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Thermalengineeringbyrkrajput chapter19 steam turbine

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Thermalengineeringbyrkrajput chapter19 steam turbine

  1. 1. Steam ; '. .EE“ I" I lnir-. »«In. .- '. z- I‘ I” ‘_' III. rm illi-III’ ‘ll irnrmi-irio lit . . . .: iv. irn -. . — ll ll-. <lllI: lFIr. I‘I‘4i<-I tlii ll -in - r: ;'nI 1" '» ll | ‘I'l[)l'I Y1 --‘ r» 'I1I‘Ii-Fl i_'. pv. crflurhira» 1” 3 ‘vii '. !'n-it 4'-l' rv‘-'IL1CIfI‘, ' v lwvl :4 r<»'~ 'l"il I“ -~ l‘: I‘v-iv um In--It _. IllIl‘. ll"' in-I rv » '. --~ii ! url'vvi: ~ III '. IILIIIII“ tiir‘-iviv ‘-3 In-'1‘). (II . ‘liil’ZI 1': 7 IT ~‘. xr1,. ' hl nIi< »'~'m‘l «lnr« an the III ido Ill «vi» . « lrw‘-Y} ix-—< ll. L"- ‘II I‘. }: :~ II I7 r iv, '1llHIilI . .II. ‘- I 'll'r r Il'I| lIrll ulv -; u,rI1i; -.[. ; ~m ii. -I jifi ‘m “if, -[(1111 w'I ,1. I I‘ - : .jV ‘.4 rim. ll r l'IIlv :1‘ ‘ill. 4'. ’ n M ‘.4 Ir ‘I. ct-inf-. uv. I- zi I'M‘, -‘,1l rnrlr ii. I" ' lain . :.. ..-: 'l11l*'. Yl- '‘. ‘--l .19. ll: ,r III tar r . i'.1~n Illrluf‘ I1‘ I I')- . 'r-. ml r. i(". ml1 ‘I’. I "~mIi‘. i ii I II ! il. .-. i|tlI! lt 4.‘l ' : :'_i I" " lLlI‘III"lll1‘. .‘1ZI. 'I~. I. ” III 'I'. ;» - -it gum r III '. a rlll tux“: iw pr‘. III: I" II 741 ii rm nt ’ w v. rrri ‘II: ‘M . «tfwV-'r I” I. ‘ H: h~-: i!Irir' 3- . in‘. W I» III: I III! ‘ If‘ ll l: 'i'-. r._. l, I- II] 1. . im turi- rw l".1., r D'. r'. .IIIIUFIIZIW‘E-, -iThIII. '.1l1LI iIIlll. I'Ill I" II) 5.1.» i ‘ -l I»-rm - I «II mi ! ».: I'! ~i"i~ - -—Iii 'hli, ,l‘. '.- ~lIl iv cti~ ‘l'jv; v- (, ji. - -1. -21- . I3IiI'Il| ’ l(. l1:« '. I: II‘ "u llfx mil l rim‘. INTRODUCTION I The utealn turbine is a prinw-mom-r in which the potential energy of the ateam in trima- formed into kinetic energy, and latter in its tum is truruformed into the mechanical energy of rotation of the turbine shot}. The turbine sham directly or with the help of a reduction gearing. is connected with the driven mechanism. Depending on the type of the driven mechanism a steam turbine may be uliliard in most dit'cr. -e fields of industry, for power generation and for transport. Transformation of the potential energy of steam into the mechanical energy of rotation of the shaft is brought nbout by difleront menu. -a. a8«'2~'~ CLASSIFICATION OF STEAM 'I'URBINES I There are scvenil ways in which the . .+u= um turbines may be classified. The most important and common division being with rempc-ct in the action of the steam. as : (a) Impulse. (bl Reaction. (c) Combination of impulse and l‘L': lCI. l0t1. Other classification are : 1. According to the number of pressure stages : Ii) Single stage turbines with one or more velocity stages usually of small power capadtips ; tho. -he turbines are mostly u. ~'. cd for driving centrifugal compressor». blowem and other similar machinery. (ii) Multistage impulse and reaction turbines ; they are made in a wide range of power capacities varying from small to large. 2. According to the direction of steam flow : (i)Axial turbines in which steam flows in at direction parallel to the uxis of the turbine. 828
  2. 2. STEAM TURBINES 829 (ii) Radial turbines in which steam news in a direction perpendicular to the axis of the turbine ; one or more low-pre. s.~. ure stages in such turbines are made axial. 3. According to the number of cylinders : (i) Single cylinder turbines. (ii) Double cylinder turbines. (iii) Three cylinder turbines. (iv) Four cylinder turbines. Multi-cylinder turbines which have their rotors mounted on one and the same shaft and coupled to a single generator are known as . -mgle shaft turbines ; turbines with separate rotor shafis for each cylinder placed parallel to each other are known as multiaxial turbines. -1. According to the method of governing : (i) Turbines with throttle governing in which freah steam enters through one or more (depending on the power developed) aimultaneou. -aly operated throttle valves. (ii) Turbines with nozzle governing in which fresh steam enters through two or more consecutively opening regulators. (iii) Turbines with by pass governing in which steam turbines besides being fed to the first stage is also directly fed to one, two or even three intermediate stages of the turbine. 5. According to heat drop process : (1') Condensing turbines with generators‘ ; in these turbines steam at a pressure less than atmospheric is directed to a condenser ; besides, steam is also extracted from intermediate stages for feed water heating, the number of such extractions usually being from 2-3 to as much 8-9. The latent heat of exhaust steam during the process of condensation is completely lost in these turbines. (if) Condensing turbines with one or two intermediate stage attractions at specific pressures for industrial and heating purposes. (iii) Back pressure turbines, the exhaust steam from which is utilised for industrial or heating purposes ; to this type of turbine: can also be added (in a relative senzsc) turbines with deteriorated vacuum. the exhaust steam of which may be used for heating and process purposes. (iv) Topping turbines ; these turbines are also of the back pmsuum type with the diilerence that the exhaust steam from these turbines is further utilised in medium and low pressure condensing turbines. These turbines. in general. operate at high initial conditions of steam pressure and temperature, and are mostly used during extension of power station capacities. with a View to obtain better efficiencies. (u)Bach pressure turbines with steam extraction from intermediate stages at specific pressure ; turbines of this type are meant for supplying the consumer with : -‘team of various pressures and temperature conditions. (vi) Low pressure turbitux in which the exhaust steam from reciprocating steam engines. power hammers. pmases, etc. is utilised for power generation purposes. (vii) Mixed pressure turbines with two or three pressure stages, with supply of exhaust steam to its intermediate stages. 6. According to steam conditions: at inlet to turbine : ti) Low pressure turbines, using steam at a pressure of 1.2 to 2 ata. (ii) Medium pressure turbines, u. -sing steam at pressures of upto 40 ata.
  3. 3. 330 THERMAL ENGINEERING (iii) High pressure turbines, utilising pressures above 40 ate. (in) Turbines of wry high prae. -sums, utilising steam at pressures of 170 am and higher and temperatures ul‘ 5."r0'C and higher. (1-l Turbines of ivupcrrritical prr: .«xut1's, using steam at pressures of 225 am and above. 7. According to their usage in industry : m Stationary turbines with mnstant speed afmtotinn primarily used for driving alternators. Iii)Stntionary steam turbines with variable speed meant for driving turbo-blowers, air circulators. pumps. etc. tiiil Non-atatiunarjv turbines with variable speed ; turbines of this type are usually employed in steamers. ships and railway locomotives. “" ADVANTAGESOFSIEAMTURBINBOVERSIIAMBNGINES The following are the principal advantages of steam turbine over steam engines : l. The thermal efficiency of a steam turbine is much higher than that of a steam engine. 2. The power generation in a steam turbine in at a uniform rate. therefore necessity to use a flywheel ‘as in the case of steam engine» is not felt. 3. Much higher speeds and greater range of speed is possible than in case of a steam engine. 4. In large thermal stations where we need higher outputs. the steam turbines prove very suitable as these can be made in big . N‘l7.l‘. |i. 5. with the absence of reciprocating parts can in swam engine! the balancing problem is minimised. . No internal lubrication is required as there are no rubbing parts in the steam turbine. . In a steam turbine there is no loss due to initial condensation of steam. It can utilise high vacuum very advantageously. . Considerable overloads can he carried at the expense of slight reduction in overall elficienc_v. l" 1 DESCRIPTION OF COMLION TYPES OF TURBINIB The common types of steam turbines are 1. Simple impulse turbine. 2. Reaction turbine. The main diflbrence between these turbines lies in the way in which the steam is expanded while it moves through them. In the former type steam expands in the noz. '1es and its present? data not alter an it moves over the blades while in the latter type the steam expands continuously at; it passes over the blades and thus there is gradual fall in the pressure during expansion. @. $-13 1. Simple impulse turbine: Fig. l9.l show; a simple impulse turbine diagrammatically. The top portion of the figure exhibits a longitudinal section through the upper half of the turbine. the middle portion shows one set of nozzles which is followed by in ring of moving blades. while lower part of the diagram indicates approximately changes in pressure and velocity during the flow of steam through the turbine. This turbine is called ‘r-maple‘ impulse turbine since the expansion of the steam takes place in one set of the nozzles.
  4. 4. As the steam flows through the nozzle its prenzsure fall: from steam chest pressure to con- denser pressure (or atmospheric pressure if the turbine is non-condensing). Due to thbl relatively higher ratio of expansion of steam in the nozzles the steam leaves the nozzle with u very high velocity. Refer Fig. 19.1, it is evident that the velocity of the steam leaving the moving blades is a large portion of the maximum velocity of the steam when leaving the nozzle. The loan: of energy due to this higher exit velocity is commonly called the ‘carry over loss‘ or ‘leaving loss‘ BOIIGI’ steam "i 3 Turome blade Lost velocity _1_ Condenser : ! pressure Boiler prosstvo Initial Steam [~", u_ 19¢ Simple impul: -c turbine. The principal example of this turbine is the well known ‘De lavul turbine’ and in this turbine thc ‘exit velocity‘ or ‘leaving velocity‘ or ‘lost velocity’ may nmount to 3.3 per cent of the- nozzle outlet velocity. Also since all thu kinetic em. -r: :y i. ~ to be absorbed by one rim: of the lnuvuu:
  5. 5. 832 THERMAL ENGINEERING blades only, the velocity of wheel is too high (varying from 25000 to 30000 r. p.m. ). This wheel or rotor speed however, can be reduced by different methods (discussed in the following article). 2. Reaction turbine In this type of turbine, there is a gradual pressure drop and takes place continuously over the fixed and moving bladex. The function of the fixed blades is (the same as the nozzle) that they alter the direction of the steam as well as allow it expand to a larger velocity. As the steam passes over the moving blades its kinetic energy (obtained due to fall in pressure) is absorbed by them. Fig. 19.2. shows a three stage reaction turbine. The changes in pressure and velocity are also shown there in. / — Casing 1 ‘L ifl . ~ ‘g l W Steam ___Steam In out +1§$353vSt» 5§$N5N5h F M, jF5M F M»F M. M= Mov| ngblades- if-’-— Fixed blades , ‘K ‘W “M . 2.» §a ‘ 3 7,; - t Lost velocity .2 ‘ K‘ fi"“’ Condenser 8 »' ‘ *3 pressure K Y V 1 L I ll" ’ ‘Y’ ‘ LT’-*7‘ -i i. ‘a. ‘ " "1; l"n. :. W ‘_'. R:: action turbine (three stage). As the volume of steam increases at lower pres. -zures therefore. the diameter of the turbine must increase after each group of blade rings. It may be noted that in this turbine since: the pr'es. ~rure drop per stage is small. therefore the number of stages required is much higher than an inzpulse turbine of the some capacity. '
  6. 6. STEAM wnnmas 833 I llm’, hmTBODS OF REDUCING WHEEL OR ROTOR SPEED As already dlrecunbcd under the heudinsztwimplc impulse turbine’ that ifthc nlflum is expanded from the boiler pressure to condenser pressure in one stage the speed of the rotor becomes tremendously high which crop. -. up practical t*nmplicacir. ~‘. There are several methods of reducing thia speed to lower value ; all these method. » utilim: :25 multiple a_v: .tem of rotor in ecrlczs. keyed on u common shnfl and the steam pressure or jet velocity is ub~. .orbed in sulgt-5 m. the steam flows over the blades. This is knovm as ‘compounding’. The different method: of compounding are : 1. Velocity compounding. 2. Pressure compounding. 3. Pressure velocity compounding. Al. Reaction turbine. 1. Velocity compounding Steam is expanded through a stationary nozzle from the boiler or inlet pressure to con- denucr pressure. So the preawure in the nozzle drops. the kinetic energy of the . ~il. x"£U'n increases due to increase in velocity. A portion of this available energy is absorbed by u row of moving blades. The t l I . 1| _ l l V V V p l-. . ) : _, , Steam exhaust / ' > / V, ‘ . d« . ’ "N; . / //l . sleamln " / / ' ‘ / , ..—. —vA Nozzo Moving blades Fixed blades Movng blades "ma! Steam 'elocity ‘ ‘ — Lost velocity . g “ ~ ‘ _’ ‘j-—j‘—j—-— Condenser at/ " ‘V v pressure '-¢ ll ‘-. ~ ‘Velocity compounding:
  7. 7. 834 mean». ENGINEERING steam (whoae velocity has decrezwud while moving over the moving blades) than flows through the at-mod row of blades which are fixed. The function of (119343 fixed blades is to re~direct the steam flow without altering its velocity to the following next row moving blades where again work is done on them and steam leaves the turbine with a low velocity. Fig. 19.3 shows a cut away section of such a stage and changes in pressure and velocity as the steam passes through the nozzle, fixed and moving blades. Though this method has thv advantage that the initial cost is low due to lesser number of stages yet its efi’1cienc_v is low. 2. Pressure compounding Fig. 19.4 show: ring: of fixed nozzles incorporated between the rings of moving blades. The steam at boiler pressure enters the first . -act of nozzles and expands partially. The kinetic energ of the steam thus obtained is absorbed by the moving blades (stage 1). The steam then expands purtiully in the nt. '¢0l1d wt of nozzles where its prcssurc again falls and the velocity inC! ‘cu. scs ; the .3 3 .33 3 3 E g; r E » :2 § ; 3 ; 9 ; 3 ~ 2 an 2 . D - 2 . in I ' n : . 3| 0‘ g. , __ , A . ; , ostveocxty on 5,318} : 2 i‘ ' w| __'__ . 8'‘ gig . 5' E3‘ 2 l I, Exhaust = l '‘ 1 pressure 8; v __: / J J- ~ _, .L, _4 P”; 19; l’rr'~~: urw-eompnundin; :.
