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516 Solution Air lows in varying cross section pipe. The apesd ata specie sectoa iso be detecine. Assumptions los through te ine i sad Analysis Chapter § Mass, Bernoulli, and Energy Equations S17 Solution Airis expanded aad is accelerated as itis heated by a ais dryer of constant diameter. The percent increase inthe velocity of air s it flows through the drier isto be determined. Assumptions Flow through the nozzle a steady Properties The density of aris given tobe 1.20 kg a thet, and 1.05 kn? al the ext Analysis There isonly one inet andone exit, and thus nn = it, = mt. Then, AV. = AY = h war of py Ye _120kgim? Fp, LOS kgin’ =1.14 (or, an increase of 14%) ‘Therefore, the air velocity increases 14% as it flows through the hair drier. nosy voo(3) 30-24 mis Dincevsion —Snkes was athe velo acres sas ted ty decreases, but the unss flow sate is constant Chapter 5 Masy, Bernoull und Boergy Equations Solution Wind is Mowing stay ato ceria velocity. The machanicel _pptertion pote aed the etal lel ower pemeration aoe teins ny of ac pee unc mas, te per Assumptions 1 The wins Bowing Indepondon oF he wind peed. Properties The dansiy of ii given bo o— 128 bp ily al» eomstan arin velty. 2 The licen of the win rine is Anatysit tac ene isthe only fom of emechanicalenengy the wind possess, att canbe converted to wurk ceil", ho over ele of he wind is rai energy, ees #2 per mtr aul i" fr sven tas Hoo ie: se ) cones os me . Jas pra pr = p25 ky ys 7 Avvmpions 1 Yolen isa ae loan vf he a ule of he st. 2 he valeciy ge ie age Wg Boy ~~ OaSSugy0Ose ENG) 620K ‘eh 9 The mech enero eset hein oni et i ropes Weta dos of etl w b> 0-100 a ‘hela eos oe sweeps ass see “The acwal cece power genertion i determined by mulling che power generation portal by the efi Fash nerve “(ONES EW| = 188K ‘Thoofne 283 KW of actual posit canbe gonsrated by his wind hin He lated eoaitons. cee _ Wiod vata 0m) + A ‘|e wind thine Tome sme som pons goon pti oh ss ats st by ep ee — si = (lag 350-0) = OO —> Jt Fas ~~ EUNOD SNCS Kg) 470 = BTN Piscussion Netto ie coy of watts ole compe we esa cg alien be igaeed a Discusion the poser pereration of wind tthe is pretense eaho othe wind voy, and ths the power Uiesoas Also, tc poweeont of sot tas wl tS Goat of as oa css ‘generation wil change swooely with the wind conditions| Caapee $e ern and Ber Butions Solution We cme fom alike tna sage nara pee. Th eal fie of he prmmetae Sand epee tones ewan bra esl th pap ts ened ‘echacges one saegy ace aspighl 4 Ibe slwing tac esse uo euch Proprio he dey ms be = 00 Anais) Weta he free fice of eke ae poi | ae ee ses. ‘als hes ae sic he tence eel =~ Oca us ypc casa pls ai? ae pe Ua ‘esse The Hem eengy a pins sina band? at ope athe sre = P= Pa Pasa ‘Than cna ath pots invere Cee heya) sash wae he nsane sly satay T sf cel atl a aot sau a ual te eV — obo? 0.0717 power ste ytiny WAS Larner Loa, Wines 01 — py — FNRI =| Mbnnas~BlC oa aaa penis — 2AM yeu 06 60.6% Chapter 5 Mass, Berea an Energy Equations Boson PAE sis (Oppo we sot te pug Th cage ia emma cy Solution Ais selena, Hema harmo nef he ore oe deena “oben rug arco cag nt rruptions ow teoagh te noses any . culo econ ang pied Ue pa, ks Propres The deny of at gion 26 2.21 kp ath in and 0.762 kp athe ot Dew - Analyt (0) Themis ow ee fa deter fen ert condos be a "Manet etic = BA, ~(221 nD 209) DSR O26 gla Seti aaa tia Fi 20ve alk Memon 5 4K (nm) b- som ape J-sr7 Kea fzmen “SS e-ss0me Paani’ | (2) Theres only one inet ad ne ei nd has y= y=, — ‘Thea the exit zea othe nozzle isdeteraned to be ler in sleet by 8 2 kas : 2 Pisassion Row dat aly toads of uo eeetis eacey caused by the punprnetec i cowed w ta wa padily — dy = 02682 gap 2? -25.2 em ‘nahn eno aero eing oe ln axes snes he pongo he ml. babs (062 egies" KISOS Discussion Since isis a comptssite flow, we rst equate ass law rte, volume ow ae, Chapter Sass, Remon, and Energy Equations 30 42 Solution The vslecigy of an sire i to be measured hy 2 Pie-mase probe, For a giver diferent presse Solution ua power plant, water eutes the uozales of bydkaulic bine ate specizied presse. The maximum ‘ling, the vslosiy of He aterat s hhe detrrin sic pte velocity wte can be accelerated toy the ravztes ven he denermine Ascumprions The flow of water i steady, ncompressbl ‘emoull equation is applicable). 2 Water enters the nozzle with alow velocity. Properties Wetake the density of water to be — L000 ky We take points Vand 2 atte inlet anc exit athe nore, respectively, Noting tha: ~ andy anlieation ofthe Berl cquat.on beeen pints | and? gives ose Sastinsng he 3 fmt exit locity is determine a , —_ (am ona wen im? Vg mn) s0vte sone 2 y, [200 Aor WoO YT) vy ae ind rotons with negligible tietional effects (50 that the Diccussian This isthe macimnm nozale exit velocity, and the acm ‘velocity will be les because of friction between water and the walls ofthe nozzle. Chapter 5 Mass, Reuull and Brergy Eguatons Solution _& Pitotsatic pote ie inordinate dou af an ar suing nt pale to flow an the fers fof he wacrtokim seminal he fo eclosian eoane ue ope Pcs pee he Feats Assumptions 4 Tae low tough te duct is sendy. snomprcsible bd erocaooa! wih neglgble Sitional effects (so thatthe ero equation iseppicale) 2 Air isan oes as Property —Wetahs de density of water tae 1000 Byn?, The gus constant oat R= 0287 KPa. Amatp’s We eke pint |e he se one probe whee ths entrance yall Now seme! he tac {be Pius probe nl pool 2 whe ip uf he probe bere te eazabe somal w fos az» cumneed (oe ‘Syne arn othe Pestatic probes Noting tha pin isa agai point an hus F>—O ands ~ 2p. opiction [hte Rem eatin betes pts Td 2 ve fw shes the prose ie tthe tp athe Pitot ai pi 200k 0.8 024 Pepa = ° ite a 258 Nin” -296 Pa e ken 2 Lrg AD (O28TAPE mig Kas . A, ae Subang fesse Yorgi Ma 2 mis IN ‘iscussion Not tat he flow velcity i pipe a dt cam be reseed ey by