 3 through a valve is a turbine followed by a small initially evacuated tank with a volume of 0.6 m . When emergency po\\er is required. the valve is opened and the tank fills with steam until the pressure is 15 bar. The temperature in the lank is then 400"C The filling process takes place adiabaticaUy and kinetic and potential energy effects are negligible. Determine the amounl of work developed by the turbine. in kJ.

SolutionKnown: Steam contained in a large vessel at a known state flows (rom the vessel through a turbine into a smaU tank of known volume until a specified final condition is attained in the tank.Find:

Determine the work developed by the turbine.

Schematic and Given Data:

Steam at i15 b:ll'.

P--j::(::l the Tallo ",. . OF of the 31T \\1 I III h!!lble. Plot the pressure. in atm. and the temperature. In ' . the tank at time t > O. Also. plot the comprc\',()r \\ herc ml IS the imtlal mass in the tank and m IS the mass 10 work input. in Btu, versus m/ml' Let mlml vary from 1 10 3.

xample 4.13

. . I e The initial state of the all' in the tank and An air comprc!>SOr rapidly fills a tank havlOg a known vo urn . the statc of the entering air are known. . . k d plot the air compressor work input eadl Find: Plot the prcssure and temperature of the air wlthm the tan . an . ,"'ersus mlmJ ranging (rom 1 to 3.Known: Schematic and Given Data:

~lution

Engineering Model:A"---:T,=70F

IAircom~

V=lOfl J

, , , , , , , , , ,

:p,'" I afm

1. The control volume is defined by tbe dashed line on the accompanying diagram. 2. Because the tank is filled rapidly. Ot> is ignored.

3. Kinetic and potential energy effects are negligible.

Tj

'"

701= c:onsltJnr - - - - - - - - - - -

PI"" I aIm- - - - - - - - - - pVI4

Fig. 4.130

4. The state of the air entering the control volume remains constant. 5. The air stored within the air compressor and mter connecting pipes can be ignored. 06. The relationship between pressure and specific volume for the air in the tank is pul.4 constant. 7. The ideal gas model applies for the air.=:0

Analysis: The required plots are developed using lnreracth'e Thermodynamics: IT. The IT program is based on the following analysis. The pressure p in the tank at time t > 0 is determined frompUlA

=

Plul 4

where the corresponding specific volume v is obtained using the known tank volume V and the mas~ m in the tam. at that time. ThaI is, u = VIm. Thc specific volume of the air in the tank initially. VI. is calculated from the illtal gas equation of state and the known initial temperature, TJ, and pressurc. Pl' That is

u,

~

RT,PI

~ ( 28,971bOR (530"R)I~1 =(14.7Ibf/m.2)144 in. 2

1545ft'lbl)

13.35ftllb

Once the pressure p is known, the corresponding temperature T can be found from the ideal gas equation of ~tate. T = pvlR. To determine the work. begin with the mass rale balance Eq. 4.2. which reduces for the single-inlel control volume to--=:onl

elmt>.dl

.

'

Then, with assumptions 2 and 3, the energy rate balance Eg. 4.15 reduces to- - = -W dl

dUC' + tillt ,

4.12 Transient Analysis ',nin" the ma0

I SpecIfy the mass and mass ratio rV1 ,. Vimt f . mimi

r ,

190

cJr-ptn-4 (ontrol Volume Analysis Using Energy

Calculatt' tht' work using Eq. (a) Win" m' u-ml' ul-hi' Cm-ml) Ul .. u T("Air Tl) u .. u TC"Aif 1)R , R ,

hi .. h""-r(-Aif-, Ti) , _ 2 dercd aoove to \alidate the program. I - mImI consl Gsinn the Sohe button. oblam a solution [or the samp e case , ' . I'dated use the Explore button to "al) the d Good agreement is oblained. as can be ven fi e. 0 nee the pro through a 2in.-diameter pipe with a \elocil~ of../ hIS 3t a panicular location. Determine the mas:,> no\Oo ratc. in Ib s.. if the substance i:,> (a) water at SO lbbn.~. 80 F. (b) nilTogen as an Ideal gas at 50 Ibf/in.:. S'TF. (c) Refrigerant 22 at SO Ibflin. 2, soeF. 4-15 Air enten; a compressor operating at steady state wilh a pressure of 1../.7Ibf/in'. a temperature of 70'F, and a lolumetric no\Oo rate of 5OOft1tmin. The air \'elocitl in the exit pipe is 700 hiS and the exit pressure is 120 IbfJin. 1 If each unll mass of air passing from inlet to exit undergoes a process described by pv l ~ = CQ/Istalll. detcOlline the exit temperature. in F. and the diameter of the cxit pipe, in inches.

