How Steam Turbines work to Generate Electricity

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Published Date:25-10-2017
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6 SteamTurbines 6.1 INTRODUCTION In a steam turbine, high-pressure steam from the boiler expands in a set of stationarybladesorvanes(ornozzles).Thehigh-velocitysteamfromthenozzles strikesthesetofmovingblades(orbuckets).Herethekineticenergyofthesteam is utilized to produce work on the turbine rotor. Low-pressure steam then exhauststothecondenser.Therearetwoclassicaltypesofturbinestagedesigns: the impulse stage and the reaction stage. Steam turbines can be noncondensing or condensing. In noncondensing turbines (or backpressure turbines), steam exhausts at a pressure greater than atmospheric. Steam then leaves the turbine and is utilized in other parts of the plantthatusetheheatofthesteamforotherprocesses.Thebackpressureturbines have very high efficiencies (range from 67% to 75%). A multi-stage condensing turbine is a turbine in which steam exhausts to a condenser and is condensed by air-cooled condensers. The exhaust pressure from the turbine is less than the atmospheric. In this turbine, cycle efficiency is low because a large part of the steam energy is lost in the condenser. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved238 Chapter6 6.2 STEAMNOZZLES g The pressure and volume are related by the simple expression, PV ¼ constant, for a perfect gas. Steam deviates from the laws of perfect gases. The P-V relationship is given by: n PV ¼ constant where: n ¼ 1:135 for saturated steam n ¼ 1:3 for superheated steam For wet steam, the Zeuner relation,  x n ¼ 1:035þ 10 (where x is the initial dryness fraction of the steam) may be used. All nozzles consist of an inlet section, a throat, and an exit. The velocity through a nozzle is a function of the pressure-differential across the nozzle. Consider a nozzle as shown in Fig. 6.1. Assume that the flow occurs adiabatically under steady conditions. Since noworkistransferred,thevelocityofthefluidatthenozzleentryisusuallyvery smallanditskineticenergyisnegligiblecomparedwiththatattheoutlet.Hence, the equation reduces to: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼ 2ðÞ h 2h ð6:1Þ fg 2 1 2 where h and h are the enthalpies at the inlet and outlet of the nozzle, 1 2 respectively. As the outlet pressure decreases, the velocity increases. Eventually, a point is reached called the critical pressure ratio, where the velocity is equal to the velocity of sound in steam. Any further reduction in pressure will not produce any further increases in the velocity. The temperature, pressure,anddensityarecalledcriticaltemperature,criticalpressure,andcritical Figure6.1 Nozzle. Copyright 2003 by Marcel Dekker, Inc. All Rights ReservedSteamTurbines 239 density, respectively. The ratio between nozzle inlet temperature and critical temperature is given by: T 2 1 ¼ ð6:2Þ T nþ1 c where T is the critical temperature at which section M ¼ 1. Assuming c isentropic flow in the nozzle, the critical pressure ratio is:  n n21 P T 1 1 ¼ ð6:3Þ 0 P T c c 0 where T is the temperature, which would have been reached after an isentropic c expansion in the nozzle. The critical pressure ratio is approximately 0.55 for superheated steam. When the outlet pressure is designed to be higher than the criticalpressure,asimpleconvergentnozzlemaybeused.Inaconvergentnozzle, shown in Fig. 6.2, the outlet cross-sectional area and the throat cross-sectional areas are equal. The operation of a convergent nozzle is not practical in high- pressure applications. In this case, steam tends to expand in all directions and is veryturbulent.Thiswillcauseincreasedfrictionlossesasthesteamflowsthrough the moving blades. To allow the steam to expand without turbulence, the