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Termodinâmica, Notas de estudo de Engenharia Mecânica

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Tipologia: Notas de estudo

Antes de 2010

Compartilhado em 08/03/2008

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Baixe Termodinâmica e outras Notas de estudo em PDF para Engenharia Mecânica, somente na Docsity! — CHAPTER ONE "GETTING STARTED: INTRODUCTORY CONCEPTS AND DEFINITIONS PROBLEM Li mass enters when . j : mass extts when exvoust take valve 1s ogen Velho te open hot sugfnces aiston exerts force ongas , VAZ COMPrESSIM, gas surroundinas Pads Con om eiatê á durma expansum hot-surtuces iuteract vitta suvroundingS aj im hat Aves porq snafh torque 3 transmits force PROBLEM I.4 | 1 A control volume encisses the valve. o! vive and turkine. Bt e Sicam enters ar) and exits at O . omiis o A torqueis transmaitled +erough tha E a Antabing Shajh. o . * lyarm surfaces of the turbine interact with the turrovndrngos. toithia tha control volume, Shaw flows acrost vhs valve andthmdugi ja lmrbine lolades, When tha generator is includad wi tia control volvma, . Slam enter at O and exite at BD. luarm turfaces of tha turbina aus tia generator interact uti tha, Surround . + Eleciroa current fiowo from tha generator, Node Had 4a trantwlhd drque does Vet crose tha boondary of Aba ateegas cmol volume p PROBLEM 1.57 A control volume enclesças tha engine- devem pomp. * later endere at O) and aut at E) * Aw fer Combushos of tha on-board fuel enters; and comburhim gases exit. + Carma surfeces of ABS no intevraet with tus surroundings, Goitia tha pump, à piston ix kept ui moon witun a counder own to combushiom of tha on-looard fuel. The prston motor 3 Narmassed te pump tra Jigurd, The amount of fuel wibhiu the system olecreases wi fume. loben +ha Lose and mosale ama metudad, a Wsh-sprnd adam joel emits the auetendad cobol Uoiuna ab tis noggle ext, I-4 PROBLEM Llo system boundary * tuo phases are present (liquid and gas). - not à pure substance because composition is different in each phase. system boundury - three phases are present ( Solel, liquid and qu), “not a pure subtunce because composition ef gas phdses is dt erent tan at ol Pre solid amá liquid phases. The system 15 a pure substance, AlMhough the liguid is Vaporized, te system reméius fixed tu chem- ical composition qnd is chemica ly homoganesus. The system 15 à pure substance, Alhough + Ave phases change, Lhe. system remains of fixed k chemical composttina and is chemically homogencous. The System 18 net à pure subo stance du g te process since the com- : osition of +ue qas phase changes db o water exaporates into the doa. E l | | Once all of the water evaporates, | 1 the gas phase comes ta equilibrium and'+He composition becomes = esses el homogencous. At Hhis point phase am be trealed os a gure Ps 1-5 ACE. PROBLEM (to - = = ! tt Abe 101b Ejs Tma=(t0 beto a pftre | GENOA E =Qu3 o, e Eae Fara PROBLEM LÊ gra» [5 EN 000 NI || legum/5* 3 A m/s UN Fgrav* AS kn PROBLEM LIZ 2a . Fyeay [4580] lg s> é m=10 leg (a) Goal det (E) OT , a7s m/s* Ilocal , cb) Ymassis unchanged. m LN Farav = 95 N Fray > ma «(10 kata m/ç2) Uaumisa = 98IN Faray PROBLEM 113 «tolb (6) — Ega «(2 to) 32.2 Ib-£Hs2 Greca * Tm “Vick à be = 30.4 tHs% Sogal (b) massisuncanged am o - + | Ibt Egas * MI e (10 lb)(02.2 É HEM Ei = 0 PR PROBLEM LIB aa a & For a linear spring , Fpri ing = ELMO, where & AX à the spring extension. Since Fprias = w Bray = mg » we have Kiax)=mg. Sina ; m and kK are independent of Aecatron, Fgmy tha oca accelerahom of Gravity à propor tona titho deftetim. Thus mars, Qmars . (ARmers Ortlbin Ta = Eee me Torto (AN) eoth e-zalin art = 1% gasáo a — meent Prooa CAN 2 (Ay) pson = Ge (Om) edi Rent (ax) esrth SAY o agrin)= 0.049in 32174 PROBLEM 114 Decelevatum ocuws from 5 milh to vestiu 015. The average osceler atum megnitudo is la | . Ar. (Imlnco) 5280 || th aEo 0.15 Imi |la600s ="73.33 Hh/s” or, in g's (al -[/73.33 a) - 2.28 a's deceleratim ing 32.2 Hist J lin 9's) Thus, +he magnitude of the average force applied is . < (601 & Hs lia m lala (Go b93.33 to) 32.2 Wo: fifsà =/139 bt dia PROBLEM 1.20% a.= 5 m/s% (up) g=48 m/s? Fg rav app” Fgm Md Fapp 2=ma+rmg -m tara = (Akg)(S+ 2.80) 28 pre | kg-m/32 = AGA NL Papo PROBLEM 1,21 *€ Fapp= 45 Ih£ é m=3S 9=32.2 Bis Fgrav app” Fara : Fogo Form m ma as a age (as Hb (EE Er lb Hjs? Il Tb Fogo - ma m - 32.2 Hs = 424 €t/s* (downward) e Al PROBLEM 1.2 2 Fgrav 4:30:5 Hs” From Table AE : dd = oras É MG E nHg =(0,5 Ibmol (18.02 ES TeD(so.s = 8.53% Ibf Mo. (0,5)(18.02) Save E (0.145) 18.02 Ib/Ibmol 3] Lib Sara Parar <br Ib/gã Save PROGLEM 1.23 o ; =sa Fopnce =m Ispace Espace Feortla 4 t E .m aa q ; carta dem ) q space Teor a =. 3 £ Eme MN Ra F êpce (quant « (42 (BE! det) weight Ispucé b ms? F. a8 m/s2 =68.61 Ne earth, Tea I-to ProgLEM 1.2.4 q La.si- (33 vab/adz mist, 4 ; T= tom -— where 2 o km. -" " me — — + É CA PA E = = 22 Suer weight io Wzmg, the percent Clhangeii weight landins Co % changez Lhe 3 Jem) mw = | Cagimd - [caem (33x05%/6º) (0% )] Je £ | soe? ce o [ (B.8- GaxWM)] “GIF of PROBLEM 1.25 les 30H [Sua 02 bip 7 G= 324 Hs” go uohere Vis Me Fra The mass of walker is mz Thus volume of Je spherical tam V= Em? (Em(3o = Ii3ixto Se? and the mass is a = 3 m= gV=(024 Bo ttstuos) A =706K0ºlb cm The weight is º e Hs VE = mg =(T.Obxtp (3215) ae a G vav = 7.04 X10º Ib£ Egray PROBLEM 1,29 o) Ar à temperature 0$ 240%, He Specified pressuve of 125Mpa falls between The tabe values of 10 amd LS MPa. To determine Jhe specific volume correspondiug to 1.25 MPa, we think of the slope of a straight live joming "se adjaceut +able stades, as follows: Y =0,.2275 ms Z2. (UV p=1,25) uv, 18 :P 3 meg JY Vs DINB3 my leg Ay ; ' P, MPa to 125 AS similar triang les : v- 01463 o.2215-0.I483 . 025 a | slopel * Te-tãas "sas P VeomBst e (0. 