~ , ,,,

Coohng IU\Ooer Warm ",aler mlet na l '" 4000 lblb

TI ", 120"F

T1 '" goOF m2",m. \1.u.eup ..~ter

+

Fig. P4.19.,20 A pipe carrying an incompres.sible liquid contains all expansion chamber as illustrated in Fig. P4.20.

1t,16 Ammonia enters a control volume operating at steady state at PI = 14 har. T I = 28~c. with a mass now rate of 0.5 kg/s. Saturated vapor at " bar leales through one exit.

wilh a \olumetric flow rate of 1.036 m"min, and salurated liquid ill 4 bar lea\'l:S through a second exit. Determine (a) the minimum diameter of the inlet pipe. in cm. so the ammonia vclocit~ docs not exceed 20 mls. (b) the volumetric now rate of lhe second exit stream, in ml/min.4,17 Liquid v.uter ill 70"F enters a pump though an inlet pipe

(a) DC\'elop an expression for the time rate of change ofliq. uid le\'el in the chamber, dUdl, in terms of the diame ters D I. D1, and D. and the velocities VI and V2. (b) Compare the relative magnitudes of Ihe mass flow rales Inl and 1;12 when dLldr > 0, dLldr = 0, and dLldr < 0, respectively.

having a diameter of 6 in. The pump operates at steady state and ~upplies water to two exit pipes having diameters of 3 in. and 4 in" respectively. The velocity of the water exiling thc 3-in. pipe is 1.31 ftls. At the exit of the 4-in. pipe the velocity is 0.74 ft/s. 'llIe temperature of the water in each exit pipe is 72' F. Determine (a) the mass flow rate, in IbIs, in the in lei pipe and each of the exit pipes. and (b) the volumetric now rate at the inlet, in ft1tmin. 4,18 At steady state. a Slream of liquid water at 20"e, I bar is mixed \11th a Stream of ethylene glycol (M = 62.07) to form a refrigerant mixture that is 50% glycol by mass. The \later mola.r f1o\Oo rale is 4.2 kmol/min. The density of ethylene glycolls 1.115 tImes lhat of water. Determine fa) the molar flo\Oo rate, in kmollmin. and volumetric flow r3le, III m' mill. of the entering elhylene gl~col. fb) the diameters.. III em, of each of the supply pipes if the \dOCII~ III each I:'> 2.5 mt's.1t19 Fill-urc PU9 ~ho\l!lo a cooling tower operating al steady

D

Expansion chamber

1~':=111L--'>J-,tr-~

-l

D,

D,

Fig. P4.204,21 Velocity distributions for laminar and lIIrbu/~nt flo~ tn I R . . C < Circular pipe of rad . IUS cart)tng an tncompresslbl~ Iiql.lJU of densil~ pare gi\en, respectively, b)

tah::, \\drm water from an air condItIOning umt enters at 1;2(, f .... Ith a md !Io flow rate of 4(XX) Iblh. Dry air enters the t~c:r dt 7U r I atm .... ilh a volumetric flo\\ rate of

VIVo = [I - (r/Rfl VIVo

= II

- (riR)]'"

Problems: Developing Engineering Sk Us.tn:~ tho: ... ",[lledln" ve "'-'Ity ror e;)l;h \elOC1I~ I,hstnhullun "

197

", the radl I dlqao.:e rom the pipe u:nto.:rhn" and\eT"oU , R

r','1 \ \(t-)

J"nH:~ e:>;rrt:...... lnn~ for thcm3~" flo\l r according 10 ~ c=, '. \lhere: 1 9ixl kg of Yoaler.