convergent–divergentnozzleisused.Inthistypeofnozzle,theareaofthesection fromthethroattotheexitgraduallyincreases,asshowninFig.6.1. The increase in area causes the steam to emerge in a uniform steady flow. Thesizeofthethroatandthelengthofthedivergentsectionofeverynozzlemust be specifically designed for the pressure ratio for which the nozzle will be used. If a nozzle is designed to operate so that it is just choked, any other operating condition is an off-design condition. In this respect, the behavior of convergent and convergent–divergent nozzles is different. The temperature at the throat, i.e.,thecriticaltemperature,canbefoundfromsteamtablesatthevalueofP and c s ¼ s . The critical velocity is given by the equation: c 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼ 2ðÞ h 2h ð6:4Þ c fg 1 c where h is read from tables or the h–s chart at P and s . c c c Figure 6.2 Convergent nozzle. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved240 Chapter6 6.3 NOZZLEEFFICIENCY The expansion process is irreversible due to friction between the fluid and walls ofthenozzle,andfrictionwithinthefluiditself.However,itisstillapproximately adiabatic as shown in Fig. 6.3. 0 1–2 is the isentropicenthalpy drop and 1–2is the actual enthalpydrop in the nozzle. Then the nozzle efficiency is defined as h 2h 1 2 h ¼ n 0 h 2h 1 2 6.4 THEREHEATFACTOR Consider a multi-stage turbine as shown by the Mollier diagram, Fig. 6.4. The reheat factor is defined by: Cumulative stage isentropic enthalpy drop R:F: ¼ Turbine isentropic enthalpy drop P 0 Dh stage ¼ 0 ½ Dh turbine  0 0 0 h 2h þ h 2h þ h 2h 1 2 3 2 3 4 ð6:5Þ  ¼ " h 2h 1 4 Figure6.3 Nozzle expansion process for a vapor. Copyright 2003 by Marcel Dekker, Inc. All Rights ReservedSteamTurbines 241 Figure6.4 Mollier chart for a multi-stage turbine. Since the isobars diverge, R.F.. 1. Thereheatfactor maybeusedtorelatethestage efficiency andtheturbine efficiency. Turbine isentropic efficiency is given by: Dh h ¼ ð6:6Þ t 0 Dh 0 where Dh is the actual enthalpy drop and Dh is the isentropic enthalpy drop. From diagram 6.4 it is clear that: P Dh ¼ ½ Dh stage Dh ¼ðÞ h 2hþðÞ h 2hþðÞ h 2h 1–4 1 2 2 3 3 4 if h (stage efficiency) is constant, then: s P P 0 0 h Dh h Dh s s stage stage h ¼ ¼ t 0 0 ½ Dh ½ Dh turbine turbine ð6:7Þ or h ¼h £ðR:FÞ: t s Equation 6.7 indicates that the turbine efficiency is greater than the stage efficiency. The reheat factor is usually of the order of 1.03–1.04. 6.5 METASTABLEEQUILIBRIUM As shown in Fig. 6.5, slightly superheated steam at point 1 is expanded in a convergent–divergentnozzle.Assumereversibleandadiabaticprocesses.1–2is Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved242 Chapter6 Figure 6.5 Phenomenon of supersaturation on T–S diagram. thepathfollowedbythesteamandpoint2isonthesaturatedvaporline.Here,we can expect condensation to occur. But, if point 2 is reached in the divergent sectionof thenozzle,then condensation could notoccur until point 3is reached. At this point, condensation occurs very rapidly. Although the steam between points 2–3 is in the vapor state, the temperature is below the saturation temperatureforthegivenpressure.Thisisknownasthemetastablestate.Infact, the change of temperature and pressure in the nozzle is faster than the condensation process under such conditions. The condensation lags behind the expansion process. Steam does not condense at the saturation temperature corresponding to the pressure. Degree of undercooling is the difference between the saturation temperature corresponding to pressure at point 3 and the actual temperature of the superheatedvapor at point 3. Degree of supersaturation is the actualpressure atpoint 3divided bythesaturationpressure corresponding tothe actual temperature of the superheated vapor at point 3. Illustrative Example 6.1: Dry saturated