Homework 10 Solutions F11

Homework 10 Solutions F11 - PRO 3am 8.17 Eggs“): WOA‘GF...

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Unformatted text preview: PRO 3am 8.17 Eggs“): WOA‘GF us +92 working fluid. m o. Randome c. as w‘tHa Super beam! vapor enmrinfl +th Wbihe. The s+a+es a+ 44M. Wino. In Id- QM CUM-“5W 9“ ‘3‘ 0".“ Wed/“£61 Mud Wflm‘vae am Pump mama 6%16Lwcies an: ngen. END.) Dmxrmme Cg) We head {rm/uh rm for Ha; Sf'eamjmemw‘m (05“ k3 5* Shawn flow 1;:ng MMavI/mlomq'w Jamaich 4448 rode m” heed “frwnrfierfor mdemrerlperlcg 010 3mm Wsma. 1 [V "EgGmgamLJG- MODS; '. same as Example 6.10'1'i‘ems k—uaexceer 34+: v.8 am! We}. AMMsls: From iDrobl-cmazJ L“: 3321.4 LJ/lcf, and MB: 15!.53 far/1:5. 11a 5‘26ch CwHlepj w+$+rd£2 is {-uum wing 'Hae Aurbl‘ue c-CC—iciemij km 2. K4: = :5 k2":- ‘m- “tunr [415) ths From Probe? 81“: his = 200165113 - 'flws , has: 22124 karllcs. Smxilarlfl) = ws“ 3 _‘ “P \mrha ‘15”“1 ‘ “3+ (“MAO/mt, FNMA Pmb‘em 8.2, \nqs =1H.‘S‘1.Tku.s, 14” = [(953 kulkfi, (a) For 44% comm! emoiosmg ~H40. $+€am genemJ-or 04M 1 WWI-1AA '3) (9).“ = [Al—L“l =332L4_ 165.0 131555 k3”ch 0"“ M m ' \ Cb) The. +herm¢l cfi-{cfcmqj Ls h = 1W . QIBIM \.- “‘13 ® an/m 1 Chrkt\“CL\rl/\a\ “- 1044.3- 14.31 = {0391.9 w/Lj K: : 0.319(32.8_%) VL K93 For: We comolenser I. ‘ T . 61%,; m ( hL—Mfiwtp 9% = Mil/‘3 = 2212.1— ISI.53= 112049 fié—MM LTkese. “ism/b am be 00 not RAH/k We 0? Prablemm 8.2.413 5e 5 me P fie diet/+5 041 imvwsila'slim ivaxe. firbine am pump on ~Hu. {demo 6 Rmkmauide,‘ 8.28 Water is the working fluid in an ideal Rankine cycle with superheat and reheat. Steam enters the first—stage turbine at 1400 lbf/in.2 and 1000°F, expands to a pressure of 350 lbffin.2, and is reheated to 900°}: before entering the second-stage turbine. The condenser pressure is 2 lbf/in.2 The net power output of the cycle is l><109 Btu/h. Determine for the cycle (a) the mass flow rate of steam, in lb/h. (b) the rate of heat transfer, in Btu/h, to the working fluid passing through the steam generator. (0) the rate of heat transfer, in Btu/h, to the working fluid passing through the reheater. (d) the thermal efficiency. KNOWN: An ideal Rankine cycle with superheat and reheat operates with water as the working fluid. The net power output of the cycle