osc poe hy sing ho probe ist he pips or duct eral! fo lov, aad reading the difreatalpresue height. Also sore hu tis isthe Veoh et 1 Toctioa ote ube, Sever readngs a ova leetons a a eraseaceio nny be rood re dst sme fw velo, (Chaptor 5 Mass, Hemoul, and Energy Equations S46 Solution The sate and stagnate pressures in &hocizntal pipe ane eased The velit at he center ofthe pipe ‘stohe deerminad zumotions Th Dow i ey, ise sad totaal with ag unl es inthe st date between Ine tie pressure mea lions 0 ut ne Hern ua s applicable Analyse Wo take points | and 2 along the eentotine ofthe pipe, wi poi vies under he piszonetar and ‘nt? atthe cnvanse ofthe Pitot tatie probs hs stagnation point. “This a steal flaw with steht snd paral] steaines, aad thus the stalic pressure al any puul is equal 0 Ine ydrossic prsure at thal pono pu san pi adh =O a = .cupphtatve ofthe Beraculi equstion between puns und? gives 266 ce a Water Pee ae te me Subecuting the Pend Ps expressions give WE eth 2) aon RY 8a te) a e = ah Solving for 7 and subsiutng, M1 Palka Tem) ROR IOSS 02 m7 1.39m6 Biscussion Note tbe Pitta probe. of the excose hid column in tat ve dete th flow velocity, alle need isto measure the is Astampelons 1 The aie Now over th ster steady, incompressible, __V ~~ ‘aad eotaivnal wah negligible irictonal eLiels (oo tbal tie Bernoulli cquaion f epphteabe). 2 Standard tuwspheric vidi exist 3 / ‘ind effets renege Sagat! Properties The density ofthe aumospher tau elevation of 3000 mis p UD in Anaiysis_—_Wotaks point atthe canoe ofthe tbe whose opaning is anal Mow, and pots 2 af The entrance of He tae whose ena i ‘Tosate presse mete: «——$ formal to los, Nong tas point 8x stagnation poinc ard thas =U and 17 he application of Ge Beroult sustion between pes Vand 2 To stagnation presse meter Solving for J un substiutng faa) DY AM kan [azn an) since 17 im? and 1 ms: Sékm, Discwssion uve tka the volovity uf wa wren com be dolemained by szaply mnsuring te eileen pressure em Potato probe (Chupter 5 Alas, Bemoull and Energy Equations 45, @ 5 Solution The drikieg varer noo af an oc are met hy lange water bales with a plastic hose incre ei, The ‘minimum illng cme of an eon an is obo determina when the boss fll nd whan it near empty Assumptions | The flow 3 toad, incompeessible, end notational wid negligible etonal ‘gustan is eppliable). 2 losses are neplecte to obtain the minim ine tne Analysis Tees (a ha he Bern We take point 1 tobe atshe tee surface af water in che Dole and poi 2 atthe exit ofthe tube so that) = Fe Hog ibe Soll is open lo he almespbere und yates discharges ny te aumosphere, Py” (ae bu to heruhe dialer) sng ~O (we take pint? he refeyene evel), The the Brnonl eqation sy Subeusng, the iacharge veloc ot water and te filing ime are detenined ae fll (@) Pu bore, ~ 3.5 Is. yaa sarcio vv ones? a : 188 Vay Gai EER fy Sr 1a v oa AP, GALSLO" WML 3B) Disexssion The siphoning time ie datermined sssuning itaeless lo, at Ags his sheen seed. eit, the tne ail be Tanger besa of ‘ction seoreen water and te tube suface Sottion Asoo pane wi fas ci lone hay copy: acres tak hrough aa ovise, The eight ho water wi seine ake eqiia is 0 be Asmpiims 1 the flow is seals. soompi, abd iotsonl oh negigise Gta. ees tha the Beto agua i failichey 2. Both the unk ante texto ae ape Me fshzoephese, 3 The nner Deve of te servo ein aoe, Amass Steere 2 DnB Bee Bee the ‘Nong that th lamers ofthe ibe ae once ae hese, ew tesa we ro ad ut of eek be te te when te water soces inte tbe a the ae sre equal ace Seung he wo velo equal teach ther gee > Rete ‘eevee, se note evel i tne wl sale whe he we ‘Thisral ‘ohare suring nex sian, The aul wel bomen dine fe sein ‘spipwandoifec wateoowrideed, Chapter 5 Sass, Hermon, and Linergy Lquations 525 Solution A hydraulic tubine-generator is generating electicity from the water of « large reservoir. The combined tuubine genorsorallcieacy aud the bine elficiency suv tobe determined. Assumptions 1 The elevation of the reservoir remains constant. 2 The mechanical energy of water atthe wrhi eligible Analysis We take the free surface of the eservir to be point 1 and the turbine ex: to be point 2. We also take the turbine ext asthe referenes level (= and thas the potential energy at points 1 and 2 are pe, = 27. mad pe: =0. The flow energy Pip at hoth points is zero since bath 1 and 2 are open tthe atmosghere (Py p,.). Further, the kinetic energy 1 both points i zeco (key ~ ke, ~ 0) since The water at point | is essentially motionless, and the kinetic energy of water at turbine exi is assured tobe negligible. The potential enerey af water at pont 1 is mies atmo) ME) cama thea the rate at which the tothe turbine become hanicel energy of the Fuid is supplied [Fncnene|=*MCusrn Caeereue)= tale, Y= nape (900 kgN.079 eg) on2kw The combined wurbine-yenetator and te turbine efviency ate determined fos their definitions, Geasrator Ws Ta Laren Taig —ettet— SOORW 9.504 on 82.4% [ME gmucna| STLEEW “Therefor, the reservoir sopplies 971.2 EW of mechanical encrgy 10 the thine, which eamverrs ROM KAW of it shart wee Uhl drives the genenaio, which yenerascs 750 KW of electric ower. Discussion This seoblem can also be solved by taking point 10 be tthe turbine inet, and using Dow energy instead of ‘pocential enemy. It would give the sume result since ie flow enengy ot the turbine inlet is equal 10 the poteaia! energy at the Tree surface of the reservar Chapter 5 Mass, Bermoull, and Energy Equations pr Solution “the water in an shove the ground swing pool sta be epi by srpluggighe otfiee ofa hori pipe lached te ube Bot: uf te pool The maximus charge cl ib be determine le 2. The flow is stay, Tighe ition effets fo thar He Bernall equacon ie applic} ‘Anramptions 1 The crifce has 4 smooth enltnce, and all Fictional Iottes are ten roomier and rrosionl with -Anaysis We aga 1a he es srs ofthe oo. sod pont 2a the ei a pipe, Wo