  8. 8. STEAM nmnnces 835 kinetic energy so obtained is absorbed by the second ring of moving blades (stage 2). This is repeated in stage 3 and steam finally leaves the turbine at low velocity and pressure. The number of stage: (or prcazsurc reductions) depends on the number of rows of nozzles through which the Iatcam must pass. This method of compounding is used in Rafrau and Zoelly turbine. This is most eflicim! turbine since the xpeed ratio remains constant but it is expcwsive owing to a large number of N! ag¢'. '. 8. Pressure velocity compounding This method of compounding is the combination of two previously discussed method. The total drop in steam pressure is divided into stages and the velocity obtained in each stage is nlso compounded. The rings of nozzles are fixed at the beginning of each stage and pressure remains constant during each stage. The changes in pressure and velocity are shown in Fig. 19.5. D“ V) 90 M 3 8 W 8 3 8 I" «S ‘E 2 -8 2 I3 2 3 2 O «D S Q Q 5 D C 3 D 03 2 3 D U! B C '3 .9 *§ .9 1: .5 5 5 )_c 5 _ 3 § § LI. 2 IL E LL 2 la. E I - V I I I I V ‘. . ' _ .4 V -‘I I I I I I , T_, __/ _. . _. . .,. »-” 5 I _. _ . . , r ‘. - I I I ‘l”"'7’ . v , W--’ ; . . . Steam “I/ I’ — , Steam T? g V : V I 1 j". ’ s —o‘—n— , _ ‘I . _/$9’ ’ _ ' ‘ -T-J’ . -J , I V 7 l I I .4 , .. v:{. ' I : _ U _: I ‘ I I 7 ‘ A, 1 V: " I’ ‘ . I r ‘ -'1 I’ 1 1 I ' ‘. . ,4 I I I I I ' .1 I I ' I ' ' 7:5’ I ' ‘ " :1" ' : I ' : : I I : i I ‘ ' : I I ' I ‘ I I ' , I I ‘ ’ , I | fl i I ' E I I ' , I I ‘ ‘l 1 I | ‘l ‘ I ‘ r I ‘ I I ' I I « I 1 I I ' I I 1 ’: v-: I: -I I :1- , I I I I I I I K - I ; ‘ ‘ I I I I | ‘I I ‘ i ' ' I‘ i ‘ 1 ‘ I . I ’ I— —I-1 ' I ' I. .. -4 J ' I I I l ‘ , ‘ I I ' I i ' I ‘ ' ' I I ' I I " ‘ E’, I I I I I I ‘ ' ' , ' I ' ' I ‘ fig: I I I I | I : ' V‘ I I I E I ' I I I I I ' I - I ‘ I , I I I I ‘ El EVE" | ‘ : —.s. _._t. ,_. I' ' " ‘ ‘ ’ I ' V: LoslveIocifY 5. : I ‘ I I ‘f I ’ : , I ‘M Y 3 TI-I‘§I I I ' : I ' ' ‘; _' ' ' I - ' -. —_ I I I I 13 I5‘ : ‘ : . -I I If , I I [I I Condenser -— A l I . I : j . 2 I : I f. k [ I - I f pressure *‘L; ' : I ' : I I | , I Y cw , - a: J: ,: I ‘I -' _. *-. -I I. H. II. ,-, Pr: -.~. ~'ure velocity compounding. This method of compounding is used in Curits and Moor-r turbine.
  9. 9. llli RH ‘ll. 1 "-Ull l, Rl1_» 4. Reaction turbine It has been discussed in Article 19.4. ¢. :., ;;}, i DIFFERENCE BETWEEN IMPULSE AND REACTION mnnmzs I ' Pro'. uvunr drop Only in nozzle: and not in moving lnfimadhladc: «(norz1JcsluiI well blade. » as in moving blade’. Arru o/ ‘blade CIIGIIIICIX Con. -ctunt. Vnrying (converging type l. Hladrs Profile typ--. A4-mfoil typv. Admicsian o/ .-«team Not all mund or complete. All round or complete. i Nozzles / furrd blade» Dinphrum contain. -« the nozzle. Fixed blades . -«imilor to moving bl. u:lr. ~' uttnchc-d to the cmsing nerve .1.-4 nozzles and guide the steam. I Pom-r Not much power can be dcwrlopod. Much power can be developed. Space Rt-quims lo. -as space for same power. Require. » more space for some power. Efficiency Low. High. Suitability Suitable for xrnnll power n~quim- Suitaiblv for medium and high: -r mo-nu. power requirements. Blade man ufadurr Not diflicult. Difficult. K’ . '{. ,«, y IMPULSE TURBINES 1 19.7.1. Vt-locity Diugrnm for Moving blade Fig. 19.6 xhows the velocity diagram of a . ~u'n, r,'le stage impulse turbine‘. C, ,, = Linear velocity of moving bludc (m/5) C , Absolute velocity of steam entering moving blade (m/5) C0 2 Absolute velocity of steam leaving moving blade (m/3) C, ,_ — Velocity of whirl nt the entrance of moving blade. : tangential component of Cl. C“ y = Velocity of whirl at exit of the moving blade. = tangential component of C0. Cr I Velocity of flow at entrance of moving blade. = axial component of C‘. Cf" = Velocity of flow at exit of the moving blade. : axial component of Co. Cr, A-V Relative velocity of steam to moving blade at entrance. C, “ : Relative velocity of steam to moving blade at exit. (1 = Angle with the tangent of the wheel at which the steam with velocity C, enters. This is also called nozzle angle.
  10. 10. B = Angle which the discharging steam makes with the tangent of the wheel at the exit of moving blade. 6 = Entranm angle of moving blade. o = Exit angle of moving blade. Intel triangle 5: ch C’: a 0 . .77?" *0. >1 J g7 z: ._ ~—c_I - on IE? Fug. l9.6.Velocitydiagram for moving blade. The swam jet issuing from the nozzle at a velocity of C, impinges on the blade at an angle a. The tangential component of this jet (Cw, ) perform: work on the blade. the axial component (cf, ) however does no work but causes the steam to flow tluough the turbine. As the blades move with a tangential velocity of C“, the entering ateam jet. has a relative velocity Cr, (with respect to blade) which makes an angle 8 with the wheel tangent. The steam then glides over the blade without any shock and discharges at a relative velocity of C, __ at an angle to with the tangent of the blades. The relative velocity at the inlet (0., lie the me as the relative velocity at the outlet (0.. ) if there is no fi-ictional loss at the blade. The absolute velocity (Co) of leaving steam make an angle 3 to the tangent at the wheel. To have convenience in solving the problems on turbines it is a common practice to combine the two vector velocity diogram. -s on in common hltso which represents the blade velocity (CM) ms shown in Fig. 19.7. This diagram has been obtained by superimposing the inlet velocity diagram on the outlet diagram in order that the blade velocity lines C5, coincide.
  11. 11. 838 THERMAL ENGINEERING C. - , , C_ ; .L(. _. c __ I l ‘*9 F’ °” fl P . ___'r, __, H o / __, /< . CIy, // / /X ‘ ' C Y ct‘ "I, ' / / / —u‘ ° Y C10 1 __ __/ /01 c'o _ _~‘ : /» ‘>Z/ _ x_ g 'f_. It _ _ _ _ _ _ . _ _ . _ _ _ _ _ _ _ _ _ _ _ _ . _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ S N My 107 19.7.2. Work done on the Blude The work done on the blade may be found out from the change of momentum of the steam jet during its flow over the blade. A5 curlicr discussed, it is only the velocity of whirl which performs work on the blade since it acts in its (blade) direction of motion. I-‘mm Newton's second law of motion. Force (tangential) Oil the wheel = hlmuz of xlI‘sum x nctu‘-lvrnlion = Mass of steam/ .-sec. x change of velocity = rh. (Cx. ,-C. .,J 4419.1) The value of C. A is actually negative me the steam is discharged in the oppoeitc direction to the blade motion, therefore due consideration should be given to the fact that valurs of C. .. and Cm, are to be added while doing the solution of the problem. (i. c.. when I} < 90") Work done on blades/ sec. = Force at distance travelled/ .~ec. = n’z, (C“l +C“")>-(Cu Power per wheel = "3.(Cu», * C. . , , l X Cu woo kw . ,,tl9.‘.2» <~. - cu, ecu_ +c, ,“) Work done the blade Energy supplied to the blade fl'l, (C“. l +C_. ‘ H LCM m C, ’ T 2 2C, ,,(C, “ «C U) = 419.3) Blade or diagram cfliclency ;
  12. 12. STEAM TURBINES 8l3 Hit, and A, be tho total honts before and alter expansion through the nozzles. then (It, - 11,) in the heat drop through a stage of fixed blades ring and moving blades ring. Work done on hlndo per kg of steam ffl = jj. .__. ._j stage G den”. um“ Total energy supplied per kg of steam = . ..(19.-4) ~ , _¢-‘i’_ Now. nozzle clllcicncy 2 M1 __ ha) Also nu”, = Blade elllciency x nozzle ellicicncy = 2CN(C, , +C, ,V_) X __c 9 = C“(C +C. _ ) c, 201, - ha) 0:; - hg) The axial thrust on the wheel is due to di/ fen-nee between the velocities of flow at entrance and outlet. Axial force on the wheel = Mass of steam at axial acceleration = n'1,(C, ' -05) . ..(19.5) Tho axial face on the wheel must be balanced or must be taken by a thrust bearing. Energy converted to heat by blade friction = loss of kinetic energy during flow over blades = ri-. (c, ,’ -0,9’) . ..<19.s) 19.7.3. Blade Velocity Co-emolent In an impulse turbine, if friction is neglected the relative velocity will remain unaltered as it passes over blades. In prncfim tho flow of steam ovor the blades is resisted by friction. The cficct of the friction is to reduce the relative velocity of steam as It passes over the blades. In general there is :1 loss of 10 to 15 per coat in the relative velocity. Owing to friction in the blades, C, “ is loss than C“ and we may write C, ‘ : K. C, ‘ . ..(l9.7) where K is termed a blade velocity co-efficient. 19.7.-l. Expression for optimum Value of the Ratio of Blade Speed to Steam Speed (Ior maximum emoiency) for a Single Stage Impulse Turbine Refer Fig. 19.7. C_= PQ= MP+MQ= C, ‘ 0050+ Cm ms ¢ = c, , cos 9 coco = ., cos9(1+lC. Z)wheroZ= me . ..(I') Generally. the angles 6 and ot are nearly equal for impulse turbine and hence it can safely be assumed that Z is a constant. But. C, .oosO= MP= LP-LM= C,cosu—C_. ,, From eqn. (0. C, ,=(C, coeo-C, ,,)(1+K. Zl
  13. 13. 840 nnzmax. ENGINEERING We know that. Blade cfiicicncy, uh. = . ..(1'i) I XTMCI U35“ -C, .:¥l+KZl '11.: = “n*--““""‘ C‘- : 2(p cor. a—p‘Xl + IQ) = 2p(coea a — pl-fl + ((2') . ..(i1'i) where p = Cf‘ is the ratio of blade sperd lo sleanx speed and is commonly called as ‘Blade speed («1 ratio’. For particular impulse turbine 11. K and 2 may assumed to be constant and from equation (iii) it can be seen clearly that n, _. depends on the vnlur of p only. Hence diflemntinting (iii ). ‘ma. ’ . .. . do - 2Ico. ~a~2p¥l -1» [(2) For :1 minimum or minimum value of nk. (his should be zero oosa-2p: o, p: °'’‘‘"“ 2 2 Now, =2(—2Y. l+KZI; —-l(l+KZl 0- which is n nrgativr quanti! _v nnd thus the value so obtained is tho mnxim1m1 Optimum value of ratio of blade speed to steam speed is cos 1: , pup, : 2 . .1 19.151 Substituting this. value of p in cqn. (iii). we get tI1,u. I,,4”—2x m§a(oosu—c3_: "— tl+K. '/.1 : mg " «1 + KZ) 419.91 It is . ~1ufl'icicntl_v tl(‘l‘Ul'. ‘llt‘ to usaumc . symmc. -trical bludcxa (8 ( c») and no friction in fluid pasangze for the purpow of ;1n;1Iy». ir. . 7. : l and K : 1 . . lnulhu = cos‘ (I . ..( 19.10) The work done per kg: of steam is given by W — ‘Cu’ 4-C, “ D C, , Substituting the value of C, ‘ +C_, ‘ 1: C‘ I W = (C, cos a — Cut‘! + KZWCJ = 2C, _.(C, cos (1 -C, _.) when K = 1 :1ndZ =1 The maximum value of W can be obtained by substituting: the value of cos a from equation (19.8). . .92. co. -.u=2p=2C 1 1. <26", — C, _.» = 2c, ,~' . ..c 19.11)