2275-0148 =0./879 mêla (a) (1) Ma pressure of 1SMpa re quer specific volume et oisss mk falls between tue tule values of 240 and 280%. To delevmine le temperature correspording to +he ques specific volume , we Hinle fi He sibpe 58 a shroag ul Vime jon ing ge adjaceut table states, às follows: 280 (0.1621,280) ç Slope = T-2Zã4o . zpO-Z4o ” Pe ASSS-4B3 1627-1483 > T=240+ ad] Ho [res (to) d WOT ciyBs,2do) = 260C Ch) q Mas dC 47 0 mYg (€) Indhis cause, he specrfiêd pressure falls between He tuble values of Lo and 1.5 MPa and the specified temperature fuls between the tale values of 200 and 240%. Thus, double interpolatin is reguived. . Hi z20%€, tha specific Volume od eaci pressure is Simply He average over the infervol : zoo 4.2245 4 O +.Z2 at 1.0 MPa 220%; E > = o 21615 wº/kg 1325+, at 15 MPa, 220% = 1325+ 1483 z = 0.404 Veg . Thus, with the same approach as iu Ca) W-OtãoY O.21675 - o. 40% - PN os LS- 14 > 5-6 > UV =otoy “(85 )to.erers toy) = 0/5567 fra co t-Id PROBLEM |. 30 ca) At a temperature of 120ºC , tra specifred pasauwre of 54 Ibf/in2 faiis beto the table values of 50 and 60 lbf/in? To determine. the specific volume correspondina to 54 Wflint, tua thank of tha slope of à straight Aine j raina tua adjacent a bs, states, as follows: 1, 3 seo v=sga ff 1 4a. 50 = Te sp Py tbflin? Similar Arrangles: |stope) = N=88a1 = IHo- 5.881 = vz 6.841 + &-(auo-s89)= corr co-s4 69-50 Rota to) At o pressure cf 60 Hoffin?, the given speafic volume, ef catrttih falis betuwcen the table values of 120 and ISO F. To determine the temperature corres pondens To tia given spesifi voloma , we thninn of tha slope ota eirarsht Line jormima Te asjacent table states, às fottows: 140 (et, 140) sbpe= T-izo - 140-120 “age o Pe comia Guz- 8.8] E 8.982. 8.841 Je » panot [SE ) zo ) AL 1 Ls =74ºF s8 54 co é! 5, gtêilo () Tn this case | the specified prasruve fails between ha table values of Go and bo lbf/in? and tha spewfved temperature falls bekweem the table values of 100 and 20ºF, Thus, double unterpe lah 3 required. e A UOSF, due specific volume at each prsture JE empty tia average ouartia intervol: as Sof ao*F; va quo det = 6.973 1º/th ut) ? 3 at colá, np; v> ssh = EM . Then, unth tus Same approach 09 Un tas V-SH€ CMS yo gar t Etr sm) c-sg 6o- sv = Cos AY ato t-15 PRogLEM Lspk m=5kg From the pressure - Specific votume relation é 13 Na ! ( E pu = const, pis LATE N 1 4, = [Ab =|-—— 3/k [x] A (555) (ora mthg) paibar ,U,=0,2 m/kg = 05810 mYbg P,:0.25 bar — MawmazgIOos mi Mo 1.2 - 14 1 0.8 + Eos] 04 a d Te 02d z od et o 0.5 1 15 2 25 3 V-m PROBLEM 1.32. e wmm=l db From Te. pressure - volume relation + +; pV" = const. PAR 209º Í EP VE (&) v = (55) 1044) £ Fut . 3 na; Vocz HH p,=20 Ibflin) Vi= 10H ne; Vi IS AE? Va p= 100 lb£/in? naLz; Ei aoo ns €&————— 0 ————+& nete, Voz 3.168. 44º 100 80 4 cabo 2255 nara suvo 60 p-Ibfinz 40 20 + PROBLEM 135" Thermo dgmanic cycle! t-a: pVa constant = B= ibar, Va lmi, Voz 0.200 3.1 V= constant rr, pV= constant. The constant cam be Fr process usas data at stater: P v= constani =» Nr = (bar )C im) = E bar-m? Accordinglys on à pressure - Volume plot process I-2 is deserihed b + lbar. ms Tal Tm portiuslas, ben Voz olm3 , p= bar, PE: Tue therredynamic cycle +apes the forma 2-3 peconstat, VW >V Lexpansion), NV = Omo ewlvated Pa=Sbars = constant 5 O Vs 0.2m LL > É Bys Sber Va= tm + V= constant p (ue) 3 2 PVE constant ! As lbar vw Im3 opta to 14 2 oa 04 os ob of og oq Lo (wi) 1-19 PrRogLem 1.36% Pam = VOL 33 PA, Prlitm 5891 q=13.54 Ya 9:4.81 mist P-Ram 2 [1028 tg tNjm VkPa $3 = (io4,O = 101,33) ka . Pleg-mifs* (13.59 9/ems)c7 Si mis?) EN vm? 108 cm =002m = 2Zemgç L (b) Page * P 7 Futm = (04.0 = [01335 2.67 LA Foge PROBLEM 1.37 PE (0) F,vac = Pe 9 L ! ! - db Hp ha LIbg HZ = (ua, af Y32.2 EM qa A TD 4 ( ps X $ tz l= tb -M/6?] | tuq jut Po vaç Melia. Ho = Odair dbgfive Pato é nes bela t 4=3.a Hs? R,abs * Em” Rvac b - 3 $ç MAM lol o = 14.5 OU = (429 Tão Brabs PROBLEM 1.38% +27) Pb * Bbm > vacdum Tank, Bum AS efa Proc = Ram” Pabs AB-dO=SEkhe Pg Pas E OM bar =40 L Pa I-zo PROGLEM 1.34 - 1b$ Fête e |APopge] = 8,41 9=320 FH/sz v:o.oou Hb Sho? Lsuã 62.344 1b/gº [Bope | a(e2.30u à lb Y32.0 fix E) 32,2 Ib-Ss2] | 144 uz 3 0.39H4 Ibf/ivt LPyago (decreases) PROBLEM L4O Pagto efe L=20cm =60,2m Tau 6 Foge" Sug gh T =(13.59 Téo o.2m) E=20 tm A |Jogea rode Um? | Ilkgmis2 ] 26, ot N/m? =0,26bb bar wi Tgnge p= LM bars Mercury (g = 19/54 9/emê) - Gage a 959.81 mjs2 bs,6 Gurrt Gogesa = LOl+Ze6b=L21b bar e Pos, = lc gt Fraga A =L2T66 HLH=267b bar e PROBLEM LH Bim E Su Lug AIR «(3.54 SL )(9.81 me) GES)m Praç*02 bar ess Vig | 10 em IN Ibar Pet 750 vom Hg 1034 im || egemjstl io? Njm? Su; 213.59 9/em3 2) bar =4.81 z - 359.8! mis Pass * Bem Prue = Ibw-02bw=08barç Fabs t- 21 PROGLEM 1.485 L/x= Sin 6 L x sine oe XL inclined manometer gives greater resolution PROGLEM | » daçS sp=e9d «(62.2 m 32.2 EX 8) Libg 32.2 Ib-fils; gr teu in? g=b22 Ice unter =2.89 Ibglivê à 9=32.2 Bjs 1-2% PROBLEM 1.47 Consider Jue variation of pressure owing to fra effect of gravita. br an element Am a motion less gas or ligurd, due foras acting are tia forces ef pressure on tua upper and lower surface! and tha argh of tha au atem Jose Z+HAz Zz [=] * Í Toma 2-0 O= plerazjAà tmg — pl A hero m= pp V da a Vraaz] = Lad) = Rearcam e Y A (ptrsaz)— pexj=(422) 4 or plsor)- Pl) dg. As au TA Fe Amit ao AZ» O. TUOs gives de. 2 Ge) 3z Ar . Bi . (4) Atmosphere. TE wa C/p tuhere Cs 72 435(mVkg (kPa = Ta gas No, is insovfed us Es (8), ue hmm RR = -4P» dinbo plo = eps RE + ink cl om p= Kemp(- 42) lonen 200, P=h (latm), giving p= Remp(-92/c). Tnsertim erewn Valado Gises pomatm when 21 taum o | IM - (conssz) [Se] = ep dia This telakonshmp is siown ui tra Accompargins plot. Gestmfs2) (x em) Tay SE pés exp ( |-25 PROBLEM 1.47 (Conta) 08 = 08 s É =. - 0.4 1 = = 02. 