Homework 14 Soltuions F11

Homework 14 Soltuions F11 - PMSLEM lz.22- new”: . . L—...

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Unformatted text preview: PMSLEM lz.22- new”: . . L— A 5“ "“HW‘ “"4“ "‘ ’Pec-‘J‘W‘ “""W' °"“"5:I': 1-H"; a. Lek-Pffl‘of‘ «t a :Pon'fi‘aJ Hut Jim.» safe. and sin-JG. And “Mr qt 0.. apgufq‘ed Prcflurf. Th2. lsen-h-opuc “mp'fi‘er ‘Ffiwqwnry n, “no gnaw“. FtND'. Dcicrmme +H£ +1: upgrai-urc 41" "Ha: can't NM. (1&0 0F eodvopy pradue-h‘dfl' c... HA'HC 3‘ GIVEN DATA: “t: / l 'an power- "quit-ed, and fig! hr pfighe 5,...— "L T.‘-2-O°C. / v'n = 0-5 “5/3 EMGIUEERIUG- MQDELzfl} The compressor is well .145“!me and sperms ai- steady S+a§c. (1) Khnch'c and Poitn‘h'll gang), eff-och Can be «'auorod. (5)7719. fawn": cultures 'h: “a: .'d¢a[,'.r_A+‘~°nr a; 4‘“ Dalia“ Model And “MP5 - StHDn (chum: conghamt. ANALYSIS: Ting 4cmPgn-‘urt T1 an be denim-mined USI'KD “Hue. I's¢nhapn‘c comvres:w efi-Eu‘enc‘d J bu Er“; 1-“ +ellapqrg+ur¢ n+ shin as, Tulhunb‘ ana. 9mm S‘s—Eluol Eq.I1-3(. can. be mvaked hurri- ‘1o,L?=s-El)ol+ finJgu-EI)”; = 19‘(§;-3:- E 'n%)o-: fip‘( EL-sT-Eihé: )N‘. :O Recon-Mnya’ Wu.“ v.1...“ e"... +£41.14“: gal-hale: %U;GL)O.& " tapgtia)”; = %’1(.E?)O‘_+ Sadr)»; + fl“ % a -.- (mountsafi-q-(oAXI10.flt'\+ mm A. 5.4-: 2.3.13 :1 ‘ - ' Klimt-K. ® Solvm5 dcmfwabj usl‘ug &a,+c\’ T5,: 470 K . 'ln accord uHH lustctl assumph‘onx, 4h: Noni-op”. compn333r effi'a'oury “15- if Bi... - 'k. {S “C: «i. an?” 383- h! 2 3:“ (Wrath. 4- Britain-ii)!“ = 0’6 ('38‘1'z"5”)+O-'+U3”3' rs“) V"; VI“ 9.?8 -.- G'QEEJLJIAMl Accorunal} ‘3" L EPEJOI + %~‘(Ill.-[H]ut = 6?3L ‘11!an Of ‘30-.G:3o,_‘+ 5N:t;‘)flt = “61(30'1 " SDI-GO‘H. +‘ “:3:- __ (0.“(K3fi+(9.4\{u20 2. 5131.: nsrLLQ- main»: SO‘IVHW?) 'T'L : SICK-(14?”C) The. pa.er VCfiUI'N‘l "5 was. (ht—h.) =--'~ ( ‘1‘”) 14. Luke". M f: +kc Mitt!“qu holocuur uu‘S “1'- M: gal-Hal". Bul Hul£<g‘bnn’+(o'4)wo :- 30-4. ThereforL PROBLEM 12 .2 z (emu. ) - Pagez m i :31: We. Va ,_ - (o.sr3\[5739“a"““ Jlfl']: —uo.e.=vd 4———. c "—3 30.4 Isa/me ‘ MM; and enaij mic (nuances reduce. .1 3+enst-I-1‘e 'I-o awe. . o ' O=% YACSl—SIJ + 02v J Ugo: WLSI‘SI) . ( Fl -?l) (gI-EI)O‘ + Slutty-Sh. 1 h 3 W M M Wm“ 3° dado... Fm— “rw A-as 30.1. Ins-r— “ca/s (rt-V KI )1 “WI 5 = 0.050? 5E. K [, “4n Ii'ev’xz.:h.{m2f‘ohd-LLr-urx~ (-01% fibk dab. Cam. In onlch 5 £45.33, (T. The read-s for T13; T1. J ch , M41 0'“ obi-“heal Mag 11' name milk #11 Val-Utes qwm Lara. 12.31 Two kg of N2 at 450 K, 7 bar is contained in a rigid tank connected by a valve to another rigid tank holding 1 kg of O; at 300 K, 3 bar. The valve is opened and gases are allowed to mix, achieving an equilibrium state at 370 K. Determine (a) the volume of each tank, in ma. (b) the final pressure, in bar. (0) the heat transfer to or from the gases during the process, in 1d. (d) the entropy change of each gas, in kJ/K. KNOWN: Nitrogen and oxygen