  14. 14. Tl l T17RH| ‘l‘. . ii-ll It is obvious from the equadou (19.6) that the blade velocity should be approximately lmlfof absolute volocity of steam jot. coming out from the nozzle (fixed blade) for the maximum work devuloped per kg of steam or for maximum efliciency. For the other values: of blade speed the absolute velocity at outlet from the blade will increase, consequently, more energy will be carried away by the steam and efficiency will demmsc. I-‘or oquiangular blades with no friction lossea, optimum value of %N- corresponds to the case. when the outlet absolute velocity is axial as shown in Fig. 19.8. x : ~ c. , r - -—— ~>+ P» * M °°'_ * — ' L. O V V V V V ' 7' -"V777~ ‘V77; . __ ‘Iii: .—‘ . / _/ "Vi ' cl: 2/ Z #2’. c&_ Y C°= cgo 1 , /~ . s ix‘ l".8 N Since the is uxiul B : 90°, Co : C, ‘ and 0., “ - 0. The variations of 11,, or work developed per kg of steam with %3- is shown in Fig. 19.9. This 1 figure shown that : ti) Whcn’E, “- = 0, the work done becomes zero as the distance travelled by the blade (CH) ia zero. even though due torque on the blade is maximuin. nu orwork Fm. l9_9
  15. 15. H6 HHRMALENGNHRMG (ii) The maximum eficiency is coo‘ a and maximum work done per kg of steam is 26'3" C when -3 r one 0/2. 0: (iii) When -E, “ = 1, the work done is zero as the torque acting on the blade becomes zero oven 1 though the distance travelled by the blade is maximum. When the high pressure steam is expanded from the boiler pressure to condenser pressure in a single stage ofnoule, the absolute velocity ofsteam becomes maximum and blade velocity also becomes tremendously high. In such a onse, a velocity compounded stage is used to give lower blade speed ratio and better utilization of the kinetic energy of the steam. The arrangement of velocity compounding has already been dealt with. c j I‘ C‘: c ___1 "J P . .-. ... --. ... -.. -.4.------_. .. -3" H L. ----p. -.. ..-- B O up . ... .--. ... ... ... (.. ... ... ... -.. . --. ._--. ..(. .-. ..----- c. .'% § 3' Fig. 19.10 Fig. 19.10 shown the velocity diagrams for the first and second row of moving blades of velocity compounded unit. The speed and angles are such that the final absolute velocity of the steam leaving the second row in axial. With this arrangement. the ICE. carried by the steam is minimum, therefore, the efficiency becomes maximum. The velocity of blades (Cu) is same for both the rows since they are mounted on the same shaft - Consider first row of moving blades : Work done per kg of steam, W, = 0,, (CW +0.. ..) = CHIC, ‘ oosova-C, ‘ oos ol
  16. 16. S'| 'l. 'MM TURBINES 3 I7 Ifthexe is no friction loss and symmetrical binding is used, then C, ‘ = C, ’ and 0 = 9 W, = cu 3: 2c, ’ oos o : 2cb, (c, cos 11- cu; .. .(19.12) The magnitude of absolute velocity of steam leaving the first row and entering into the second row of moving blades is some and its direction only is changed. . . C, ‘ = Co Consider second row of moving blades : Work done per kg. W, = C“ . C’, ,| on C3”. 2 0 because discharge is axial and B’ = 90' Alto: -nately. W, ’ »_ Cu lC', | cos 8’ *C', . eon <5’) For symmetrical blades 6’ : o‘ and. if there is no friction less, then C‘, ‘ = C', ., W, -= 20,, 6'', ‘ cos 6' an 20,; (C, ' cos a’ - C”) . ..(19.13) Now of may be equal to B. - C, 'ooeo’= C.ooeB= c,. oos¢-C5, = C" can 0—Cu = (C, oos a—Cu)-Cu = Cl one cc - 2C“ Substituting the value of 0,’ cos tx’ in eqn. (19.13). we get W, = 2C, ,, l(C', cos a - 2C”) — C“) = -. 2C” (C, cos a -- 3C, ,,) . ..(19.l»i) Total work done per kg of steam passing through both stages is given by W‘ L W‘ + W: = 20,, (C, cos a - Cu] + 20,, I0‘ 003 a — 305,] = 2C, ,, (20, cos a - 4C“) = 40,, (C, cos a -— 2C“) . ..(19.15) The blade efliciency for two stage impulse turbine is given by W 2 “N: ‘'2’ =4cu[Ci°“"‘°1"2cul" —2 91. C1 2 BC C C 2 (0, cos 13- 2c, ,) = 8 —c*‘1(eosa—2.—C*lL] = - 89 (cos on - 2p) . ..(19.l6) where P (velocity ratio) = -' -2"-. I The blade cfiicicncy for two stage turbine will be maximum when $151 = 0 d 5 [8poosa-16p‘)-. =O
  17. 17. ;| li TliER. iAi. ENGINEERING 8 cos (1 - 32p 2 0 mm which. p -. - °°: “ . ..u9.m Substituting this value in oqn. (19.16). we got (nu)m_-:8. cow-2.°°‘°‘] -mu: .. .(19.1s) The maximum work done pcr kc of steam is obtained bv substituting the value of __ C _ cos a " " c, ” 4 «SC , or C, -= ;s'% in the oqn. (19.15). 4C a‘. ‘ o 3' G “ = ecu’ . ..(19.19) The praent analysis is: done for two stages only. 'l"he similar procedure is adopted for ane- lysing the problem with three or four stages. In general, optimum blade speed ratio for maximum blade efficiency or maximum work . ’.-we is given by p 2 — —- . ..( 19.20) 2.n and work done in the last new 2 -2‘; of total work . ..( 19.21) where n is the number of moving/ rotating blade rows in series. As the number of rows increases. the utility of last row decreases. In practice. more than two rows are hardly preferred. 19.7.5; Advantages of Velocity compounded impulse Turbine 1. Owing to relatively large heat drop. a velocity—eompounded impulse turbine requires a comparatively small number of stages. 2. Due to number of stages being small, its cost is less. 8. As the number of moving blades’ Iowa in a wheel increases. the maximum stage effi- eiency and optlmuxnvnluc of p decreases. 4. Since the steam temperature is suillciently low in a two or three row wheel. therefore. cast iron cylinder may be used. This will cause saving in material cost. Disadvantages of velocity-compounded impulse turbine : 1. It high steam consumption and low cfliciency (Fig. 29.1). on next page). 2. In a single row wheel, the steam temperature is high so cast iron cylinder cannot be used due to phenomenon of growth ; cast steel cylinder is used which is costlier than cost iron. Example 19.1.11 stage of a steam turbine is supplied with steam at a prvcuurc of 50 bar and 350‘C. and about: at a pnruure of5 bar. The isentropic r/ ficimzcy of the stage is 0.82 and the steam consumption is 2270 hglmin. Determine the power output of the stage.
  18. 18. STEAM TURBINES / — Sxngka stage A Tsta / —wo go Three stage ‘ .5 pi C1 I~~. ~ I” 11 Solution. Strum supply prm-zuro. pl 7 50 bar, 350'(' Exhaust pressure p. _, : 5 bar Iurntropic cfficicncy of the stage. I1,, ___“ = 0.82 Ste. -un consumption. m_ : 2270 kgfmin Power output of the stage, P : Refer Fig. 19.12. h A. i 1,‘ 3500 K ‘I Saluralvon " ‘ "u “ / Z ‘M38 4 ‘ ‘ '1 , 1 L9» ~2- 2 7‘ r_. ,b'j“, , _ - 5 F4: 1*‘ 1.’ 845
  19. 19. 820 HIERMAL ENGINEERING From Molllcr chart . - lz‘ : - 3130.7 ldlkg of steam Ix, : 2640 l<J/ kg of steam Isentropic heat drop 15- It, = 3130.7 - 2840 = -190.7 lul/ kg Actual heat drop = ll‘ — It, ’ h _ o 3*“: ‘IL. .. once) = 5‘ .4,’ 032 = 5}; ;&'—'7?- = or It, - It; = 0.32 x 490.7 = 402.4 l<J/ kg Power developed = in, (h, — h, ') 2270 60 x 402.4 kw = 15224 kw. (Ans) Example 19.2. In a De Laval turbine steam issues from the nozzle with a velocity of 1200 tn/1. The nozzle angle ll 20‘. the mean blade velocity to 400 mix, and the inlet and outlet anglu of blade: on equal. The man of clean flowing through the turbine per hour b.1000 lg. wlculate 0') Blade angles. (ii) Relative velocity of steam entering the blades. (iii) Tangential force on the blades. (iv) Power developed. (ti) Blade efficiency. Take blade velocity co-efficient as 0.8. Solution. Absolute velocity of steam entering the blade. C, = 1200 lnls Nozzle blade, ‘. o = 20‘ Mean blade velocity. c, , = 400 m/ s Inlet blade angle. 0 —: Outlet blndo nnglo, 4) Blade velocity co-efliclent. K = 0.8 Mass of steam flowing through the turbine, rn, = 1000 kgfh. Refer F‘g. 19.13. Procedure of drawing tho inlet and outlet triangles (LMS and LMN) mapcctively is no follow: : C, ,== C, “+C_. = t3l0nVs .0 . ..-. .-.4.. .___ 1713.19.13
  20. 20. Cr. = KC, ‘ = ma x 330 - 664 m/ . 'g A: point M make an angle of 30“ (M and cut the length MN tn represent C5 lo 664 uh). Join LN. Produce L to meet the perpendicular drawn from N at Q. Thus outlet triangle is mmpleu. -d. (:1 Blade angles 8. ¢ : As the blades are symmetrical lgivcn) 9 = G = 30°. (Anna) (ii) Relative velocity of steam entering the blades. Cr. : Cr, = M8 = 830 mls. IAns. > (iii) Tangential force on the blades : 1000 Tgnggnfial fume = nl, ((. "‘| 0 ("H1 = 60x60 413”)! : 1153.8 N. tAns. l (iv) Power developed, 1’ : _10Q0 1:110 x 400 2: gm, I’. m = w: , , I I", I I + II‘, M 60x X k (vl Blade efficiency, n, _‘. : 2(‘, ,,«(‘, ._ + l‘, , I 2x40()x 1:110 ‘ 11“ = (, ._, W = 12'm)»__. = 72.84. tAn. ~a.1 1 lllllple 19.3. The’ u-lm-l! _v n/ '.s~mun uxlting (hr I1Il: :l¢' nftlw impuI. -«- stage of a turbine i; 3 All. The blades‘ operate vlusr In (hr maximum bladmg a-/7?: -1. : :v_. *. Thc nozzle angle is 20°. idcring cquiangular bladrs and rwglccting blade [ru-(inn. calculate for a steam flow of IGATE) lglc, the diagram power and My diagram efliriency. |. |.“on_ (; ,'u. ,, ; (" = -100 mls. (1 = 20’, 0 = 0 ; m_ = 0.6 kg. /5, ( ', .1 cos (1 For maximum blade em. -1.-nay. p = (: - , , "011 = 5”“ W or ("N = 137.9 mi: 400 2 (‘W = y‘ con u = 400 can 20“ - 375.9 mm -1‘ 3 (" gin u 3 400 sin 20“ - 136.8 ml: (2 was . .!a__ . 0.727 tan 0 - Cw‘ _¢-u 3-mo - 137.9 ' o . tan" (0.721) - 36‘
  21. 21. 1’ -----4.. ---- Fig. 19.14 C, ‘ 136.8 ‘ - __ ‘ - - = 2 232.7 In/3 NOW. C11 BID 9 — cf‘ 01’ Cr‘ - sine» “ C’: sin 36:: Now neglecting friction. C, ” = C. .. = 232-7 111/5 fince the blades are equiangular, therefore, 0 = 0 = 36° CW0 = Cro C05 36‘: — Chg = 237.7 cos 36" - 187.9 = 0.36 m/ s cw = cw, +c, ,n = 375.9 + 0.36 = 376.26 m/3 Diagram power, P = 'l1s(C. .», +Cw, )"C1,x ’<10"3 kW = 0.6 x 376.26 x 187.9 x 10-3 = 42.4 kw, (An3,) 2Cu(Cu. ) +Cwo) &Ile or diagram efliciency, nu = C 2 1 2 x 187.9 x 376.26 "‘ Hoop = 0.884 or 88.4% (AIIIJ II or IDA. A single stage steam turbine is supplied with steam at 5 bar, 200°C at tlll III d'50 hglmirl. It expands into a condenser at a pressure of 0.2 bar, nu; ugde gr‘; 5, m M]; (U. P.S. C.. 1301* fisfliuen Zp, -6bur.200°C ; p,-0.2bar, m_ =50 k¢/ m§n_¢u.4oom; .;¢. .fi! ’ cg"-'5 (boclunlblction louunrenoglected) ms 'ali.1iiIutcanuuu: Aubar. soa°c: *1 ' 3355-4 “/53: I, as 7.0593 fir. '-351-5|! -FM. 1-, , 1 If. I ; i »- ,