steam at 2MPa enters a steam nozzle and leaves at 0.2MPa. Find the exit velocity of the steam and dryness fraction. Assume isentropic expansion and neglect inlet velocity. Solution: From saturated steam tables, enthalpy of saturated vapor at 2MPa: h ¼ h ¼ 2799:5kJ/kg and entropy s ¼ s ¼ 6:3409kJ/kgK 1 g 1 g Copyright 2003 by Marcel Dekker, Inc. All Rights ReservedSteamTurbines 243 Sincetheexpansionisisentropic,s ¼ s :i.e.,s ¼ s ¼ 6.3409 ¼ s þ 1 2 1 2 f2 x s ,wherex isthedrynessfractionafterisentropicexpansion,s isthe 2 fg2 2 f2 entropyofsaturatedliquidat0.2MPa,s istheentropyofvaporizationat fg2 0.2MPa.Usingtables: 6:340921:5301 x ¼ ¼ 0:8595 2 5:5970 Therefore, h ¼ h þ x h ¼ 504.7 þ 0.8595£ 2201.9 ¼ 2397.233kJ/kg 2 f2 2 fg2 Using the energy equation: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼fg 2ðh 2h Þ 2 1 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ð2Þ£ð1000Þ£ð2799:522397:233Þ fg or: C ¼ 897m/s 2 Illustrative Example 6.2: Dry saturated steam is expanded in a nozzle from 1.3MPa to 0.1MPa. Assume friction loss in the nozzle is equal to 10% of the total enthalpy drop; calculate the mass of steam discharged when the nozzle exit diameter is 10mm. Solution: Enthalpy of dry saturated steam at 1.3MPa, using steam tables, h ¼ 2787:6kJ/kg; and entropy s ¼ 6:4953kJ/kgK: 1 1 Since theexpansionprocessisisentropic, s ¼ s ¼ s þx s ,hence 1 2 f2 2 fg2 dryness fraction after expansion: 6:495321:3026 x ¼ ¼ 0:857 2 6:0568 Now, the enthalpy at the exit: h ¼ h þx h ¼ 417:46þð0:857Þ£ð2258Þ 2 f2 2 fg2 ¼ 2352:566kJ/kg Therefore enthalpy drop from 1.3MPa to 0.1MPa ¼ h –h ¼ 2787:6–2352:566 ¼ 435:034kJ/kg 1 2 Actual enthalpy drop due to friction loss in the nozzle ¼ 0:90£435:034 ¼ 391:531kJ/kg Hence, the velocity of steam at the nozzle exit: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼fg ð2Þ£ð1000Þ£ð391:531Þ ¼ 884:908m/s 2 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved244 Chapter6 Specific volume of steam at 0.1MPa: 3 ¼ x v ¼ð0:857Þ£ð1:694Þ¼ 1:4517m =kg 2 g 2 (since the volume of the liquid is usually negligible compared to the volumeofdrysaturatedvapor,henceformostpracticalproblems,v ¼ xv ) g Mass flow rate of steam at the nozzle exit: 2 AC ðpÞ£ð0:01Þ £ð884:908Þ£ð3600Þ 2 ¼ ¼ ¼ 172:42kg=h: x v ð4Þ£ð1:4517Þ 2 g 2 Illustrative Example 6.3: Steam at 7.5MPa and 5008C expands through anidealnozzletoapressureof5MPa.Whatexitareaisrequiredtoaccommodate a flow of 2.8kg/s? Neglect initial velocity of steam and assume isentropic expansion in the nozzle. Solution: Initial conditions: P ¼ 7.5MPa, 5008C 1 h ¼ 3404.3kJ/kg 1 s ¼ 6.7598kJ/kg K 1 (h and s from superheated steam tables) 1 1 At the exit state, P . P ¼ð0:545Þ£ð7:5Þ¼ 4:0875MPa; and 2 c therefore the nozzle is convergent. State 2 is fixed by P ¼5MPa, s ¼ 2 1 s ¼ 6.7598kJ/kgK 2 3 T ¼ 4358K, v ¼ 0.06152m /kg, h ¼ 3277.9kJ/kg (from the super- 2 2 2 heated steam tables or the Mollier Chart). The exit velocity: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼ ð2Þ£ð1000Þ£ðh 2h Þ fg 2 1 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ð2Þ£ð1000Þ£ð3404:323277:9Þ ¼ 502:8m/s fg Using the continuity equation, the exit area is mv ð2:8Þ£ð0:06152Þ 2 24 2 A ¼ ¼ ¼ð3:42Þ£ð10 Þm 2 C 502:8 2 Illustrative Example 6.4: Consider a convergent–divergent nozzle in which steam enters at 0.8MPa and leaves the nozzle at 0.15MPa. Assuming isentropic expansion and index n ¼ 1.135, find the ratio of cross-sectional area, the area at the exit, and the area at the throat for choked conditions (i. e. , for maximum mass flow). Copyright 2003 by Marcel Dekker, Inc. All Rights ReservedSteamTurbines 245 Solution: Critical pressure for maximum mass flow is given by Fig. 6.6:  n  8:41 n21 2 2 P ¼ P ¼ P ¼ 0:8 ¼ 0:462MPa c 2 1 nþ1 2:135 From the Mollier chart: h ¼ 2769kJ/kg 1 h ¼ 2659kJ/kg 2 h ¼ 2452kJ/kg 3 Enthalpy