are contained initially in separate tanks, each at specified states. A valve between the two tanks is opened and the gases mix to form an equilibrium mixture at a known temperature. Find: Determine (a) the volume of each gas initially, (b) the final pressure, (c) the heat transfer, and (d) the entropy change of each gas. Schematic and Known Data: Engineering Mode]: (1) For the system shown in the accompanying figure, there is no change in kinetic or potential energy between the initial and final states. (2) The tanks are rigid, and W= O for the process. (3) Each gas and the overall mixture behave as ideal gases. The Dalton mixture model applies. Analysis: (a) The volume of each tank can be determined using the ideal gas equation of state together with known initial data for each gas: Problem 12.31 (Continued) — Page 2 8314 N-m if 2kg _—]450K V _M-( {me-U )_o3816ms «— N1— —_3‘*2—‘ ' lbar 8314 N-m _, (1kg ———-](300K) RM ‘ V0? =M__aw_=msms .— m 5 2 p02 (3 barfo N/m 1 bar where pNz and pg2 represent the initial pressures .of the N2 and 02, respectively, and are not partial pressures since the gases are in separate tanks. (b) The final pressure, pf, can also be determined using the ideal gas equation of state: 0.6414 tn3 105 N/tn3 HEY} Pr = V where V = 0.3816 m3 + 0.2593 m3 = 0.6414 m3 and n = nN +no = i+—1k§— = 0.0714 kmoi+0.0313kmol=0.1027 kmol 2 2 28.01 kg/krnol 32.00 kg/kmol Finally pf = (0.1027km01X8314Nm/1moi-K)(370K)‘ lbar y: 4931)“ (c) Considering the contents of both tanks to be a closed system, as shown on the system sketch, the energy balance reduces to give AU = Q - E , or =0 Q = AU = AU|N2 + AU|02 = n“: [27(1} )— 17(er + no: [27(Tf)— 5(a)} with if data from Table A—23 Q = (0.07411qnot){7687 — 9363]% + (0.0313 kmollms — 5242] k] 0 kmol =—73kJ ' 4‘— (d) The N; is initially at 450 K, 7 bar, and finally at 370 K. The final partial pressure is yNz pf = (HM: /n)pf = (0.0714/0.1027)(4.93 bar): 3.4275 bar Thus, with data from Table A-23 and the ideal gas model Problem 12.31 (Continued) — Page 3 (A3)“ = nN: E; — E,” — E [n p”? 101341 = (o.0741mm{197.794 — 204523— 3.31411: 34275] 1" k3 =0.01484—~ 4— 7 kmol - K K The 02 is initially at 300 K, 3 bar, and finally at 370 K and the final partial pressure is yo} pf = (no! /n)pf = (0.0313/0.1027)(4.93 bar)=1.5025bar Thus, with data from Table A-23 and the ideal gas model (AS)02 =n02 [5; ~ 5: — Em mm] pm: =(0.0313l<mol{211.423—205213—8314111fl] kl =0.37432£ 4—— 3 kmol-K K Noze: E van though the entropy i‘mnsfir accompanying heat transfer is out of the sysrem. (he entropy increases because of entropy production due to irreversibilries during the mixing process. 