  22. 22. '. : . _: . . ,03II - 0.8391 9 3,‘: 7.0778 . 7.0693 - 0.6321 , *0 ’ 7.0713 ' °‘” datum at 0.2 bar. . hi ' hf: * ‘'2 hi: - 251.5 + 0.88 at 2353.4 - 2326.9 ml/ kg Bthalpy dmp '= h, - h, = 2855.4 — 2326.9 = 628.5 k-J/ K8 Velocity of steam entering the blades, 0, = 44.7 , f;. , -52 = 44.7 ‘/5235 Q1028 m/ s _' ‘flue velocity diagram is shown in Fig. 19.15 , . NOW. Cu, ‘ = 1028 cos 20° = 966 m/ s C, = 1028 sin 20° = 351.6 mls C- 351.6 ' 9 = jg-j---— = -"'—"j‘—""‘j" '—' mm C, cos '20“ 7 C5, 1028 cos 20° — 400 9 = tan" (05212) = 31.85“ ' I —— — — ~>g<—— C, “ an vo—— ——‘c. ., l"u. : I9 I. ’- Now. c, ..s. . 31.35“ = 0,, = 361.6 351.8 «-=9’? ‘ch ' . Tn 31,3» ' 668 ‘‘‘’' C“ -fllI6I: oI30"-400 -177mb
  23. 23. THY FA! l I"-. 't'»li’l"Rl‘'(} Power developed. 9 = n‘a. <Cu, +C. .,) ch. = :—3 (966 + 177) X 400 X 10*‘ kw = 38] WW. (, ;1.. _) 2CMlCM +C“l ) Blade efficiency, 1}” = l : £: M__"‘£‘i = 0,335 , ,. 35.5% WM 0028)" Since there are no losses, therefore. Stage efficiency : blade efficiency = 88.5%. ‘: ,n. ,) Exnmplr 19,5. The following data relate to a single stage: t'mpuL~>e turbine : Steam velocity = 600 m/ s ; Blade speed = 250 m/ s Nozzle angle = 20“ ; Blade outlet angle = Neglecting the effect of friction, calculate the work dcr. -clopcd by the turbine for the steam flow rate of 20 kg/ .«. Also calculatr the axial thrust on the bearings. solufion, Absolute velocity of steam entering the blades, C, = 600 m/ s Blade speed. CH : 250 In/ S ; Nozzle angle, a = 20“ Blade outlet angle, (9 = 25' ; Steam [low rate. rh, = 20 kgl. -4 Refer I-‘ig. !9.l6. -3 Triangle LMS is drawn with the abovu data . » Then angle LMN Leu. 0 = 25" is druwn such that NM : MS (because efiect of friction is to be neglected i. v., K = 1). , Join LN by vector Co which represents the velocity of steam at outlet from the wheel, This completes both inlet and outlet triangles. c, c_, + c, _ .655 m/ s Fr. ‘ ‘R16 By mva. -mremvn! : C“ : Cu] + C, ” : 655 min ; C, ‘ : 200 m/ s ; C, " : 160 mln. Work developed, W : W -T n'1,(Cu) +C, ,”) C, _, ~ 20 x 655 x 250 3276000 Nm/3. ( Am”)
  24. 24. smw TURBINES . 8SI Axial Thrust : Axial thrust = n'x, (C, ‘ ~ C, J = 20(2G0 — 160) = 800 N. (; n_. .) Example 19.6. A single row impulse turbinc dc-Lv: lop. s 132.4 kW at a blade‘ speed of 175 m/ s, using 2 kg of steam per sec. Steam leaves the nozzle at 400 mls. Velocity coefficivnl of the blades is 0.9. Steam lt"(1Ut'. ' the turbine blade: axially. Determine nozzle angle, blade angles at entry and exit, assuming no shock. Solution. Power developed, P = 132.4 kW Blade speed. C, _. = 175 tr: /5: Steam used. 77-1. : 2 kg/ s Velocity of steam leaving the nozzle, C, : 400 m/ s. Blade velocity co-elficient, K : 0.9 (CV + C“ n x Cu ll Power developed, P = "3. 1060 kW 2<C. +C H175 1324x1000 _ = L C‘ ) T _j___ __ I324 woo or (Ca1+ .1.‘ 2x175 378 m/ s ' CS. ” , 0, since the discharge is axial. Construct the velocity diagram as shovm in Fig. 19.17. 1,. ) in this diagram C - 0.9, B ~ 90°, since the discharge is axial ; and '1 C. _., 4. CM — PL = PQ : 378 mm. H — L1 C. ,=C. .,+C. r—378 INS ——— Ti’ V»; a cm 175 In/ s , P . M L. O 4 , - «— e<— — . , . . ‘O 0:‘ “Ix fl go C‘. Y 0.“. .. _ 7 c, — on , r C: ‘fl *~ c. ' ’rc; . A J ‘A A N S 1- 1-. » 1 I From the diagram (by measurement) : Nozzle angle. a = 21‘. l/ ns. l Blade inlet angle. 8 : 38°. t. n-. .) Blade outlet angle, 0 = 32°. (. »n. ~.> I-Ixumplc 19.7. A . -rina, ':. 'r impulse turbine has a mean blade speed of 200 mi». The nozzles are inclined at 20”‘ to the plane of rotation of flu‘ blades. The stvam velocity from nozzles is 600 mls. The turbine u. ~ms 3500 kg/ II of steam. The absolute velocity at exit is along lllt‘ axits of the turbine. Determine :
  25. 25. 826 msnwu. ENGINEERING (iii) The diagram cflicicncy. (iv) The axial thrust (per kg steam per second). Assulnc inlet and outlet angles to be equal. ('U. P.8.C. . 1998) 8olution. Cv‘ivcn: C5,=200m/ s;u=2O’; C,=800mls; m.=3500kg/ h'; B=90‘;0=¢. (I) Inlet and exit angles of the blades, 0. ¢ : Refer Fig. 19.18 -----+----- 5 . Fig. 19 18. 0,, = c, sin 20- = coo sin 20- = 205.2 mls SP _ C‘, 2052 ‘““ °" PM ” c, cos 20°-cu ’ G00cos20° -2oo °“°‘ e = tan“ (0550 = 294-. Mm. ) Also 8 = 4) . ..(Givcn) . . Q = - 29.4’. (Ana) (ii) The power output of the turbine, P : p = ri: ,(C. ,‘ +C, _.°)cu = 3% (800cos20'+0)x200x mam: 1o9.8kW. (Ania) (C, ,_ = 0. since the discharge is axial) (iii) '11:: blade or diagram edicioncy, nu : “N: 2C, ,(c5 +c, .) C13 - fl”J1 - 0.329 0: 32.3%. mm. ) (in) The axial thrust (per kg steam per second) : The axial thrust per kg per second = I X ‘CA’ N when C, ‘ = C,sin20‘=600sin 20'=205.2 unis.
  26. 26. where (. ‘., ‘ = C, .~in 20' = 600 sin 20" = 205.2 In/ s. C , ‘ Now, -‘— : (an 29.4 («H (. ‘, = 200 x tnn 29.4 = ll2.69 m/5 Sulxslilulinu : l1«- v.1lum, wr gel Axial thrust (per kg steam per second) , 1 x 1205.2 ~ 112.69) 2 92.51 N. mm’, ‘= "1.; ,m, ,,p]¢. 19.3‘ Ste, -am with absolute (‘eloci! _' of 300 n1.f. < is supplied through a nozzle to a slrxglc stage ilnpubzr rurbmv. The nozzle angle is 25‘. The mean diameter of the blade rotor is I mu-Irv and at has a . ~.p«-rd of 2000 r. p.m. Find . -mmblc blade: angles for zcm axial thrust. If the blade velocity co—e/ ficient i. ~ 0.9 and (hr steam flow rate is 10 kg/ s. calculate the power dewloped. s(,1u(;0,, _ Absolutm velocity of an-am entering the blade. C, = 300 m/ s Noule angle. (1 : 25' Mean diameter of the rotor blade. D = 1 m Speed of the mmr, N = 2000 r. p.m. Blade velocity co-ellicicnl. K : 0.9 Steam flow rate. ri1,_ = 10 kg/9: Blade angles : nu V . ‘. Blade spec. -(I. C, _. : r = 105 m/ s. I" In With the above data (i. e-. , C, : 300 m’. <. (LN. : 105 m/ s and a = 2:'1"rdrnw triangle LMS ll"ig. l9.l9). From S drnw perpendicular SP on LM produced. Mozisuro Cr,
  27. 27. 854 " 'm1zx. wu. ENGINEERING _ From S draw a line parallel to LP C‘-' Cf, =C{fl and from point M draw an are equal to Cr‘, (:o‘9cr')hO get the point of intersection N. Complete the triangle LJJN. From N draw perpendicular NQ on PL produced to get C/ h . Measure 0 and o (the blade angles) from the velocity diagram. 0 = 87" and 9 = -12’. (, n-. ».) Power developed, P : ra. <c, , +c. ,t lxcu 10:: 306x105 P: m = ——'fir = 321.3 kw. (Anya) ‘: “Exnmple 19.9, In an impulse turbine (with a single row wheel) the mean diameter of the bladm is 1.05 rn and the speed is 3000 r. p.m. The nozzle angle is 18°, the ratio of blade speed to steam speed is 0.42 and the ratio of the relative velocity at outlet from the blades to that at inlet is 0.84. The outlct angle of the blade is to be made 3° less than the inlet angle. The steam / low is 10 kg/ s. Draw the velocity diagram for the blades and derive the following : (i) Tangential thrust on the blades (ii); -Lu'al thrust on the blades (iii) Resultant thrust on the blades (iv) Power developed in the blades (U) Binding eflicicncy. (P, U_) so| u¢ion_ Mean diameter of the blades, D 1.05 m Speed of the turbine, N ~ 3000 r. p.m. Nozzle angle. a = 18° Ratio of blade RP4‘('d to nu-um . -4p<-ed. p —' 0.42 Ratio : 0 84 . C’! - Outlet blade angle. o = 0 - 3" Steam flow rate ; i;_ = 10 kg/ .- Blade speed. c, , - —"élo"1 = . — 164.5 ml. -2 But I) - %g- -I 0.42 (given) 1 C, = F$‘bL2= = 392 In/ S With the data. C, : 392 min ; a = 18“. complete ALMS 0 = 30" (on measurement) . . 0=30”-3:27‘. Now complete the ALMN by taking 9 = 27" and C, ” = 0.84 C, ‘ . Finally complete the whole diagram a. ~. shown in Fig. 19.20. (1') Tangential thrust on the blades : Tangential thrust = "'1.(Cw, *C. .,) = 10 x 390 = 3900 N. (Anm)
  28. 28. Fug. 29.26) (iii Axial thrust : Axial thrust (iii) Resultant thrust : n‘z5lC, l — Ch» = 10 r120 — 95) = 250 N. Mus. ) Resultant thrust = ~'(: !900_. r2 +4250)” = 3908 N. (Ans) (iv! Power developed. 1-’ : m. '(«'. .; + (3,. .. .' X 61,- 10 x 390 x 164.5 J ,7“ _ 1. mm 1000 _64l.55kW. (Ans) (wBlading efficiency, n, .,, - : 2C', ,,- '61. vC. ,. I 2x16-1.5><390 . . = — ‘—: ’;—"~' = . « =83.5°’. mm. » "’" C. " 392-’ Example 19.10. In a SUI}, -‘I’ n/ 'im. pul. w rmrtirnn turhmv prm'r'r1('(! with single mm wheel_ the mean dwmele, of the ; ,[m[, .,, . 5,, 1 m_ I! runs u! .'m()() r. p.m. The srmm r’. -‘stuns’ from the nozzle at a velocity of 350 m/ s and fhv Imzzlo angle‘ is :20 ‘. Thu rnlur hlmlc-s are equiangular. The blade friction factor is 0.86’. Determine the pulrvr 1/¢'m'Iu]n—'d i/ ' flu‘ axial thrust on the end bearing of a rotor is 118 N. Solution. Mean diameter of the l)lzul(-5, D = I m Speed of the turbine, N = 3000 r. p.m. Velocity of steam issuing from UK‘ nuzzle. ,C', = 350 this Nozzle angle, I1 = ‘-30" Blade angles, 9 = '1! Blade friction factor, K = 0.86 Axial thrust = H8 N Power developed, P : 1 aooo cu=2;§’. =""; ‘0 =157m/3
  29. 29. 856 THERMAL ENGINEILRINO T‘ * cm on C“ ‘; c_ c, ,+ (-c_") 320:1’! /s"“—€*‘| *—c_n P l J‘ -c, ,. = 157 rnls '1 ‘* " "‘ ‘ ‘AM 5 9’ ‘K E ‘ 5 ! _, ,»——‘’' :1 c. .' 0.. A “ X C, Yr °° / ‘V - _ u 31 51 ch — it xi ’ N s I*’r, :. I9/ll “ With the dam. CM : 157 ll'1’8. C, : 350 mls, a : 20‘. draw the ALMS (Fig. 19.21). By measurement. 0 : 35" Since the blades are equiangular. 0 = 0 = 85' ” Now with A as _ 35“ and C, “ = 0.86 C, ‘ . complete the ALAIN. On measurement ; Cf. * 120 mfs. C, " A 102.5 m/8 A150. axial thrust m, (C, »_ -CA! = 118 m, ’ 0:185‘; ‘ $5 = 6.74 km. Further in this case. C“ = C“. +0”, C“! + '(—C", ) = 320 ml»: ('3 > 90°) Now, power dovclopcd. P : -? *—-—-j-"MC" .0" ‘How kW l000 = ‘S-7“_"l‘(’)f)_“’)_”«‘_57 ; 335,3 kw, mm. ) ' EXHEPW '9-IL A ximple impul. -cc lurbinr has one ring of moving blados running at 150 mix. The absolute velocity of steam at exit from the stage is 85 mix at an angle of 80' from the tangential direction. Blade u-lncity co-eflicirnt iv 0.82 and tlw rate of strum / lowing through the stage is 2.5 kg/ s. If the blades one equiangular. determine . - 0‘) Blade angle. -«' : (ii) Nozzle angle ; (L'ii)Absolutc velocity of steam issuing from the nozzle ; (iU)Axial thrust. Solution Blade velocity. c = 150 um. Absolute velocity of steam at exit from the stage, C0 = 85 31/5 Angle, B -. 80° C, Blade velocity ctrcfficient. K = = 0.82 '1 Rate of steam flowing through the stage. m, = 2.5 legs Blades are equiangulnr. i. e.. 8 = O.