ake th eee i pp est 0), Noting hath dss pois pon ato art has Py Ps Pg) al a {Ind velocity al the tee sutce i very low (=), the Bernall equation between these "vo pent ump iss 0 The asian discharge tale wocas wo tho wate gh in the pow! fs « maior, wel is he ease a he Begin ig and sus ~ 1, Suhstinting, the masini fw velocity and discharge rate heen 542s Discussion The rst shove is obtain by dstcgardng al ictienneffers. The acral ow rte wl bo Ls boowuse of eden fects dung ov: aad ie sul presse op, Aso, tbe rte wll wradually decrease othe wet ‘ove inthe pip decrsss. Chapter 5 Mass, Bernoulli, and Energy Equations 554 Solution Ths water in an above roe ground swimming pool is tobe emptied by unphigging te orifice of horizontal pipe ettache tothe bottom ofthe pool, The ime it wil take to emply the tn is a e determine Assumptions: 1 The orifice has w smooth entcance, and all ‘ictioral losses are negligible. 2 The flow is steady, ‘ncompressisle, and trotational with negligible Fritiunal ees (a0 thal fe emoull equation is applicable). Analysis We lake pint | al the free surface of water inthe pool, and point 2 a the exif pipe. We take the reference level at tho pipe ext (2, = 0. Noting thatthe fui at both pints in open ta the atmosphere (and tas P, a) 284 that the ud velocity at the Gee surtiveis very low (= 0), the Bernoulli ecuation between these (wo posts simplifies to salty, we expres the water height in che pal at anytime r hy s, an che discharge velocity hy V, = Dg Note that water su'ace im the poo! moves dawn as the pool drains, and thus 218 @varible whose value changes frm a the beginning te 8 when the pools emptied completely. ‘We donots the diamecer of the orifice by D, and the diameter ofthe poo! by D,, The flow rae of water from the poo is ‘obtained by mulilying the discharge velocity bythe orifice cross-sectional aca, DE V = Apis = 28 ‘Then the amount of water thal lows drough the orifice during a diferenial time intervel dei av Vie Paget ) which, frem conservation of mass, ust be equal to the de ease in the volume of water inte pool AV = Aga te) hae ° ? we the change in he wae level in he pool dng, Note tdi egatvequsiy sinc he posve Wesson ofc ute, Town ean acto gets pesos Guay for mca ot ater dacargad Stig op i)unl@) pu aca sa scan 2 i, oi, feat Pe ae rr) D? Yee The last relation canbe intarated eas: fiom != Owhen =~ hts = fwhen2 Dy rine the variabloe are separated, Lating be the discharge time and integrating it 0 (completely dnuined suo gives Substituting, the drsining time ofthe poo! willbe Bm [29 scons at8.4h ‘oosmy Vaxt rv ae Swimming Foo! Discussion ‘This is the minimurn discharging time ssnce itis obtained by neglecting all tieion; the actual discharging time wil 3¢ longer. Note that the discharging time is inversely proportional to the square o° the otifice diameter. Therefore, the ‘ischarging time can be reduced to one-fourth By doubling the diameter of he rice. (Chapter ass, arnold Energy Equations 56 Solution Air Sows upwanl at specified ste tough an inclined pipe whose diumstrisreduced trough «reduc. ‘The diferent height etncen Guid teres of the bo me oa wate manometer atached acnns the redueris k Be ees Acumptions | The Now throu the duct ie stead; incompressible and iottional with nopigible ition fects (50 na Be Bernoulli cgi is applicable). 2 Ait isan ls! gas. 3 Tho eles of st coluran on the gresare change is opie because fis ly deny 4 Tho ar Howe peel to earenes of cae arm af the randmetr, and tus £9 dynam efets acs inva Properties Weahe the demsty of water tobe p= 1000 kg’, Te gas constant of i's = 287 KPa ne K. Inaysis__We te points | and 2 a the lowes and upper connection point, respesive'y, of the evo arms of the ot (ong that th effet of sevstion oe the saree Paw P oskre RT (0287 APa oy KYGT= 279K) 1180kgin! 22.991 Substinig PP iL. 18D kgm’) (17m)? -(2299 mi { 1267 in? 1267 Ps 2 ig-nw ‘The cibrental height of water inthe manometer eoesgonding to this pressure change i detemined Tom AP = Pay gh tobe 1267 Ni (1hg-m" ) Tom kp yo.8Im Pome Diseussion When he effet uf a coluran or peessre change msider te presi chug bees ae ow PO FN gai, a) SIP mie) 12 99my* is 1 asp and 1269 Pa) less San 1%. Thaefore, the effet of ar slum. on sresrure "hor words, the prosure eterge o° BF the cuet i ablrstembrey das 0 iy change, al the effet of elevation change is alge. Aly ie were 0 account for ths woud be mre press aeoumt fr tse Ac uF rin ke ream ety ting Pg joe can whe clog ‘Te add nal air column ine manus lends eee ou the cage in ema on dTereace i ibe Now inthis eae Chapter § Mass, Bernoulli, and Energy Equations 5-58 Solution Water is siphoned fom a reservoir. ‘The minimum flow rate that can be achieved without cavitation ‘occurring in the piping system and the maximum elevation of the highest point of the piping system to avoid cavitation are to be determined, Assumptions 1 The flow through the pipe is steady, incompressible and irrotational with negligible frictional effects (50. that the Bernoulli equation is applicable) Properties We lake the density of water to be = 1000 ky, Analysis «the velocity V, cannot he 2 5 ews. Applying Rernoulli Tg, #tom (1) 10(2) 2308.10" | 13 * 9810 2g Vere 21309 For T= 20°C , Py 2.338 x 10! Pa (abs) “ . n d= 100m, D=I6em. an 161s. Therefore th veloc wll ever exon 16 ms. Accondagys 0 nee nado oe 5 Ve datz 690.128 he hepa Bn ite nossa re 7a, the abso Prog = 208 Pe y lg le 9) ora maxima 7th aon pes Puy = 29 Pala) sane constant -tberefoxe (On the other hand, trom the continuity, and} goust'be minimum. %® Pig 2338 yao Vy =2.56¥y =2.56x9904=25.35 mis We should check if these velocities would be possible, Bernoulli Eq. from (1) to (2) yields plying Bemonlli Ng. from (1) t0 (3) 101325 y 15.0 Zp nye 10.2381 2 9810 . 