drop from 0.8MPa to 0.15MPa: Dh ¼ h 2h ¼ 276922452 ¼ 317kJ/kg 123 1 3 Enthalpy drop from 0.8MPa to 0.462MPa: Dh ¼ h 2h ¼ 276922659 ¼ 110kJ/kg 1–2 1 2 Dryness fraction: x ¼ 0.954 2 Dryness fraction: x ¼ 0.902 3 The velocity at the exit, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼fg ð2Þ£ð1000Þ£ðDh Þ 3 1–3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ð2Þ£ð1000Þ£ð317Þ ¼ 796m/s fg The velocity at the throat pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼fg ð2Þ£ð1000Þ£ðDh Þ ¼fg ð2Þ£ð1000Þ£ð110Þ 2 1–2 ¼ 469m/s Figure 6.6 Convergent–divergent nozzle. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved246 Chapter6 Mass discharged at the throat: A C 2 2 m_ ¼ 2 x v 2 g 2 Mass discharged at the exit A C 3 3 m _ ¼ 3 x v 3 g 3 Therefore A C A C 3 3 2 2 ¼ x v x v 3 g 2 g 3 2 Hence, A C x v 469 ð0:902Þð1:1593Þ 3 2 3 g3 ¼ ¼ ¼ 1:599 A C x v 796 ð0:954Þð0:4038Þ 2 3 2 g2 Illustrative Example 6.5: Dry saturated steam enters the convergent– divergentnozzleandleavesthenozzleat0.1MPa;thedrynessfractionattheexit is 0.85. Find the supply pressure of steam. Assume isentropic expansion (see Fig. 6.7). Figure 6.7 h–s diagram for Example 6.5. Copyright 2003 by Marcel Dekker, Inc. All Rights ReservedSteamTurbines 247 Solution: Atthestatepoint2,thedrynessfractionis0.85andthepressureis0.1MPa. This problem can be solved easily by the Mollier chart or by calculations. Enthalpy and entropy may be determined using the following equations: h ¼ h þx h and s ¼ s þx s ; 2 f2 2 fg2 2 f2 2 fg2 i.e.: h ¼ 417.46 þ (0.85)£ (2258) ¼ 2336.76kJ/kg 2 and s ¼ 1.3026 þ (0.85)£ (6.0568) ¼ 6.451kJ/kgK 2 Sinces ¼ s ,thestate1isfixedbys ¼ 6.451kJ/kgK,andpoint1isatthe 1 2 1 drysaturatedline.ThereforepressureP maybedeterminedbytheMollier 1 chart or by calculations: i.e.: P ¼ 1.474MPa. 1 6.6 STAGEDESIGN Aturbinestage isdefinedas asetof stationaryblades (ornozzles) followed bya setofmovingblades(orbucketsorrotor).Together,thetwosetsofbladesallow the steam to perform work on the turbine rotor. This work is then transmitted to the driven load by the shaft on which the rotor assembly is carried. Two turbine stage designs in use are: the impulse stage and reaction stage. The first turbine, designatedbyDeLaval in1889, was asingle-stageimpulseturbine,whichranat 30,000rpm. Because of its high speed, this type of turbine has very limited applicationsinpractice.Highspeedsareextremelyundesirableduetohighblade tipstressesandlargelossesduetodiscfriction,whichcannotbeavoided.Inlarge power plants, the single-stage impulse turbine is ruled out, since alternators usually run speeds around 3000rpm. Photographs of actual steam turbines are reproduced in Figs. 6.8–6.10. Figure 6.8 Steam turbine. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved248 Chapter6 Figure 6.9 Pressure velocity-compounded impulse turbine. 6.7 IMPULSESTAGE In the impulse stage, the total pressure drop occurs across the stationary blades (or nozzles). This pressure drop increases the velocity of the steam. However, in the reaction stage,the totalpressure drop isdivided equallyacrossthestationary bladesandthemovingblades.Thepressuredropagainresultsinacorresponding increase in the velocity of the steam flow. As shown in Figs. 6.10 and 6.11, the shape of the stationary blades or nozzles in both stage designs is very similar. However, a big difference exists in the shapes of the moving blades. In an impulse stage, the shape of the moving blades or buckets is like a cup. The shape of the moving blades in a reaction stage is more like that of an airfoil. These blades look very similar to the stationary blades or nozzles. 