  30. 30. — With the above given data velocity triangle for exit can be drawn to a suitable scale. From that. value 0 : 8 can be obtained. Alao the value of C, can be obtained which helps to get the value of C. I with the help of given value ol"K'. With these values having being known the inlet velocity triangle of the velocity diagram can be completed to get ll"u. - value of C, . the absolute velocity of steam issuing from the nozzle and value of axial thrust can also be calculated. The Fig. 1922 gives the velocity diagram of the turbine stage to a suitable scale. — From the outlet velocity ALMN '0 l I I ' c C. ‘ V 1. l l c l S Flt‘ 19 22 By measurement, Cr, = 186 mi: C, 135 C“ — — U-52 — 226.8 m/ .~ (H Blades angles 8. 0 : By measurement 0 c blude angles 27°. (Any-, ~.) (ii) Nozxlc angle, a : By measurement ; nozzle angle, a : 16°. (Am. .) (iii) Absolute velocity, C, : Absolute velocity of steam issuing from the nozzle. C, = 386 m/ s (by measurement). (Ans. ) (iv) Axial thrust : C, _ = 34 m/ s Also. C, ‘ _1o2 m/5 By measurement. Axial thrust —- "MC/1 “C; l = 2.5 ‘I02 - 84) = 45 N. (Ans. ) Example 19.12. One stage of an impulse turbine con. <ist. «: of a converging nozzle ring and one ring of moving blades. The nozzles are inclined at 22“ to the blades whose tip angles are both 35‘. If the 1.-/ ocity of steam at exit from nozzle is 660 m/ s, find the blade . s'p¢. '€d so that the steam passes on without shock. Find the diagram efficiency neglecting losses if the blades are run at this speed. (U. P.S. C.)
  31. 31. 853 THERMAL ENGINEERING S-Ollll-loll. Given : (1 = 22" ; 0 = 0 : 35° ; C1 = 660 m/ s. Fm I9 23 In case of impulse turbine. maximum blade eflicie-nc_y. cos! (1 (nM), “,, , = 2 (l+KZ) . ..[I-: qn. (19.19)l when: Kt». blude wlocily co-efficient) - 1. ('3 Los. -sea are neglected) cos 0 Z : : 1 (‘I Bllxdcs arc cquiangular) flu co: G (1 + 1) — cos’ a — (con 22")’ — 038 or 86%. (An: -‘. ) Also. 9”" : co‘-Zn . ..n: qn. (19.3): or 9_». ;_ : cos 22° C, 2 CM * C, x 0.4636 v 860 x 0.4636 = 308 min. (An. -.l 175'-‘I-Ixaunplo 19.13. In a single stage irnpulae turbine the mean diameter of the blade ring is 1 math: and the rotational spccd is 3000 r. p.m. The steam is issued fiom the nozzle at 300 m/ s and nozzle angle is 20‘. The blades are equiangular. If the friction law: in the blade channel is 19% of the kinetic energy cornexponding to the relative velocity at the inlet to the bladex, what is the power drveloped in the blading when the axial thrust on the blades is 98 N . ’ : 0.4636 Solution. Mean diameter of the blade ring, D : 1 In Speed of the turbine. N = 3000 r. p.m. Absolute velocity of steam ieuuing from the nozzle, Cl : 300 ml: Nozzle angle; (1 = 20“ Blade angles are equiangular. 0 : ¢ Friction loss in the blade channel = 19% : ‘.e. . C, _ = (1 — 0.19) C, ‘ = 0.81 C, ‘ Axial thrust on the blades = 98 N
  32. 32. Bhdeupeed. » _ nlw n I 3000 ‘w H" ‘v ' ‘:3 -u57.1m/ - M” _ _ 9 = ‘l (given! Now. 'l'l0cll_' dmpzmm is drawn ha :1 nuilnlxlu M-nlv my slmwn in l"ig 1924 l ‘l n: x Axial thrust ‘— ”‘4('; (I! ‘ _ 95 us _ m_ ~: |m). n H1: ur "I. ’ ~—~ = 5‘0-_; _5 kg l(lll.5 — Sl m. I('l_‘ . ‘*1 (‘Mil Puusrr tluu'/ v/[u'rl, ’' = ‘Hum (‘pa (‘[5 3- 90°] * '5 2 ': .5 r. 5" 2.1.0.3)! hll )hl), <l = ‘SL2 k“'. ”~n"‘ lllllll Einnple MLH . S'Imu' I/ ml I/ u’ mu. unmm pn~', ~'lhl. ' v/ 'firn'rujs ufu Dr Lurul strain turbine is 83.3% when nozzle unglu as 2!! I): -uluw (hr / iormulu usual Xian. Maximum pumsnlolv t'llll‘lt‘l| (‘’, 11%" : RH IN Nozzle angle. :1 = 20‘ Maximum possible ellicnmcy of u ll: --I mule angle. - 1| . mi 20' 8 10.9396!’ = 0.883 = 88.3%. «Pu-ow-db. . uvu| lurlmm umpulsu turbine! = cos” (1 where 0. is the so For derivation of the formula used refer Article 19.7.
  33. 33. In a aimpl: impala: turbine the nozzle: an inclined at 20° to the direction of motion of Mr moving blades. The steam leaves the nozzle at 375 mil. The blade velocity is I65 m/ .1. Calculate xuilable inlet and outlet angles hr the blades in order that the ¢I. u'al thrust in arm. The rrlatizv velocity at’ strum as it flows over the blades is tainted by 15% by fiidiou. Aha. deter-nu'ne the power developed fir a flow ml: of IO lag/3. Solution. Nozzles. n = 20’ Velncityol's1eaminsningfroInthcnozzIeu. C,s375m/ a Bladespoed C‘, = 166 ml: Axial thrust as zero i. c.. C, ‘ = C,_ C 59- an -0.15: = o.s5. i. e.. was loss due to friction. man new rate. fin, - to km. '1 Iuldaudoutbtaaglu: Withthaahovegivandan, dnwvdodtydin¢rIm0onoIdhbhIuleud1owninFig.19.26. ? F’; 19.5 By measurement (fmrn velocity discrun). 0 =85“ O-48° . Man. ) B I 1”’ Power developed. P : Also. C, “ x354m/ a:C, ~ -24:1:/ s lfiymouuntnant) n': ,lC, ,.‘ QC“. Ixcu Power developed. P - moo . ."Lc2;_’__". .92¢_"LC£ . l° . u4.5|; w. mg) 1000 1000
  34. 34. -33"‘ 1933- I" 0 -‘Will! ’ “flit? impulse turbine nozzle angle is 20" and blade angle: . ‘gel. The velocity m-rfllcirnt / hr hlmlv an 0.85. Fiml maximum blade e/ ficienqy po; ;;‘b1¢_ 1/'3)” . I Ol'. ‘¢ll¢l£l: :l; is 92'? of Mr mnrimum lalmlr c-fllrio-nry, find the possible ratio of blade Solution. Nozzle nnglv. ct = 20' Blade angles an‘ vqnnl i. c.. H = o . . . f (7 Blade ‘l‘lm‘Il)' (‘o—ofl1c| mu_ K . _. (L85 ' "H I (}r ' I ) ’°“‘‘‘l N“d" “m°l“'“‘~. V' = 92"} of maximum blade efficiency Ratio of blade speed to steam speed, p = (""*' (. ', Maximum blade ollicit-ncy is given by : (‘D52 (1. _, , | “|. _~. '.H, ‘ 3 ' O’ ’ ll 4' . ..[hqn. : W3: 21 - K~ as Z = C932 = l 2 cos 9 (‘oral 20 "1e. .~'. ... .. = Z ‘ ~'1 + n. ,~<5.- = (we or 81.69? The actual ell'wicnc_' of the turbine : 0.92 X 0.813 = 0.7:’) The blade vlliciumgv u[' u . ~'nu_: lL- . ~'lu}. ',c impulse turbine is ggivun by be relation. 1], __. : 2 II 1» Km x cos (1 — p'~’; 0.7.") : 21] + (0.8.-'z~Ip x cm‘ 20 p5' ().7."> — 2 x l. H.5I().94 n ,1-‘I 19.20:: = 0.9-: p p~‘ p‘-’ _ 0.94 p + 0.203 —: 0 (UV: : 'U’. ‘.l-l If - -l . « 0.203 0.94 + 0.267 p 2 _) —’ ~ 3 W or p = 0.603 or 0.336 Hence pu. ~~zl; lc- r«. ::; .. . ‘ 2 0.603 0!‘ 0336- (Alli-l fir B‘. Qpl¢ l9.l7. 'l'lu- fillluwmg (lulu rc'[£'r tn (1 siriglr . ~'tu_A. 'r' impulse’ turbine‘ . ' [&, n,, .u, u~‘. "Uzzllf lwut «Imp a 2:3! kJ-leg ; nnzzlu n-/ 'fivi¢-In-_v = 90“; ; nozzle angle = 20° ; ratio of blade speed to whirl mmpunvnr u['. s'tcuIu spent - 0.5 : hlmlv velocity co«z'/ flclvrlt = 0.9 . ' tlw velocity of steam t'IIlt'l'Ilt[. [ tlw unulv : 20 In/5. Determine : ft) 'l'lu- hlmlr uuglrs ut mlr! um! uutlrt I/ ‘the sh-urn c-titers Into the blades without Ihoclc and lruuvu tlw Ialmlva in am u. I;ml tlu'c'4'lluIl. llllfllade efIirlo'Iu'_y. (ill) Power developed uml uxlul thrust 1/tho stmm / low as 8 kg/ s. _ Solution. llentroplc hunt drup -. 25] lulfikg Nozzle etficiency. n, ,,, ,,, , = 90'! » Nozzle angle, in = 20“ Ratio ofhlnda speed to whirl component. of steam speed as 0.5
  35. 35. co-efliclont. K ' °-9 Vdlocity ofnmm entering the nozzle - 20 " '1 Blade an en 2 0 " lIscf1:| ‘l_1vraIflr{rP_ Natale emcioncy is gdvm by : '1u. .m. » "' l5c, ,m, pic heat drop 0 9 _ UE&1'E¥‘E1'"r3 °' “ ‘ ’ 251 . . Useful heat drop = 0-9 X 25' = 225-9 "Mg Applying the energy equation to the muzzle. Wt‘ 89' 2 2 9. :20 = 225.9 x 1000 . . (-13 = 2 x 225.9 x I000 + 400 = 452200 1.0.. C, = 672.4 mis O ‘ -- Ch‘ - r C’-: tht-r data gwcn : a = 20, 6‘ = 0.0. Ix = (, " = 0 9 and Cu-, , = 0. us llw stmm Ivaws Hm blade. -; axially. C0 = Cf , Cw ___j_‘ r: a P . - — - ~ — ~ 4 » — - _ ‘ __ i L. Q j of ' “= n ‘V ' I /1 ' ~» I3 = 90; C. ‘ C. ’ ‘ - ” - Ca: Ca, / __ 4"‘ is '. .-’/ /I I ’rC=672VAm S With the above data cunalruct the velucitv triangles as follows ' 0 Select an Mutable scale. any 1 em = 50 "1/3_ 0 Draw a horwmlal line through a point I. and angle (1 a 30-'_ Mark the point dough; “ " . . . 672.4 L5 5 (.1 3 III a ___ l3_5 cl“. ODrnwulimthrough8whihis di| cuts at the point P. Manna; thapc: i?I: :wc; |.I‘: r-u;2t_': (’¢l, :,o_'h°nt‘l "”‘ ”"°"¢""”. CI-a ‘C, -12.7cnn. cu'°‘5c. '0.5x12.7-e, aA5 ¢n_ . .".
