practice problems - PRDGLEM 9,1 RNng ‘- A Pipe CQWCVfi...

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Unformatted text preview: PRDGLEM 9,1 RNng ‘- A Pipe CQWCVfi (1w “muohfrcssR‘o‘c “$1.14 Cmfim‘ks aw praunzm ’ _ chambu‘. ‘ ELNQJ- Devclop 0M ezpnessz‘on 40¢ Hie, $03!: 0'? (Laurie 04 H M The. expaws'wu ‘chawkw- Cu kvms d;- cerhu'u 6W3. . Veiocihés.. , r , ' . ' SC | [van ‘DA—m: AéSUMPIlOMS in) The con+ml vomme is as skown cm +Me accompamding diagmm.(2) The :5 via is Mam; Prccs'xblc,(3) The {jaw (1+ Hue {Md- aw! QXH' is one -d\'meusebno.\. { fiNAL151§ '. “We; mass take \oct‘quce, becomes ‘ TC'DE' dL 1 L 5 (14:1; ’-’- §(’%)vr - §(%)W W}S§‘v{ng ~For dLch‘ anti SiMpic‘flGfifl 4L war: pl \r‘ a? = 132 i“ in! [cvd may“? ' FROELEM 4.58 KNOQM: Ammcmia Hows ‘H‘mufik ox con-m; +Y I b [A ‘ ‘ J fiiume. M‘ fleade sinks. Tke w“ 0 V0 «Ame at: we M 61" an o QXH'JWCXM am know“ 0d“ team How bmmdan . (ei- diamekr— so "Hug ammonm Vale“)ch does ym+ exceed 2.0 urn/5(5) we vommemc How rwfe of’ ’ h " ‘ “ ‘ “ ‘” " Pzfl! bar _ , (H , , SM‘. Unix)?" I? “1” bar -—--— (AV) 2 L636 rug/m3n 7‘, :28“: 2 MA =O.5L: {s Par-451$ ' 3 m M m ‘ m .___ ___J (3) sad“ (Wul'd Aggggmelméggs (I) The comm! v0! tune is at Shady sfwéeil) The flow at +116 inla+ is 0b£~o€tmen$£bnd- W: (A) To mlm‘e velaciwfy Md 1:599 «:1?sz 91%er 1%: {Me v-a MFG} .___ MI“? I At not: i q. 7‘; use ‘ANE W3, Vet‘scflg \Joxfes (mmrsezlg w'cHA dxdmefer, The mfms‘rmm ohazmwicr Corresponds in \n =20 M(s. To avf’u’; I nah {rain Table A~ W i‘kod‘ Ta = 28°C is less flam'fiw, at What: H2an {1mm Table A43J ‘1)“ 2: 'Q‘FQJdaflC-a [.671‘1‘ Ho‘s” M‘s/{<3 me! (at) = iYfiflfi “ (“(0.3Icy/fic’Le‘iwadmvlcg) ‘ W W: “ J “WWW-m 7" 6.00729 "‘8 = 0.7lqcmwmm __ @‘lmfi‘ (b) To flhd@r)3, begin DUE-HI: Hm Mags rafe. éalamce d V0 ~ ~ m —. ~ 22ft“: mlmmvms ==§> m3: mrmz KNEW via a CW”)/’u* (Av-)3 1% E with m. (ng/mj TEES»; TableAulH art: Mom» 5 1x1 2 0.30% hag/Pei] amt uaml.‘53?7xw~3WB/Ij. .. «3 '3 S ,. 3 ' K.“ (AV)3-Q.Sst-cho 7%)E03 @Xfiiflu goaam/M 9,“; 1M1» £0.30?” @903 mm it PROBLEM 4. 25 mam: Skam Hows Harth a. wall— {now/a fed nozzle «1% known condfihéru (1+er Mid“ and txi‘f'. FINQ'. Deferm‘ne ‘Hae czii—Wmhm. ‘T W: ~ I l 5:300 lva/mz' ,W a z ‘40 MM" I 77: (0009’: ~Tl+ hozzle -—%—-—> V: = {900 His : H .. V.— (OOF-Hs ( W0.) 1,, W10) The confro/ volume 25 af: shady 51-01212) There. is no 5623‘ MPG“, and ch=0, L46 MIM- ‘ .U. (3) P04914121] energy eFFed-s are negligible. Aflktjmfit The pres5ure '1: known «f ‘Hm exit We’d-Me. is flied by ddcrmuhmj hz usMg +ke SW-Sfm‘e energg balance _ O , O - O 0331(574, + m [(h,—L.,,)+(V77_ Lyrgcy/Lfl when; Ml :Mlavh. Soleg {or L1; raw: h; = [(1.4— C ‘ 22) Rom TableA-lfaJ 1m: 131%.Sm/Ho. TM; {OGZJQOOE 2 a (L4 mm. ‘41 :( BILLS awn.) + (M; 32.2 pkg/52 7734M“! = £250.03 ERR/Ho lnfurpoia4-{n3 [m "Table A“ WEE. (1f y}: qO1M/Ihl} {Air-123.0133 BEL/{b 7';L '3" (981mb 0F— T2. WO'BLE‘: M 4. 3:3 Knowm than we provfd-k" for a+w~bm¢ o.