Lecture_Notes_Rankine_Cycle - L I ' ’ 'Ww AND 0N...

Lecture_Notes_Rankine_Cycle
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Mk # Mama” Mia. ...———-—__._._..._.\_. “mum. a mam» wmgaufiaé 'fb {95563: in 'z‘wfomeg Jr Pum PS _ a 0&7 ms 6 wdfng, Mow Samzxifm'?‘ a M tit-flow! NW M35?!“ is {wfifigfib MEAT We +0 Tsar. m _ 9 f I1 + “AMEN—am... {km M [0m «F M6 06 WW “WE‘VE? La m9}; akmwifie‘kv gm? w Rama “5% Waflngjg Wham MUWhQ- Pan‘s w swam 4M Edam am? w?" 6W Wm? Em; m Ufif‘é‘ff‘ a fbw_ Cmqwmfia. aUXihgggfi’ ggwgfmgfif’” gR‘fi‘iS Qfiflégfv 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 400°C. " The condenser pressure is 10 kPa. Determine the cycle efliciency. ' ' ' - To determine the cycle efiiciency, we must calculate the turbine work, the pump work= and the heat transfer to the steam in the bofler._ 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 fixed; ._ ; ' 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:” Izlwb—VVIII'I PHI-nit LHANL‘II: 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 fixed. E'ttt: . xzsae P4lmown . flfififiyi 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 filled; Exitstate: State-3 fixed (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 flat work could also be determined by calculating the heat rejected in the c.'.on<'1el'1SOS-iii.j . fl, and noting, from the first law, that the net work for the cycle is equal-to- fimnet 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 ( psicolég'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 3efiicie'ncy. ._ _ To determine the cycle efficiency, we must calculate the nirbine wo _ the pinup work, and the heat transfer to the steam in'fl1e_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' 'da‘nd 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, rim—1813i“qu ' 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'fln - - 7" '- ' DEVIATION OF ACTUAL CYCLES FROM IDEAL CYCLES n 443 - URE11.16 T—s 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:- n‘ansferted 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 . ailabili’qr 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 firepg-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 influence 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 efficiency of the turbine _is.86%, and the-efieie'ncy of thepumpsssosar Determine the thernylefljciency ofthis cycle. _ '- ' ' 3 ' _ '- '_ " ' . - up“; Maelsawegewmws mm an: 3.3 MPa: F5 ' 380w 1’: ©4.8 MPa= P3 its 5 MPa 40°C $5 10'kPa = F: - 42°C #11 -FEGURE 11.17 4 '- quematic diagram for n F: €99!” 444 a CHAPTER ELEVEN POWER AND REFRIGERA'I'ION SYSTEMS—WITH PHASE CHANGE As in previous exampless, for each control volume the model used is the steam tableé and each pfécess 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 fixed. Em Stare: P5 known. Analysis . EnergyEq.: w, 2 [15 _ ha - I Entropy Eq.: 36: — sir? (Mahat- S tnfbifffla -. 1 ' . The efficiency 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 ._ . __ (mimfiwr (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 fixed. Exit Stare: P2 known. Analysis EnergyEq.: wp=h2~h1 __ - .I - . _ Bum-op)! Eq.: ES The pump efiieiency is ' ' _.0W8'-an{'idwi Wmf. _h21_h1__h2.s“'hl _ WP _ 62—h] 77? FIGURE 11.18 T—s 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 “@gflowfi -. 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 894.1”963 ——_' 8873:ng I. I I. Finally, for the boiler: - ' ' - ' . Control volume: Boiler. Inlet state: P3, T3 known; state fixed. Exit state: P4, T4 known, state fixed. 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 efiiciency ofi<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 efficiency of the turbine is 86%, and the efficiencyofthe is_-80%L- Determine the thermal efficiency 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 T—S diagram of Fig. 11.18.- ' ' Control volume: Turbine. Inlet state: P5, T5 known; state fixed. . _ E'...'£ fi‘u‘rv—A n 1...... am... ;_ [pfldflx‘i‘zm’g {3% 546%? yam/fig!” flag??? Ué-fiafiiwf DEW’MU kl”) der Ping/{€541 Weswfi i \0U+ “WINS an \tmsf'cm 1wa law Jfle Lomigfléyzf F? (“will £96, 5—; w [mm-v Eag/ tower “Wary fig Whom P tofffigfpaflghhgn 41; 1%", “g” 91:“— ~%—% wagéafifl mgaw’g ma; W (m yo {pokfi UL? Wadi/W 9%Wfl/ a (WWW? HM" M- ISBO , 0M «mumpng a; m 0—9 {0% {w ewww: mi? 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