is given. FIND: Determine the mass flow rate of steam, the rate of heat transfer to the working fluid passing through the steam generator and the reheater, and the thermal efficiency. SCHEMATIC AND GIVEN DATA: Q-m pl= i400ibfi’i112 T,=1000°F e 1 Steam Generator 3 p3 =p2 = 350 Ibrnn2 T3 = 900°1= p6 =p1 = 1400 Ibf/an2 Wcycle =lx109 Btu/h 5 ps =p4 = 2 lbf'lin2 x5 = 0 (saturated liquid) Problem 8.28 (Continued) — Page 2 T—S diagram p = 1400 Ibffinz ENGINEERING MODEL: 1. Each component is analyzed as a control volume at steady state. The centrol volumes are shown on the accompanying sketch by dashed lines. 2. All processes of the working fluid are internally reversible. 3. The turbine and pump Operate adiabatically. 4. Kinetic and potential energy effects are negligible. 5. Condensate exits the condenser as saturated liquid. 6. The steam generatOr and the compressor operate at constant pressure. ANALYSIS: First, fix each of the principal states. State 1-. p1 : 1400 lbf/in.2, r. = 1000°F —» From Table A-4E: h. = 1493.5 Btu/lb and 51 2 1.6094 Btuf(lb-°R) State 2: p2 = 350 [bf/in}, $2 = 51 = 1.6094 Btu/(lb-R) —> From Table A-4E (interpolated): kg = 1313.48 Btu/lb State 3: p3 = p; = 350 111171112, T3 : 900‘? —* From Table A-4E: ha : 1471.8 Btu/lb and 31 = 1.7409 Btu/(lb-°R) State 4: p4 = 2 lbf/in.2, s4 = s; = 1.7409 Btu/(lb-0R) —> From Table A-3E: x4 = (1.7409 — 0.1750)/l.7448 = 0.8975 and 114 = hfq. + x4hfg4 = 94.02 + (0.8975)(1022.1) = 1011.35 Btu/lb Problem 8.28 (Continued) —- Page 3 State 5: p5 =p4 = 2 lbf/in.2, sat liq. —> From Table A-3E: kg = kg = 94.02 Btu/lb and v5 = 125 2 0.01623 ft3/1b State 6: [15: h5+ V5(p6—-p5) 144 in2 fi2 3 he = 94.02 Btu/lb + 0.01623[f1t—b](1400 — 2(E’i] “3“” ‘= 98.22 Btu/1b in2 778 ft - lbf (a) The mass flow rate of steam is found as follows. Mass and energy rate balances for control volumes enclosing the turbine and pump give Wusz’i—hz): W12=m(h3‘h4)w and Wp:m(h6—h5) The net power of the cycle is chcle = Wll+W12 _Wp = mflhl ‘h2)+(h3 "h4)—(h6 415)] Solving for m m =—W°Y‘“~‘__— [(171 ‘ hz)+(h3 _ h4)_(h6 415)] Inserting values 1x109 % m = —.._____mh—__—_n [1493.5fl—1313.48§E]+[1471.8§E~1011.35PE)—[98.22% 44.0213“) lb lb 1b lb 1b lb m =1.s7x10‘1b/hr (b) The rate of heat transfer to the working fluid