6.8 THEIMPULSESTEAMTURBINE Mostofthesteamturbineplantsuseimpulsesteamturbines,whereasgasturbine plantsseldomdo.Thegeneralprinciplesarethesamewhethersteamorgasisthe working substance. AsshowninFig.6.12,thesteamsuppliedtoasingle-wheelimpulseturbine expands completely in the nozzles and leaves with absolute velocity C at an 1 angle a , and by subtracting the blade velocity vector U, the relative velocity 1 vector atentry to therotor V can bedetermined.Therelative velocity V makes 1 1 an angle ofb with respect to U. The increasein valueofa decreasesthe value 1 1 of the useful component, C cosa and increases the value of the axial or flow 1 1 componentC sina .Thetwopointsofparticularinterestaretheinletandexitof a 1 the blades. As shown in Fig. 6.12, these velocities are V and V , respectively. 1 2 Copyright 2003 by Marcel Dekker, Inc. All Rights ReservedSteamTurbines 249 Figure 6.10 Steam turbine cross-sectional view. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved250 Chapter6 Figure 6.11 Impulse and reaction stage design. VectoriallysubtractingthebladespeedresultsinabsolutevelocityC .Thesteam 2 leavestangentiallyatanangleb withrelativevelocityV .Sincethetwovelocity 2 2 triangleshavethesamecommonsideU,thesetrianglescanbecombinedtogivea single diagram as shown in Fig. 6.13. Figure 6.12 Velocity triangles for turbine stage. Copyright 2003 by Marcel Dekker, Inc. All Rights ReservedSteamTurbines 251 Figure6.13 Combined velocity diagram. Ifthebladeissymmetricalthenb ¼b andneglectingthefrictioneffects 1 2 of blades on the steam, V ¼ V . In the actual case, the relative velocity is 1 2 reduced by friction and expressed by a blade velocity coefficient k. That is: V 2 k ¼ V 1 From Euler’s equation, work done by the steam is given by: W ¼ UðC þC Þ t w1 w2 Since C is in the negative r direction, the work done per unit mass flow is w2 given by: W ¼ UðC þCÞð6:9Þ t w1 w2 IfC – C ,therewillbeanaxialthrustintheflowdirection.AssumethatC is a1 a2 a constant. Then: W ¼ UC ðtana þtanaÞð6:10Þ t a 1 2 W ¼ UC ðtanb þtanbÞð6:11Þ t a 1 2 Equation (6.11) is often referred to as the diagram work per unit mass flow and hence the diagram efficiency is defined as: Diagram work done per unit mass flow h ¼ ð6:12Þ d Work available per unit mass flow ReferringtothecombineddiagramofFig.6.13:DC isthechangeinthevelocity w of whirl. Therefore: _ The driving force on the wheel ¼ mC ð6:13Þ w Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved252 Chapter6 The product of the driving force and the blade velocity gives the rate at which work is done on the wheel. From Eq. (6.13): _ Power output ¼ mUDC ð6:14Þ w If C 2 C ¼DC , the axial thrust is given by: a1 a2 a _ Axial thrust : F ¼ mDC ð6:15Þ a a The maximum velocity of the steam striking the blades pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C ¼ 2ðh 2h Þ ð6:16Þ fg 1 0 1 where h is the enthalpy at the entry to the nozzle and h is the enthalpy at the 0 1 nozzleexit,neglectingthevelocityattheinlettothenozzle.Theenergysupplied 2 to the blades is the kinetic energy of the jet, C =2 and the blading efficiency or 1 diagram efficiency: Rate of work performed per unit mass flow h ¼ d Energy supplied per unit mass of steam 2 2UDC w h ¼ðUDC Þ£ ¼ ð6:17Þ d w 2 2 C C 1 1  V 2 Using the blade velocity coefficient k ¼ and symmetrical blades V 1 (i.e.,b ¼b ), then: 1 2 DC ¼ 2V cosa 2U w 1 1 Hence DC ¼ 2ðÞ C cosa 2U ð6:18Þ w 1 1 And the rate of work performed per unit mass ¼ 2(C cosa 2 U)U 1 1 Therefore: 2 h ¼ 2ðÞ C cosa 2U U£ d 1 1 2 C 1 4ðÞ C cosa 2U U 1 1 h ¼ d 2 C 1  4U U h ¼ cosa 2 d 1 ð6:19Þ C C1 1 U where is called the blade speed ratio. C 1 Differentiating Eq. (6.19) and equating it to zero provides the maximum diagram efficiency:  d h 8U d  ¼ 4cosa 2 ¼ 0 1 U C 1 d C 1 Copyright 2003 by Marcel Dekker, Inc. All Rights ReservedSteamTurbines 253 or U cosa 1 ¼ ð6:20Þ C 2 1 i.e., maximum diagram efficiency  4cosa cosa 1 1 ¼ cosa 2 1 2 2 or: 2 h ¼ cos a ð6:21Þ d 1 SubstitutingthisvalueintoEq.