  36. 36. STEAM TURBINES 863 Mark the point M as LM = CM : 6.35 cm Join point MS and complete the inlet velocity triangle LMS. Measure MS (0,) A 7.7 cm. 0, = 0.9 x 7.7 = 6.93 cm. Draw 1: perpendicular line through point L to the line LM. From M cut an arc of rudius 6.93 cm to cut the vertical line through L and mark the point N nnd join JWN which completes the outlet triangle LMN. Now find out velocities converting lengths into velocities : C‘ : 672.4 m/ s C“ = 12.7 x 50 = 635 m/ s CA‘. = 0.5 CV, ‘ = 0.5 x 835 : 317.5 mix (3,, 7.7 cm 2 7.7 at so = 335 mm C, ” = 0.9 C, , = 0.9 x 385 = 346.5 m/ s C, , < 4.45 cm —. 4.45 x 50 = 222.5 ml» C, , = 2.6 cm = 2.6 x 50 = 130 m/ s. Blade angles measured from the diagram : 9 = 35", as 22". (. -ns. .l (ii) Blade efliciency, 71,, : 2C , 2x3l7.5x635 Y1,‘ = = = 0.89 or 89%. (. ~n. s.l (iii) Power developed. P and axial thrust : _ n'v. ¢Cu7 IXC5, fl 8x(635+0'): -(317.5 P — moo W2 : 1612.9 kW. l. ns. l Axial thrust = m, .tC‘, , -C, » = 8222.5 — 130» 740 N. (An. -.l I-Ixumplc l9.lH. In a . -u'ng! ¢' stagr stvam Iurbinv salurated swam at 10 bar abs. Ls" supplied through a conu-rgen! -dizre-rgcn! st. -am nozzle The rm: :!. > angle ix 20“ and the mean blade . s'pr: cd is -100 m/ s. The Mean! pn. -ssurr I. -am/ zg lI'u- Im: :I0 i. ~ I bar abs. Find . (1') The best blade? anglu if black-. ~. an: cqmangular. (ii) Thc maximum powvr d4'L': 'lup1'd by thv turbinv if n numbvr ofno. -:1r. ~' u. -md arr 5 and area at the throat of each nozzle is 0.6 cm". Assume nozzlr c-[7icu-ncy 68“; and blade / n‘cn‘un m~¢~/ [must uf 0.87. Solution. Supply ata. -um promote no nozzles! = 10 bar abs. Nozzle angle, on = 20‘ Mean blade speed. CM 400 mi: St; -um prcasurc leaving the nozzle: 1 bar nb. -. Number of nozzles used 5 Area of throat at each nozzle 0.6 cm‘ Nozzle efliciency. 71" W, 7 88’? C, Bludc friction co-ellicient. K - "" : 0.87. (: ,i
  37. 37. The velocity of stcuxn .1! tin. outlt‘-1 of nozzle la found reprt-:1-nting the expansion! through no/ zlo on in-~ chart .1.» shown In Fig. 19.127. I'll’ f1'. K.L'l9l + 3 an ‘ . ‘ or e er -av <9‘ i U 9 « ‘O ‘ch ~ . ‘ 4. 1.. Y / I’ , .. / ’ ,1 #1: 2 _/ ‘Z / / Saturanon ,2 ’ 4/ hm. - . V . . 3. *3 — 7 — p 5(kJI. kg Fit. 1!) L’? l"mm Ia». ~ chart. Ii, — h‘ - 402 k. l!l-:1: __ - 0.88 “VII If I‘ ' hl ha II, - /11' = 0.88 x 402 2 353.76 it-llltg (3.- 9 Also = Ii, — II_, ' C1’; V911,, 41,’: _ Vi‘: - 3:'.3.76>r fix? ) _ 341,14 mi. in Blade angles : Con: ~truct the ~elv. >cit_v triunglt-. ~ as per data given as shown in Fig. 19.27. Bv nu-usurumi-nt. 01: m s 35.5‘. (A, ,._) m» htuxiuauni power developed, P : For Finding out the maximum power developed by the turbine let us first find out the maximum muss of steam passing through the nozzle. The required condition for the maximum mass flow through the nozzle is given by V; 2.: _[ 2 pl nol where, p, = PI‘: -Hsure of steam at inlet of the nozzle. p. _. = Pressure of steam at the throat of the nozzle. and n I index of expansion) = 1.135 as steam is saturated.
  38. 38. STEKM TURBINBS 839 .9 -. ... .-. ... -.g. .-. .-. ._. .__ 3 “$1923 12‘. a . 2 0.115 I A (1.l35+-1] 0'68 p, =10x0.58-=5.8bar From 5-: chart (Fig. 19.27) h‘ - It, a 105 kJ/ kg and h, —h, '=0.88xl05=92.6kJ/ kg 0,’ (specific vohune at point 2') = 0.32 In’/ kg Tho maximum velocity ofstcnxn at the throat ofthe nozzle is given by C= - J2(h; -l_I; ' = J2x92.4x 1000 = 429.88 111/8 Using the continuity equation at the throat of the nozzle, we can write In . u,'= Ax CwheroAisthenrenofthonoz: .le. m x 0.82 = 0.6 x 10* x 429.88 0.6 x 10" x 429.88 m - —-—m-—- - 0.0303 km. Total mass of steam passing through 6 noules per second is given by rn, =0.0806x5=0.-i03kd3 Power developed by the turbine = ’l’1;‘T°u0°§M kW From velocity diagram, C“, = 750 min (by menmrexnent) . Power developed: E; -'5": -ELM = 120.9 kw. MM. ) '3'‘ Example 19.19. The first stage of an impulse turbine is compounded for velocity and has two rows ofmouing blades and one ring affixed Mada. The nozzle angle is 16’ and the leoving angle: of blades are respectively, first-moving 80', fixed 20‘ ; second-moving 30‘. The
  39. 39. ofuqam leaving the name in 640 ml». 'I'hr friction km in each blade now 0&1’ N10330: volnvilflv. Slmm (mow the no-mud mm of mmrhul hlmlwl Mi“! !!- flnd : (ii Blade wlnrily . ' W’ W044’ “/7954"“-‘. '>' 3 (iii? Symon’/74' xlmm rmum nmlinn. Inluflnu. Rofvr Fin. 19.29. Noulo nuglo. u = 15'' ; 0.’ = 20" [Y = 90" I since the uuvurn Imnwm the blades aid“ 0 = 0' = 30" . . (3, oIont_v of steam leaving the nozzle, CI = 540 mm and (, ' = 0.9 (v . (",4 ' —'- - 0.9 and , , = 0.9. '0 (rr >Q— — _ Z , Cw _ 4 ‘»+ 0-"- _ —c, _. -- c, _ M ‘Z G” ’ L D C . T H I ‘Is 5 V C" t . C, ’ . C‘ . A C, , J 1 ‘ S W 1 CV . 1 < C‘! .| - ‘ cx , P ‘ . . I . ‘ ‘ "in". A O . c ' 0 ‘ 79. " ‘ l" : 90 Furs! row at C ‘ I ‘ 0 ‘ C. ‘ movung blades c C, ' l W N: Second row oi movung blades Second row of moving blades : The velunly Inamglnm slnmld hr drawn slurlmg from 1). ,» _, .,. W,, ,‘[ m“. 0’-, ,u, ‘., ,.‘. wad“, The pro: -c-dare is an fulluum : Refer I-‘fig. 19.29. 0 Draw LM In may mlwu-nwul r+rulc- lam)’ 2! cm! mi (‘N is m-I kumvn. 0 Draw 0' a Jill“ and draw pt. ‘l'pt‘IIllit'llltll' Ihruugh llw point 1. mt Q’) to LA! u fl’ s 90'. This meet» the line MN‘ «I. N’. Tim: rmnplx-tea the outlet triangle LMN‘. Q Mgagure Cm’ : MN’ = 3.5 cm L“ - Cr, ‘ = -9- - 35 -3.9cm. . 0.9 0.9 I Q|
  40. 40. . Ilitu’-20'omldn t‘ dl .9 anus ~ . ' at 8’. Join M8’. Thin eom7:I: t': aa. t'lcieoln‘l: t Magnum"; now of moving blade: : “IO lhllowlng ntopu am lrwulvucl in drawing velocity trlunglo for flrnl mu: nfmovin‘ una", 0 Draw M! = (r, ,, t: 3 cm) , C’ 65 , (-0 = = = 7.22 cm. 0 Draw . :'O - 30“. through the point M to the line LM. 9 Draw an arc of radius 7.22 cm with centre L. This an: cuts the line LN at point N. Join . I. ’. This completes the outlet ALMN. ' 0 Measure C, _ = MN = 9;; cm (I " = C, -I : ‘ = cm -— Draw an . _’(t = 15". through 21 point L. Draw an arc of radius of 10.8 cm with centre at M. This art‘ cuts the line LS at 8. Join MS. This completes the inlet velocity triangle. Measure LS from the velocity triangle LS : 13.8 cm = (7. = 540 mfs. The scale is now calclllulcd from the above. ‘ 40 33“; Zn Smlo 1 cm = = 39.] m. u'. ~'. an: x m Blade velocity, Ch, : Measure the l'nllnwing. _: distzmce: -' from tho -'vlucit_; ' diagram and cmwert into velocities : (. ‘,_‘. = 1.. »AI = 3 em = 3 x 39.] = 117.3 m/ s. (A1-us. ) (iii Blade c-fi'i(-. ienc_v, 11,“. (. .'U_ : P! ) = 114.8 cm : 18.8 x 39.1 = 735.1 mls C ' : I"Q' 4 6.2 cm : 6.2 x 39.1 = 242.4 m"s . a . J". i": :"‘(('1z_ ‘ (7.-. i'_' _ (‘I2 = 3” '”"‘5'_7“': ("-; ‘., ' ' 2423' = 0.786 or 73.5%. (Aun) iv’) l- um Specific ate-um consumption, m‘ : IIlV. }lV(;3 0 ('h_' g 3600 . x lmm Iilillll ‘ I000 _ 7 = ' .8 In kWh. (Anal '"~ ' 4735.1. 242.4n117.3 3‘ 9 ‘V M: IEWJ. 'I'Iw / iilluwmg pulfli-ulurs rrlulv In a lwo«mu' t't'lt'-‘W’ W"‘P0“"d"d “"9"! ” mxl o ll7.3 3600 1000 wheel : Steam Ut‘ll)(‘l! y at rmulv mule! = 650 m/ s Mean blade an-locily = I’-’5 "W Nozzle outlet angle = 16"‘ _> angle, firs! row of moving blades = ll? " I U I ll -
  41. 41. 868 THERMAL ENGINEERING ' -Outlet angle. fixed guide blades r 22° Outlet angle. second row of moving blades : 36° Steam flow = 2.5 kgls The ratio of the: relative velocity of outlet to that at inlet is 0.8»! for all blades. Determine the following : (1') The axial thrust on the blade‘: : (ii) The power developed, (iii) The vfl7ciency of the whevl. (AMIE Winter, 2001) solution. With given data. C“ = 125 m/ s. C, = 650 m/ s. 0 = 16°. firs! row inlet velocity diagram is drawn. C, _'+C_L: 924 mls °" ‘ 23: Second row of 9' 35 moving blades Fuf 1930 Now with given, C, = 0.8-SC. -, . ¢ = 18°. first row exit diagram be drawn. With c, ' = C0 ; a’ = 22°. second row inlet velocity diagram is drawn With cm’ = 0.34 c, _' :9’ = 36°. second row exit diagram is drawn. The values rcud form the diagram an: an followa : C, = 180 m/5. C, ‘ = 138 mix. I C/ ' = 122 In/3. C/ . ' = 107 m/ s. C, ” + C“. = 924 m/ s. Cu, ’ + C. .,' = 324 m/ s.