0 T DA 0 2 4 6 8 19 z(km) (o) Ocean. Lething 2É-Z ) denote dont, Eq. (K) reads dp. 4 d2 v- Than, AStumina Ar constant paG) ak o = (294 ' lim 2=6, pe fp (lama) quina ps (2)2 +B. Tasertiny trsum valiis aives Pi atm atuam Zé tm Cast mist) EC) ida] atm p= bot 0-4 S6 x1073 m3Tkg [Tum || poisaspio” Mm ameno] =p+ios Z Ts celakincoup is shewn x the acto vepamenina plot. 200 N 160 120 p (atm) so 40 A PROBLEM 1.5 Using Eq. 1.22 T(F) = 1.8 T(c)+ 32 (o Tltej=21 Ter) =(Le)(z)+32 =69.8 T(R) = T(E)+ 45267 = 529.47 (b) TELE) = 40 T(r)=(h8)M-40 )+32=-40 T(R) =40 + 459.67 = 414.67 ce) T(C)= 500 T(e) = (1.8Hs00)+32 = A3A TOR) =Ã0A +4459.67 = 13ULT7 cet) TCC)= O Tltp)cÔLECO +52= 3A T(ºR) = 3A + 959.67 = UM te) Tlf) =100 T(ºF) = C1.8HX100)+ 32 =212 TC R)= 2121945967 =67.67 (4) T(Q=-273145 T(ej= (1.8 -273.15)+32 =-459.67 TOR) =-45SA6THUSIGI=O PROBLEM 1528 Usmg Eq. 1.27 <a) e) e) (43 eo G) TÊc) = TCE) - 32 T.8 18 = T(E) 178 -õ tise)= 68 (0 = SE -1276 = RO T(K = 20+27318- 223.15 (0) = HO T(ºe)= e «278 =-40 T Tte)= -40+ 2735 = 23315 T(ºp)= Seo e)= SOS 17.782 AGO Tt e) - 17.:*8 ae T(r) =abo+ 273155 633.15 TCE) O me fa -l78 = 1278 T(K)= “ABAS: 255.37 T(er) = Sa = BUM 17:28 = 100 T(c) TB TO: 100 +27: 373.15 T(CE)= —48747 T(C)= — 48% -1A78= -27315 .B T(x) = O !-30 ProBCLEM 1.53 Tel TCC) = [nte +agais] -[m, (CO+2741S] = Talk)- Tick) Tal CP) TF) El Tel CP) 450] TCA) ESET] E TCRI = TR) PROBLEM 1.54 The expression Tor resistance is R=Reexpl (+ 40] 3 where Roz 2.70 and To=30k 25) | Since R= 0.31. ab T=422K 1] À Bis (az)emp Lo Cais - Lo] E | Solvihg for 2 1 . In *(.31/2,2) é 44 e: Ti dy + t286AK 4iz 3o 054 me Thus 01 mm TEA . 1 250 300 350 400 450 500 R=2.2 exp |r2084 (L- 45] Ta PROBLEM 1.55% GIVEN DATA: From the data ot T=32ºE TRE) R(O) 5139 =Re[t+a(a-5)] > Ro= 51.370 At TEAG'F, R=S/.72. Thus - Testl 32 51.39 3 s Tes2 146 5172 staz = BL8I[i+ (IDT x ss a Finally, R= SIA [+ 34USS HO? (T-33)] RE BoLitaCr1] * Ras326 + dotre x lo TT Let To=32º E T=406.A1(R-SL.326) T TIRESIMTO) = UF «——— PROBLEM 1.5to Using Eg.1.22, TIP) =I.8 TCC) +32, For T(P)= TC) = A A T=:187+32 4 Solving 4 e-40 Foca T ink: Tl(k)= TÍC)+2ZIZIS = -VOJZTI3IS = 233. Be Tt) TR: TERETUEASGOT = -dOrESIbIE MIT TER 1-31 PROBLEM 2.1 Known: Am automobile of knovn mass acelerates from a given veloci to another. Eron: Determine tira inrbal kinetre eneray and the change ra Kmeho tnerqu. ScHampTic ACIvEsM DATA: ma 200 9 mf= 50 km/h. No = (00 em/h ASSOMPT.ONS: (1) Tha autommobrle is ts cloçed SyStemm. (2) The veloutes ond tinche energies are felative to tia, Aoad. ANALESIE! The initiot kinefic energa Casturmplio 2) is Ke, bw + El im/am | |ixT pin 2 so em VI TT) | Ha z aU eo( e) [ie | to%Mem Tea) KE, = usP4 KI “1 Tua Cla vi Junetre enerys va Ktkeço Su LE V] e E] - à (200) [geo (se | | ie] | El = 34% KJ a %KEacks PROBLEM 2,2 KMOwWN: An ooject of knowa mass is located at a specified elevakica relative tote surface of the earth. FIND: Determine gravitational potential energy of lhe object. SCHEMATIC É GAUEN DATA: =400 kg sa Mm ASSUMPTIDNS: (1) The Object IT 92418 vis isa closefcustem. (2) The ac - celeration o gravity is Constomf. 2.30 LA Á (e L ANALISIS: The gruvitational potential enerau ig PE = ma z UN LRN (ts = (400lkg (9:18 a Xas libelo MN TEN TA | =92 kJ « A PROBLEM 7.3 * KNOWN: An object of known wetgkt undergoes à specified change ru its kinetic energy while its potential energy inereases . FIND: Delêrmne the final velocity. Rs 3=32a0/8“ SCHEMATIC É GIVEN DATA: s , vça40 Sy/s ASSUMPTION: The object is à closed system. “object AKE = -S00 4k-bF AnáLysis: The change tn linekic energu is Foray 2 GO be=m3 O axe a Sontri ENT m= feto ja go |= 100 lb Thus solving for N7 and insertihs values and à unit comiension Factor . 2 ACE GS [om | - - ) A bs 2z(- 500 St-Ibf [A E.d/s . (40 Hs)? (100 1b) 4 4 = 3598 Bs Na 1. The analysis makes no use of he Information related to potential energy. 2-2 PROBLEM 2.4 KNDWA: | A bricla of Knogon volume and densiky expenênces a given decrease. ua grawiratinal potential energy. V=2.5X3.Sxlo vê FIND: Determine the change im elevation. pras SLHEMATIC É GIVEN DATA: LEY sro by | 9=32.0 M/8* ASSLMETLONS : (1) The brickis a closed Ape =- Soo fHIbf system. (2) The aceeleration E Gravity z às constant. (3) The density of thê. body às Uniform throughout. ANALYSIS: (e? ; V=(0S SB) Es =0.030384 = 5=5517 Based on assumplion (3) ma=gV =(noblm Vo 030381) = 3.6S1b The change in pobential energy and the elevation are related by Ape = mgy4% Thus, solving for 4z . AP (-s00 &.ibe) 32.2 W.fh/sa SE = 9 * Guião Mis | IbE D =-138 He dz I The negative sign denotes a decrease, in elevaton - 2-3 PROBLEM 2.8 Knowa: Anobject of known mass moves with a given velocity. Einp; Determine (o) the final velocity for a given change iu kinetic enevad »andíb) the change iu elevation for a given change im potential energy. . Pes melho DeHEMATIC À GIVEN DATA: CM wtotis ASsUMETIONS: (1) The objectis a closed g=32.0 Hs? System.(2) The acceleration of gravity z (a) AXE = -100 H-tbg is constant. | (by 4pE = 100 Pt- bg ANALYSIS: Ca) The change in kinehic energy is related to the initial and final velocities ly Ares dm bu] Thus, selving fer +he Final velocity Vi 2 4kE Va = TA sv, Tnserhiug values and converhng units accordimgly der . | (2)C100 H-IbE) [32.2 Mo-Skls? 2 Vs (1) | ribe |! Os D =54,67 Hs « Nz fb) The change in potential energy is related fo the change tu elevation ly APE = mg AZ Thus, He change in elevakon is . APE Az a m 2 Moo He tbe) 32.2 Wo. fHs* (1 IbY 32.0 Hs2) vibe o = 100.6 HH « Az 1. The velscity decreases, as expected. 2. The elevation increases , as expected. 2-6 PROBLEM dA KNOWN: Am object 0€ known mass accelerates from a. given initial velocity to a given final velocity due do He action of a cesultant force. EiND: Determine the work done by the resultant force. ET &Iv : 28 me à bg a v h N5=200m]s Rá Vizsoomis ASSUMETLONS: 4) The object is à closed system. (2) The resultant force. is the onty interaction betucen the object ab its surroundings. ANAIYSLS: By assumphon(2), the work of +he resultant force must equal Te change iw Kinetic energy. Thus, using Eg 2.6 O work = dm a 4 (aka) (So0t-2.00* IN Ikgm/s2 =aio ES work, E 102 Nm t. The increase in lunekic energy of the Object is the result of energy +ransferred to it by EM e work of the resultawt force. — PROBLEM Ao Known: Anobject of known mass undergoes a change of leinebic energy due to the action of a. resultuut force. Te final velôciy and the work. done by the force are given, END: Determine the initial velocity. 