2 Bn gy atl Pom ye Yr 2g or frst past, 2 101325, 9810 14.461, Zou 13m 560 Solution Water discharges to the atmosphere from the orifice at the bottom of a pressurized tank, Assuming Irictionless Now, the discharge rate of water from the lank is to be determined Assumptions 1 ‘the orifice bas a smooth entrance, and thus the Grictional losses are negligible, 2 The flow is steady, incompressible, and irtotational with negligible frictional ellects ($0 that the Bernoulli equation is applicable). Properties We take the density of water to be 1000 ky’? Analysis We take point | at the fice surface of the tank, and point 2 at the exit of orifice, which is also taken to be the reference level (2: = 0). Noting that the fluid velocity at the free surface is very law (¥; = 0) and water discharges into the atmosphere (and. thus 2:= Pay), the Bernoulli equation simplifies to IP 2g ae 2g Pe +3 Solving for ¥, and substituting, the discharge velocity is determined to mR PEP) age, = [PC Ve Y 1000 gm’ 0-100) kPa [1000 Nim? | 1kg.-mis? LkPa IN Je 201m 25m 02 mi ‘Then the initial rate of discharge of water becomes V ~Agitee¥, ie 0. voy (18.7 mis) — 0.147 mP/s Discussion Note that this is the maximum flow rate since the [rictional eflecls are ignored. Also, the velocity and the flow rate will decrease as the water level in the tank decreases, SSE ‘Solution Water flows through a hosizoneal pipe that eonsiss of to ssorons ata specfiod rate, The differential ‘eight of « marcary manometer placed between the tro pipe setions ita he detersine Assumptions Uke flow thcough the pipe is steady, incompressible, aad ivoetinal with negligcble rctonalelfet (90 ‘hat he Bemoul i squatan ig apatieabte. 2 The Teset inthe reducing section are negligible Properties 847 Ibm and, = 62.4 bo Anaiysis We tee points | tnd 2 long the ceatcline of the pipe over le two lubes of the manometer, Nut ‘he demstes of mencury snd weler ue 2, = ke emul yuisonbeveen pos 1 td 2 awk mee oe ‘We let the diferenial height ofthe mercury manometer be hand the ditarce between the centre and the mercury level inthe tbe where mercury raised b>. Thea the pressure dfererce P;— Py ean lec ke expmessed as Ae Ve VE VE i-ve PAE on ri yoh na Pe! DE Palen — Pad Phen 2. —Y Satan ae 2|© —OE 220i" wear.62.4—1) ‘Tretofore, the differential eight ofthe mercury eaturnn will Be. Pisearsion Yn ealily theres ction Toses inthe pe, and Ss pressure at Tire 2 laa he sale a That estimated here, aad Career willbe lara Usa tha calculated bere oF Solution lurbine-generelrellicieney ave given. The flow rate required ist be delenmined, Assumptions 1 The low is steady and incompressible, ‘water levels at Ge renorvoir and che discharge sie remain causal — Prupertn We she te deisity 0° water ww be p= OA a Ube = we Anais We take point 1 at the fee sunce of be — M0 Teserenir ané poinr ar ie Fee surface nf the dschenge warce stream, whic isa 30 taken asthe reference level @) 0}. Also, puta Land 2 are open to the almosphere (P= Ps Pa), the velocities are regligib e at bath poins (V, = Vs = 0). Then the energy equution ix terms of heads for sleady incompressible ‘owe fwngh cone wolnras herween these rn pices that includes Ue turbine unl dk pines recuces to Ave Fira Hessen, pee gsm “Bs Pin Sebettunag and noting “hat Hrs cn ~oetn plih tes of water ae detained o be Ingen =~ hy = AOD 36 = 364 Bint 0 ‘oso nts? Voaere ase me ook 8 OR asp Termesethanne 085022 02)(364 Hh atom JTW &_28Tms 2 Alf ‘hetefor, the flowrate of water must beat eest 3.82 fs to generate the desired electric power while evercaming fiction losses in pipes. anon Discussion Note that he effet of trietonal losses in the pipes isto neraase the required low rare of water ta generat ses ied aunt oP electric power. 530 ‘Solution Water is purnped from a large tmochanicaleificeney nthe puny 5 bd cto a higher rservot. The head loss of he piping 53 Assumptions (The flow is steady and sncompressile, 2 The ‘elevation difference beweer the lake aad te reservoir i coms, We lake the density of water tbe 7 1000 kei Analysis We unas puis | and a th Gas safes of ts Take and he sservois, respectively, and take the surface of te lake as the esterence level (=; ~ 0) Hoth peints are npn to the atmosphere (, 1) and the yelacitis a bah locators ar> negtigihe (Y = 0}, Then the energy equation Tor sleady incompressibe ow trough 4 conte volume berwezn ‘hese tro poims that insludes the pump and Ihe pipes reduces Properties - fon bey ser se Bantam 2%, since, i he bence of turbine, Las tne ~Lascion umm “Emectecsinae #4 Woeapa Fre ~Leachtneum Noting bat Z, mnie use pump power is Weg ties rah ges Hh) a 7264 mvs 7361 ‘Then the mechanical efficiency ofthe pur hecowes ¥, 7368 png — =0736-73.6% Fug 10W Discusion A wore proctval measure of performance 0! the pump is the overall efficiency, which can be obtained by ‘multipvine the pump efficiency by the metar efficiency. Ima hydroelertc powerplant, the elevation diference, the heed lot, ths power goneraton, avd the overall sonaprer snes, nernonn anu ergy quan 566 Solution A water pipe bursts as a result pressure of woter in the ipe is tobe detersined, ‘reering, né water shoots up into the air a certain height. The gage Assumptions 1 The flow is slealy, incompressible, und imolatoaal with negligible SctionalelTets (ou th the Bernoulli ‘oqution is applicable). 2 The water pressure inthe pipe af he burst section is equal (e tas waler main prossu. 3 Friction belieer ‘he ovaler and air negligible. 