passing through the steam generator can be determined by applying mass and energy balances to a control volume around the steam generator to give QinSG = meal — h6)=(1.57><1061b/h)(1493.5 Btu/lb — 93.22 Btu/1b)= 2.19x109 Btu/h (c) The rate of heat transfer to the working fluid passing through the reheater can be determined by applying mass and energy balances to a control volume around the reheater to give 7 = m h —h =(1.57><1061b/h)(1471.8 Btu/lb — 1313.43 Btu/lb) z 0.249x10’ Btu/h InRH 3 2 .— (c) The thermal efficiency is Problem 8.28 (Continued) — Page 4 W. ,7 = W = (1x109 Btu/h)/(2.19><109 Btu/h + 0.249 x109 Btu/h) = _(_10.4100 41-00% Q“ The results of Problem 8.28 can be compared to the results of Problem 8.10 to see some of the eflects of reheat or: the performance of a Rankine cycle for cycles with the same net power output. In this case, reheat results in higher thermal efitciency, lower steam flow rate, and lower heat addition. PROéLEM 8. 24 gugwg: Wm‘w 1‘s Hoe workm M {m an [deal RaHLfMe Odo/[e wW‘ Katmai“. The 54-04% (22% 4M, Mei: +0 94% Whine gnaw curd 4W2. condenser ch+ are Specified. Fle‘. Memme. (a) +ke rake eat hm? mihim r L30? Mm -Howfm JCb) H46 Jr‘kcrmal efiCicicu ,(c) 'Hne. WW. 0% tear}- WM-Fcr-I’he condenser Pa“ 1:5 9 S+mm Wmtwd. a = 10 0 bar Turbine | Turbine l C’ C. C N'D'TA‘. -. fl \ . \ lump ,‘ caka = 0- 6: bar Egalweglwfi; MoPEg : Sea Example 8.3. ANAWSIS'. 95+) ‘Cbc all o-F-an Principal SW89. m" gig [I {3.7.}00 “:4809c mi? ht“: 3311.4 b?[kfi)5\=6.8287_ i :1 : = 3 "3 m— FL 1 bar) 3; 51 > x12. :SJ’ZZJ: OQQIQ J hi: 2984's Sim:- 123:7 19a», T3 =4 80°C =5 k3: 3438.9 Erika) 53218123 LJUfi-K SW94= pchob bm5q=sg=> x4154-‘51'r4 = 0.614913) M4: 1423.6 E $34r--S,$fli yfi SW51 136:0.05 bar’baklfiwid =b la S—JL'N‘” h; *1 L‘S+’U§CPG'~P5) S=rs|.ss hCI‘fkj M3 IOSNIVH“ “=3- :IS'LSB %+(J.ooeno’3)f§(Loo—omeflow W loam“! :tsi.53+;o.o(o = 191.5? [J‘ch Cod F‘or-f’lae Ccm+ml vo [ALMA enclosing 5mm 54nwa+ow Qt.“ = m [OH—1%) +(h3—hg] 31> awn.“ = (L‘rhbnmfhl) 6mm =(3321.4 atolsqnnesaa—zasqa) ‘ 1 Odd :: 3mm k311ch . m ‘ ‘2 ' _. We clef - Us) Tirte War aJ efihcuwmj ,5, yL_ Wmvoc‘fi/V“ = mm W lag-m —cm—m = 63&.b+lol3.3—-ro.ocp :waMg Wikg W3 mass 1 ; ‘41 4:32” W ma 0 «c /o>6_____________.