22223 m=º300lb é | V5=200 Hs SCHEMATIC É GIVEN DATA; FO ENA work done by fone dor todos = 1008ty (Odd Cera ASSUMPTIQNS: (1) The object is a closed system. (2) There is no change in elevatia. (3) The resultaut force ie the only interaction between the object ondits sunromadiags. AnAN Sis : By assumphon (3), the work. of He. resultaut force must equal The change iu kmetic energy, “Thus, using Eq. 2.6 work = bm (V5*-vr2) Solving for VT; inserhing values and converking units 2W 2 Vis a tz tur 2 (00 Blu) | 778 A-ib$ [e Mo. fls? 2 pé + 100 (B001b) | Ba | VE 5 = 23300 7 ç> er o VN, = 152.0 se Nr !. The ihcrease m velocity reflects the increase in livetic energy of the object as a result of energy Iransterred te ih by +he usork 6F He vesultaut force. PROBLEM 2.12 ENDwa: Am object of Enown mass moves along a stroigut Ive worth a Known velocity. End: Determine He e otutiônal speel sta WU wheel whose votatronal kineho energy is equal ta mag nitude to Ye, objects Incar kúetic energy. SCHBMATIC É GAVEN DATA: KEo; * KE py AE i f===- meto db TAN | Losv-10Hls ED 4 , = object + r=-150 b-H AssompTON : (1) The Objeck and the Flyuheel are both closed sstems, ANMNySiS: The kmekHo enerqu of the object is ke. dmy? ob í (1) 2 13 UE «= 4 (Golb(ivo 32 l5za Wenos| * 1853 HH ibf For the fuwbeel (See the soluhtewu to rroblem 2.11 ) Ko ET? ee to « [2KEtu . [esses H-ibf) [32.2 tb fhjs? T (so lh. 42) pot = 25.82 4 Tn terms of RPM +) vevi | tos =. 24 ve : - (25. sz ) | E | mat 246.b revimin Q aii PROBLEM 213 + Two objects fall freely under le infueuce ef gravity feom restand He same initiod elevaton. FIND: Show that Abe nitudes 6€ He velocitie are equal at the momeat just before They stilo fe earth. 8 SCHEMATIC É GIV : 7 A o, | vºo z E fa Ze pl Ed 6 cp Tur s:(1) Mn object in free fall is a closed system 2.) The acceleration of GTONUM AS constawt. (3) There, is no affect 0€ au resistance . (4) The only forde octing is trai dus do gravity. ANAaNAISA For an object falling freely under fhe infuence of gravity + ER Ogg les imivà +mg(z-2,)=0 For N,=O aud E,=0 w. mn? = Mgz, Thus We 2 9z Súnce the Ema! velocity doesn't depend on mass, both objects vil have identical velocittes at fhe moment just before Shey strike Ihe car. PROBLEM Z. It KNOWN: An object of known mass is projected upward from the surface 0f the carta with à lenewn initial velocity. The only force acting on the object is the force 0€ Gravity. Em: Plot the velocity of the object versus elevation and dekermine He elevation when its velDeity reaches zero. To 13 W20 SCHEMATIC É GV. ETR ês 9=31.5 Hs? AssUMETIONS: (1) The object is a closed v7=200 Hs system. (2) The occelevratiow of Gravity is 4 -meBolb Cônstaut (3) The only force aching on Fiz ES objectis the force o gravity. q ——— ANALNSIS: Since he only force oching ou the loody is He force cf Quavi E 210 aqplies. Tuas the velocity and eleva are related do da E conditima o im Ceniga = Ly tea Tuus, solving for NT Ex om for N” a/Nç-agã inecms of E When V5=O, Z,1S ONO 200 Roz 29 (gaste) Plotting the above velat onship ted 2 =bmaizHtç EE 200 4 160 1 - <<. 120 À ve (fis) 80 + 40 | T T T T T T 0 100 200 300 400 500 600 700 =(t) a-i3 PROBLEM 2.17 enown: A box sides doum a pamp. Themass of tin box and ctg veload amd the. top of tm Lam p oa Eram. The Aa jeometry vs alto Speatfiod. FioD* (4) Tx jue oba ! feccdtim , deternana tha velocity oftia hex at Aa ba à 4 and the clum wu trafo que potentras certa ta box. Ch) Determuva qua Clanges ui kinchrc aud potes Aus qr box when frrehm Lã acting and +ou Veto city ato +ta bate qto dam à msm: Compor uatia dado aaqulth of part la), seHemaTIC É CiVT DATA: m= 25 1b st = ato AsgsomproNa: |. The loox is the etored system. 2. Tn part(a), fecchon is negligibie. 3. Tha acceloratia ef gravo 3a.oftist, ANALTSIS: (4) Tr lue absence 0f any resultont force acting on tira system, including fricham, E4.2.3 veduces to dm (NEN?) + mg (20-25) =0 68) Sovrna, Vez Wizg (2-2e) 2 7 Ve =) (24%) +2 (22 É) (SH) = (8; As shown bu Eq (Rr), tu ineo and potenhal euergy Cumget [a ta same ma gua tuda bro dra opposite A Sima. Thus Uh APE= mg(Ze-20)= Qsw (32.0 (ct É) = 12427 E-lbf 4 APE and ARG = + 4 hE. th) luhether thure à frchon or not; tha potential eue ni que case vê Ae tame ao dejermined va qurt (a): cIZAT pr DE. TF Vez 4HIS, à bei ; i ade? tira €lhanas wi kincht Curso vo smaller in mag ni d : nn ave = dm[WiNH] sp (a) - (287) | aonde ame |= 218 Hlbf 4xE Reterrina again to E.29, the decreaes wi potenhal €nar ui URL se Cam dor ccounted dor wi Jerma of wicrease um kuatic ensrgs opta box and cuark t+ overmme Lfrichm, a-tb PROBLEM zag KNdWN: Asystem of known mass anda given mitialvelocity experi- ences à constant deceleratiow due to the action of a resultaut force and comes to rest. END: Determine the lengkh cf time lhe force is applied and +he work. SCHEMATIC É GIVEN DATA : vs, m=5kg Fe v) Nicd0 mis + wo x ax =—amjs> LU tt dt dC td A AssuUMPTIONS: (1) The system is closed. (2) The horizontal deceleration is constant. ANAINSIS: To trhd 4he time, use the fuct that the acceleration is constant, as follows av = 4: q > am = asdt Va tz Siav= | at Y É o or NEN, = Qx st Thus - o at= DE COMO sos. a Ay E amis?) The work of Me force Fi is found feom Eq. 2. oO Work = LmpÃEwo) — IN VkT | kgmis? 102 N-m 6) =-4 ki work. =- Ss ka)(40) E 1. The negative denotes energy transfer out of lhe system, PROBLEM qd Enouon: An object of known mass, fone for a speufred time, experience a constant agelera bon from o Eron initial Velocrta, FiND: Deteruune tu work of fha Agultuut force. açted on ba a resuHant scHtMaTic A GIVE DATA! ? /Z Sw 4: 0 2 esto) az fts Lo. x No sm-WSS 1 Tha objecf tia closed system. 