4 The Freversihilies that may acura the burst section of the pine cue io abrupt expansion are negligible, fied poteatil energies to each other sataoul invelving any pumps, tures and wascfal components with are fetonal loses and thas a0 Ais auluble for ike use of the Beco equton. The water helght ‘ll be maximum under the slated asarpions. The velocity inside the fan) tone relatively low (F 20) and we ke the burt secon of he pipe Gs he eleenc level 21 ~0), Atte po the war Cajestry tnd ampheris prewure pecaims "Then the Mera euaton cua Pipe plies Ak ae 28 Solving for Pu: unl sbettuting ea Lain nw 1000 Kg mi Figs 088: nom oan a2m{ anzkPa tm erefore, dhe pressure in the min must be ul east 412 KPa above the stmospherie pressure, Discussion ‘The result cbiined by the Bemouli equation represents a Lm, since Lrictivnal losses are neglected, nd sheuld be interpreted accordingly. Ic cells us that te water pressure (gage) eaanot possibly be less than 334 KPa giving us & lower limit), and in al likelihood, tke pcessare will be mich higher. Chapter & Mass, Rernoull, and Energy Equant Solution A pump is punping ei ca spoefisd mm. Ths pres 0 inthe pump is meas, an tae mote: lien sine. Tv oneatl ich uf pn oe ern Arumpliney Uh fn sey wd ioomessibe 2 The eleva difeconce ans ip hg 9A te our inthe pnp are accounted fry the pum eMeney an huss = Dod The Kirti anergy comet los are the extracted turbine head and the mass and volume flow Properties Ths Gas of ois gives ote 8D gn ‘equation forthe pum reduces to Fi gay M ae chgne = 10 Esty ane be not ws hee aww © oy "a OVA we A \ - \ om Ye Io) ‘Subsimrin, the useful pump head and te onmrpondng nef pumping porter SS dearbcé bie & 256,000 Nit A884 my? — 09.49" suas (860 cain 9.8 on) ‘UVST ins?) Fe = Mies = 8a Wms Hs WIR) HN nme wee 1 1000kg- mist ATEN as , “Thea te so paying pores a he chm cine of fe pp becone Wrepatse mene near ~ WAIKES KW) 22.5 WW oan = fm 05 gr arty a ‘Tee ova cine of is zampincar a te oda of he mca ad marci, 0 which 09 0.471 Chapter 5 Mass, Hernouli and Energy Equations 82 Solution A pmmp with a spasifid shaft power and efFciency is nso to raise water 30 higher elevation “The maximum florets of water iso be determined. Assumptions ‘We assume the Mow in the pipes lo be rctonless since the masimum low ral i o be determined, E. 1 The flow is steady and incompressible. 2'Tho elevation differance between the reservoirs is constant. 3 ~0, Properties We ke ie density of water be p= 1000 kg Analysis We choose points 1 and 2 al (ae live surfaces of the lower an upper zeservits, eapectively, and ke the furfac ofthe loterrese-voir a the reference level (== 0) Bota points are open to the atmosphere (P, = Py = Py.) snd the velovties uu Locations are negligible (V = V2 0) Then the enerey equation for seady incompressible low tough 2 ‘control volume between thee se points that icludss the purnp ard the pipes reduces "0 ff a | lee pone H Tis case and FY, J aa Enc Fos Fi the una pumping power is Wyse = Tam any. =O Substituting, the vo ume tow rate of water in dotermincd ta be y Boos 230 saw isin PE. MOU VON mR VASMNL Lee LW 0.0208 m/s Discussion ‘Tis ke minimum Now rte sce te tional eels are ignced. Tan actual syste, the flow rate of water will be less eeause of friction in pipes, Chapter 5 Mass, Beroull, and Energy Equations Chupter §Mfass, Bernoulli, and Energy Equations sas se Solution Wa fv ais syed tia vin pge wos dines dead by ect, Te press Solution Was ows ais oe el a benz ipe wane anee doses by » Solution ne cones to he oto wink eee with noe heen pining sgh ip. The wwtler fs prossuived by a pump, ce the eight ofthe water jet fs metsured. The minimum pretsure rise supped by the are given ww be a= c= = LD Assumptions 1 ‘The flow is steady and incompressible, 2 Friction bet the water and air as well as friction in the hase Analysis We take points 1 and 2 along the cenerine ofthe pipe hefore and after the reducer, respectively, Notng Properties Wen ie ety ofan p= 100g ia BEE acta eat halt Manto uelbs not ons miso ome et + ay Fete tes ras +H ooh low (Vi 20), the energy equation for steady incompressible Flaw through a control volume between these two points that thier the al fo) es pains @) tow a 4 mst aO0Rme Reducer hy 20-39 1 (480445) kPa (1000 ky-mus? 1.051.981 ms)? (6.968 mi] iy amiga yoataw tage as out) More eg ea Weim nal —) L “Tris hand lees corespords to & power porta loss of IN Tg. mi? iw TN mis = gh, ~ (1000 kg/m? (0.038 m5} 0.81 ens mI }-a6 wo ® Solution to al AA fan is vensTate & bazhmom by replacing the entire volume of air once velowiy cemmins below w specified value, The walle uf Un fn-tmor uni, te Usmter ode Can easing, and he pues Aliference aoross the fan ate to be determine ‘nuptial The'tow say nt compete 2 Pl bv — along the flow (other than those due t the fan-rontor inefficiency) ere Ac ~ megs Site ons hatowela a= center dry ew, @) Ss »® ceeeec cee eet ucieedestse ee © e crest sry coments ie — Prapenes These ene 28 dnrtnis (6) Te whee ante alos V2 203 3 = TH ‘Hav vos a as Snes of neh east ats y-% tm sms eins m= p= (1 254g (D315) =DUTTS k's We lace points | and 2 =P, ‘and the Mowe veloiy is negky Vo = 0) AlS®, Pr= Pye. Then the energy eguatn for ths congo volume ‘bomscon the points I end 2 reduces to (: (3 } wf Bisa, ip =H PE, ha |Win Lge PM \e pt sINC€ Fete =Potionpay MUMS ERSE AME Hy. =H — Fei an SUSIE cams)" (_ Nn y_1W Wana = mas” = @0375kg1.0) 2 Hina mes He = OUST LOL Amis)! Hone 128 and = Yo2aw awe 05 Theteore, the electric power ating af the far unit stb (6) Forairmean velocity tw remain helow the spose valu, ihe diameter ofthe fan casing should be Vn Ah =002 147, 0069 -68¢m {ey To dstormine tho presere difference seross the far nit, we tke points ¥ and ta he on the ro sis othe fan on horizanal ine. Nosng that ~2 and Py since the fan a narrow cross-setion and the loses of the fan unit, whichis aecounted for by the efficiency), he energy equation fr the Can section seduces to fa es mp 12 0am"is (ne lw Thereore the fan wil ise the pressure of tr by 40 Ps before discharging it Discussion Note that only half ofthe elec energy consumed by the fan-tor uit is eonveried to the mechanical ‘nerzy a ar while the remaining hat is converted 9 eat because oF imperfections Sobstiuing, = doin? ~40Pa Cipla ermal am Energy uations 5.088 Solution A hore comstad mth boron af a prescarind tan fcc wth a nme aie en pting sie