12.41 Hydrogen (H2) at 77°C, 4 bar enters an insulated chamber at steady state where it mixes with nitrogen (N2) entering as a separate stream at 277°C, 4 bar. The mixture exits at 3.8 bar with the molar analysis 75% H2, 25% N2. Kinetic and potential energy effects can be ignored. Determine (a) the temperature of the exiting mixture, in (’C. (b) the rate at which entropy is produced, in kJ/K per krnol of mixture exiting. KNOWN: A stream of H2 at a specified temperature and pressure enters an insulated chamber at steady state and mixes with a separate stream of N2 entering at a specified temperature and pressure. A mixture exits at a known pressure and molar analysis. FIND: Determine (a) the temperature of the exiting mixture and (b) the rate of entropy production per kmol of mixture exiting. SCHEMATIC AND GIVEN DATA: ENGINEERING MODEL: (1) The control volume is at steady state. (2) For the control volume, l/lfW = O and ch = O. (3) Kinetic and potential energy effects can be ignored. (4) Each gas can be modeled as an ideal gas and the exiting mixture adheres to the Dalton model. ANALYSIS: (a) An energy rate balance at steady state, expressed On a molar basis, reduces with assumptions 1, 2, and 3 to give 0 = ~ch wit? me at 2» a = a]; + ija (1) 2'0" 30‘ "a "3 where 1'11, 1’22 , a, and are the molar flow rates of H2, N2 and mixture, respectively. Since Ma, 2 0.75 and hz/fi3 = 0.25, Eq. (1) becomes Problem 12.41 (Continuing) — Page 2 A = 0.75;?] + 0.25;?2 = 0.7SEH1(TI)+ 0.2575413) (2) The enthalpy of the mixture per kmol of mixture, is E, a 0.75%“? (r; )+ 0255,1112) (3) Substitute Eq. (3) into Eq. (2) and use data from Table A-23 0.755% (73 )+ 0.25%,“ (r3 ) = 0.755“: (11)+ 0.25m; (r2) =(0.75{9971 1“! ]+(0.25{16064 k‘] ]=11494.2s 1“ kmol kmol kmol Solving this iteratively with data from Table A—23; T3 2 400.5 K 4— (b) An entropy rate balance at steady state, expressed on a molar basis, reads ‘- . _ . — _ . — . 0 - Z T +rtisl +11232 H383 +crcv J. 0 Thus av=§3-[g)§.+[2zt] "3 "3 "3 From the given data, til /r'23 = 0.75 and fig/1&2 = 0.25. The specific entropy of the exiting mixture on a molar basis is 53 = (0.75FH2(T3,yH2p3)+(0.25)§N1(T3,yN2p3) (5) where yH2 = 0.75 and yN] = 0.25. Substituting Eq. (5) into Eq. (4) and rearranging :1“ = 0-75 EH; (T3,.VH2P3)' EH1 =Pt)]+0'25[§N, (T3:}"N2P3)_ EN, (Tzapz 3 Using the ideal gas model and data from Table A-23 Problem 12.41 (Continued) — Page 3 0:cv = 075'???“ (T3)... EH1)_ Elm yH2p3 ] + 0.25[:§1:1(T3)_ EN: (T2 )_ fem yN1P3] "3 pl P1 =0.75 (139.1416—1352085) k3 —[8.314 1" jmw kmol-K kmoI-K 4 bar +0.25 (20011072—209461) H —[8.314 1‘3 Jlnméfl kmol - K kmol.K 4 bar = 5.7131 1“ 4—.— kmol The rate of entropy production per kmol of exiting mixture is positive for the adiabatic mixing process. as expected. PKocBusM :2 .90 [Lg-12.83.: A caA'n-Lr.‘¢..1 “flat lawns, Spar—«fwd. da‘mens'u‘cmv r'm'h‘aflq Cami-new w-o.r\‘ am A W cong‘h'ong. TIMI. hm; confinfi arc cooled, Frui)‘. Mkqun-n. La.) wkeHur- Conunra ham Occurs) Ch) 1”“ eca‘vu—t, pussy-we, cc) +6... Lunar fianrf-fir‘l and (d) +“-' (Ema/c In; Cuh'Ong. km:th A 4mm: th a. vol-me 0-! ’90 Ff: "NH-"y «mum-n: am- at 300°F} 8° "If/ful- Jtfi‘ : [Ob/o, The hunk. Can-feu'l': are taglfll "lo 170°}; FIND: Dcl‘eruu-sc +k: +1.