2. Metrom is lusrigontas, ASSUMPTIONS: um potenhtas ementa. s. Tha soma system ex periencos NO Cum horgomtat acceleratim «3 coustant ANAL TESES: Artigas Eq24 wite astumphm 2, tha mork % ha force Rd gistn hy va Vgorte m o LV, = MO) Tha finas vetocihy Vo Câm be detarmnes irma astonjh'm 3 dV = à (constant) dx Tums dve ad E mm=attito) o or q att) =026 (05s)=18 H/s be mes uork= 4 (ao) (19 83 | AE, | -ASSR Elbç < toorie (6 log) Blu ; = (1558 Bebe) [araacta | = 471 to iria PROBLEM 7.72. (Contd) (b) Letim N range fem Oto TOwmalh, Aus Ae 00 tgp trair plots” co boo developed, 40 30 4 —— Wiotal — 1 e Wr E —- Wa 5 20 + z 5 a 1 a 10 4 a 1 AT a so AE TT 044 FATITITIT IO IDO 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 Velocity (min) Wa par frnndia plote 4d up fo amous SEua/h | fia poser sagured + ovarmuma Aotlina Aosistunce vi mena sig fi cmd Hm tia poa to ouer coma dvas, “a lucia specdo, fo cdyas sf los comas damanant becunse pj 5 peedo, 3 . Va VE dava ua Ta Jim prossima Pre Va. Sina tos quo E ouartma +hair aff Se ve td 6 Ana onane from toe fuar stoved em looard tua velecha, baço pasa elyrvins bus gu sapo alla Sijuficno e fbot om Lab emsurmphm. arai PROBLEM 2.23 Known: Mensured data for pressuve versus volume duving +he compressica of a refrigeraut ave given. ElND: (ay Determine n for a fit of the data by pV"= constant. Cb) Use fhe result of part (a) to evaluate the woric dône during the compression. (e) Evaluate the porte ustug graphical er mumerical integ vretum of +he data, (d) compare au clisuss paris Cb) aud (e). SCHEMA É GIVEN DATA: Data Point p (bf/in?) V (Gn) 1 112 13.0 2 131 1.0 3 157 9.0 4 197 70 5 270 5.0 6 424 3,0 assumprtons: |. The refrigeraut in the piston-cylinder assembly form a closed system. 2. The pressure values provided appromimate” Hhe pressure of the piston face. AN MYSIS; cay One approach to fiud mn. is fo begiu with pV "= constant. Taking He º log o Toth sides ok Vis equation P mt. Metag Log p+ n £egV = Loge bo ow Log p = (-n)logv tloge Tuus, (-n) covresponds lo Ha slope of a plot gp vs. logv. Us Da spreadsheet program to obtaiu to. Plor a + aqua es bes Sil qurve: 2.7 2.6 º . 25 Prom the wurve fit 241 Cn) =-04688] 2234 8 F 22 “ acDIOBBI Mm 21 Thus 0.40887 1 py = tonstant 1.8 0.4 0.6 0.8 1 1.2 log(V) a-àa PROBLEM 2.23 CCont'id) (b) Using He results ok gorilas and Ha procedure of Example 2.1, jhe work. is Vz . BNE cp V We dave Ee Cum bene (3.0 03) - (uD)(13.0) [1 | Blu o U- DaoLEBND 12 mf | 78 Hibe =- 0,21b3 Su 4 w o «o 4 graphical evaluatim of he wore involves a plot of He tabulated data and & smook urve druwn Hurough Ha data points : 450 = 400 | 350 «300 E 250 E 2450 100 50 0 0 5 10 15 dd Vin.) Each elemental rectungle in Hae plot contribules He following Jo Ho area umber Ve curve: be Vos ins )jtRI 1 Su Is ot6 (to SÉ NO. iu pls EE) * 5-3 56 x10'! oh The numbor of vetangles is approximately 40, Hus Wa(moLi)(S.356xt07“)=- 0.2148 Blu NM (Ay The results oblalued in parts Cb) awd Cc) gre iu g00A agreement. Each should be consideredia plausibleestimale for He reasons presented ou age 40 iu tre discussion oF atual expansum amd conpressioa processes. |. The software IT could be used to oblaiu He lensk Squaves Curve fit by rogramm tas the equatisas for curve Lithing. His custer to use & spreadsheeF program in Huis Mshance, however. 2. The outy measured cata are We tabulated data paints shown as filled cirdles , The smooHh cuyve does not necessavily represent He actual pressure af Yhe piston fare for tha corespondai volume. a-a3 PROBLEM 7.2 5* Known: A known amount of gas undargoes à constaut- pressure process in a piston- culinder assembly besiuning af a speertied Specifi. volune. The Dorle is Enovsnd Gun tag p pecific guyp: Determine the final volume. SCHEM é Ge TA! Pl | ya wa [pdl=-ISKT vis 0,2 mig 7 o Je ES we=-lsty 77 O LO pesar 7 Assumpr ins: (1) The gas is a cloged system. Vo (2) Pressure is cmstaut duvins Ha process. ANMNSS: Usmg Eg.2.17 Va w= S pdv = pl(Ve-V) = Pp(Y4- mu) Solving for V> and mserting values Va & É tm, CISET) ! bay 10ºN-m É (Sbar) TES ear! + (02Skg( 0: Mk) = 002 mº4 Va 2-2 PROBLEM Z. 26 KNONAS Ogg a piston-aqlindew assembly undevgoes a process for thich pv b3 = constant he vb is bene y ' Finn: Determine the final volume aud pressure. 2 ACHEMATIC É GIVEN DATA: p Va Wa Jpar- -4S Blu v as HW? AssuMeTIONS.: (1) The Oq is q closed system .(2) The process is Pelytroçio. Anasis: To determine V,, substitute He pN velatim into Eq. 2.17 and integrate va vv? a a constNay = 2 MN wi = d pax 4 (Sm ont. 25 Now, Const = PV"? With +his expression , amd solving for A 30 (=-3)W —3 Vo pv! Us Y . C3)(-4S Bh) vs bel 4 HT? Co bis po? Bla ||iiu* +ês = 440 PV,<Ig3S Ha Yz Now, we can use +he p-V relation to get p 1.3 .3 - V - bev/25 . PR. - (4) =(30 Tê (RS = 44,85 btw Pa 1, The shaded avea on +ue p-N diagram represents the work for the polytropic process. PROBLEM 2.27 * KNOWN: À gas undergoes a compression process. Pressure and votume are given at Ie in" Rial and Pitol sines! Pressure and vo lume are related linearty during he process. EIND: Defermine the work. E SCHEMATIC É GIVEN DATA: 8 2 p=ibar,Vi=03 m? ; a fe=3bar ,V,=0.| m? GAS | Log r Ir o ol 0% 03 - mê ASSUMPTIDNS: (1) The gas is a closed system. (2) The compression is a quasi - equilibrium process. with a linear relation between pressure and volume, ANANSIS: Based on the given data, He p-V relation cam be expressed GA p=4-10V where pis in bars und Vis inmê The work is delermined using Fg. 247 . w=[pdV ' ê ' Inserting He p-V relation and integrating Vos Olm3 w=[" [4- 10] totiim er lay V:03 m3 — dear PO oa V,204 — =[ev-(2)v']] fiel vz 03 = fu(o1-03)-S (12-08 |too| O =40kT W 1 The negative sign for work denotes energy frausfer to -+he System. a-a8 PROBLEM 2.30% KNôwWN: Ar undergoes a polytropic process belween two specified states. FIND: Determine the work. Schema É GIVEN DATA: n=0, lomol p= zo Wf/in3 vu; siso A/ib p= GO Ibetin3 v,=3984ib pv”= constant ASSUMETIONS: (1) The air is a closed system.(2) The system undergoes a polytropic process. AntiMSIS: From te pressure - volume relahon for a polytropic process pu = pv > pulp” Solving for n logtr/rO — Yog(2o/go) log Cu) loq (328/1,50) Now, using 45% 2.17 te defermine work. and cora tu. molecular Weight = [,307 of air from Táble UE “ár =? W= fear o m [7 pv e mens 1 . (Gr ju, rm. Pura AV PY «mf mo ml Bu) Nov am = (0,2 bmol) (284 1b/Iomol) = 5194 Ho and o (80 Ibshu2)(3AS 2214) -(2(11,80)] has mf] Bu W =(SAM o (1- 1,307) | EIS? | [TS Skdbf o =-308,8 Blur w L The Negative sign or vsork denotes energy transfer sinto the 3yste M. &-3! PROBLEM 2.31 kNouN: Warm aim cools slowly im a piston cylinder assembly from a Enoun initial volume Jo a Evown Sinal volume. During the process, à spring exerts a foree on the piston Hat varies Linearly from a nona intrial value Jo a fulad value of zevo. 3 Emo: Determine Jhe initial ud Linol pressure of He air, and He work. SCHEMAME É GWEN DATA: A=0.018mº v=0.003m? V -D002m? Pass = 100 kPa . a Spring force varies linearly from 900 N when 003 m? to zero when Va = 0.002 mP ET ASSUMPTLONS : (1) e ar is a closed system.