__ fROGLEM 8‘26? ((5201613) LC) *"Fo-r fie condenser ‘ GEWZMCL‘4"L\5) :5 %= L14_.|45 ‘ = 2415.5Hr5;.33 : 2214M Lé‘, om M+er+0veLj )8 0mm = @I‘wlm - Wade/m = 3613.61 — 193% = 2214 .1 £311.13 iv The rcsul’rs 041+th and '5 c‘. ' \m +0 see Some vP‘HxQ Whiz? M figmiomgarffiw? SE: egrflaz ovate. G) PROISLEM 8.37 [(51:9th W +6 ' 44.4 [0' MM } ; ' " ‘ on: Wfim kw. 3m. 5m mmgmzw EH99! "Dakarmn'ym («B-{‘th 91; head' W‘fim luv k up 3+urm cnhiwj 'Hng +1%s+- s Mme) (lo)-Hae-Hnerml GMCiel-flcfi; 3% Ca) 'Hpe. me 5% hard- +mms{'€r r—Wa WWW per kg 8% Mm eW-I‘bc 'clf5+‘5+u§'£ Mine. 1 I (1—)) ff 3 IOOIOGI" _......D 77‘480‘6 , r — H — — -‘| '- . a izmkfim‘afié “CW W“ 5W“ W I Wva W: FIM,-Piac each principai 5M2. fl: Ioobwrl'r‘ : 480%: =5 h£135m|+§<3f|¢jJ 51:65282 15!vi aim F1: '7 Ioer s1 :51 o> «a: 81—541 20%”. 3 L12: 2984.8 Elk: W flab b Shiraz ‘. 249.0 M ‘3 =5 - 3 _ F? I 3 1 ’KB= 7‘ S43 :03qu In =zeofi’.8 Wiles) 363"3«Fs J 3 W4: fifooebm 3.1+. “some: nah =15L53 1:5?ch ,j _ 4 m hsw 4+U4CP5~PQ 3 ? msle \kT .— Yh =15153+CL00MXID mg (VD-05W” mom» 103nm = 51.33 + 0.918 = V5223 tr»?! 33$}: 6'. 19b: '7 bar, 5a)}.[icéuid a 145 =: {99712. Erl'lcg E silagwhwh ’U'C— a —3'° ~ 525-r - 1~ (9+ 0 P7 PL)=b?'7.22+(l.[080X£D )(100'7 103 ~707- k3 Ca) For +911. dream June/mkr at‘u [M] = h." M =(332L4 —-Io*z.52) = 2913.61 mij @cnlw'n LbJApp Enj was And (Marga \oMaMces 'I'o 'HAC cowhvlvohJLme endosn‘hjm 0% ww’fir hell/WV“ __ mth _ 64722—15213 _ O 2 52 3- hi— 5 h ZbB‘i-a—L‘SZLB “ ' _l For :Hrua wwhoi VOWWLQ, mmwsl‘uj m Wbme sjmfias i“: = (“Fl/‘1) +CI’LJ\Q/\zf {AB} = (332L4—Zb84.8)+ CL~O.2151)(2b84.8 —2009.8)= mobs Lctrl [c5 PRDBLEM 8.57 (Comf'd) cw . “19(ka (hawks-MM (kl—m 2 Cr— 0.2lsz\(152.’23r-i‘Sl.53')+(70'I.S'2-é‘i'f.223 =10.85 Lorna Efigfl: nefi“ fewer dbve/(DPQAJ par LUIH- mass en-kwfmsr ‘HLQ firfihsfigt Wade [m1 = Wefvgk-WP/m =be.3-—1D.as = 1155.3 lair/k3 Md Wflcrmic-ijciMcj is. n W‘flfl M1 11155.3 0 h“ @m/vk‘ : 26133 = 0'442 C44“?- A’) h- (c) For 45kg. Mensa/r“ Qo-ufi' : ml(1_-Lj3ck3fi\"4) (pm/W = [tr—Lj\C\/13-1¢q)=Cl—D.2]S'2)(zoo‘f.8-—13!.63) ‘ ‘ =1438A 15:] kg WWI 0V" rrzwbam 8&8 m: The} “Meal (flerrOVi-‘wve Paulame 04¢ 6}? Problem 8.3"] is modfjtyed +0 mutual; Whine. sfmge Mad 53M 00 [sem+ro[3‘\c, emicichcies 0.9 0.5, Emu: Answer -Hne same %\L£8+L0ns as in Prob m 8.3”] ‘JEOV‘ +he Momma! adds. swam/m $ mug“ 5&1 ._ s 5M2!ng mg; Mgpeg: Same as Examroif‘. 