Lecture_notes_rankine_cycle

Lecture_Notes_Rankine_Cycle
Download Document
Showing pages : 1 - 14 of 14
This preview has blurred sections. Sign up to view the full version! View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: L I ' 'Ww AND 0N 3%Elfixs bum Wm? am: we W _. \fxiffi 3mg??? +,m.gs3' 2 {UN burg} {wad LS almig K % (MM/{MA ( {Pmm J 9 WW W??? 6W?) VM*%mwawige@ + W mawg 04. Vagamgizaga , _ $905: F-(zWEV Pimw : wad mefs : najagmr FEM 3 w was? 6M mm #2 1;? 0F WM {0 gawggw EM .l CM {10% cgcje mam/2 FY1051 ewa' 63-day.) igvga 1 M4 WWW-(:2 {HMS 80+ Hf: WT m SUHZMK/ mode: far Fuauef' Lag/less = 09%.; 4~1JLEMQ T 3 jzffmi km? -"EV&%:/(m he awsazgaci <r EWC ' 'r/9\ @m mmhgn g 1%: 9% 9 ?"'w"-?i%r%wr WWW"??? Wir- (an m; 5.256 6a in vi. Qdf (-mmtq' T7: 'a'ifrc 5}. Thu hma - 3 {4% Ji'V'C-fmr ghtcwn a f 5 23 ?s=%"-r=aa, y\;2:e.zvz:;m {2351} CW 3% UKWMWWK bi? a Wig 3'?! 1 WT WWW: {mil Wynn f... _! 1a,, T WWW W wax??? kw L 3:33?- mmmm Lgz') mU (40555 UUPM ugl- g 'hjfi. 4 40% {S hf 3) Ixrmm Compresgm (1H) LS harzi aha mirme Eswgm, Hg; 3.33. +9 mef M wdwwam' i J ' If - .' ! P433358 Eb am My $353552? 5%; 5th Lg- { PM ML 3 mag 4}} {''kft mmfgzwf 7LP-Jr handles 'lihmgaii =79 Qk?'fx0( " Csrzigci} Lg. g... Lamkg {Mag 52gb UMQQM Wax Cu THE 9mm FAME 5%(35 #7312 Jim? ~* Moritz? WW agaw- (ma/WM? H mgmmg i 4% con (an m: I acex W Vapor WW6? Osgdf $354.6 m 1%; mum CanJMgm? ~ EH; (m _1f5{%337 Vaififfd 05456 1 QM _ 4-1 MMFK/ ComYrggg/m m Fwy Mi w -~%-* I? L33 f Emuw 3-5 (mmm mama M admm m Me? My. cam? g: P) 54* imam, megawatt: m {1&9ng LkW FrsW) {edim lnwxhg M in, , \ii/ gsvf .u \JQ/Hi Figs ER Pgr gawk Enigma"? IS meiiub, WEHMEE due, 40 a Sigh? decrease: in 39:97am ((37; I I . . gm m m} walla was Wig, iwmfwmwa mum Wag? (gig? {mm 7 5T Carm- 055E552}. (96/ no _ . I In \mt" i Wwf E'ig as oompfggsmj WQULA (Sm?!) 3< Saves 5% $1196 Maa WW {5W6 33 , U3an 16mm Mme; as a @qu WWI/r0 my AW??? {Emirm (5%) w {Wu P? if; mum/La a+ cum-mi $3 in am Macs-M 42- Ema 6% gamifg 1 luggath gg 6&9? PUMP Wk I , Boiler 15 bawtaiitg a. iwsx Mai? {figif (mm WV? 1cm (WWW Wm; mam yQ-QWS {6%, I Conmm ts a.th m WM W5? dwgw-Wi YQJMS hembio ax WWW WNW Sum as cit 5am! W22 0: Wosyh-etx- " - 9% TN. miza @932 (455 Law wig $45) { 14} '17:: O f mg" a: 359: 3: (MEI; VJ? 0? (1332:": PF?) Ug U1 : F4 Pumps, haaauman maltd MW?! ' Mmmw; 4 a - r: _. r Wdefed BMW {5 a) r) Lug 0 J 9 Q3 22 (2%: 9W) Inmwmsgxwa (f%@ _VTwam (35%) .4; 3ing i xyaf (3%: m5) _ n , ,. m x I I coma/grggp My {3 ? gr/K (gig 9; r; , FD! W WW!" 2}} RC wq-Wm 9%) or MM, ram" iWZx a) " W WE....?YEE1L Wm W In 3 4M 4* 7, --v-~-H""' W "M"; W EM wgwii FE>~mreJ {9W 55 W391 " Jug? ivynmsg lHQJLd wig {(U' 3125 Marga} xzxmmmewloig ( Ly: wg _ for a (my; gig-316m (Pig/ro @Indi F # w,th ............................................. N . I! II 0% g wwww we W . WWI _ Sigh: \~ Wu my? " "" 3V; 1 m s 3 ' . d. - t A K kkkk W. \// . am. WW area under ' q1x 1'73 G: qt 0: 9m . Wnzrmirhy mgrm boiif> @Whvr} " Digrid 15; m 1W6meshes {in We; meoW-{s : gm {nem if kgf [3g {0 gUP{YJUa~;i%E3%i N6 We MOST (Lomim a uid fdri a Causes Frgggw dmfl m +he bolder} We? Wmgm 2 H5, gag}??? *6 MForiivE C(wm mm {mm mgs- 151.3%? 793m gear gummdm as SiQam ows wsrik .W MEWS {B'ifak (WM hag-f M T5 beMfcrm! JID 5km m Emmi) " - This iMnm'aiii WWfo 1(1er Wm 9m damage; 0% 43%;: 3dza atezmsag . .1 :2. 2; / irrelrfimbt Vh w {M'm'wimm , Wan-ex fasgeg " WWSWWE WW5? ham??? 1% gawk Ceex e 3:3!" a 7 g z r m I U v E 7)? mg?) em? 1001' harm-r WT M Ma? R3, . .: z ":29? A - / 3:" Q gammy 5mg f4fjl 5 meaudmng +516 {mfvogcm egazmg, r11 : " 1 T we * Pam? lOSges; a boSeig ma g/EW'EUK ES mg/ xa'khyigmg (Sq/rtmleiF mag,- {Jim-a; 3;; Wait) o KM ermai HYMWEWS mg ugmwa Idrggfg wag-{WW 4' Davmhoa {mm Uifamegffjmfi ,? Mk # Mama Mia. ...-__._._..._.\_. mum. a mam wmgaua 'fb {95563: in 'zwfomeg Jr Pum PS _ a 0&7 ms 6 wdfng, Mow Samzxifm'? a M tit-ow! NW M35?! is {wgb MEAT We +0 Tsar. m _ 9 f I1 + AMENam... {km M [0m F M6 06 WW WEVE? La m9}; akmwiekv gm? w Rama 5% Wangjg Wham MUWhQ- Pans w swam 4M Edam am? w?" 6W Wm? Em; m Ufff a fbw_ Cmqwma. aUXihggg ggwgfmgf gRiS Qgfv m 1") me) VEme \ m m Egg-1.49% arm/Mes. UrHI EFFECT OF PRESSURE AND TEMPERATURE ON THE RANKiNE CYCLE I 429 FIGURE 1 1 .7 Effect of pressure and v temperature on Rankine-cyde work. FIGURE 11.8 Effect of pressure and temperature on Rankine-cycle 5 efficiency. I' EXAMPLE 2 In a Rankine cycle, steam leaves the boiler and enters the turbine at 4; MP2: and 400C. " The condenser pressure is 10 kPa. Determine the cycle eiciency. ' ' ' - To determine the cycle eiciency, we must calculate the turbine work, the pump work= and the heat transfer to the steam in the boer._ we do by considering a "ccntr'cl 3 surface around each of these components in In each case the thermodynamic model is the steam tables, and the process is steady state'with negligible changes'fin'l'cinetie- Dutential energies. ' ' - ' - - = I- ' '- ' Control volume: Pump. _ - I - ' I I' - - .L Inlet stare: P1 known, mutated-liquid; state xed; ._ ; ' I - I Exit stare: P2 known. I ' . Analysis _ . '. - EnergyEq.: w? = hg [11 _ _ . ._ m 90th:? EntEOPY-ECI-I .5: .=.51-:.'1 wuu u...- u-u- n. ELEVEN r uvvcn HIVU ncrmulzrv-u [UN 31: IzlwbVVIII'I PHI-nit LHANLII: 1101015102612 {"4 Elli/2100 3mm limit. 