“! Fri-“are; 'Hu. hen-f figngfgr- Jud“! +1.9 inhopy change. SLfiEMA'HC T GlVEN DATA: H516“ 1-D: “I 1‘; :3oooF ‘ 111-",an pl _-_ an Muffin" ¢r -_ w% T T; a I70 °F T’- ‘ no”: V EMGmEEIEIMG- MODEL: (I) As Show». by 1*“: actowpand 1H3 {0°3urc, 1“! .1):me Cour-r“ of ‘HNI. +Ahk- ¢°“+'H+S For Huck +h'yc Art no ck‘nje“ [K tf'uffit arpaifuf'l"l' euevy. (L) The Mmth Ac” “3 I’m-I'd“! 3‘5 and 9g¢h cowronruf adheres-1L0th Dk‘hu Model. AUALTSIS‘.(fl-]Tke firrf sier r; 'h, 62+"me I'f (Onfiflfla h'on °C¢U’r- ‘5'“‘1 'h‘r-fi volume 0-f- -H1? “at. and 4h! +O+Al MAJ: 'F 0-!"er pun-19.1 d'! math-T, Hat Loo-Hr Undtr on a coax-hr! rprcffi'c. vafume procfl‘! 0‘5 “‘1 +W‘L Camden-H; ar-e cooled (lee 'T-v dl‘gaquL on.“ +9.: {dud 3M 01vn'h'0h 9-} 8+4“. V" : (ii/“v31; ulnar: Pyle. ¢, 5 :' (amuse.an ) : (r593 (bf/,4ql- . Fa“,t I Illa“ TIM-.3 (genrexrw) Ibf/FT‘ T_;.,r,[t4.,-nj .v,‘ 11”, A.1E “Mk Vy‘;V3 3H9: T2109o39F as “at ‘HnPcra-hlrt A'f Lou-"ch. condrnrah‘oq would beJF-u. A€C°"-"*3"JI +“"‘3 "1' "0 q“d'"”h.o“ "" “'4 P"""1 +—" CA):- (b)Trea-h'nj +h¢ oucrnlr IM::I-ure as a“ 4'dtnl ,1: n1‘ £4119: (Andz = “RT: P=hRTL = 35-3. P=6°ltf M—°")-a an P‘ Tr” ‘ T;— v ?.'T."’ " LW‘Y‘mofifi ' “LEE 2. (arm WWWK a? warden-r La “\f=(E41LL)CH): WW) 9 13.8599 W “A” “h”? Air is m: aqv = [(So—s.cqs)tl+43C'3-°‘): 4.4! Ma) (imam 51-51163?) (“5‘60 D A“ emu-3y bfl-IAH‘Q "dun: +0 AU: @‘fll 0" G): 4v' Th“: oneoem rune Ccvn+td)—Pagaz Q: M‘LMCTLJ-utt-rl)]+ “‘vLuvt-rl] ‘uVL-r')] U) L=V/v, In “M‘s can.) when. +0...“ u‘: no candfhfa‘h‘lln) +h¢ clam-«5c I'M. :penf-p; whim-9. mug?!“ © Of- ‘Hflt wait-f Vo'or an.» be. EanI-agitd DIV-‘3 311a.“ {whit dad-L and mum; “1 or. “u .36..“ a“ +..y:._"|"o.l-{¢ A-Z'S. Mow-.3 fay} Tab“ A-1‘3E' give: F, +ht pvevfaut Qqu. h'm waald be mrfiHeu Q = “a E “am — him] + Cm: V wan 66.41. g... "nun n-225423E C0: L49”) [ noun. [2.9.qu wags-fin“ )1: :75'5.‘i— 45574,] ‘EIOL = —uo.+3—-I'2.41= "11.2.3.4. Eh. F________Q (A) The Change m enhupy “For 1”" proton is AS: MQL$¢CHI $1.)“3atTu Pad] 4' "v [5" (T‘JPu)—sv Where “'00 5F9¢"f'lt. ln'h-upr of. hulth hupcrnhue Acurh‘nala Campy] hck “rum: camraneui is evahonfcd, 1+ 144: "W "H"- P‘V'h'fil prcnurc 04' 1H: tamponqufi in 4|“ “NIH”- . _ ‘01- _-9 - j’fi AS" m‘[an(TK)‘s:(TI)-fi; [H’fiEJQ- wmh dover {rm 115m A-Z‘ZE. M #135 2337 T o 617;; ‘° @ 12.03, — ‘OJS‘iS— ——o.0l'«f¢)=~o.17:;i,l (Sh/cg d————____AS AS: (43:) 63777- .5331: - '-"_3‘-. 1-. «.31 +(‘3'355 44.353542?“— mm. in _________________ L A Soluh‘on usl‘ng shaw- +o.bro_ d4“. cam AlSo b! developed. LIZ-Hut: cue .Eq-tn Lao-ll be aspen-ed as Q -_ MA LEAVE 3 -— Maud] + LV/v,][ Luca)- ugh-ID tutu-u- Ma 4;, alg‘hu'ne-l “FYMM SHAH-h- +kblog. .2. Enh‘DP‘1 chorus»: Wm“ ““W'I'W “'5 “"36 ‘Frh 41mg. M E “j M War-“Pv- FROfithM [2.92 maggN.