Lecture_Notes_Rankine_Cycle - L I ' 'Ww AND 0N 3%Elfixs bum...

Lecture_Notes_Rankine_Cycle
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UrHI EFFECT OF PRESSURE AND TEMPERATURE ON THE RANKiNE CYCLE I 429 FIGURE 1 1 .7 Effect of pressure andv temperature on Rankine-cyde work. FIGURE 11.8 Effect of pressureand temperature on Rankine-cycle5 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 pumpwork= and the heat transfer to the steam in the boer._ we do by considering a "ccntr'cl 3surface around each of these components in In each case the thermodynamic modelis the steam tables, and the process is steady state'with negligible changes'fin'l'cinetie- Dutential energies. ' ' - ' - - = I- ' '- 'Control volume: Pump. _ - I - ' I I' - - .LInlet 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 obtainWP = MP; P1) = (0.001 01)(4000 - 10) = 4.0kJ/kg '_h] = 191.8 kJ/kg 'h; = 191.8 +4.0 #1958ngr; U{@100 Lt?For the turbine we have:Control volume: Turbine.Inlet state: P3, T3 known; state xed.E'ttt: .xzsae P4lmown . yiAnalysisEnergyEq.: 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 " ISolutlon .' 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 _. _ _ IThe 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. rAinveslrigmpsicoldgi.Sen.u.~ .\JC's"9.5:-r?\dImam 1.: yfamiliar c:Per 10 gelvestirse, ctratamrentcorreserlos accid:EstadistirFamilypersonas (psicolg'ildjferenmIrene HM(UL/US pmcome In thum um. .E3? EFFECT OF PRESSURE AND TEMPERATURE ON THE RANKIN CYCLE I 431transfer. Considering a control surface around the cendeuser, We haveq; = [24 h; = 2144.1 191.8 .= 1952.3 kJ/kgTherefore, _ . -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 pinupwork, and the heat transfer to the steam in'1e_boi1'e_r_;_We_do this by cons _ a controlsuiace around each of these components in turn. In'eac'h case the the ' odynainic'niodel -' ' ges'in'kin' 'dandEXAM 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 ' AnalysisSince 53 = 31, SolutionSubstituting, rim1813iqu ' w, T. van, P1) = 0.01614(600in = 69.70112 = 69.7 +1.8 ..= 71-5 Btu/11mi- or the turbine we have Control volume: Turbine. _ ._Inlet state: P3, T3 known; state - - 3Exit state: P4 known. ' I Analysis .Energy_Eq.: .713 - [I4nah-navr'n - - 7" '- ' DEVIATION OF ACTUAL CYCLES FROM IDEAL CYCLES n 443 - URE11.16 Tsgram showing effectasses between theer and turbine.steam leaving the boiler and u - u g the turbine, respectively. Note that the frictional effectscause an increase in entrop- ta:- nansferted to the surroundings at constant pressure canbe represented by proce : c. This ect decreases entropy. Both the pressure drop and heattransfer decrease the . ailabiliqr of the earn entering the turbine. The irreversibility of thisprocess 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 waterentering th oiler must be pumped to a highe s essure than the desired steam pressureleaving .1 boiler= which requires additional pump \ rk.' ondenser LossesThe losses in the condenser are relatively small. One of these I: u or losses is the coolingbelow the saturation temperature of the liquid leaving the condenser. is represents a lossbecause 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 thepumpsssosarDetermine the thernylejciency ofthis cycle. _ '- ' ' 3 ' _ '- '_" ' . - up; Maelsawegewmwsmman: 3.3 MPa: F5 '380w 1: 4.8 MPa= P3 its 5 MPa40C $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 CHANGEAs in previous exampless, for each control volume the model used is the steam tableand each pfcess is steady state with no changes in kinetic or potential Energy. This cycleis 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 - IEntropy Eq.: 36: sir? (Mahat- S tnfbiffa-. 1 ' .The efciency is *" UWJW? 1v} M 5/w, [15 h63?: = lam h = --h M- hSolution 5 65 5 6:From the steam tables, we get115 =_ 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: IControl volume: Pump.Inlet state: P}, 1; known; state xed.Exit Stare: P2 known.AnalysisEnergyEq.: 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 Tsdiagram for Example11.5. DEVFATION OF ACTUAL CYCLES FROM IDEAL CYCLES E , hits; a. who? asSince 525 = S], / _ awd] 6r! he then goes?312: h1= V(P2 Pll/ __ -_ -_Therefore, - I I "1/."_ h2s h1 _*-V(P2 '. P1) IWP . . - - '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. AnalysisEnergyEq.: 93': la ' 11;Solution . .Substitution gives I ._ I I _ . :._ IqH = 114 - 123 = 3213.6 _ 171.8 '= 3041.8 kJ/kg837.8 ' ' '= --- = 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 cycleis 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/{541Wesw i \0U+ WINS an \tmsf'cm 1wa law Je Lomigyzf F? (will 96,5; w [mm-v Eag/ tower Wary g Whom P toffgfpaghhgn41; 1%", g 91: ~%% waga mgawg ma; W (m yo{pok UL? Wadi/W 9%W/ a (WWW? HM" M- ISBO , 0Mmumpng a; m 09 {0% {w ewww: mi? WNW? mgleam @053? F l j 'fhug QM? iiim Phm)ng by 7 3,; Wm (Simm f M" 555551;)I @ if {@me 0 g 6%de P I0) Posmn'hr? 0(- mr leaf-am Wm mam-my(2) *Ewfmlb momzm aim-EM 0? 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