1 ? _. xm- Since S2 =S1= 4% {r . Solution Substituting, we obtain WP = MP; P1) = (0.001 01)(4000 - 10) = 4.0kJ/kg '_ h] = 191.8 kJ/kg ' h; = 191.8 +4.0 #1958ng r; U{@100 Lt? For the turbine we have: Control volume: Turbine. Inlet state: P3, T3 known; state xed. E'ttt: . xzsae P4lmown . yi Analysis EnergyEq.: w, = 123 ~ In; Entropy_Eq.:_ 1-34 :53 _ I ' Solution I I '- Upon substitution we get ' .523 = 3213.6 kJ/kg, 33' _=' 6.7690lekgK: _ __ . _ ._ ._ - .3: .wame LS. e_-r@ 6.7690 = 0.6493 47.5009, .154 : 0.81529 - h4 = 191.8 + 0315903918) 2 2121411;ng I _ __. I. I W Liar Wt = 113 "414 = 3213.6 _ 2144.1 2 10595 ' _ w, = w, wp =.1069.5 - 4.0 = 1065.5 git/kg __ -' ;_ Plate {.00 Finally, for the boiler we have: Control volume: Boiler. '- _ Inlet state: Pg, kg known; State lled; Exitstate: State-3 xed (as given). I i _ _ _ - . , Analysis - - I - ; 5M6 " I Solutlon .' gE-wm __ I_ . I _.__ _. Substituting gives I I :- . em = h3. lz'; = 3213.16}. 1953.: 3017.;3kjfk 'Wnet . I - . I :.' .- = = ' = I ._ . m 4.9 3017.8 _. _ _ I The at work could also be determined by calculating the heat rejected in the c.'.on<'1el'1SOS-iii.j . , and noting, from the rst law, that the net work for the cycle is equal-to- mnet 11w,F I. . , ._ P :0 @nhgml till??? ., 1. r A inveslrigm psicoldgi. Se n. u. ~ . \J C's "9 .5:- r? \d Imam 1.: y familiar c: Per 10 gel vestirse, c tratamrent correser los accid: Estadistir Family personas ( psicolg'il djferenm Irene HM (UL/US pm come In t hum um. . E3? EFFECT OF PRESSURE AND TEMPERATURE ON THE RANKIN CYCLE I 431 transfer. Considering a control surface around the cendeuser, We have q; = [24 h; = 2144.1 191.8 .= 1952.3 kJ/kg Therefore, _ . - W = 9H qr =-3017_.s .1952-.3-= "1065,5'kJ/kg' . i ' The condenser pressure is 1 lbf/in.2. Determine the cycle 3eicie'ncy. ._ _ To determine the cycle efciency, we must calculate the nirbine wo _ the pinup work, and the heat transfer to the steam in'1e_boi1'e_r_;_We_do this by cons _ a control suiace around each of these components in turn. In'eac'h case the the ' odynainic'niodel - ' ' ges'in'kin' 'dand EXAM PLE 1 1 .2 E In a Rankine cycle, steam leaves the boiler and enters the turbine at 600 1 volume: Pump. . . - In tstare: Pl known, saturated'liquid; Exit are: P2 known. I I ' Analysis Since 53 = 31, Solution Substituting, rim1813iqu ' w, T. van, P1) = 0.01614(600 in = 69.70 112 = 69.7 +1.8 ..= 71-5 Btu/11mi- or the turbine we have Control volume: Turbine. _ ._ Inlet state: P3, T3 known; state - - 3 Exit state: P4 known. ' I Analysis .Energy_Eq.: .713 - [I4 nah-navr'n - - 7" '- ' DEVIATION OF ACTUAL CYCLES FROM IDEAL CYCLES n 443 - URE11.16 Ts gram showing effect asses between the er and turbine. steam leaving the boiler and u - u g the turbine, respectively. Note that the frictional effects cause an increase in entrop- ta:- nansferted to the