Fig. P4.25_.26 Carbon dioxide gas IS h constant at g = 9.81 lIlI~. If the pD",er output at steady stale IS 500 kW. "hat is the mass (10'" rate of \\aler. in Kg/57..50 Steam enters the first-stage turbine shown in Fig. P-t50 ~I bar and 500 C "llh a volumetric flow rate of 9Om' min. Steam exits the turbine at 20 bar and 400"c. The steam is then reheated at constant pressure to 500"C before entering the second-stage turbine. Steam leaves tbe second stal!:e as saturated vapor at 0.6 bar. For operation at steady siale. and ignoring stray heat transfer and kinetic and pC of the coolin\! \later in K. if the cooling water ma~ flo\\- rate IS X2 kg/mm (b) Plot the temperature increase of the cool 109 water. In K "ersus the cooling \\-ater ma,,~ flo\\ rate ranglOg from 75 10 90 kglmin. 6 Air enlers a compressor operating at steady stale at 4'1~\ Ibf/in.2 and 6O'F and is compressed to a preSSure of 150 Ibf/in. 2. As the air passes through the compreSSor. It i~ cooled at a rate of 10 Btu per Ib of air flowmg by \\oater cir. culated through the compressor casing. The \'olumetric flov. rate of the air at lhe inlet is 5000 ftl/miO. and the po.... er input to the compressor is 700 hp. The air bcha\'cs as an ideal gas. there is no stray heat transfer. and kinellC and polential effects are negligible. Determine (a) the mass nOli rate of the air. IbIs, and (b) thc telllperature of the air allhe compressor exit, in OF.4.65 A pump steadily draws water through a pipe from a resef\ioir at a volumetric flo", rate of 20 gal min. At the ptpe inlet. the pressure is 14.7Ibf./in. 2. the temperature is lH F. and the velocity is 10 fus.. At the pump exit. the pre!>Sure is 35 Ibf'in.!, the temperature is 64 F. and the velocity is 40 ft!l. The pump exit is located 40 ft above the pipe inlet. Ignor. ing heat transfer. dctcrmine the power required by the pump. in Btuls and horsepower. The local acceleralion of gravity is 32.0 ftJs 2.f 4011 "me {X1\\Cr Input to th tcm~rature 0 t> heat from th J'i'i kW :.nerg) tran,ft;r) .

4.56 (;lrhon dIll Ide- ga., l oompre-,~d al stcady state from a pro;: 'tire of 101M In.: and a temperaturt:: of 32 f to a pressure ()j '\I) 1M in. and a temperature of 1~O F. The ga~ enters the C\lmprt::. The mass flow r:lte IS 0.98 Ibis. The m:lgniludc of the he.ll tran~fcr rate from the compressor to its surroundings is 5% of the compre~sor I>o"cr input. U~ing the ideal gas model "ith cr = 0.21 Btu/lb R and neglecting potential energy dlt..>sor to the surroundings at a rate of 5(X)Q Btuth. Kinetic and potential energy effeclS are negligible. Determine the lOkt "olumetric now rate. in fr'/min, first using data from Tahle A-15E. and then assuming ideal gas behavior for the ammonIa. DISCUSs. 4.6] Air enters a water-jacketed air COmpressor operating at stcady state wllh a volumetnc flow rate of 37 m\/min at IV) kPa. 305 K and exits With a pressure of 6RO kPa and a

4.66 A ",ater pump operating at steady state has 3-in.--diameter inlet and exit pipes. each at the same elevation. The "'".lter can be modeled as incompressible and its temperalure remains constant at 70 F. For a power input of 2 horse power. plOI the pressure rise from inlet to exit. in Ibf/in.:. versus the volumetric flow rate ranging from 4 to:) galls. 4. 67 A pump steadily delivers waler through a hose termi nated by a nozzle. The exit of the nozzle has a diameter of 2.5 cm and is located 4 m above the pump inlet pipe..... hich has a diameter of 5.0 em. The pressure is equal to I bar at both the mlet and the exit. and the temperature is constant at 20T: The magnitude of the power input required b} tilt pump IS kW, and the acceleration of gravit} is g = 9.81 m/s'. Determine the mass flow rale delivered b} lhe pump. in kg/so

8.?