tap. Thema tank ei prosers (esas) aes sonding age height foresee nd Aramis hh ni ania 2 i waa yl Finn i pei Woste be dnyatenaioh Si? Ona: Anatysiv We take perro the Fre sur ice of eter ne ak, ana pair m ke ds tcacs lov hs bot eink, Not tat = SE Saf, ~ fo gate mimanmn eave for che rose a prem sed Tat th Mul veloiy athe sua of te tn wy He OF = the eneray on the inet and ext sides of the fan espectvely. Pint | is sufiienty far rom the fan so that scmg Fe loss fther than TheceZoe, the pau ust supply s siniuans pressure tise vf 68.7 KPa Discussion The rorut obtained shove represents the minimum vale, and should be intopretd accordingly. In ealiey lexger pressure rise will ood te be supplied lo overcome Getion sry 10 minutes while 587 Solution Water News through a horizontal pipe al specified rate. The pressure deep weross a valve inthe pipe is saecasured, he earesponding head loss andthe pow: nexced 0 overcome it arto be detemined 1© Assumptions 1 The ow is steady and incompressible. 2 The pipe is given to bo horizontal (ocherwise the elevation sifference across the vale is negligible). 3 The can dow velocities a: the inet andthe ext of tac valve are oqual since ‘the pipe diameters constant Properties We the te density of walsto be = 1000 kg Analysis Werks the valve a3 the conto volun, ad points Yan 2 at: ine and exit ofthe valve, respective ‘Noting thet 2, ‘Vs, the energy equation for steady incompressible flow through this control volume reduces to nie nw Fv ee, geage aot A tes Hay Hy PS 2g is 2g Sohtiuing Vee Was (1) @ aon OE 004m Tog ORTMIL Ik ma “The useful pumping power need in veccome this head Toss is foun Wrenn ~ igh, — Meh 4 ¥ iN y_iw = (000 kin" yo 020m'iy9.8 m9? 40 264m] ‘ies nonetigen otal SA ‘Toerefor, this valve would ease «head loss of V.204 mn, and it would take 4D W of useful pumping power to overcome it Discussion ‘The required useful pumping power could also be determined ftom Fy tarragon) nem] 589 Solution A water ark open othe atmosphere is iiilly Filled with water, Asharp-edged orifice ot ube botora drains to the atmasphere The intial discharge voleity Fm the tank st he eterrined Assumpatons 1 The flow is steady ard incompressible, oe is given tale oy = 12, 2 The tank is open to the acnosghere, 3 The kinetic energy Analyste Wo take pine | at he froe surface af the tank, and pet? a he exit ofthe orifice, Netng tar the nid at both points is epen te the armosphere (and thus = P= Ryn) and thatthe fad velocity atthe free surface of the tank i very low (7; =0), the energy equine between thes to pois (in ters oFheads) simplifies to Fla ve Z 2g whic: yields 240, 2 ars Solvang fr Ps and subs:taing, Therefore, tre wape a1 pressure on tyref the ste tank must le a Kass 1A pe Fe = 2a (9.81 mis? (5 ~0.3 mb1.2 = 8.77 mis Deana Tse chine ahve css imma Shoe sehen. sey Tab bevels wl preva al hs easing The can low vel wil dora he wee level m he nk cecreases, 94 ‘Solution Underground water is pursped toa pool at given elevaton, The maxim flow rate and the pressutes st ‘the inler and outlet af the pump are to he determined, Assumptions 1 Ve tow is steady and incompressible, 2 The elevation difference netween the inlet andthe aut of the pump is negligible. 3 We ussume Ce itil effects in piping lo he negligible since the maximum Mw nate is to he eteumined, Envios ying ~0- 4 The fleet f de kinetic energy vovsection Laci is mepligible, c= ‘We ake the density of water tobe L hgil~ 1000 ky! W of power, and is 78% efficient, Then tae useful mechanical (shal} ower Propertis Anutysis (a) The pump-moter dows 5 itelivers to the Mui is 8, (Us9OKW)—39EW ‘We rake pit | atthe fice surface of undergrad weer, which is alan taken as he reference: level (2) =f), ané pont? at the fice surface of tre pool. Als, hath and 2 are open to che atmosphere (P= P= Py), the velocities are negligible at Ith ponte (Fi = Fs = 6), and frictional lasses i piping are diseganted, Teen the erergy elation far steady inccrapressible or through a eanral volume between thess twin points the includes the pump avd te pipes reduces to {En vs ig af pap» —Mnarpamsel drone 2 Pati + Ena Tm he absence of amaine, Enns = Foe pn +s ng 4 Fama Fang Easier Tras, Fans = M822 ‘Toon he aus and volume Now sles of water becerne oO) ®, ves —_ftonom? yt mn 3.28kgis - ao a th UB2SKEN gyysasmn'he = 0.0133m"18 ~ toe agin? (8) We ake points 3 and 4 ut he inlet ond te ext ofthe pump, respectively, where te low velovies are Ve _ ons mis VV _ooisisatis A DEA ROOT FA 4 ‘We take the pump as the contol volume, Noting that =,=24, the energy equation fr this sonra vo'ume reduces to ae ) (a a es) +823 | rag HL eS 884 aie | ay > ea 2, Substituting pp = MODERN GAIN! TAM! |! TN) 38s cs ™ % 2 Lrnookg mis? |" norsasm¥al 1s» (18842983) kNim! 2Fh Sea 2 27BKPA Chapter § Mass, Bernout, and Energy Hgwarions S961 Solution Woler is puraped from lake toa neacby pool by 8 pump with specified power and eliceney. The heed Toes ofthe piping system and the mechanical power used lo overcome il are lobe delermined, Assumptions 1 Ths Cow is steady and incomoressile. 2 Ths elevation diffrence between the lake ané the fee surface of {he pou is causal, 3 Alle losses ia the pap axe accouated far by the puuap elficicacy aad thus fh eepeeeats Use lasses in piping. Properties Anais We tice the density of ware to be = 62.4 Ibm The use puuping power sad the couespotding useful puapiag bead awe =a pay ~(O:73X02 bp) = 8.6 3H 32.2 Ibm -fts® S50 tis bt Thp {We choose points 1 and 2 et tho free surfaces ofthe lake and che poo, apoctively, Hoth points are npen to the atmoaphare (P,~ P,~ Pal and the velocities ct both locations are negligible (P; ~ P ~ 0). Then the exergy equation for steady incompressible How Uueugh a conte] voluue belwoen these (eo pots iat ielades Hae puaup aba Ue pipes raluoes to (2A mR? -2 88). Fw pg ‘Substituting, the head logs is determined t0 be ae 24) 643-35- 203 Then the power used to avercome ie becomes Vihy Eesha 990 -(anmm ys rating aie e9 af = -4.0hp Discussion Note tha: the pamp must raise she water ah additional heigh: of 29.3 ft to overzome the frictional lasses in Pipes, which requires aa additonal useful pumping power of ebout 4 be. Solution Cideipround water is panyeal te ¢ pool ata given elevation, Fors given heal oss, the Mow tate and the oresures at Ee ile and eusle 9" he pump ato be determined Assumpinas 1 The Sis scl ad veampesii. 