surroundings at constant pressure can be represented by proce : c. This ect decreases entropy. Both the pressure drop and heat transfer decrease the . ailabiliqr of the earn entering the turbine. The irreversibility of this process can be .. uiated by the methods a. tlined in Chapter 10'. A simil - 053 is the pressure drop in the i c iler. Because of this pressure repg-the water entering th oiler must be pumped to a highe s essure than the desired steam pressure leaving .1 boiler= which requires additional pump \ rk. ' ondenser Losses The losses in the condenser are relatively small. One of these I: u or losses is the cooling below the saturation temperature of the liquid leaving the condenser. is represents a loss because additional heat transfer is necessary to bring the water to its sa ' tion temperature. The inuence of these losses on the cycle is illustrated in the foil m 'g example, which should be compared to Example 11.2. ~ EXAM PLE 1 l .5 A steam power plant operates on a cycle with pressures and temperatures asdes'ignated in _ ' Fig. 11.17. The efciency of the turbine _is.86%, and the-eeie'ncy of thepumpsssosar Determine the thernylejciency ofthis cycle. _ '- ' ' 3 ' _ '- '_ " ' . - up; Maelsawegewmws mm an: 3.3 MPa: F5 ' 380w 1: 4.8 MPa= P3 its 5 MPa 40C $5 10'kPa = F: - 42C #11 -FEGURE 11.17 4 '- quematic diagram for n F: 99! 444 a CHAPTER ELEVEN POWER AND REFRIGERA'I'ION SYSTEMSWITH PHASE CHANGE As in previous exampless, for each control volume the model used is the steam table and each pfcess is steady state with no changes in kinetic or potential Energy. This cycle is show bu the Tws diang ofFig. 11.18. ' Control volume: Turbine. Inlet state: P5, T5 known; state xed. Em Stare: P5 known. Analysis . EnergyEq.: w, 2 [15 _ ha - I Entropy Eq.: 36: sir? (Mahat- S tnfbiffa -. 1 ' . The efciency is *" UWJW? 1v} M 5/ w, [15 h6 3?: = lam h = --h M- h Solution 5 65 5 6: From the steam tables, we get 115 =_ 3169.1 kIfkg, 35 .._ 6.7235 ._ . __ (mimwr (J <___ @ P5491: (0% 55.. = 35 = 6.7235 = 0.6493 +x117.500'9, - ' 1530.8098- 1:6. = 191.8 + 0.8098(2392.8) 2 2129.5 kJIkg ._ w. = 21,025 65.) = 0.86(3169.1 2129.5) = 894.1ng ' _. . For the pump, we have: I Control volume: Pump. Inlet state: P}, 1; known; state xed. Exit Stare: P2 known. Analysis EnergyEq.: wp=h2~h1 __ - .I - . _ Bum-op)! Eq.: ES The pump eieiency is ' ' _.0W8'-an{'idwi Wmf. _h21_h1__h2.s'hl _ WP _ 62h] 77? FIGURE 11.18 Ts diagram for Example 11.5. DEVFATION OF ACTUAL CYCLES FROM IDEAL CYCLES E , hits; a. who? as Since 525 = S], / _ aw d] 6r! he then goes? 312: h1= V(P2 Pll/ __ -_ -_ Therefore, - I I "1/." _ h2s h1 _*-V(P2 '. P1) I WP . . - - ' 73p '77? Solution @gow -. Substituting, we obtain u 1\ ' - . _______ _/__L_\ _ P ~ P N 0001009 5000 ~ 10 ' ' w = v( Emmi) = w =- 5,3 lekg . . 