prob.vi-2.ver2

# prob.vi-2.ver2 - Problem 6-2 531133311 1 OCT 96 Part...

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Unformatted text preview: Problem 6-2 531133311 1 OCT 96 Part 3'. {3:33:13 #1 33 23303, 33333333 [3.33 13333) - bi = 1333.34 kJ3kg ' 3r — 3.1333r kagDK 3fE =1433.1 1:323 3135 = 2.3313 kakgﬂK 11I = 2331 4 333kg 3! = 3.3532 33333593 @ 33%;, Saturation (0.035034 P131113) 11r = 133.33 31% sf = 3.4331 3112,3331: 13\$: 133332-33 m333g hfl = 24234 31333 31% = 3.3143 3.133% 3E = 2531.3 133323 3: = 3.3323 UfkgoK State 1; 3l =3ﬂm = 2331.4 3.1333 State 21 32 I 31:- again = 3.3532 kLI'kgoK .5'2 — J! .‘1'; +I2Sg- = \$2 I} I: = 333 331 w 3.3333 h: = h:- + 33-133 112 =1?55.37 Ufkg 5333.3 3: h3 = 11,5333 =13E.33 kakg State 3" I13. = hJ + W? where: WP= 3311;333:313 WF= {LDDSBE-UI} mJIkgHT “1'6 MP3—. Dﬂ5ﬂ34MP3H 11353113111?! lhﬂ’aﬂlﬁkgﬂ-Imkgﬂlkﬂ IDDDI} WF= 13102 kag Page] 3H {13-2 SULN} State: 4: 114 = hﬂam = 13135.34 klfkg Wm: whabhr: ' wpump Wm = ht ‘ h: ' WP Wm = 2251.5 11.9155 - 1255.32 urkg + 2.592 kJﬂ-tg Wm = 992.22 1ng Qin=hl ‘h35=hl '(hg Qt: 2515 259 1:11:53 11H: ﬂﬁ= _993.23 klﬂtg = [LEE]? Qm 2512.255 kag I I 3500 Steam rats = — = = 3.694 kg 5mm 95] 82 3M 1" kg [hr J kWh“ Wad r Cycle #1 Same as cycle I, accept 111:1 stab: 3' and a different state 3 State 3: 53 = 54 = 55,319, = 2.1524 kJr'kgoK S1 — .5} SI+I33¢ =51=>xs= 5'1: £3 = {1.3423 h3 = hI+ xjhfz he, = 961% ILTJ'kg w11ﬂ=(h1‘hJ)-{h4-h3} ' Wm = {2251.4 klxkg - 1255.92 k11'kg)-{1295.24 151193 - 952.99 kJr'kg) Wm: 559.59 111115; Q51: 1‘1 ' I14 1}: 2251.4 1111153 - 12115.24 151223 or 1455.115 kag 111,: %= 665.59 I_1 159g 9 11.4592 in 1455.ﬂ5 kmcg Staam rate = 1 = [;][3mﬂseﬁ = 5.38 kg mag. H"... 563.595.! 1" kg [hr J kWhn Page 1 of 4 {6-2 SOLN} CHI: #3 State 1. ‘11I = hﬂ5m= 2?94.3 kJﬂcg State 1': hr = huwhnma = 2913135 klﬂcg ' State 2'. 32 = 3L. = swam:ng = 3.13334 3111:3013 Sat—£3 SI+X=SJ= 31:51.1:— 3. x2 = 3.11911 he = 131+ M2111: h1 = 1881155 kchg State 3: Same as cycle #1 State 4: 11., = hﬁsm = 1154.23 kIﬂ-rg Wm E wlurhinn ' Wm Wm= h]. - h1 - WP W: {l .UUSEE-UE m‘fkgﬂﬁ WEI-.UUSDHIil-hif’aﬂlﬂmfmif lhﬁaﬂﬂkga‘Nm’kgH lkHlﬂﬂﬂI} wp= 5.33 3133a . wmt = 2331.33 3131;; - 1333.35 11mg - 5.113 kag Wm= 111113.113 1313113 (1:11: hr ‘ hJ' = 1'11" (113 ‘E' WP] Qt: 2901315 kﬂkg - {133.33 1c]ka +5132 k]!ng Q1“: 2753.41 kJJ'kg Th3: Em; 1616.139 kakg = 0.3634 Qi" 27534131133 Steamm =_l_:[__l_][36ﬂﬂscc]=354kg3tﬁam 13.. 1111333333133 1hr kWhr. Page 3 31f 4 {En-l SDLN} Pan 1:: Ca.th Cycle eiﬁciency Tut 3U|5 game: ———=1——=.4594 1 Tram 555 11m "‘ “\$137M: 3’ “an i" "II-L1 PM a: I _I f 111111.69 1456.1 1455.? 1T45ﬁ ' 26” £455.? 1125.29 +1 {Ir-W3) TOTAL {mug} Part (I and summary: BEL. LIT! u I all " HE . _._.- —”=::—.—Bﬁa ﬂ" JEE Ja—I—nduna.‘.a..av|.u. jﬁmnpgp—I-ﬁ-ﬁ'I-n—FIII- uh! a- w I“ .11-u-r-r EIIII E 1 J Luz-Ia - Lu n."“‘"ﬂ':'£1'1‘ Jutﬁﬂn PH _.._._. _._ n..- _._ 3,“ H '- a»: u - - :_ ma-rw . 1w: . . . . . . . d-I a..1 3" .-.- 1.- . mimiﬁg+ girl—HJIIIILL}. m— +' mini? H 'Eﬂlmimlﬂ': LJuﬂii . .. F"... ' :P r' in“. 11"“ .. l ' b n; .T ' i - H-..“ * _._.I-II HaﬁrﬁnéJﬁé—él—n [mm . ‘* 'mm lawmﬁm Ami: ml. Til—LIL! - nnmx 1—r HIT . n . regal-9i WI?- ' Page 4 nfé {5-2 SﬂLN} ...
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## This note was uploaded on 02/02/2012 for the course PHY 4049 taught by Professor Acosta during the Fall '08 term at University of Florida.

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prob.vi-2.ver2 - Problem 6-2 531133311 1 OCT 96 Part...

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