(2) The process oxurs slowly, so Here is no aceeleration of Ha piston (3) There is ho frichom hetweck He piston aud He ey lui der wall.(4) The spring force varjes Vinearty LutHy volume, ANALYSIS: Tue initial aud ftual pressures of he air are defermu ed roma free -toody diagrum of Fa pidon,as follows. Thafis, ZF=0,so Lnitialla : Fspriiso = 9)o0N Pp = Pam + Brig - 100 bp, 4 COND | IkPa | .ssokp (0.018m2) to? n/m? N Fiada: Fprim =0 >p,=100 kP h A. F, Va + Nou, the worlk is determined usina Eg.2.17, W=f páv, but from above p= Pet + Fspring Since the spring force varies linear from 4300N to zero as volume ques +Arom Viz0.803 Mm? to Vo 20.002 M 400 N spring = Uooor, X V-0.002) and wW «Vips Eprina) dv Soo «(ge V-0.002) Tu , & y O.b0I] [D.oI8) lie? 1 Vo 0. 062m? = st IPS + S0000V - Io JAY = (ssgoo)yz A v,:0.003 m? HT D 3 2 10“ Nim mia) | mg) = - O,125 ES, w VPa JON -m OizSbda + t. The negative sign denotes that the prston does worle on Ha Gir 06 the air cools. Also, tua cotmosphere aud Yha spring do more ow Ya piston, 2-32 - 0125 kPa-m? PROBLEM Z. 33. Known: Air undergoes wo processes in series. FND: Sketch lhe processes ona p-u diagram and determine +he work, per umt mass of aur. SCHEMATIC AND GIVEN DTA.: pReessi-2: pulPiconst A=icokfa,v; 20.04 mYhg Des Vz= 0.02 mY kg PROCESS 2-3! p=const, ve ASSUMPT NS: (1) The air is à closed system. (2) Bom processes are quasi- equilibrium processes. 1.3 ANALISIS : For process l-2, R.=R ) = 24b.23 kPa. Thus, je pur rágram is as * Z 3 q Zoo a 100 o OL O 04 vw - m?/kg The worke for eada process is determined usting Eq. 2.17 wW=[pavemípdu > Wa (pdu Thus as Wa ( “entido. (Pur mw L 50 = a ) º to co bo XD mt) - Ctooicoomo | to Nim | HT Q-1.3) Via lioFN-m =-3.082 KS/kg W. Va e = Sp páv = PalVg- ta) 3 No? = (246.23 LAY 0.04-0.027) E es =4. 4246 k3/eg Finally o Wa - Wiz, Wa 4 | 842 ps Wialm E 1 The result is postiive, denot mg rod Ha net energg transfer by worke is from the system to He surroundiugs. 2-33 ProBLEM 2.35” knows: Aun around à ED: +he object of krown mass is puiley atiichad fo a tepe (uound and falls at constant speed. Delternune fina pomem transuultad to tia Pullea and totatronel spres ef ha pultey. Sepematic Leves DATA: ue m=50b bos W=3+&s F=mg AssumPrONS: |. Tue object falis at constant Speed. 2. Tue acceleratim of Yravita wu constant) ANA LTSIS: una, Since E 2i3 W= FO = 32.0 friçã Tha power tranenaies do tr pule com be determuned = ma Vo 16$ Colo Yao gt (38) || (14% 4. EMP oraith | 0.27% hp w n VzReo , we have. to = o 38ts rRey eos RO CGldt | 02% eua |) Iuin = us RPM at 2-3 PROBLEM Z.3b KNowN: The rotational speed and diameter of a drive shafk pull are known. The net force aphl red by the beit on the pulley isalso giver, FIND: Determine +he applied torque and +he power transmitted. SCHEMATIC É GIVEN DATA: -SEHEMATIC 4 GIVEN DATA —s] | [— R=D/2 D=03m 300 RPM k z E- F, =2000N Lo eps (net tangential Sorce due A to but 'tension) E F. >F, ANAmSio: The torque is calculated using He tangoutial force amd te radius ot which Lt acts S=ER = (zoc0 N)J( 23 wo) = 300 Nm « dd Thus, with Fq 2.20 Me power transmitted is Wma * Í:0O Uma] OT fu bos ||t N-milid! ais = 42 kWe W H ! (00 Nem ( 300 Enf Fo) 2:37 ProgLEM 2.3€ EDU: Operating data ate provided for an elechit motor ar Stendy state. ED: Deternuna the elechit power required by the metor and tia pruor developed ba tha outpul Shuatá - Deternune E La net power mput teta motor. Also, eterna tha amounk of energy transtar by electricar «vork and by the Slug dorvns 24 of operam , SeHematic À GIVEN DATA : pes==0==. fOamp h uoy - 4 t AssuMPTONS |. As Shown inths Schomant., the motor is the cloged sygim. 2. Tha system is at Steady state, ANDALTSIS: (4) Using E 22! Wetectrae =- CVolinge) (current) al | warn fai KW = — (loves) amp) | TvolF TE» -l100 veria “9 ati 4 Wejeghic Using Egirto Wit = CGequea ngular peloch) La Lew = (oz 19:m pos Rev | enem] ="))| eu |) = LotkW 400 Wa (6) Nuk Welechic + Wokoft g = (ewjt (roteW) = -0.03kW q we (6) Tategrahm, + fina enersy transfy Awovats 2h , - w i Weleckic = É Wetecdt = Cottew)(2h) = -22 kwh q Welectric bo, Guopwo)teh)= anétieih + ——Wistass Wishga = [ Wiugrdt o 1.º The nuimus sign is needed because energy is dransttrced to doa Wa tor elechicathy 2. This vou mapusoenti +iu porhon of th electric pomer rapto Ahad «q vt obtocved 9 a shoga poser output because fe! effecia witiua tia mejor such sq eleciricas netishuro aut frich'on. 2-38 PROBLEM 2.4 KNOWN; À Wire suspended vertically is sivelched by an applied force, FIND: Determine (ay the work done, amd Cl) the Young's modulus. SCHEMATIC É GIVEN DATA: datas ASSUMPTIONS: (1) The wive is a closed sustem, (2) The moving boundary is the only Work made (3) The appled force vártes Linenriy with X. (4) Thechange “in area À is neglecttd. — x = 10 (unstvetched) — x, = 10.01 $5+ ANALYSIS !fThe applied force varies vita y : F varies from O to x ALCOrÁINo, 25800 Wf F = (500 É (x IO M) = 2,5x108 X- 2.5x10€ there x às in Hr ond E de im Ibf, Thus, apply ing Eq. aiz to evaluate the ore of Fha torce E Xz w=-frdx x; 10.01 4+ =- J (z.sx1002x-2,5x108) dx 0 Ft = [232408 (30,01 10*)— 2.5 x10º (10.01- 10 3] 212,5 Hibfa W (b) From Problem zm0, He Youngis modulus € ca be expressed ou Ko e = o (ão) where q =E/A is He normal strecs,aud X, is Fhe unshretched lewgth. Thus E( ko ) cs El K-Xo . (2500 bg) 40%) * Corn?) (o. =10) E . Young!s 2.5x10" Ibt luz ng a modulas ad PROBLEM 2.42 KNowN: À wire 0€ constaut cvoss- sectional area and à a gue initial length is stretched. The stress - straiu relation is leno EIND: Derive au expression for the work done onthe toire as a functum of Straiu. SCHEMATIC À (IVEM DATA: 5] c WWEM D Z x À-area g qn G&- normal slress 4 Xo————] £s “ão (strara) AssUMpTONS:(1) The Wire is à closed system. (23 Stress and straiu are related linearty . 