8.5. ewédg 515; pang.me am principal 5%. MN 12‘: Ioobcw,"n = 480%: =5 h(=3321.+ 143!ng 5, :(95282 Erika-K ufifnjpflm mew-few cmlciemfl 0+ Tha—FWngbme 3+6?- w;+u \ntg = 29:84.8 Lamas 41mm Pm‘olem8.3'l)h7_=2.812.l mug Sig—£93; P3=O\QG \owr, S35 2 52 .To 39.1“ SZJ-IMjK‘BY'PO1Qi‘Q{hTableA-3191:b_8[0%< ‘5 -— 3 1m 7‘35: Sis—Silt: 0.80M; Kass: 2049.0 Ella: 53— 433 . . WSMS -Haa saccmi— S‘l‘afie. Wrbx‘ho. efihumcg V13: \nL— whim—lass) = 2241.5 talk-5 Sm= gummy bar) Saul-.1? mid =5 Inqr-ISLSBKIHQ ~ ~ r - ~ . =Ltnss— Mm 4‘ m ugLMQ—Hm Lsme u.er ewe.wa M ( S 4) m v.5 = k4+Chgg—k4)/hf\:192.4l tang 333%; “Pb='1 bear) sodr. Lifiwld =‘> ML=GQ7.1Z1¢THC5 M]; W‘L‘HA MP1: OWN“ he} [Ch—rka 3h1=hL+CM15-L\5Vy\?:7lo.l Q") C‘le‘ c IMAM =2tp1t3 \cT(k54—fifijlk| M 003 5 —_ bit—TV. = 0.2048 mot ch/Waqr‘rkqm—gkaks) =%23b Jaflj wwiuigs, Mar/r9“ =(175)(h5*hq)+(k1—\\6\=y3.58 5&5 PM NW (mg thmvam‘ =q4mg mks (D W: :wfjlfis =o.3b4 C36.4°/o)‘——-_‘ M L~ = cx—gwn —kq3=1bw wlkg QW’W QC.) 9.005% ‘M mus resuvcs who. tomyar‘efikw 2 09 km 8.8"? +0 342 5W9, D’G‘HAE rifled: (NS! 1. o? in: w pun/up Ewe, \H-ies on mania PerOrmmmce . PraogpeM 8-7 gmowgg: Wed-er is +010. modems JRde in am icicoJ Ranku‘me (Adele Comnw PressWe QM “We Whit/1Q iuld fiat-e are Specr‘acied. The mass How ma Sham 15 5025“. m: DWmine (cu +he. Me}!- Power) Ga] We. mice 0? {Mai- Way/Later +0 Hue Sfeum @assiu +LWOLL-jk-er lomcrfim -H«e Warm! efiiciemuj, and ((L) M mow: u.) Panic of CODWMG wakw (m;st ~Hnrou5h ‘Hne condenser. flgwaMgfl 3; ngEN 9352.: Plzlbo bow rzo kjls Bofierl ___ __ Cuoimgwutcr ATCW;I'BOC- L----- Condenser F -- 0.08501‘ ‘\. I s I ENQINEERWG M0251: See Example SJ. Twp 3 50* “3"“4 AN&L15153 Fz'rs'i‘J 41m each of He {Princian sinks. T S+a+€t= 19,: {(90 Law) Sail/643m" $ hl=2saoxo lax/kg, s, = 5.2935 EEK m: F;'-‘-O.DB bad/L ‘5'135'.Lb xix—5%?- = 0.506” J kl: [(0315 \g/kj _ _ 1 7. 5+a+e3: g: 0.08%») 5d". Wand =3> kg: [7359 [(3ij Mi: Pq=H90 baP)L\4eaL13+’1§(fit—f33 3 105“ In WI. =I73.58 5+ (Looawmfigglfabow‘oefi [oar l W logN-J = £73.38+:b.|3 = no.0! Lair/1:3 (GO; The 1nd“ foam dweleeal Ls waadflz w; WP : «wring-w flag] k? fl 1;? 1 kw “201:5 “1350-!”l937-Q-(l‘iDpl-4‘TSBBD :3 m _ =1.112 mos kw WW“ KB} For 1W1 Shown fMSl‘VASWmmSK We ladder . (3% = W‘H’H“L\u¢\“0103(1580.