2 The eleva ifsc ele sta the elf Te ‘pump is negligible. 3 The effet oF ts kinetic energy corecion factors i negligible, a= We eke the density of water to be I kL = 1000 ken! Analyst 4a) The pan Un daivrsto the Sidi Properties tr rae SV of power, ane 878% tte.‘ the wal mechanic (bat) power ‘We ake point ta he fee sures of undesground water, whichis also ken ws the seerenee vel (= O, ad point 2 et the fie surfce of the poo, Also, bth t and? ee open tte atmosphere (= P= Pah ad te velocities ae negligible alt both points (7, ), Then the ensgy equation for stacy incompressible ow sacouga a contol volume Bativeca these we pots ht inlacs the pur wade pipes reduces to Inthe sbeonoo of athe, &, ‘china +Bntrce AM Mp Wyong Es engomy 84 thos Wi, Nosing that Baas ion =Fihy he mass and volume Mow eles oF weler Bacon Five Reh asks oto soa Oatms}G04m | Lm Mek \ = W016i! fs 0.0117 mils P1000" {) We tke points 3 and 4 al te inlet and ie ei ofthe pump respectively where te low velocities are novieom ss VY _ onnisom's reat 8 sunems, YM " A> abd isos? 4 Say apt 14” moor? 4 ss {We ake he pumps the cunt wel. Noting a 5 ~ 4. the erengy enim or his cont ye reds ee Ge 7a sey | Hine e a Subic am imei Naat? =Gostmsy ar), seid (unn) 7 2 sotoig: mis" J oztesan ee 3.1) S88 61hNn? 320.9 ©924KPa Disession Nove that Cio losses ke pipes ease he Nw tte af wae a derese. Ao the ei: 0° ow weloclies or the presse cage azo he pump ineligible in hic about 158) ad ca be igure 5.98 Solution A tank with two discharge pipes accelerates to the JeA, The dameter of the inclined pipe is to be determined, Assumptions | The Mlow is steady and incompressible, 2 The effect of the kinetic energy correction facton: ix negligible, a= 1 Analysis c Se ut us Applying Bernoulli Equation between the points 4 and B we) Ls Negligible lasses ve o a 0 2g ‘ Let's divide (1) by 2) 5-99 Solution A firebual is fighting fires by drawing sea water and discharging it through a nozale. The head loss of the system and the elevation of the nozzle are given. ‘The shaft power input to the pump and the water discharge velocity are to bbe determined. Assumptions 1 The flow iy steady and incompressible. 2 The effect of the kinelic energy correction factors is negligible, aml. Properties ‘The density of sea water is given to be 2=1030 kg/m’. Analysis. We take point | at the free surface of the sea and point 2 at the nuzdle exit, Noting thal P)= P:= Pye and Vi ~0 (point 1 is at the free surface; not at the pipe inlet), the energy equation for the control volume between | and 2 that includes the pump and the piping system reduces to +a,2 at 422 tMcbiaee TH, where the water discharge velocity is Vv 0.04 m3/s —s = 0.37 mvs = 20.4 mis ADE 14 2(0.05 m)? 14 Substituting, the useful pump head and the corresponding useful pump power are determined ta be 2037 mi Ayano = Gn) + (1) 22378 2(9.81m/s?) +Qm)=27.15m Ww Youmpu = PVBh, tump.at 2 1kN kw = (1030 kg/m? (0.1 m?/s)(9.8 Lm/s? )(27.15 m)) ———— mal em % x 000 kas? TKN ans =27.43KW Then the required shaft power input to the pump hecomes Woop 27.43 KW 0.70 oem shat ~ — 39.2 kW pom Discussion Note that the pump power is used primarily to increase the kinetic energy of water. S104 Solution Air flows Uhrough a pipe that eonssis of two sections alu specified rae, The dilTerenial height of w water ‘manometer placed hetween the two pipe sections is 10 be determine Assumptions 1The How through the pipe is steady, incompressible, und izotational with negligible Grietion (so thatthe ‘Bemoulli equation is applicable). 2 The losses in the reducing section are negligible. 3‘The pressure difference across an sir column is negligible because ofthe low density of air, and thus the air column in the manometer can be ignored. Properties Analysis We take points | and 2 along the centerline of the pipe aver the bwa lubes of the manometer, Noting that 2, +» (ot, the elevation effects are negligible for gases), the Bernoulli equation between points 1 and 2 gives The denvity of air is given 1 bo u,= 1.20 ke. We take the density of water to be 2, = 1000 kim! > Pe AM Pee By 2g 0) We let she differential eight ofthe water manometer be h. Thes the pressure dilfirence P,—P can also be expressed as @ ‘Combining Fags, (1) and (2) and solving for h, RP =pyak 2em Pac OP? Fi _ eal We 2205 spa © Of ir —> 120 Ls © 226! Pa Calculating the velocitics and substituting, yh 00m 5 stay War Aaa m022m) 4 y=. Om ois.anms DEA wl meM py (S28miS*- 1ST 5 o367ma4.87em 2(9.81 mis)=(1000/1.20) ‘Therefore, the differential height of the water column will be 1.37 em, Discussion Note that the differential heigh: ofthe manometer is inversely proportional to the density of the manometer uid. Therefore, heavy Mids such as mercury are used when measuring large pressure dilTerences S107 Solution Water is Towing through a vealuri meter wit known diameters und meusured pressures. The Mow rate 02 ‘weler is w be determined fur tbe caso of Sictioness flow. Assumpfions 1'The How through the venturi is steady, incompressible, and rotational wilh negligible eetion (u thatthe ermouli equation is applisble). 