4,68 As shown in Fig. N.68. a power washer used 10 dean ~~~ SIding of a house has water entering through a hO"C at C. I atm and a \'eloctt} of 0.2 mJs. A jel of water eXit' with' . a veloclI~ of 20 m.. 's at an a\erage ele ..:atl0n of:'l m Volt , . . } no slgmficant cha nge III temperature or pressure, At ,teJJ \ . . Slate, the magntt ud e 0 r t h e heat transfer rate Jrolll t, ~ flow rates. NeglecllOg klOelic and ~tlal energy cffcclIream of air entef'> al 22 C ",ith a rna..., flo\\ rale of 188 ke mm and e,it, .l\ 17 C A".. uming tht' Idcal !!3 model f~r aIr and ignoring klOetic and potential ent:r~\ effect'\, delernune (a) the ma.... flo .... rate 01 Ihe Reln~~r ant D-I.I.IO ke min. and (1'0) tht' heat Ifan..ler bet\\e~'n th,; heat c\l:han~~r _l11d II\. \urroundm!'! 10 lJ film

202

cJu.pfU'4 Control Volume Analysis Using Energy

s.nn1~V'"WI\'"

o IX"" Left1 10'

..............1'1",0 1bIr

,

Rtturn

lnulo.e

M~?~_~

L:"~-='~'~-=:"

":""""'"'"

T_15~

Fig. P4.78

4_78 As sketched in Fig. P4.78. a condenser using river water 10 condense steam with a milSS flow rate of 2 x 10' kglh from salurated vapor 10 saturated liquid al a pressure of 0.1 bar is proposed for an industrial plant. Measurements indicate thaI several hundred meters upstream of lhe plant. Ihe rher has a \olumetric now rale of 2 x lOS m1/h and a lemperalure of 15~C. For operalion at steady state and ignoring changes in kinetic and polential energy. delermine the river-water temperature rise. in C. downstream of the planl traceable to use of such a condenser. and comment. "79 Figure P4.79 shows a solar collector panel embedded in a roof. The panel. which has a surface area of 24 ftl. recel\es

energy from the sun al a rate of 200 Bluth per fll of col. lector surface. Twenty-five percent of the incoming energj' is lost to the surroundings. The remaining energy is used to heal domestic hot water from 90 to 120F. The waler pas.sts through Ihe solar collector with a negligible pressure drop. Neglecting kinetic and potential effects. determine at steady state how many gallons of water at 120"F the col

lector generates per hour.4.80 A feedwalcr heater in a \'apor po....-er plant operates II steady state .... ith liquid entering at inlet 1 ....ith T l = 45 CIDd PI = 3.0 bar. WOller vapor at T1 = 320~C and P2 = lOIw enlers al mlet 2. Saturated liquid water exits with a pressurt of Pl =- 3.0 bar. Ignore heat transfer with thc surroundmgs and all kinetic and potential energy effects. If the mass DOl rate of the liquid entering at inlet I is lil l "" 3.2 X 101kgih. determine lhe mass flow rale al inlet 2. lil~, in kg/h.

481 A fcedwaler heater operates at steady state with liquid water entering at inlet I at 7 bar. 42"C and a mass floll rale of 70 kg/s.. A separale stream of waler enters at inlet 2 as a two-phase liquid-vapor mixture al 7 bar with a quahl~ of 98%. Saturated liquid at 7 bar exits the feedwater heale! at 3. Ignoring heal transfer With the surroundmts a1!ol neglecting kinetic and potential energy effecl~ det;~ Ihe mass flow rale. in kgls.. al inlet 2. ,..82 For the desllpuhemer shown in Fig. N.82. liquid watn". state I IS Injected into a stream of superheated \apor ento:f' mg al stale 2. As a result. saturaled vapor eXll~ al state .1 Data for steady Slale operation are shown on the figure Ignonng stray heat transfer and kinetic and potentIal en~rg.:. effects, dclt:rminc the mass flo," rale of lhe inconllng. 'U~f healed Vllpor, in kg/min. ,..83 As shO\\n in Fig. P4.83. 15 kg/s of steam ente.... a tic" paJ/rUler operating at ~lcad\ Slale at 30 bar..'~O C. llhe~ :1 IS mlxcd \\ith liqUid walc; at 25 har and temre-'r;ltll!. (bl Plot m~. in kgls.. versus T 1 ranging from 20 to 22ralure in Ihe tank. at the time ~hen the \Jh located at x = 0 and the ~pring e"ens no force on the piston. The atmospheric pre~~ure i~ 14.7 Ihf, tn.!. and Ihe area of the piston face: is 0.22 ft! The valve i) opened. and air is admilted ~Io.... h' until the volume of the air inside the cylinder is 0.4 ft', D'uring the process. the spnng exerts a force on the piston that varies accord1I1g to F = h. The ideal gas model applic) for the air. and there is no friclion between the pi!>lon and the cylinder wall. ~or the air within the cylinder. plot the final pressure. l tn Ibl-tn. . and the fmal temperature. m F. \ersus k rangmg from 650 to 750 lbf!ft.