  42. 42. STEDWI mxsmes 869 (1') Axial thrust on the blades _-. ‘ Cr 3+ (Cf! I — Cf’ = 2.5 H180 - 138) + (122 -107Il= 142.5 N. (An~.1 (ii) Power dcvclopvbd = '“c‘“' + C“ 3 ’ ‘CM’ + C“ ""0" 1000 = = 3.. kw. (An3.) Um Emden _ g_924 +_: ;2:g;5_1g§ ‘ (92.1 + 32-: n x 125 W ' C. ‘:2 ’ 650*‘/2 = 0.738 or 73.8%. (An. -,. .) Example 19.21. The fir. -ct stage of an impul. -.-c turbine is compounded for vclocit_y and ha. »- two rings of moving blades and our ring of fixrd blades. The nozzlv angle is 20‘ and tho leaving angles of the blades are re. »-poctivcly as follows : First moving 20''. fixed 25” and second moving 30". Velocity of steam leaving the nozzles is 600 m/ .-. -cc and the steam velocity rclativc to the blade is rrducrd by l0’. ~’r during the pasxagc‘ through each ring. Find the diagmm efliciency and power developed for a steam / low of 4 leg per second. Bladc . -speed may be taken am‘ 125 m/ svc. (. VI. U.) Solution. Ch, = 125 m’5, C1 : 600 rules a = 2o= , o - 2o= u’ 25". o‘ 305 10 ' : 1 " j’ : K 100) °'9 Vll. : 4 kg/5 Vfith these values velocity triangles can be drawn (Fig. 19.81). c, =c_, «c_ =850mfs C, ,r$5m‘s C_ -285rn. ‘s Cu — 125 mfs I-- L F‘ . . l 3 —c. ' = C. » c- ' = 2&0 m-s—! >1/ c. ) ~ 20 «vs .13 n"§'L«'Za. . P{ 5' ‘ Second row oi movung blades
  43. 43. 844 nnmuu. mamwuno From diagram (By measurement) : Co, = 555 min, Co. =- 235 min 0.. .,’ = 260 min, C. ,u' r 20 min Now. Ow= C,, ,+C, ,‘, =685+285=850mIs 0,’: Cw, ’ + C. ..‘ = 260 + 20 = 280 Info ‘, (C C, .' Power developed 2-. fl——¥31%0—6’—-x—: N- = 4x(850+280)x125 1000 _ _ C ‘(Cw +C', , ) Daagram efiicxency - -3573’? -— = “#23539 = o. 'm7 or 7844793. mas. ) Example 19.22. The following data relate to a compound impulse turblne havlng two rows of moving blade: and one row affixed blade: in between them. The velocity of steam leaving the noule 600 ml: 3 kw! Blade speed = 125 ml. -: Nozzle angle :2 20‘ First moving blade discharge angle = 20‘ First fixed blade discharge angle = 25‘ Second moving blade discharge angle. -— 30' Friction loss in each ring = 10% of relative velocity. Find : (I) Diagram efiiciency , - (il) Power developed for a steam flow of 6 kgls. Solution. Refer Fig. 19.82. Flat now of moving blade: : To draw volodty triangles for first row of moving blades the following procedure may be followed: Select a suitable scale. Draw LM = blade velocity (C, ,,) = 125 min. Make (ML? 2 noulo angle; a = 20'. Draw L5 = velocity of steam leaving the nozzle = = 600 lnls. Join MS to complete the inlet triangle L248. 0 Make AL! -flV = outlet angle of first moving blades = 20'. and cut IDUV = 0.9 MS, xinoo K = 0.9_. — Join LN to complete the outlet. ALAHV. Second row of moving blade: : Tho velocity triangles for second row of moving blades may be drawn as follows 2 o Draw LM , = blade velocity (05,) = 125 mls. 4 Make AMLS = outlet angle of fixed blade = 25‘ and an L3‘ = 0.9 LN. (‘. ' K -=0.9) §-tbs?
  44. 44. m -. ... -4--- -. -_----. .‘. ------n N First row of moving blades Second row of mowng blades Fig, !sn. :s2 g Jnin .118’. The inlt-I 'v| m'it_' lri: mgl(~ l. .l. “ is mmplvtc-(I. 0 Make / I.. 'lI. " : nutor unglv uf . ~¢V-mm! nlmiru I-1.-ule-:4 : iii! and cut M. " : 4). !) Ms’ K = 0,9. . Juin I. ;". 'l'h(- null: -I ‘(~l<)<'iL tt'i: m,«_ch- i. ~ cumplvl-‘ll. The ihlluwim: rt-quirul dntu unny mm‘ he . <('uI(-(I MT frum the dingrmn : (j : (V. -(1. : H} = 8-1:’) m. -.- ('_, " — I"()' _. ‘,2.‘~T(l H15. . (i_ '- 'u Diagram effit-it-ncy. HM I ‘ (H-- 2 - 1:154 1-54:’ < am». A . , * 0.781 or 78.l‘"u. (A113,) Itimh“ III"('” 9 ('__ " HNMI _ lilH~| ."n v ZMH - 113:’! : ‘HSJ5 k“, . ‘ ’ I000 I8’ $1,“ ".33 17,‘. [mg , ..mg. - "/1: Iurhuw m u (mu ruw l'¢'Im'II)’ mmpounded impube W581. The steam velocity at ml: -I an (500 m In mu! Nw uwuu him! » m-lur: !,‘ Is 130 mill. The M332’! .. .ongl¢ E3 16'' and (he 9;-it angle; for the fir. -«I-ruw nf mm'm; .' blades. the fixed blades. and "I8 now of moving blades are 18”. 21° and 35" re. -sperm-rly. ' 0) Calculate the blade inlet angles Ibr we-h mu‘- (m Power dvvu-lapped. I’ 2 w_. J
  45. 45. Z. ~ W -dnlinaah mvflfnuouinl 5104“ """""""‘k"'"‘“h' " a m. .¢: ::u: gm'n’ vrfmmuw flu‘ the wlm-I and (In! dlflflmm P°“""'P""‘ ' ‘ . - ‘ “m cm Who! would be the maximum posulihle diaxfflm P/77¢". """-7 I” M‘ U‘'’‘’' '“°fl| ‘h &i(y and mule nngic ! Thlv tht blade wlomtv mveflicient as 0.9 / ‘W’ “[1 ’’’“d'~''''‘ fihlhu. Refer Fig. 19.33. (1 = 16“, 0 = 18". c, = 600 mm. Cu = '20 ""5 11': 21". ¢' = 35°, #1,, = 1 kg/ s. Blade velocity co-omcient. K = 0.9 With the above data velocity triangles can be drawn. fivm the diagram thy measurement) Cw = (, ',, I «Ch = 875 m. ~’s ; Q": C "7! r C J. " = 294 In/ S cf 3 168 m"s, c, . = 1:45 mss ; (7.; = 106 m/ s. = 97 m/ s. u‘! Blade inlet angles : First mm : 0 = 20 Imm-int; blade‘ [3 : 24.51‘ “fixed blade- Smrmd mu' : 9' = 34.5 «moving hlndex (1,: c, ,‘ + ch: = 875 m. -‘s
  46. 46. :7 * iolvufmaoing Mada-n - m, ,((? ,,. _ o(, ‘“, ") - m, r:, ,, . 1 , . pm; . "5 N_ (E, ml! of mo: -drug blaulwn - u‘: ,((. .', ,," 4 (‘. ',, ,"') — m, (.', ,,' - I x 294 - 204 N. M Axial thrust : new mw av’ mvvinu N-vulva = m, «(. ',_' + (. ',"'l= 1 x mes - mm = :53 N Samnd mcc- nfrnnmm: blade-s = m, (C, ' ' — Ch‘) = I 3: (I06 — 97; . -. 9 N Tlwol axial thrust = 33 + 9 = 42 N per kg/8. (All! -l uiil Power developed = ''l~-1C«-- ‘* Cu ' ’ CM, IOOO = lx(8 z9 = 140.28 kW per kys. 4AM-D I000 Diagram cfllciency = 2C""C" . ‘ C“ " = 2 X '20 787". ’ ' 292 C, ’ c600.’- = -— 0.7793 or 77.93%. (Ans-l n‘x'I Maximum diagram cfficic-ncy : (‘(153 u. : cos! I6 1 0.924 or 92.4%. (AIIBJ Example 19.24. An irnpuI. ~'r' s! a,4.:4' n/ ' a lurbinv / um In-n maze of mmring blades separated 53' fixed blades". The sfcmn Ivr1m': ~' Ihr Iuazzlm u! an rmglv n/ '20’ wifh I/ Iv «IiI'u'(. -(inn of motion. of the blades. Thu‘ blade’ exit un; :Ivs arr : Isl IIml‘iIu, ' 30: : fi. n'r/ 22' ;2nd running 30“. If flu‘ adiabatic he-at I/ rnp / hr Ilw rmzzlv is 186.2 I: -l"I. ',u and (hr H022/P ¢’fficirnz*_v 90%, find Uh! blade spcvd Izcn-. ~'. s'(zr‘_' ti" (hr / Fmri M-/ m'i! _v u/ '1/w sfmm / ls‘ in he axial, .—‘l. ~:. s'umv a loss of 15% in mlatlre velocity fur all hlmlr / )(. '.~'. ~'u; ,'r'. ~:. l"r'na1' ru'. w Hz. » blmfr 4-/ ‘/.7r~i: -In‘) am! the stagr e ciency. (P. U.l Solution. St(-um ‘l'll)(‘ll_'. ('1 = 'l‘l-7'-3 ‘l~ "1 = ‘H-73 0«9 "'1 '8“-2 = 579 mls. The v9](, cn}- di, -ggrggln / ‘ur uxznl a/ i.w~Imr_: .,'¢- Iurhinv is drawn in rmwr. -:9 direction Isee Fig. 19 34:. . The Nude -(. |,, (-jgy | (', H,l LA! is (lmwn In any cum‘: -uic-nt . ~‘c: Ilv. 0 As discharge is uxiznl I. ." is (lrmm pt-rpondicular tn 1.3!. O Knuwinfl the outlet : ml. 'l¢‘ "l. W" -"*""’"d "‘""'l"$! "l"l~' ‘O’ = 3””. IV" 58 located. MN’ represents relalivv vulurity ut uullct I(", " I. O Relntnve v(-lucity ut ml: -t tn the ewcund mm-in}! l‘l*| d*‘ is MN’ r CL, ’ . ‘ ’ r S’ —' ‘ —‘ ' ( M 0.35 l K The triangle at lulu-1 In thus nunund nmvifltl l"“‘l"5 "‘“5' LMSI i‘ ‘mmined by drawn‘ the n . _ _ discharge angle / Ml. .S" Iu‘ : 22“), wlu-no l. ." 1(', 'l 1?‘ “W 9”“ ""—'l“‘“. V °“h" 5“-'°“d bud‘ ring. N C LN cl’ 3 13"" gm. -unung a loan of 1506 (given). ° °"' ° ' ‘ 035 0.85 '
  47. 47. 848 THERMAL avomannlna -9 With campus: at centre L and radius LN an is drawn and the velocity triangle at tho axit of the fir-st moving blade ringLMN la: completed. knowing the exit angle of the first moving binds ring (9 = 30"). I I I I t C». I I I » I 8 9* First row 0! moving blades I-fig. 19.34 a With a= 20° and C = M8 = 5- = 51- the inlet velocity triangle i< completed ’* 0.35 035 ‘ ‘ Now 1.8 is the absolute velocity from the nozzle. Since this velocity is known, scale can be calculated. It la __ Absolute velocity at exit from the nozzle bonxthls and then the blade velocity. etc. can be calculated. From the diagram : Cu (2: LM) = 117 min 0., + C. .. = 752 we _( . + '): < _’. 'I62+23~l)x117 Blade efflclency, nu - cf’: — -fir - 0.6952 or 89.52%. Mas. )
  48. 48. STEAt TURBINES 875 En ' ‘Z "* Ch: - C"§E3E’: _‘_’ " ‘,1? 1._, 186.2 x T1660" = 0.626 or 82.6%. «. n~. > Stgige efficiency, n‘w_, '_ * 1 1:; -,: , nmcnon mnnmss The reaction turbines which are used these days are really impulse-reaction turbine. Pure reaction turbines are not in general use. The expansion of steam nnd heat drop occur both in fixed and moving blade: -.. l9.3.l. Vt-lucit_' l)iugrum for Rt-uctiun Turbine l3l. idt- Fig. 19.35 . -~how. ~a the velocity diagram for reaction turbine blade. In case of nn impulse turbine blade the relative velocity of steam either remains constant as the steam glides over the blade: or is reduced slightly due to friction. ln rcuction turbine blades. the steam continuously expands at it flows over the blades. Thr effect of thr continuous expansion of steam during the flow oucr the blade is to incrca. <¢' Utv nrlatirr vs-lucit_ of . ~.-train. . ... c_ _. ,0 --__-‘_-___ U) l-'2: 190 . . 3 Velocity dl. |)fT1IlTl for l'‘: IC! l0n turbine bludc. C, >C, _ for reaction turbine blade. (C, SC, ‘ for lmpuls-" turbine blade). 19.8.2. Degree of Reaction (Rd) The degree of reaction of reaction turbine same is defined as the ratio of heat drop over moving blades to thr total Itrat drop in the stogv. Thus the degree of reaction of reaction turbine is given by. R _ flout drop in moving bla(lt~: ~' ‘’ > Heat drop in the stage ‘Sh; -1 . , = MI ‘A nee shown In Fig. 19.35. The heat drop in moving blades is equal to increase in relative L'('lOCll_Y of . -»tcam passing through the blade. am: L
  49. 49. 876 THERMAL ENGINEERING >3‘ Saturation line (0) (6) F1;-. 19.35 The total heat drop in the stage (Alt, + AIL") is equal to the work done by the amxun in the stage and it is given by . ..u9.22) Referring to Fig. 19.36, C, " C, " coscc () and C, and (Cm! A» C“) = C, ‘ cot 6 + C, ” cot o The velocity of flow generally remain constant through the blades. C, = C‘ = C] - Substituting the values of C, ', C, ‘ and (CHI +C, _.. ‘) in oqn. (19.22), we get (IDS-(‘Ce 20” COIO c = ac; (eot¢—cot9) . ..<19.23) If the turbine is designed for 50‘? reaction (M, ._ M”). then the cqn. (19.23) can be written as C! I 2 = ac” (coto—cot0)