77p 0-80 . _ Therefore, WnEt = W; W1p 894.1963 _' 8873:ng I. I I. Finally, for the boiler: - ' ' - ' . Control volume: Boiler. Inlet state: P3, T3 known; state xed. Exit state: P4, T4 known, state xed. Analysis EnergyEq.: 93': la ' 11; Solution . . Substitution gives I ._ I I _ . :._ I qH = 114 - 123 = 3213.6 _ 171.8 '= 3041.8 kJ/kg 837.8 ' ' ' = --- = 2929 -- h 3041.8 . -' _ it . . . This result compares to the Rankine eiciency o<35.3%_. for the cycle .of' Example 11.2. ' ~ ' ' 3 ' - - ' EXAMPLE 1 1.5 E A steam power plant operates on a cycle with pressure and temperatures - Fig. 11.1713. The efciency of the turbine is 86%, and the efciencyofthe is_-80%L- Determine the thermal efciency of this cycle. ; . . , As in previous examples, for each control volume the model used is the steam tables, and each process is steady state with no changes in kinetic or potential energy, This cycle is shown on the TS diagram of Fig. 11.18.- ' ' Control volume: Turbine. Inlet state: P5, T5 known; state fixed. . _ E'...' urvA n 1...... am... ;_ [pdxizmg {3% 546%? yam/g! ag??? U-aiwf DEWMU kl) der Ping/{541 Wesw i \0U+ WINS an \tmsf'cm 1wa law Je Lomigyzf F? (will 96, 5; w [mm-v Eag/ tower Wary g Whom P toffgfpaghhgn 41; 1%", g 91: ~%% waga mgawg ma; W (m yo {pok UL? Wadi/W 9%W/ a (WWW? HM" M- ISBO , 0M mumpng a; m 09 {0% {w ewww: mi? WNW? m gleam @053? F l j 'fhug QM? iiim P hm)ng by 7 3,; Wm (Simm f M" 555551;) I @ if {@me 0 g 6%de P I 0) Posmn'hr? 0(- mr leaf-am Wm mam-my (2) *Ewfmlb momzm aim-EM 0? Wm 6d 40W" Smfs 01L WWW} {gmw m m-LWMWX w has Emmi Qualmr") This (Haggis, Wizama exthuanvgw 4? 64/0565 Miami? 19mm M...- M_ (7/) Sum_i,mhgam 12;, my T-mpe/gbwg (mam Vmh) WWI-"V a? wmh hag-E" 63 Wmi {a (z-Wm m baiir can go mama \Q\o kefmwg gaggle? SUPU/Waht *h/ Siam +0 A hwrar T (Tef- lmtad {if T . F "sham arm FWEci-M a-ngrmaa \1\ WON-x myuj Mfr may? am mamas; (area mm 33'?) 73) Maa away e; Eaax W\ HQWc-A EVQ Mange}- r 0M, MMMa Wok fa demeam imravwm comm mg Sf-if W? (Quaxw m U? :3 WWW L?) l him-K, 76" Mam-h 93. J: YM:=E~z:-alm%mi (=17 Hi3 Mair T awawa af Mbie/ emr XS v 513% . ' W (EMQM WM / . \ {x m Oremo ) In @3131 83; ODmd" '3 P of m MR f/ \\\ ,3? 3 JEN, WM? a-U40i'm~%em%mr i h mamas m T adv Whip-5v 100M? I f s Wage; (whadn macs, TM" mhxOn my \5 1mm; xrer' in 964m c mus Hm, +~amm MM war) For & erxm WWW, mm- T . TE? 96?! Shif T QM Wig 3v maiwc/ LonHM; 0% - {>4er int/ream (am (an 00 wcccoi ECU/3"" mmma, #er SW} . DWhd$z E? CAh W WEQMWE (wwh h) {97111965441000 MW) I 1 w Wait} mm Omar?th- (1+ Sibzigzzrs-mi V U? 7 a? w; WM OH 1 W" r T % Vow? T ~ \ xix. ...
View Full Document