2 The Ilow is herietial so tat elevation slong the cxnterline is comaant,3 The pressure {is unif el a given crontsecton athe venta! meter for ‘he elevation elects em pressure meastremen are negligible). Properties We tke the density oP water ta he 7= 1000 ky Analysis Noting dat ‘We take point 1 at the main flow section and point 2 at che thoat along the eencerline of the vnturi moter the aplication uf the Berruull squsion between points 1 and? gives 2 7 oe 2g A me 7 {he flow in artnet he inenprsesile and thus the density is constant "200 the eanservation of masa relation for thie single steam steady flow device can be expressed ae 5.106 Solution _Atcpiscpened on che wal of very age tn that cents ait. The marinara lew te f tough he lupstohedeerrined ant let algerie lead esr be ee Asap Fh omg ti tp ately, pei, ational with eligi iti (9 Ba thew aa ‘ra naa the Baul equation a bleh Anais The deny af rin the an * wake ae byte? Rr (a2e7ePa-mkg KOSH) Wie tae pour 1 inthe tuk, ad point 2 ‘lection aft are egies) and cet ofthe tp along te sas horzotl ine. Noting hat 2 = 2m the +o AUR sg) -t0tm le Ths i hme rl ic ind by tg ‘lose he sau se ‘Adding 22m ong larger diameer ea section wil have no effet on te How rts sae the Hm ie Hictoness ‘by wing the Demoul equation cao e som thet he velo ia is setter isrsee, mee prowetedecreae, 236 eesti ewalyrnd wits spss eM thay si v Vy Solution Water Faws though the eslargemert section ofa hor zontal pipe ata specified cate, Por a given head les Viv ae wt ne Seen chan acre eae ean hececramed sting ine Assunpions ("The flow through the pipe i steady and incompressine.2 The pipe Is horizons. 3 The Kinetic energy Suietcting ine Fa.) asthe correction factors are given i be = 8 0 1.0 104s 7 Properis —Wetake the density of water te = 1000 kg Anaysis We take points | and 2 atte inlet aud exit of the enlargement section along te cotarline ofthe pipe “~ Noting thal 2), the energy equation fr acon volume betwee nee Gwe sim emo eas lies Solving fox Vives the dested relate fr he Dow rate, +) | re = aren 6 so hag te \ Povey Hae thane Ved | AAA) ® pe '28 Vell (ay ay ‘ee the inet and ex veloc re The Dow ste fre give case er be dled by sustain he ven vals iat hs elation b> VV oot’ yay > = 14” mia a.06my? 8 water wi PHA) _athotm® [2680-15 kNin® [000K | 9 9968 mize ES eteony HS tata 4 Yat-myiny) 4 Yaonokemsit-@m 1s oh Yt permis GIA eon a Discussion _Vertar meters are commonly uscd as flow meters to measure the flow rate of gases and liquids by simpy measuring the pressure difference P, - Pz by amacinmeler ne pressure transducers. The actual flow rate will less than the value obtained trom Bq, (3) because ofthe friction losses along te wall sur"aces im actual flow. Hut thi differance can be as Litle as 1% ina well-designed venturi meter. The effi of deviation from che idealized Iernoulli flor can b> accounted Ir by expressing Ly. (3) a8 (R= Subsea img, dhe chun in state pressure acrows the erlargement set ix determined be kes 1.05{3.890 mis)? — 0.157 w5)*] 2 1000 Nin? B f -coonien?f osmyosse| N | hg ms i 0.865KkPa V = Cys | MAB) ‘Therfors, tho weer presse reas hy ORS Pa aoe the enlargemen:seotem Vota ms i Discusion Noto tha the pressure nereasx deri the hea losin th enlargement ection This i ot dynamic whee i isthe ver’ dchorgecosfiin! woe value i ess tan iis lange 28 199 for wellsgnee ventrt pssze bong courte sali presse. Bu Us plessae (aie > Uy) deicass by 65 anor 638 MPa 4 rites in crtir ranges of flow. For He 10°, the vale o° verti discharge cooffsient is usually renter than 0,96, resale of: Sictneal fats ss Solution box paravetors is rahe bain Assunprions.—1"The fl is steady and incompressible Analysis ‘Chapter > atusy Bernowtt, and Energy Exuations type flow meters are used wo measure flow rales relation fer the flow rat a a fancton of given Herrall equation in the direction nora‘ sieambnes lor steady, neampressile Now is given by (Fg. 5-43) ripe Pf aes Kosi ‘where ris the locel mis of eurvature. Substiusing Y= Cir 2 K,(enohercontss) (1 pe Appling pins and 2 end aking y= ges yo PP AWE ga Hi w sobeeaing ¥ = and Fy = © andsting [apm oof gy . Vo jean wun Gp = 2 f VE But om the figure, r= 2—Rand ry = 2-4 R , Substnsing, Fa 2 becomes wae =u i-v#e— ‘te © cf “he flow re pen by Q~ [VHA Le sate what celbow below we ean wits (21)ar atts Solation inthe ark-wer water snp fling out of he Fly open valve Assumptions 1 Tha flow is incomes Anaya Applying Berl Fy fom Fe sur Pon Fat Ub 1) Pag DD 5-4)= Py (SH) 0000 re ask lagging Py the Beroul Fgescon 0000 “soup 10800 seus) S27 sows $2 14.093ir4 40.7730 1-103 371m when the water evel necnmes h=3. 7m the flow woul (0 pessise} Ritts WE ps a.7hh D668 0? and per‘omning ntsgration gives Chapter § Mass, Bernoulli, and Energy 1 From the sketch, =? —2—rF dd =2yR? -(A-rP dr = friar = 2f Rar ae AA would be, Considering the eross.setionsl area of he Combining with Tig 3, we oblain the Now rate to be (Chapter § Mas, Remoull, and Enargy Bquations A cylinrial water tank witha valve a the hottom contains ara the tp pat and water. The vstr height to be detained. tthe point A would sive b-0am (Chapter § Maxx, Bernoulli, and Energy Equations stk ital pw Solution A wind tunnel drs aimospheric wir by a large Lin, Fora given air velocily, the presse in the tune! i 10 i” he determined Assumptions 1Txe flow through the pips is steady, incompressible, and irrational with negligible Irieton (so tha the ‘Bernoulli equation is applicable). 2 Airis and idea gas. 287 kPam'ke K. Analysis We lake point | ia almospacric air belo il enlers the wind (uenel and thus 2) ~ Pg ated Vj £0), a poiat 2 in tke vind tunnel. Noting tht 21 ~ z+ (or, the elevation effects are negligible for gases), the Bernouli equation between points 1 and2 gives Properties The ans constant of sir is R wate Hatin He awe aL aka tak ~ RT (0.287 kPa-m'kg-K)293 k) 1.209 kgm? Subotiuting the pesne inthe vid wages eae obe measure, azsiems M9" [IN LY 28) arene sp. Thedchared ter at een 2 Lng ms? | ooo in Diseussion Note that the velocity ‘na wind tunnel inreazes at the expense of p ‘even liner heeause of Tosses sure. In realty, the pressure will be 