£~140.Dl)=2.8Mxrosl«w¢ Le) K: W =Q388 [saw/QM [01) For HAL cowfid datum enchjsmj m ngwcassum‘na AL: =8 AT W v.“ : mum—M0 5”" ”° “° , ‘W C» ATM WqulA aw=4nq mar/Egg 49mm TabltA-lfi , ~ _§\20)§ \(a37.b—173.85) :2 14 mm» MW? CALI-HM [EM 335 "35" PKOfloLEM 6.3 kMoww: Waz’rer is +149. working jqu/M Eu aCarno+ Vmpor Powerajle. 30.4mm \i uid wka boiler mad mum vulgar 2% +44 Winefim Loo‘derfloscwe Lg spacfiliedfl’ke WWW preng is sPacCaCfedJmf W was 0w make 09 a+eam Lsgwm. FIND: bmvmue Ca) fihe-ervmd eflFicie )Ual +he. Mic mark who) (CW2 h£+ Fewer) aMCd) ‘l’iae Wait 5?- VIGO=+ Mm¥m~ ‘flr‘om “HAL Luau-{aha ‘ ass‘lpfl “through +Le condenser. Ecuadch e? envao Dag: Pr “501°” T .. if»; ~‘ - Cooling water Condenser 5d.13§""1d\ '3 PfPs-“O'OSW * pr ENGLNEEKIIOG MODEL '- U) Each cowPDnngr 15 analgcea as 0L cm+mi Volumm; 61+ 5+ s—{vivalz} Mt W5 am flaw/mat wevwslble. (3)1WLWLLM aMA +442 u Oferuk mj‘awamttfla) mafia am 994% {out energj)eFF-cck m negligible. P “'0 flflgflfilg; fag/5+) fig cash 019 +Ledar1'meipaf wales. 5mg l: 3 “90 bar; 54+.va =,'> L1,: 2580.L 1:1?!ng s}: 5.2%‘5 m/Lg-K 5+6”er : 0081004? s =5 =a’> 51‘542 ' ‘J 1 I x = 30.60 :1 {9145 -—~— F2, ms _5 ‘H, In: lb37. “:5 S+gia3: EB: 0.08Ear, 5325? = 5+)Ci “mm film“ Km?“ 5 -— 5 :5 “a: 3 "c3 =o.4I30J h3=11bu.4 [CG-HE 353'" $st 5+a/(e4: P? 2190 bear, Sat: “Sud ,cb [ALI fies“ k3“? Lax} The fixerml afiicicm.j ofi-Hae Carnaf— 0.3018 13 ‘3ch b5 SH. wiHa duh hm Tab‘e A35 TH =qu+ereobqr=b3053 K Mn: TM“, 0.085”: mm, K. m CD VL=L— 21—32:; =0.4G:3 (43.3%) ‘4 OD] The WOYLL ‘ra/‘rb 31 : yflhvug} 1 “950.141be W4: yum—M1,) 2550.5-1L37Ja Cc) The ne‘l' 'Fomr devdwed is ® Quack = wick—kg {LN—Lg] —_- (120‘553)[Czsao.b—Lang—(twat — mmfl . = 55,120 W H d.) For “We WSW: Qouu— = ‘5” (l4 {\«fiflioXngmb-n may r- 59» ENE) 1mr <——QM [.mszeuj ,vkawmywm a fivwwcwwm 5M) =0.Lmto- Z.'Tlnese results Can be coward @444 fig, Rankin: (Icicle, 9Q ’frobem 8. ‘Z. Lwr‘ 20.5I3 5”" PEOWEM 8.57 mggw: am isn‘mworlcm M I'm r mam}ch 5r (war de with 0% c 0364 «Pasteuer gin L3 5!) P w The, WA {law «we 0+ 9+ka Wx‘m Wél‘rs'i'vS‘l“ Mint LS 5‘!ng- ‘fkekmpzmfiure 99% Emma wMav-gassm Wrong Arkawwiow Kid-Wh- ‘ 09mm Am;wa (Lg-We raj-e 9-?