43 The cross- “Sectonal? area remeins constant. ANMMSIS: The voor dome on the wire is gem by E. 2.18 We -J SA dx Xy From the given stress - straiu relation o) lohere CG is à constant (Young's modulus). From His expression e dx de = Ke > dx= x de Substituting info the work expression E £ w=- | (cejA (x de) = — cAx [ade o o Finally Wa - CAX E voork . z e Cap ressior A-HA PROBLEM 2. 43 know: A soap Bim ow a wire frame is siretched. EinD: Delermine the work. done. SCHEMATIC É GIVEN DATA: te -s ASSUMPTIONS.! (1) The Lilmis a closed T=2SxiO “Nom System. (Z) Tha moving boundary is desu The only Work mode. (3) The suctace tenston is constant actina on both = sides ot the Lim. 3 Ira ANMNSIS: The work Ás determined usina Rg. 2.14 w=- Staa = - [caga For constaut surface fension W=- TA Ax == (osn Pa (Sem) (ie tom | MT totem! mem a -5 D =-2,5 40º Tá vw I. Thenegatie sign denotes work done on the film. Note Ha small magnitude of the wovk reguired to stretch He film. PROBLEM Z.4U knows: A liquid Film ona we frume is stretched, FLND: Determine the york done. SCHEMANEO & GIVEN DATA: ASSUMPTiONS :(1) The Film is a T closed system. 23 The moving bowadary às He only sorte mode. Qezin. GB) The surface Fenton is constant, I é acting on bolha sides of the Film. — pax = ANANSIS, The vosrlo is determined using Eq uu e xa wW=-S 2a <- Jo tata For constaut surtace tension W=-TZRAXx .- -4 WE : ' VA =-(2.5x10 TE) ZCM tm) EM O =-C53u0P Hds q w I.The negative sign denotes work dene en jhe film. note He small magnitude of ha work required Jo streteh the £ilm. 243 PROBLEM 2.17 known: Energy transfer by conduction occuvs at Steady stute +urough a pane wall. dimensions, termal conductivity, and surface temperatures are specified. cinn: Determine the rate of energy transfer by conduchon, SCHEMATIC É GIVEN DATA: |. K=0.038 Blu/h.fºE TE 1ÓF ASSUMPTIONS:(1) The wall is af steady state .(2) The temperature varies linearly Lhrough the wall. (3) The thermal conductivity is uniform and heat transfer is by conduchom - ANAINSIS: Using Eg.2.3! together with assumption 2 Mn ar dr , To-N Qt qro ae Ta Thus decr Inserting values . 30-71 o Ás o (0.0318 pese too 6) — º) = =407 Bulhe Ou |. The values given ave for a typical insulated frame wall. alo PROGLEM 7.4 8 Know: À surfuce of Gwen diameter emits Hhurmal radiatum at a known vate ab a Specifizd temperature. ED: Determine the emissivity of the surface and plot Jhe rate of radiant emission vs. Su temperature, SLHEMATIC É GWEN DATA: . CI) The Steaua — GeaisW Boltamamm law applies.GlThe. iafasa” To * 000 emissiviky is constant. o 3 d=2cm =o.0ozm T= 567 wimik! ANAINSAS» Using Eq 2.33 & * EGAT, GE Setyring for & e. Be da SATA Ta? “4 xa) Us W) (sem x108 (=0.92m!oso KR” wk! 4 =0.842. « E Using the emissivity just calculatel, and varging 7; from o fo 2000k, the folowing plot can be constructed based em Eq. tw) above 250 + 200 1 / 150 -) Go tw) 100 4 so 1 0d == —— 7 0 400 800 1200 1600 2000 TLCk) Nokicu fa ropid growth in de with Th. 247 PROBLEM 2. 49% KnNdwn: A sphere of known surfaga area, temperature emiisivi emits mermél radiatia . > TEMP . aMef emissi ty emp, Determine the rate of thermal emision . SCHEMATIC É GAEN DATA: q Ra ASSUME TION: The Stefan - fl E=0b Te ormiquo? , Boltzmann law applies, N A=tag 6hu/h 8 0p To= |000ºF = (4O?R ANAVÓIS: Using Eq 2.23 : “4 Qe = ECAT, «(06 Hp my xiS* saia H(Idoo ce)! = Sb07 Btu/h de 2-48 PRobuem 2.51 (Cont'd) Thus »— IZb0- SO = —>—— = 97,19 btu q (9.535 46 bbT) 1719 Bralh Now, inserkha values in (4%) T = 1260 — (347119)(0.5357) = 208 R Tz L The form of (kk%) illustrates the analogy between heat conduction through a composite wall and electric curreut flovo-Hrough a series of resistances. The temperature difference iu He numerator is analogous to à veltage difference , and tin vuluo ap Ry ind: are “termos resislonces amalogous to -he electrical resistamees. Rs , SI PROBLEM 2. 52 Known: Energy transfer ocours from the inside air to tue nulside air Hhrough am musulated frame wall. EnD: Determine the stendy-state heat trausfey rate Hrrough he wall, ÔCHEMATC É GIVEN DATA: rnside air outside au Tj-70ºE To 2 10º F Acts uu/h Hier hos bBm/h HR disome non (1) The system is at Sendo To; Tio stute.(2) Newlows law of cooling applês dx em. tor heat from jue air to the voalt. (3) The temperature distribution is linear Hmrough +ue wall. (4) The tuermal kk: 0.038 Bhy/h. HR wall conductwity sf the wall is uniforra. ex ANMNSIS: For energy hanster between the inside air awi the wal Q 3 h,A CT; = Tui) <m For conduettow Hurough rue voall : dr Twi "Two (2) d=KA de = KA (Tie) ' Fially, for energy transfer belweeu the wall aud the outside air Bo =h A (Tue-Te) CG) Now, from (1) — O A +) T 8: (qa) : Is And, From (3) . , Toe * To + ãc( a) (5) Com bin g (2), (4), and (5), and nokiug Hat A: 0, QE A á a A (To = TO) L A O (x + ze + ES) Thsertig values Ci6o Hi) (530 -450)ºR , Q = Í 4 ER + 1 ) E E Bio (= Pa E 0318 rãs “e b PICTTA Gto)tBo) = CAES) = 772.7. Blu/ha à l The quant in te denominader is called Fe overall reciriance, or “Rvalue ! d-52 PROBLEM 7.53 Known: Á hot surfuce ie coveved uitA musulation, Enevgy Jrauster occurs from nuter surfuce of the msulatiza and The surround air. END: Determine the minimum tinickmess of insulatiom to mata the outer surfuce temperature below & dest red value at steady stude., CAE A TA Tair = 30º f=/0W/mix kK=0.08 W/m-K insulatiom qurt surface ASSUMPTIONS: (4) The system is at ate state. (2) Newton's law cf ecoling Gpplies for heaf hunster from He mulato to the air. (3) The temperature distributiom Through the iu sulatiow às linear. (4) Tre thermal conduchvity of the insulatiom is UA tfora, ANAMSIS: For emergu Irançter bu conduciim Harousk the nda AS ME CMruis TO) “9 Quz-KA dx * KA E Lu =300º€c Further, tor energy transfer from the insulatiow to He air ão ta. A (To Tai) A! sieady ctute, Que As ,5S KA tia TO E BACGTao Solving for L. K CTouee — To) PAO Tao) Wit &,=0.08 w/m-x , À 210 w/mix,T «300% avá'T. 230% the plor below can be Lo Sur , air e 108 Ê Luva qm A 1 -bb q oY .ez 2-53 PROBLEM 2.576 nown: A System of Krewn mass Undergoes o proces fer toluceh the hect transfer and work art Exowr, Tha inihol speufc internet enem is also kreun, EroD: Determine tha Anel speupc internal energy schemaTic £ GIVEN DATA: scbemáric d GIVEN DATA O Q=-|000:J W =-200 87 u,= 300 IS) ASSoMPTIONT: |. A closed Syçhum of km mass is under convderatim, 2. Changes in Einehr and potenhrs enersy com ba heslectd. 