1'4 T"""Val~c

lOOkP

Airsupply 100 tbfJm.~r::'1..;J-='

SO"F

Fig. P4.120

l

4-119 A .... cll-in~ulated piston-cylinder assembly is connected by a valvc to an air supply line at 8 bar, as shown in Fig. p.l-.! 19. Imlially. thc air in!>ide the cylinder is at I bar. 300 K, and the

P_'"Ib~

4.121 Nitrogen gas is contained in a rigid I-m tank. initially at 10 bar. JOO K. Heat transfer to the conlents of the tank occurs until the temperature has increased to ~OO K During the process. a pressure-relief vah'e allo.... s nitrogt:n to escape. maintaining con!>lant pressure in the tank. Neglecting kinetic and potential energy effects.. and using the ideal gas model with constant specific heats evaluated at 350 K. determine the mass of nitrogen that escapes. in kg. and the amount of energy transfer by heal. in kJ. 4.122 The air suppl} to a :!()OO..ft' office ha) been !>hut off ~ overnight to conserve utilitie for blood flow because they can cause damage to blood cells and are subject 10 mecbanical failure. The goal of current de\e1opment efforts is a device having sufficient long-term biocompatibihtl. performance. and reliability for 'lide~pread deploymcnl. Investigate the status of centrifugal blood pump de\e1opment. mdudmg ldentiflmg kel technical challenges and prospects ror o~'ercoming them. Summarize ~our find ings in a report, including at least tbree references. -4.60 Recent natural disasters. including major floods. hurri canes. and tsunamis have revealed the vulnerabilitv of municipal ....aler dIstribution systems 10 waler-borne ~otamllu tion. For the 'later dlslribution system of a municipalit~ of your choice. study the existing procedure for restoring lk system 10 safe use after contamination. If no suitable decontamInation procedure exists. make recommendations. Suggest easy-Io-implement. cost-effective. environmental!) responsible measures. Document your-findings in a memorandum -470 Prepare a memorandum providing guidelines for SfIC'ct 109 fans for COOling electronic components. ConSIder tile adunlages and disad~'antages of locating the fan at the inkt or the enclosure \ersus at the exit. Consider Ihe relatl'" merits of alternati\e fan types and of fixed \I~rsus \anal-le speed fans. Explain how clUlracreristic f/lnl') assist 10 !an selection. -4.80 Forced-air 'Aarming s)stems iO\ohing mflatal:ok th prOJ'.>. Extensive properties of a thermal rcscnoir wch as internal t:ncrf:\ can change In mtt.:rachons with other systems even Ihou h the rt.: 'rv . lc:m~ralure remain.. con l'M)(W) 'J h ~ ., h "... anu t e JXm er input rl~

5.1 Maximum Performance Measures for Cycles Operating Between Two Reservoirs

237

1,1 \lIXT.I1~ th.: rdrigcr.tlor i:-. .nCM} kJih. \lith th.: ~\'h:lllClent 01 pnlornhltlCC of att:rT1~rJturc,