- head” (Jul) 4m mass («who mica 0*} W. Dad-a. mst‘Fi 42L varwus )0 cmme Fl N12} :D.e—}~£1rvm a (a) ‘Hae no} addi’twni’gj Mam efiC'ICit/M J M (1001:1215 wad-w nggsiLxxj flywka ‘MQQKL' (n EQOL‘ componcxf i5 analgzed ILS' qwgm’ Vomm¢ 9+ wen-d3 5+q+¢\(2)AH processes o? 'HML work-{K3 Huikam. {wk-rle ravers'xlote, emep+f0r +4; engahséd‘m 'HArOu-al‘l 444‘- +W4P (a‘+h"°m"*\9 (OF-"0 “ma i“ N d‘m‘l ¥¢ckm+w hawk-ornfi) The +Uxbfu§cs J pufip_,o.w1 «Geelwmr 'keafir super-0.42 aiiabcfi'chngj .k‘l] Ethel—Cc. M Pow.an cuergg cHeC‘H are ht lC3the.(5? Condeuake “1+3 Hm. dosed hen-hr. and Mmdeuscr 54'- SO-gm’l'e’t “13“ 0.17 W rapedwe pressures. Amggms: F1'r3f,-C—ix Bach Prihaii’al sink- siaiglz T5 2:60 bat/0T. =SéoOC. r—b “I = 3495f} '5”ch 3 S'lZG'SIBZ 143/le 3:123:42; =10 bum" 5:5 ,Sz-Ssa _ = 9::— 1 z- » % XL__ 851,541_ 0.133% J h-z 2.745.! 6W3: P3: €3.08an 53=Sz=b x5: 53112:; 1 0.17533 L‘s: 2037.© U”? Sig/1:231 mzo‘os loW,50¢F.lqu?ok a 33 “c? “421.13.88kfl'll-j M: his '2’ kq'i'MQ-LPS" P4) m3 L05.”le 1153” 21133.3 +(:.oos4m’3)?§' (Leo’o‘ogyow 1 bar Loan-m = f'l3.88+ 10.13 2 Mom 1.5;ij Silk—6'" Pang 5“”; Te=§¢eyobwfitao°c me Tqbfe ks) kg: 77/ m1? M flaw bar, Sailqu/M --—‘> In? :TEZIBIQ‘HCS 533M TWO‘HWE; pmcess ——> lngnL—l = 75135! 1:31 cm ‘For-l'he W1 volume. “mug ~+fieoiesa4 comm mm hfa'L‘S q 171— 1610.01 -- ———-——- = (9.2.93 141—14.? 2.14s.I—%2.81 I W, -Por 44v; wrv’r‘mk Wham; cuman filo—WEN; SW we 1 m1[cL\.-\rsg+u—g3chfl,lguj I 1 1 law zgw 5:5 )Sc 34b5A-‘L’I4Sfl +(0.‘LO€9‘1](Z'14S.1“2.031031% Wis u — I : lAhasxaéLw PRDstEM 8 .57 (amino Fe 4M. mp r‘ \ (in. L {4 ICE _| kw WP _ m1 C 5"“ q) =on 3.)(Maoh113.581] [H15 ms 2 = [Os ' (E) -For‘ 44x12. Sir-Mm qujor ‘ 69;“ = m C14: “(A ‘ Q10) (3468.4 “£713 = 3133x105 kW 45—61:“ (c3 The—We;va ermich is Vk: WflkiQ/\£/®;v\ 3:445 XIO5 3.133XLDS) 7- COU Wm cam-ml volume, momma Adm. WW 0 = m {cl—@MBJr gha— Lug] 4: MW (hmw- humus M 2 WA! CIFH3M3+jhEFh41‘ 0W _‘('/lawhcw— [firn’w3 WM QAM‘WF Karmic“; : cm ATM W cm, =4J'I‘f ~FYOm TableA-l‘f 216736 law O'Y‘ fin :Q'Lok ls Conea‘Nzwmo)+(o.?~‘i3l):i.%2.81—113.88]{at W 5 (Ammc L8) \Jfkg z m ._-_ 233-“? Lfifits m 15,7.LL ...
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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 10 Solutions F11 - PRO 3am 8.17 Eggs“): WOA‘GF...

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