3. The inibiad and final pato are equiibrivm States, ANALYSIS! The el wi internet €ners cam ta determined fra me enera balanee vimg assurephm sKE + SPÉ + su = Q-w » o 4AU=Q-W Then, ida asomphm 3, AUS miW-U,), So - tus) OW o Wi = + [DM we = 3005 + (-1000) =(-200)] KT 20 leg - es ato eg uz + According to tha Sign Comvenhons for Q and WI, Qu negahua hem Amore is à net heat transfr of energy from pha tyrttm and WU & wegaive Uhen Iuire Ta net work hrans for efeneras to Are syelem. PROBLEM 2.57 knowa: Five kg of sitam undergo an expansion in a prstun -eqndom astemb from state d do stata 2. During tha proces dhere is a Enoum hear transpor +o tha Siam and à Enmm cvork tranca of en erjy to Jo sicam by e paddte trheel. Tio Chomas ui speópre interna tiraram qr team 6s alto |Emown. gb: Determune tha amount 4 energy Pranstm by work from ta. segu to Fa puston during tia prcss, senemaTiC GIVEN DATA! o=+s0H À my = 2709.9 k]g o = 2689.6 KJ/kg ASSUMPT(UNS: | The Steam is Hu closed system. 2. There ds mo change. wi fue Einehe or potenhos enerau Leoa state do to sht 2. AMALTSIS: The net vort can le determined Som an energy balanes. That aj eta &SsUmphonz GRE + pré +ou = GW ara Q- at The net User 1 tha tum of fa Work assocalia ifia fina peddlenee! Why Gud + Uoork done mta pistom Weixton: INT = Wow + Woidon From His aver informa tim Wu = -18SES , here tia penta is requirts O becanta (ho quddia what fransfers energy to tis iqstim. Gllectrng Assu da US + Upiston .Q - ot Wprtr = Q— SU— Wa = Go lt) Wu - 8043- 58 (cost -z70 DEI - (188) =+3SD lc 3 a . Was here +ha posibve sign indicates tur euersy transfevred Agra Amar Piston sperm do Ho pistom ao tha sim Supands ariny Tia proces, 2-51 PROBLEM 2. 58 kNDwaL: À system of known mass undergoes huo processes iu series. FIND: Determine the work and hect transfer for the second process. S T : Process Lx: U,=U, = H434 4P/IL pa100 Ibflin? U,= 105,8 6lu/Ib, Qua, =— 38/.36 6h Process 2-3; p =p, =bOlbfhn? va 7.828Yb Uç= 214 Btu/lb dásUM TIDAS: (1) Aclosed sústem is under consideraton. (2) kúetc vd potential energy effects can be neglected. (3) There is no work for process |-2.(4) “he pressure is contam d duvida process 2-3. ANAINSLS Bg assumphon (4), he work for process 2-3 cam be found Using Egd! Vz Wi = SJ, páV = BG) = pemlvçar) =(bo RX 2 tb.0a 4.434) 8 E mese o = 75,21 Me Was Use He energy balance to find Q;s AU,, * Qs -W3 With AU zmau Quis E MlUgUz) +Ws Ca) To get us, use the energy balance for process |-2 o MiUs-U,) = Garbo u, = Bia +U = 581.36 Blu) , 105.8 Bhullh m Po Ch) = 815.12 Ghu/lb Thus, from (%) 23 *C)lM2ig- 8ISUZ) +(75,21) = 687.8 Blu Gas a-58 PROBLEM 2.6] known: A closed system undergoes à process with a lenown hear transfer rate, and Hue power varies as a specified functim of time. END: Determine cas the vate of change of system energy ot t «0.bh and Cb) the change mm system energy after 2h. ScmEMATIC É GIVEN DIA: Qa-tobw Gem) . (-Bt oct<Ih 5 pé W [a trih == “o where tisim hand Wisin kw ASSUMPTION; The syskem is closed. AnALNSIS: (A) The Hime rate of change of energy at any Hime tis gen by dE = à O) M t=0bh de = [à - (-84) a(-10 kw) (- 8-06) kw t=0.6h €-0.eh o a-52 LW4 dEldr (b) The change in system energy is oblauned by integrating (4) over Ha hme period 0f Ah. thatis t=2h AE = J (a-wde > CO O tem €=2h = Qu E (-8t) dt + J (-8) «e] tºo t=th =(-10)2) - [co 4] Vo [ese] z -20- [C4)] -[0-8)C2-0] 2-8 kwW.h Thus t, Att=o.bh, the energy of Me system is decreasing at a rat of 5.2kW because the rate ef endrgg prons cut by heat êxceeds He vate of enemy transfer tu by work. 2. The negatwe siqu denojes a net decrease. of energy over the fume period. d-bl PROBLEM 2.62 gv A storage battery develops à known power output as a funchen “me. END: Plot he power output and the change 1h energy of the battery, each ada fun dh, of me. Í SEHEMATIC É GIVEN DATA: === IT Es Wa=12 exp Ético) Cohere Wisin kw and Eis Hime ins) Q.=0 AsSSUMPTIDANS: (1) À closed system is under consideration (2) There is no heat transfer. ANadsa: The hume rate of change of system energy is o E A. E. És ow E =w dE = —12 -t/60 TÉ I emp( / ) Integrating from +=0 to deny me E € AEs - / iIxexpl-tiogdt = E Lexplt/to) =] om = m2lexpl-t/to) -1] ud () Usihg software to plot Wand dE vs. t rangivg from Oto 2505: 15 0 1 «0 - 1 -20 À 41 1 1 a -30 À Ê 2º 40d E 05 1 4 -50 À 0 - “70 À fa 0 rem -B0 T T T T O 50 100 150 200 250 O 50 100 150 200 250 t(s) t(s) (b) Inthe limit as t +00, we have Nx O and dE + -IZkT. The plols show nat Hrese limits gre approached closely attey alsout 2255. For +mes greater than 4Mis, no more eleclvric power is developed by Pe lbatrerq amil He energy remains constant. &-bl PROBLEM 2.63% ENOWN: A gas of lenown mass expands iu a piston- cylinder from a specified initial stute to « lenown final pressure. The pressure -volume for the proces is given. Also, the apecific internal energy change is leu un, EIND: Determine the heat transfer for the process, SCHEMATIC À GAVEN DATA: m= 025 kg p= 8 bar, V= 002 m? fre 2 bar pvtãa constaut au: -SS LI/kg ASSUMPTIDN:C1) The gas isa closed sgstem. (2) kinetic and potential energy eftects are negligible. (3) The process is pelghrapic wity pv constant ANAysIS: The heat transfer cam be found Using au energy balance. First, fino Je worle using Eg. AT, as follows: Va Ve =Todv = Coust w Jp Í, mst: Ay Integrating and simphf yr . PeVa = BY w= CI=[2) The volume at Yhe final state is Vi a(P/p) = 0.0635 m?. Thus = 4Qban)(B0bIswl) — (2)(0.025 toSN/m 1kT W= G- 42) | ibar Hiofn-m Ea ha V = los bT Noto, +o get te heut transfer, begin vita the energy bulunce NEGA AU = Q-W Solving for Q and nofrisg that AU = mau Q=mau +W =(oaskgK-sS kT/k9) + C lS kT) D = 295 tT« q |. The heat transfer is posttive, denoting energy transfer by heat to ue gas as it expands. a-63
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