+ $44.41 S‘Hhti;+‘*‘~“3k which. ML 9: “Pan'h‘ma. .Eiiguz Ofihfwne +6“ Faufiir dc Vt Logicé. Scfltmmm &6106~ Dina»: N2. PfiSb‘mel 1;: ‘Wo‘Kil VT 2.2.0049?“ "Di: 1;“- Assumo‘nouc: 1- Thu con-fro! Volume S.an w; +5... rckn wah‘c (:- ad- £40.67 S‘hH- 1~ Far 4’30 CmTJmA wLuw-IJM c.th u...‘ (“Rake—l {wary-l 'FY'M 1'“ [21. “IO WH- l3 “9314‘3t'blé. 3- “(Maxn L; mudgd 0.0 a... JWJas‘ ANALYSH: R‘Jucn‘na-M mu a,“ chars-a mire bhlanu: . / ‘ " 2 1. a °=‘- ‘Qw “‘“w * “L‘w hu—F Wm: gufiu] - . t , V L ‘L “B in“ MM ‘3? Way"— CQ=v+Wth‘H-L+ éjtwgvbwgé‘? mh :12.” ' - M W ' “a w 0“ i J '1?» fwd wt (m z V . .4: ~17: L...“ “7‘ .1 (WIN? 3 ... [n(fiM/mwflimwm (P'T P. — l o I J 3 I (1182:? you woo R) = 0.51!» W; Thcmimuha. cum-1 claw. fry... 7:414 Anus 6-H ‘5WF(GE=v/&)(¥a) V4“ / [WWW w fiEfiflmfl—W 0 2963 ' .. - (534m ., 5177.7". a: . __ fl WCV " L W][O~S'7' £9] ‘4‘ 053% ( (2%!“ with!) + w K—(‘Zocfllfizatlflvlflzl V 1 61M. ' 1 31a “95*”(5‘1‘ 1‘18? ’M‘oi‘ 3‘ £912 . _ P: g = 05-;[45 4-}24 + 0.9,] E;- $16qu UMP ‘Eggos\;%g§l~? zgqymu/k K K M at)?" _ $433M?» 4—4: PROéLEM 4. 3‘? kmogggi flaws exf’qwds +hr0Wil’1 m weUH‘MSUMk‘i *“V’b‘;‘a- "ma-hm“! ex“ whdvc‘i'kom are known) and .Hne gag-H" diameflris fik’vgm. FQMQ: 1910+ +140. Pewar devetopad. b5 “Hm.” mvhfiae vavxus 62(11” itmlffil. ECflEMArTiC flan/EU "DATAX flSQQjEDQMS‘. (v) The cowl-ml Volume 35 wk 5120.020! 51-63212.) Hem} *mus-Fer /: nag/{wide .(3) mind—(c, enema ad- 'HAe. {Vile/1' and fohwfidlemerflfl (gated-:7 an: 113.31%! la. V &N&L!§l§t To dmrMMe +ke pause/r, 02,50 We emqu 1041mm a“? a1~wd~g 5% ~ 0 ‘ . 0 a Oz/Q/‘Z‘ch+ m {Um—14:3—i-«Ll Z" )+j(%2~%,)] wkewe. M, “131?. W1. (NH-k fiisump-Hbmfl) \ . 2. wwz‘uMECkplnt}... [:2 G0 EValLkM-mj +he mass ~Hc>w wax-1e I z = 4:2. 44.31 SaMplg gigglgfigm:_ From Table A—Lf art Ff 24.44/94: 40 bow; 7: = 3200;} 10ng lm': 3013.4 {LT/[45. From TableA—35 vlz'zuswm‘fi (gnaw: - bosom?) marl-$2.12 ac, WP/kj . 304421ng, 141:: 370.70 +(.M(2253.a) =2q3/,«7 07/13,. New, ficmn (an!) . 71001042901015) “4:. = 0.15 kj/s 4(21230 W/kg) Now, Mm car) ' + *2. W 411-15 53)[(sots.q—2v31.~2) £70132 Esq—JAM. Ik-T , m, cv S kg 2. 5‘ [kg-MHZ, logw‘w‘ Iky‘s a (9‘1 2 7 law 7000 J\ ‘ lTCode 5000 .. \‘\\_»_‘ p1 = 40 // bar — T1 = 320 // °C 5000 ~ \‘\\\ p2 := 0.7 // bar E 0 \ V2 2 90 // m/s é 40001 ‘ dZ==06//m 5 x2 = 0.9 ‘3 3°00 ‘4 m = h__PT("Water/Steam“, p1, T1) 2000 , h2 = hsathxC‘Water/Steam", p2, x2) . v2 = vsatmPx("Water/Steam", p2, x2) 1000 p mdot = pi * (12".? * V2 / (4 * v2) — 0 = -Wdot + mdot * ((111 - h2) - V2A2 / (2 * 1000)) 0 // Using the Explore button, sweep x2 from 0.9 0.92 0.94 0.96 0.98 1 // 0.910 1.0 in steps of 0.001 ,3 PROSLW MM! Under 1clows 'Hu’owfln Pumptiaj .5ka lcnown Iltk'f'aud exlf cwiifia‘ns. "(be p9ch Raina/ed {35 H”; wuwyo [5 also specifiéd- EIMQ'. Defievmx‘me +146. mass {Low raie. Jr Mule SQHEMAJ’K. g GWEN gm: ' fls§gmangus:(n The can+rvl Volume is at .S'l‘CAAIj Sfa+e£m Hea+ fiahsfier“ Ls 54631153512. (3] Thc wafer behave; as an Ihcompggsfb/e q "1 liquid .