D..:tcrmine the coefficient ot perlormam.;~ of the refrigerator .Intl comp.trc reversible relngeralJon cyde opcralin!! betwcen reservoirs at the ..arne 1"'-0

solution)CnOWn: A rdrigeralOJ' maintains.a Ireezer compartment al a ~JXcifietllcmperature. The ralc of heat Iran~fer from lhr.' rdrigeratctl "pace. Ihe pO\\l,:r mpUI 10 operate the refrigerator. anti the of a closed system taking place in a piston--q-linder assembl~. Figure 5.15 shows the :.chematlc and accompanying p-i.' dia gram of a Carnot cycle execuled b) ",aler steadily circulating through a series ortour interconnected componenls that ha!> features in common with the simple vapor pl)\\er plant sho",n in Fig. 4.16. As the ",ater now~ Ihrough the boiler. a cha1lgt> of phtut from liquid to vapor at constant temperature TH occurs as a result of heat transfer from the hot reservoir. Since lempcratun.: remains constant. pressure also remain~ conslant during the phase change. The Sleam exiting the boiler expands adiabalicaJl\ through the turbine and v ..ork is de\'c1oped. In this process the temperature dccrea~ to the temperature of the cold reser\'oir. T(-. and there is an accompanying decrease in pressure. As the steam pa.;ses through the condenser. a heat transfer to the cold reservoir occurs and some of the vapor condenses at constant temperature Te. Since temperature remains constant. pressure also remains constant as the water passes through the condenser. The fourth component is a pump. or compressor. Ihat receilcs a two-phase liquid-vapor mixture from the condenser and returns il adiabaticall\ to ~he state at the boiler entrance. During this process. which require.; a work inpu't to lllcrease the pressure. the temperature increClses from Tc to T H. The thermal effi ciency of this cycle also is given by Eq. 5.9.

I'

Fig. S.15 (arnot vapor power cycle.

50" Clausius Inequality

241

5.10. 2 Carnot Refrigeration and Heat Pump CyclesIf j elrno! power cycle is operated in the opposit" d'Ircellon t he magnitudes , . . .. f of all cn that ....oulJ ll\f~ the con l,:f\ilti(ln of tncrgy pnnup!e.l'>ut due nlllilctuil!h occur 10 naturc. 4- Are health n k a !oCX:latcd ",nh con'UmlO!! IOmal(lc 10duced 10 npen oy IlIn tth,lLne rra~"

6. Rclcrring to fi!!. 2..l. Yohat an: the internal irrc\cl"\it-ihtl' a""ooatcd Yollh ")"Iem A and ~}~Iem B. re~p...'cli\eh I7. For the gcarl'l(),- of E\:ample :!A, .... hal are the rnn\lf'JJ I\tcrna! and c\.lernal irrc\cr,ihilitie of the arrows. Determine (a) the intermediate temperature T. in R. and the thermal efficiency for each of the two po,,-er cycles. (b) the thennal efficiency of a si/Ig/e re.. erir at I mpc a tun~s I( and I; r..: pccti\-cl\ Incrc are no {'her heat T~ transfe". ~hol'l thai I ( (c) Rl:\cr lole and Irre\o:r'lllk heat pump C)c1es each reeei\e cnerg~ Q, from a cold resenoir at tcmpeature 1( and dl"Char~c cnerg~' QIl 10 hot reSt':nOlr-. at tern peralurcs Til and Tli rc'pecti,c1y There arc no (l r heal transfe ..... Sht,,,, that TH < T....588 Figure P5.XX "h""'5 a S) lem con j'ling of a ptno.cr CIcio: dn\ln!!- a heat pump. At sleady qate. the power C)de reccI'c Q. 0) heat transfer al r from the high-Iemperature sourco: and deli\e" QI to a d"'dhng at T,. The heat pump rCCCI\es Q, from the outdooB al T,. and dcli,ers Q~ to the d~ellmg.. U~ing Eq. 5.13. obtam an el[pre~sion for the maximum theo... relical \alue of the performance parameter IQl + Q Q In terms of thc temperature ratios 1';1'4 and 1'111'.

S-B1 Usrng Eq. 5.D.complete the follo\\ing im'oh'ing rc\enible and irrc\'':r\lhk C)'..:k"(a) Re\cr;ible and Irre\erslble pm\~r cycle! cach discharge energy Q( to a cold reser"olf at temperature Tl. and recei\'e energy QH from hOi re!\Cr\-oirs at temperatures Tit and Ti,. respccti,'el). There are no other heat trans~ fers.. Show Ihat Til > Til. (b) Re\cr;ible and irrc\ersib1e refrigeration cyclcs each di!>charge energy