(6.14),thepoweroutputperunitmassflowrateat the maximum diagram efficiency: 2 P ¼ 2U ð6:22Þ 6.9 PRESSURECOMPOUNDING(THERATEAU TURBINE) A Rateau-stage impulse turbine uses one row of nozzles and one row of moving bladesmountedonawheelorrotor,asshowninFig.6.14.Thetotalpressuredrop is divided in a series of small increments over the stages. In each stage, which consists of a nozzle and a moving blade, the steam is expanded and the kinetic energy is used in moving the rotor and useful work is obtained. The separating walls, which carry the nozzles, are known as diaphragms. Each diaphragm and the disc onto which the diaphragm discharges its steam is known as a stage of the turbine, and the combination of stages forms a pressure compounded turbine. Rateau-stage turbines are unable to extract a large amountofenergyfromthesteamand,therefore,havealowefficiency.Although theRateauturbineisinefficient,itssimplicityofdesignandconstructionmakesit well suited for small auxiliary turbines. 6.10 VELOCITYCOMPOUNDING(THECURTIS TURBINE) In this type of turbine, the whole of the pressure drop occurs in a single nozzle, andthesteampassesthroughaseriesofbladesattachedtoasinglewheelorrotor. The Curtis stage impulse turbine is shown in Fig. 6.15. Fixed blades between the rows of moving blades redirect the steam flow intothenextrowofmovingblades.Becausethereductionofvelocityoccursover two stages for the same pressure decreases, a Curtis-stage turbine can extract Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved254 Chapter6 Figure 6.14 Rateau-stage impulse turbine. more energy from the steam than a Rateau-stage turbine. As a result, a Curtis- stage turbine has a higher efficiency than a Rateau-stage turbine. 6.11 AXIALFLOWSTEAMTURBINES Sir Charles Parsons invented the reaction steam turbine. The reaction turbine stage consists of a fixed row of blades and an equal number of moving blades fixedonawheel.Inthisturbinepressuredroporexpansiontakesplacebothinthe fixed blades (or nozzles) as well as in the moving blades. Because the pressure drop from inlet to exhaust is divided into many steps through use of alternate rowsoffixedandmovingblades,reactionturbinesthathavemorethanonestage are classified as pressure-compounded turbines. In a reaction turbine, a reactive forceisproducedonthemovingbladeswhenthesteamincreasesinvelocityand when the steam changes direction. Reaction turbines are normally used as Copyright 2003 by Marcel Dekker, Inc. All Rights ReservedSteamTurbines 255 Figure 6.15 The Curtis-stage impulse turbine. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved256 Chapter6 Figure6.16 Velocity triangles for 50% reaction design. low-pressure turbines. High-pressure reaction turbines are very costly because they must be constructed from heavy and expensive materials. For a 50% reaction, the fixed and moving blades have the same shape and, therefore, the velocity diagram is symmetrical as shown in Fig. 6.16. 6.12 DEGREEOFREACTION The degree of reaction or reaction ratio (L) is a parameter that describes the relationbetweentheenergytransferduetostaticpressurechangeandtheenergy transferduetodynamicpressurechange.Thedegreeofreactionisdefinedasthe ratioofthestaticpressuredropintherotortothestaticpressuredropinthestage. It is also defined as the ratio of the static enthalpy drop in the rotor to the static enthalpydropinthestage.Ifh ,h ,andh aretheenthalpiesattheinletduetothe 0 1 2 fixed blades, at the entry to the moving blades and at the exit from the moving blades, respectively, then: h 2h 1 2 L ¼ ð6:23Þ h 2h 0 2 Thestaticenthalpyattheinlettothefixedbladesintermsofstagnationenthalpy and velocity at the inlet to the fixed blades is given by 2 C 0 h ¼ h 2 0 00 2C p Similarly, 2 C 2 h ¼ h 2 2 02 2C p Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved

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