(4) T112 accelerah'on of firm/3+5 {5 amsfiufic. «173:181 M131. (5) The. magnatme Md ymssum are neArltj (Lakshmi Waugh out. | A; w @5915“: To #Ika’ rr‘s/ 565M Lu2+h 36341141 4112:}: mass “1516;1qu rule balagces 0 ~ 0 . 1 7. O = (vww 4— via [(béz)+gr‘ "Uleézpazfl z - E = ‘ch'*‘ m 2V; 3 +ficzl’21-3] when: ML?- V‘ALE WK, and-Me Specift'c emfindm jun/m Ls ehfiu‘mafid based aw assumpfioMCS) m4 . a-zob From 53-5»3fi—,V"= ka/A,md C‘X‘) Begum , . ~ 1 ‘ 7. : «WCV + M + \ . 3 1 ~. - V“ U .L .. ‘ ~ - ww + L {A} At) + M3 ca! 27,3 (g %) From A: 7100/4; A‘zoflooqm m1 M A1: 0.001%:‘1! m1. Now J Lqu 15‘” L$Q 205C ‘ l-OOl BXlD’3 MIG/leg {:1qu Table A07.) Lug ca,“ {laser-f- VaUJLM .wfxw5 , ' ‘ R375 - 3 (LDouSXtO—B max I I I = — (*3-bkw)! ‘ kw + \m Emka—l 0.000%!" 0,001quq“ E4) I M May . i N lag-mme + Mm fiv-w! 4 {kg-“41’s w3 . . O 1' 8.5 ~ (HS-l3 ¥ (0 m3—O.03‘12‘-{\m (mixer; wk 15 {m kjls) This exudm Is: who W Via. The SokuHu—Kis mix = [3‘18 k—g/LSMMMM (k? ' " toawvm! OY‘ PROESLEM ‘fi. (a0 W: Carbon dioxide is heded 0.3 H Hows Hammqh amsfau+ area. pipe, Im1e+ and cxH— Candiv‘léns are known. £1,012: :De‘kzvmfne; Hmz ride. OF hen} Etta/mm; i Gil/EA) DMA: carbon dioxide — - ~ w ~ u - — m — a a -— n R = 2 bay- ‘jf" -+-+ P2: 0.63455 Ear .. . ._ _.. _ _.. flwl a 77=sooK w VZ~QOOM/S UT: [00 mlS cf, = 0.74 kzr/icfiwc. 6sg =Z.S¢m = 0.025m ggsgmeflougz (I) Them/1th \fmlume Ls 4+ (Shams .si-afe. (2) WW =o.C3) "The dad has Cansfanzfarea. (‘1) Fm‘ermf—(a! ewerg; CM“ 2’. from Mic?" f0 exit-can be. neglecfed.(5) The carbon dioxide can be madel a0 am Madam will/14 Cansfawt syecifi'c. (needs. Wit To {fix fine exif $+w+€ , begin mm Hue. 5+eadg~5+a+e macs balance. W1, 2 Vi: L a M XV- v 1% 7?? ‘fi’z‘f $61» (vim w‘l’rl/x fine M ed 3&5 afiuafib’h of- S'MJW, “KI£=4§3)£EL F2. W R av T1=( “ (7T(.0251m“2\(mo Wis} ( 2?; bar HIOSMW‘I " W—mmm 8.3M #23“ éfifiifi*E;KWC3OOK5 ‘3 0.5 752 kj/s Uslkfl fine 5*eaM-s¥afi enema kahuna #0 {45¢ ma EM“ arm/Mg“ mm; a = 62“— y‘d‘; w: [01,- Pig) wag”; fiwgkfl Lo/Ha 51ft” 2. 3P (1: «T1) [Ha {.05 HM m“ J 1 bar . 21.: Z ch : “4 [QPCTz“Tt¥ 1*(V; 1‘]: ] fl , _._ two —[c>o M7» 2 (01731;?)Ebfltifléifiglfblffifiock +£WW/gi‘3 77%;] """" ' mmmm Mm¥iFY1TJ 1! ki-mts‘ I mmm \1 V315 a 1 55:4 1: W 62‘" PROéLEH 4' - (p .Z W: Separm‘c Vapor and Him}! Shear“: o1“ Peén‘gemwt 13%;. pass in cam/Her How +£19qu a. mall— Msulmled («em exchanger. flaw am known a+ fixe [odds and exifs’. Emu“. Dehzrmrme “We cxH- HmpemMr’e 0? "We licéuid sham. sew/ma é vagu mm: ' / R4344 W-‘f-il’i/u/ {-15 44/9 T‘ = 00;: ’ ll ‘ I 2 T2 :ZOOF $01.1I‘apo-r F2:F| /I ' /‘ . 7}: [05¢]: F’v‘Pz fly. .I _. 7.. 7 i. I gum 1mm:- Ttk,_:'z // /// //I/ m‘zhl‘g fiSfiU/Vlfll LQMS: (I) The con+ml volume. is ad“ Sh?ng Sfufe. (2.] Heat!“ 9mmch beJ-ween ‘Hne Mew? exchanger and Has». surround M 3 am be in heated am} cha (3) Mimi-[c and poi-emfiaf energy changes {Mm IL:le 11—0 at] are hegiiadaie. mmgjxhézg : \To 2:13: 3% 4, begin wifla sfcq‘a mgs and energy baithgs e rmme k, “‘2. m1 :Vikq m ‘ _ b \ a 3‘ \r G?