HW3_solution

HW3_solution - PROBLEM 8.6 W: Wad-er 11er working Fluid...

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Unformatted text preview: PROBLEM 8.6 W: Wad-er 11er working Fluid I‘M am Ideal Rawhide cycle. The COW Pressure and 4449; Wham m La $+wfe are sfecified. EIMD‘. Bekrmfiacw) Hm ne+ work per wm- was: 04: Show: Plowfl») fke heat ‘Immcer {Mr un1+ mass oat s+eam How +hmu7h we (oedema) 4m. ‘Haermal eH-Cuevnc ,aM (0!) H46 (new!— +runr€er +0 coolmfl adder pMSMg Harawy e Condenser Fuumf mass 09 8+cam Wwyed. 5c EMA! we : 61in panama =l80ba.r u’ 3 '0 \ Assgflpmbls: Sac. Example. 8-l. W: F‘irsh fix each of We pru‘mcipM ssh-‘25, QWI: n:18MPo.,$cd-.v¢por =§> k: 2509.1N/k5, 3.6.1044 Icing-K 5mg 2; Pl: 6 lcP-kd szas, => x1: ‘13:"- : 0536‘), “I: 1569.4 «(II/<1 1. M: )7; =9 kPwlsa/fi Nina =x> k3: 151.53 kI/Eg W he: 2" I"; +1r3 (Fifi—PS) = 15!.531-31- (Loowxzo‘3 i453)(:e>o- o.oe)w [ = 51,534. (94¢ =IH.b4 511% (a) The ne+ work duocayazd per und- macs cf sham How is lkJ‘ ' Jo’M-m IOSNIVHI -lbar “1%. d< we - (250fi.l—156%4)- (lemw- 151.53) =?2/.& lekg (a The raj-e a? laud +mnr1¢er 19w (MAH- Mass 5+ Sham \Lloub ~H¢muflh e boilgr is (in? =Chf‘4q) = 29044- 159.64: 23301. 5 143-1 {L3 813“ M (a) The Hawmal eflficienud is weak/wk GALU ¢ VL nigh/m 2337.5 = 0.3614- (3a4-%)(_____£ X—u PROELEM 2.9 (6071+ ‘d\ can The, 511119er + Murat!" wo!‘ an! L. comm“ k a {0 mg to er 07455: 7 {’hrauglaflu. 6%: z hf!“ : 15b$4~15153 \ = 1417.7 WW5 QWM 947. PROBLEM 8. ’43 Km: The [dead Ranlu‘ue. aide of- ?f'oblem 8.6 is modified 40 fmclude 5W. (neat-ed v01)" en+ewhg +{4e 41rd {wrbdae We and we open Redwaier beajekr, Sameal 11'7u1'ol «ifs {Me Redwwfer heafer and Mar data. are Mum. me: Dek/mfne 7‘01" {he mom‘h'cd 01:1:th “Hue nd'worlc per kg of dam How “death Hm firs!" Wthe shige ,(b) +142 Haemal efiFicieucgJand (672 He [4 1511015961” 40 quhkg wafer 1‘14 Mmdcmer Per Lg of 5 2am Flow ewlu‘mg flue #rcflurbfne 5W. scwmc i mum D4 : ASSQMEHQN§1 Same as Example 8.5, encepf‘ Hm {4Mch singer operative: in an, I‘MWM'lg revers'wlc Munch MAR/51$: instfix each 01’» Hat Principal sf-a-fcr. 5.13134: 1», = 18m?“ = (80 low 1, T, =$éo°c w l4.=3qqq.‘/ lair/(c5, s, 4.31392 kJ—ch-K 51m: F2: 114m =Iolowr ,s‘=s, =5 76,: 0.967 J 1111:27/11, La'lkg Staci! F,= bLPa =o.oebmgs, =4; => x3: 0.7578.) 113 =1982.3 1:31ch Mai: fit = 0.0% bar, sat. lixuid => ‘1‘, = /.9'I.5‘3 EJ‘lkg J 1); =I.ooeqmo‘3m3/é5 51115-5 ks ‘ “"1519: "PH = 151.53 {- (Loafing3 :73) ( Io — 010:.) Law I :151.sa+1.eo = 152.53 kIlkf, m1 Pa = ID bar-5,501: li’iuid => ‘1“ =762.6( b71145 , ’U; = (.12731‘10‘3 ma/Lg m: {2, :3 =180bar L11 "" L1. 4' 1’}. (P749) =(752-5!) +(l.t213x16’)(190-Lo) hool =78137 U115 (A) F1813 Off) 111») macs and. energy rode, bahnces +0 fie wnfrol volume enclosfij The open hm r, +910. {rad-Cam «d ef- How exhad-ed of 7. is _ hs- ks - 76181-15253 5 “1: I"s 1711.4, #5253 The, W work devetoped b9 ~Hw. Mbfiae’ per unlf mass flow aria/1143 +1112 First {undoqu shat is losN/mz' [bur I I '53— l [0351mm = 0.2385 3'77. PROMEM 6W3 CCM‘A) Mm; (tn-m + (1-9) (Lyn) = (3qu- 2714.6) +(0.7ers)(21u.b—1qs2.3) -— 1268.7. NW5 SlMilarlj, for-Me puwps Wyn/1%,: ("SH/15 449) 4' (k7‘hb) = (away/525345153) +(76/97 — 742.81) :1992. ILU/ka W5, 4‘“; NJ. work is W dz _ Wt _ ' \ fly;— _ —— - _ 1298.5 kit/kg Wanda/m . . WM l (b1 For ng warloth fluid yassmg ~H/Lrouqk Hae S‘Feam gememfor {fix - _ WM - In; L‘1 - aquq.q-7elnv 12662.4 Lcr/kj Thus) 'Hmo Haermd eCFicievxoj is “hack/15H [2(983 :‘—‘—*‘ z—‘I—‘ 7-0.4 7.6°a H“ K cum. 1662.4 w (L! 0““ (m For +612 war-(619:9 \qwid PMSMi ‘HAr'OkglA ‘Hae Condenser" (in WI -. (www.3— In”) t = (o.7ers)(mez.3—tsx.ss) = new kJ/ch‘__________________j§:*_/i~, 8—73 ...
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HW3_solution - PROBLEM 8.6 W: Wad-er 11er working Fluid...

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