‘ is-s+a.+¢ operad-I'na defies. are provtded {or at COW-ruin" followeé EH6- M a-H-vrcoaldf. FIND: "WWW-m (a; wwa] Md ‘I'fangf-Ar- frown-ft.- Cow-Twusscrr‘ Th: H-t mmwun ,ch +k¢mu WM» 03 m “New”; (c) +‘* rat-Lt 9’] Four ‘H'M-S‘f‘r 'FI'M ‘H’H HIV} 4”; 1"H“ "' F"?""‘7- SCH-Ehlg'rtc fiGIveN 'DA'M.‘ Refrigerant circulating in a cooling coil T1 = 400°F p3 = 100 lbfl'in.1 2 Air-water vapor _ I mixture . ' T] = I00°F $3 = l00% m = sootbr/in? Saturated ‘ liquid at [00°F Powerin -' ' ' "5" Heat _> ._ trunsferto 15 hp the surroundings Atmospheric air, ‘ :00 rennin at T, = 90°F. t», = 75%.p,=14.7lbmn? QNQINEERINO- Moot; t (I) The. COMprcn‘or and nfhruatcr opcrpfc at sic-.43 sf‘f: tut-M hushsublc {Kluge} .‘n kt'uch'c- and pal-gufinl curvy. (1-) There u't “0 .lh'a)’ Inc-d 'h-II— sfu- u+unew +5". «Humour and H: Surreonh'ngg_ ANALYSIS: M “um: 541+; mu: nu. mun-tee: on curls—at volume: Cutlolfn, Hm compressor and afiercooter give kfim‘fihfi'fi‘kr ""VI'QV‘V “‘Vt: “H's-*5"- Amara-“317, w. = 0’1 am& \n‘nwl u‘n‘ : {011-033) 1 (wt. " “3) T0 Fad U: and a5 4 P": 4>, %‘=(o.1r)(O-5‘i?3) 30-51“ “if/H“ “‘4 va= $5993: 933: 03503 ‘H’hl- Th“ (Ls-LL i ——°'q“3 ‘1: 0-0097 '22? "b - cm 3'3 um) Lu"; 0.51.7. £———-"“' _J': 0‘07-3 '1’“) J “3" n+1- o-sm lb “J The. mass: ‘4'“ P0"- °F H‘Q d"; ‘-'* (-0“ ht Evalunird usins ‘H-HL Vflfumcr'P-L {Juan fur! fit‘ an‘ *kt £40...‘ 3“ e-flu‘h'o" "F S+Qf¢: dew-3. _ a. t W" I14.1.amoun-IH/n‘ltwo‘fis/”"“’ = 6.99: LL"! __— " -——-'__"' =- «1;. fit 7- “Tu mil-1733., H-‘b [lb-m {550°}! MI'H TM “[1- ar-(Ou- ffi‘l— l-F- {and fih‘fip‘t "I +hc-H h‘w'l-hkn(.t"'w£) ' h Am gnu—37 rat'c baht-mg «.1 S+£udj 3'1“ 4-..- ._ mafia; VDIUHL enautu'ns +kc COMP'GJL‘D" Div-O! O : as“ .. WW 4: think-- Arv‘w- kw] - Lug ken-t Hal-In] u. .. 6.Q(.(0-022—0.oofq7):o.ll85' a... 4 w) PfioeoeM mm (Cmd'd.) Jaye/Z °" 6%.. we, + “Nu-kmw «nun-hm] Lawns-m 4”... TM... Aux-15,226 “3" “‘3‘ m! ‘ 2515' 3+ In. .- mw: ("IVs-hp), mph/RHWI+(636L%¥)F“'fl-m'qu+0.02um'lm1) '5“) = ~53to-7—S' 4533.22,: —‘tb.03 3Wm‘n 4_—-—— (c) A“ “"97 '“Me bum ' 7 cc .1 rind: rhdc. {or a. conho| vol" he. e - H“ 'H'rwd" abut- tpd-b'na 4u¢ refinacnnt H»: “Chan”: . I a O : Gal-)(cv + Eh 1N4; +I~sz\w.,] _ [ha‘h‘4 hank") _. AWL? Thu“ wdm \nv a: hJ -ng'. uh. {(kn-g- Nu) + u‘k51_ has ksi's _ IL“, k*4 .. um. {QMLMH- - BIN.) + (o. nanny-L) .— @,g,,11)(...fi‘g_ (o'llaffl‘ur) 12.:.+ amu— 5.1.073 — 9.0:. = 543.74 MAR lbn 9-; rghc‘scrgh'oq 1150 flfn/ m’m (SEC. (LL!) . "TIA-MI ._ ' - 14% H" (Dev - Ila-'h’M'hl = +°hl ‘——_'— ...
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This note was uploaded on 04/03/2012 for the course ENES 232 taught by Professor Hines during the Spring '12 term at Maryland.

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Homework 14 Soltuions F11 - PMSLEM lz.22- new”: . . L—...

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