- c» '92 9. O zflfv'y‘fc‘: + W‘;[H’\vh1\+gé; HMjC (21% +~ Malékg-L‘Q vfi' 3 “HR '5‘, ‘4 “‘13 X [’h,’ =(hf‘ L11) + From Table. A4105; kl :. toms fii—u/(L MA ,2, : 23.2.03 lL-Hmz. Imhrpofafi}? im TableA*!2—E ;, his 105175 6M/He. Saks3aud 4 are how subcooied' iiimfai minim. 7k {0310(1)an Wraxu‘maicém‘ an: reasonable Pia ‘3 [HQ-r3 [mt r: Tc; wHk k3 = 46.0! Batu/‘9 Mm TabIeA-los but 1‘ (10!.15-I05.7e)+ 46.9; 2 42 mums ' IVIWPOIa-hfig M Table: A4052. M I“, f: 43."? ‘F¢.____.~WMWWW_WMMMB PROBLEM 'A supply line carries a two-phase liquid—vapor mixture of steam at 300 lbf/in.2 A small fraction of the flow in the ' line is diverted through a throttling calorimeter and exhausted to the atmosphere at 14.7 Ibf/in.2 The temperature of the exhaust steam is measured as 250°F. Determine the quality of the steam in the supply line. ' SOLUTlON ' Known: Steam diverted from a supply line expands in a throttling process through a throttling calorimeter and is exhausted to the atmosphere. Find: Determine the quality of the steam in the supply line. ~ Schematic and Given Data: [71 = 300 |bf/in.2 p2 = 14.7 ibmn.z r2 = 250%: Figure 54.8 Assumptions: 1. The control volume shown on the accompanying figure is at steady state. 2. There is no significant heat transfer with the surroundings, and W:v = O. 3. The changes in kinetic and potential energy across the throttling calorimeter are negligible. Analysis: In accordance with the assumptions listed, hi = [’12. In addition. for the two—phase liquid—vapor mixture hl 5 hr! + Iiihgi "“ hr!) Combining these relations and solving for x, _ h: " hn x‘ ks ~ hm From Table A315. at 300 lbf/infi, 113; = 1203.9 Btu/lb and h“ a 394.} Btu/lb. State 2 is in the superheated vapor region, so from Table A—4E at 14.7 lbf/in.2 and 250°F, h; = 1l68.8 Btu/lb. inserting values in the above expression, the quality in the line is x, = 0.957 (95.7%). Comment: The use of the throttling calorimeters is limited to qualities greater than about 94% to ensure that the Steam leaving the calorimeter is superheated. Small corrections can be made to account for the effects of unavoid= able stray heat transfer with the surroundings. WW PROBLEM A tank containing 45 kg of liquid water initially at 45°C has one inlet and one exit with equal mass flow rates. Liquid water enters at 45°C and a mass flow rate of 270 kg/h. A cooling coil immersed in the water remOVes energy at a rate of 7.6 kW. The water is well mixed by a paddle wheel so that the water temperature is uniform throughout. The wear input to the water from the paddle wheel is 0.6 kW. The pressures at the inlet and exit are equal and all Kinetic and potential energy effects can be ignored. Determine the variation of water temperature with time. SOLUTION Known: Liquid water flows into and out of a Well—stirred tank with equal mass flow rates as the water in the tank is cooled by a cooling coil. Find: Determine the variation with time of the temperature of the water within the tank. Schematic and Given Data: <}__.._ "’11 = 270 kg/h X a; t. :1 .. E 0 as E m w R» a: .. m 3 Figure E4. l 3 w Assumptions: 1- The control volume is defined by the dashed line on the accompanying diagram. 2» For the control volume, the only significant heat transfer is with the cooling coil. Kinetic and potential energy effects can be neglected. 3. The water temperature is uniform with position throughout: T = Ttt}. 4- The water in the tank is incompressible, and the change in pressure between inlet and exit can be ignored. Analysis: The energy rate balance reduces with assumption 2 to ciUCV d, = Q“ — We + nah, — hz) where #2 denotes the mass flow rate. The mass contained within the control volume remains constant with time, so the term on the left side of the energy rate balance can be expressed as dUcv d(mcvu) d" d: dt = m“ ‘21? Since the water is assumed incompressible, the specific internal energy depends on temperature only. Hence, the chain rule can be used to write where c is the specific heat. Collecting results 11.41:", C?! dz ‘" dt With Eq. 3.20 the enthalpy term of the energy rate balance can be expressed as 0 hi " h2 = C(Ti — T2) + U( "(72) where the pressure term is dropped by assumption 4. Since the water is well mixed, the temperature at the exit equals the temperature of the overall quantity of liquid in the tank, so ’21 " h2 = C(Tl “ where T represents the uniform water temperature at time t. With the foregoing considerations the energy rate balance becomes mcvc -d—t = ch ~ WCV + Inca] -— T) As can be verified by direct substitution, the solution of this first-order, ordinary differential equation is T= Cl exp<~ m I) + + T. may mc The constant C1 is evaluated using the initial condition: at t = 0, T = T]. Finally T= T, + - exp(— m t)] mc mm, Substituting given numerical values together with the specific heat c for liquid water from Table A-19 [—7.6 ~ (*0.6)] kJ/s [I _ exp< 270 — —- I (%%)(4-2 5E) 45 = 318 — 22[1 - exp(~6t)] where t is in hours. Using this expression, we can construct the accompanying plot showing the variation of temperature with time. ' T=318K+ Comments: 1. As I -+ 00, T -—> 296 K. That is, the water temperature approaches a constant value after sufficient time has elapsed. From the accompanying plot it can be seen that the temperature reaches its constant limiting value in about 1 h. 2. In this case idealizations are made about the state of the mass contained within the system and the states of the liquid entering and exiting. These idealizations make the transient analysis tractable. ...
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This note was uploaded on 09/28/2008 for the course ENGRD 2210 taught by Professor Torrance during the Spring '08 term at Cornell University (Engineering School).

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practice problems - PRDGLEM 9,1 RNng ‘- A Pipe CQWCVfi...

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