  50. 50. STEAM nrnnmrs 877 . C, _. = C, (cot o — cot 9) . r1s4,2»n Also C” can be written as C” = C, (cot 0 - cot B) .1 19.25» and CH : C, (cot (1 — not 8) . . (19.26: C, ‘ C, “ C, is uaaumcd in writing the above equations. Comparing the eqns. (19.24), (19.25). (19.26) 0 = fl and o = (x which means that moving blade and fixed blade must have the same shape if the degree of reaction is 50%. This condition gives symmetrical velocity diagrams. This type of turbine is known as “Parson's reaction turbine‘. Velocity diagram for the blades of this turbine is given in Fig. 19.37. 1"‘ . : IE". 37 Example 19.25. Dc nc the term ‘detzree of reaction‘ as applied to a steam turbine. Show that for Par. »-on'. ~ reaction turbine thr degrw of reaction is 50'}. (AMIE Summer. I998) Solution. Refer Fig. 19.38. Fufl I". C? ‘
  51. 51. 873 THERMAL ENGINEERING The pressure drop in reaction turbines takes place in both fixed and moving blades. The division generally is given in terms of enthalpy drops. The criterion used is the degree of reaction. It is defined as Enthalpy drop in motor blades _ Ah, n__ Total enthalpy drop in stage Ah, + Ah" A special case is when the degree of reaction is zero ; it means no heat drop in the moving blades. This becomes a case of impulse stage. Other common case is of Parson's turbine which has the same reason for both the fixed and moving blades. The blades are symmetrical, l. e.. the exit angle of moving blade in equal to the exit angle of the fixed blade and the inlet angle of the moving blade is equal to the inlet angle of the fixed blade. Since the blades are symmetrical the velocity diagram is also symmetrical. In such a case the degree of reaction is 50%. Applying the steady flow energy equation to the fixed blades and assuming that the velocity of steam entering the fixed blade in equal to the absolute velocity of steam leaving the previous moving row, we have (Refer Fig. 19.36) Ah L C12 " Cl): ’ 2 Similarly. for the moving blades Ah qhq" "‘ 2 mt q= qemq= q - Ah, _ Ah, " Hence degree of reaction : —fl‘1— : -1- Allm + Ah, " 2 This is a proof that Parson's reaction turbine la a 50% reaction turbine. Example 19.28. (a) Explain the functions of the blotting of a reaction turbine. (b) A certain stage of a Parson’. -‘ turbine con. -ulzsts of one row of / ixed blades and one row of moving blades. The drtaibz of the turbine are as below : ~. The mean diameter of the blades : 68 cm R. P.M. of the turbine = 3.000 The mm»: of steam passing per set = 13.5 kg Steam velocity at all from fixed blades : 143.7 mix The blade outlet angle = 20“ Calculate the power developed in the stage and gross e/ felt-ncy, assuming carry over a)- eflicient as 0.7-! and the efficiency of conversion of heat energy into kinetic energy in the blade channel as 0.92. (M. U.) Solution. la) The blades of reaction turbine has to perform two functions : 1. They change the direction of motion of steam causing change of momentum. responsible for motive force. 2. The blades also act as nozzles causing pressure drop as steam moves in the blade passage. (bl D : 0.68 m, N = 3000 r. p.m. . ,5. ; 13.5 kg/5 C. “ = 143.7 m/ s. o = 2o= . u = 0.74. n = 0.92
  52. 52. STEAM vunulm-; s 853 Blade velocity. 0“ = % = = 1053 m/ ,; C. ., +0., ‘ = 185 min mg. 19.39) c. =c, ,+c__=1s5m/ s I I I I 0 i‘ I I I I I -----+----- FmJ&M P d I , (Cu +C, _. )0“ owe: eve oped m, x -1:-'-foo --135:: L1"E‘o°i‘3 : 23739 kw. mas. ) ln Parson turbine blades are symmetrical. i. e., n= m0=B c1; cm C, .~_c° C ’ - vc ’ 143.7’ - 0.74 x 57’ :3 go? .2.’ ‘_" Enthnlpy drop 2 x 2“ 2 x 2 xo.92x moo 19.83 l: J/kg c. -0.. . . m.. .,. .¢, - _W«= rk<'<= _n«= '*ar_ - ‘_°: wL‘£efl ' Enthnlpy drop/ kg ‘ 1983x1000 l55)<l06u8 V = @3700? = 0.8880l"88.3'§>. (Allen) Example 19.27. (:1) Discuss the factors that influence the erosion of turbine blades. On a sketch mark the portions of the blades more liluily to be eroded. Sketch the methods used to prevent erosion of steam turbines blades. ’ (b) A reaction turbine running at 360 r. p.m. consume: 5 kg of steam per second. Tip leakage is 10%. Discharge blade tip angle for both moving and fixed blades is 20‘. Axial velocity of flow is 0.75 times blade velocity. The power developed by a certain pair is 4.8 kW where the pressure is 2 bar and dryness fraction is 0.95. Find the drum diameter and blade: height. (U. P.8.C. ) Solution. (o) In the high pressure and intermediate pressure stages of turbine the pros- euros and temperatures are high and the blade material should be such that it stands high pres- sures and temperatures. In the intermediate pressure stages steam is wet therefore. the material
  53. 53. ,3l! * 1 . . _mm- raglan dun tn the presence ofwator particles. In ' . . tho hlzilnzngrh also nulxjvvlc. -ii In hill. ‘ """"'. "/"lB'¢3l ""939" ll ill! " __' "um" ‘up. ".9 gonwr. 9,. ..-, rm-. ,_ um blade material and DUI design should be luchthgg 3 Ihilda corrosion. enminn nnd high m-nfri/ lfltfll 4""”"“’“' ' _ When the spend in high and mnisture exceeds 10 P0? 090'» ""3 "m’°'- "r "‘°'5_t‘"'9 ll llllllt prominent. The most 1-IT1-cu-d portion is I. he back of the inlet edge of the b'9d9- Where 9'! -he’ 8|'°WoI are formed or own ammo portion breaks away. Due tn centrifugal force the water particles tend to concentrate in the miter annulus nnd their tip speed is i. "'e¥1“"' “W” the "°°'- Weed» 1191109 efoliflll elfut is must on tips (Fig. 19.40). Worst affected areas Fix. m. 4»; Methods adapted to prevent erosion : ail By raising the ti-nmpr‘-rutnm, » nl‘. <u-am at inlet. so that an exit. nl' turbine the wetness does not exceed ll) por cont. iii! By adopting l1'l1‘m (‘_w'lv ; so that the '(31l1tr. ~’: ~‘ at uxul rmnnin. »- in linm (iii I Drainage bolts uro prnvidod Im the turbine. so that lhv -. m.. .- d, .,, p]N5 which an, on out“. P9I‘iPlu-ry. dun to i'mx1rifu; ,'ul force are (l! ‘Zllll(‘(l. Tho drzum-d :1 _ muum is wbo t 2' t of total wntcr pnrtwlq-. ~s pr: -sent. ‘ U 0 per can UN The leading 1-tlgo of the turbine is pmvidvd with :1 . -lIl(«| d M. Wlrd "nmrifl - In the method In dil'l'1cultios are the limit. -* of u-mpc-mturv :1 nmtvrial can withstand -' Rel"-‘N C)T| t' lmellmd lit '! hm il: < own mlvnntugc-. < and di~‘zul'unt'ngo-z - . . . . - Drainage belts Imothnd «Ii: '1 c'| u~w ~4lru('tur'1l chun ‘ _ ‘ - ‘ - go. -a II) [b - t ' . .' ‘ however, hlc-1-ding may help. t urblm ca°mg deflgu‘ The most suti: sl'u('lury vaululinn is‘ prnrialingv tun ‘ ' - - . gsh-n shwld Nu: ; I . . , , d. 1‘ hquggqxn “ dun, no! n. mm. . H, . , -t. .."5(a, u.. ,. ’ h, _ J - ‘ . [70 n! ?.‘». . I . i- (via 0. lff. row, ‘ a 4 s 4 4 m I A Male! ’ drnpl. -ts unp(. ,-, - 0,, 3;“. mm_, iL, n 0,-‘he (b)3l>u-‘d. N = 360 r. p.m_ _ 10 . m, - 6 1 — mituplozukzngelj = 4.5 kg/ s “ -"- 4' = 20“. 4' a 0.75 (' Povlor developed in a certain pair I ~ , ‘'43 kW atzbuu-(x. o,95; mbwwm c“'LM"i6)e¥‘”l: ;‘3°°-1s. ssom/ e C5 -o.15xms5n.14,;33D_, _
  54. 54. 5'| "EoM TURBINES and 855 Fig. 19.4) C, ‘ _ 14.13:: D C‘: sin20°_ 551120‘ In/3 LP: c, cos 20' = 14.13 ax °f”‘2°° = l4.138Deot20" 311120’ PQ=2LP-LsI= (2 x 14.138Dcot20‘-18.85D)n~Iu 2 . '."'.11.’3i‘_’: ".‘. moo _ «i.5x(2 x 14.133 Doot 20° - 13.35 D) x 18.85 D ’ 01000:- Power developed 4.8 Solving this equation, we get Drum diameter, From stonm tables at 2 bar, D 2 0.98 In. (Ans) u, = 0.885 tn’/ kg Flow are: x flow velocity Specific volunie 4.5 = Xc’ Mass flow rate = ‘"1 __ n x 0.98 x In x (H.138 x 0.98) " 0.95x0.885 h = 4.5 x 0.95 X 0.885 n x 0.98 >; 14.188 x 0.98 . . Blade height : 0.0887 In. Mun. ) Example 19.28. (0) List the advantage: of steam turbine: our gas turbines. (in) Determine the isenaopic enthalpy drop in :11: stage ofParson': rtndiou turbine which 4.5 = 0.0887 In Iuoslhefollowingparticulara: Speed: 1600rpn1;nuundiametcrofm¢or= = Im; stoge¢fl’ici¢ncy=80‘: Ia: speedm! io=0.7: blade outlet angle = 20'. Solution. (:1) Advantages o! steam tux‘-blues over 2113 turbinea : 1. The iond control in steam turbine: is oasy simply by throttle governing or cut-off govern- (B. U.) ing. In gas turbines the air-fuel, ratio becomes too high. 100 to 160 at part loads. This causes problems to sustain the flame.
  55. 55. 832 THERMAL ENGINEERING 2. The steam turbine works on Rankine cycle. In this cycle most of the heat is supplied at constant temperature in the form of latent heat of evaporation. Also the hent is rejected in the condenser isothermally. Hence the cycle is more efficient, and its efficiency is close to that of Carnot cyclo. On the other hand. the gas turbine worlus on Bruyton cycle whose ellicicncy is must less than that of Cnmot cycle working between the same maximum and minimum limits of temperatures. 8. The efficiency of steam turbine at part load is not very much reduced. In go. -5 turbines the maximum cycle u. -mperature decreases considerably at part load ; therefore its part loud ellicioncy is considerably low. -1. The blade material for otcum turbine: is cheap. For gas-turbines the blade material in costly as it is required to suwtnin considerably high temperatures. (bl For Parson's reaction turbine. the velocity triangles are symmetrical, as shown in Fig. 19.37. Given 2 N ; 1500 r. p.m. , D -_ 1 111, nmL, _ : 80'} ; Speed ratio, £3‘; = 0.7. c = u = 20“ 1 Cl = Cr and Cr] = C0 ILDN n >: 1 x 1500 . ; ——= --T — 78.54 m/ s C" 60 60 Speed runo . 0.7 — - C1 78.54 C‘ = = I122 l'l'I/ H c, ;' . c, ‘-' + ch, ” — 2c, c., cos a ; u12.2r-' + «7a. :.. n= - 2 x 1122 x 73.54 can 20'’ 2195.34 or C, ‘ = 46.86 m. /.~ M = Actual enthalpy drop for the stage : gnc, ’ — com (c, ._‘ - c, fn = gnc, ’ — c, _% + cc, ’ — c, _’n = c.’ —c, _’-‘ (‘. ‘ C“ : C,, ;C, ‘ = C,i or M = [(112.21-’ —(46.8-1)”! x 1/1000 lo]/ lag = 10.39 l~: Jfka Iscntropic enthalpy drop. wn : "J3 = 1?? = 12.99 kJ/ kg. (. «n. s.l . v'. .1;; v~ 19.8.3. Condition for Maximum lzlfficicncy The condition for maximum e/ '/iciency is derived by making the following a. ~;. «.-umptioru : ii) The degree of reaction is 50%. (ii) The moving and fixed blade. » are symmetrical. (iii; The velocity of steam at exit from the preceding stage is name as velocity of steam at the entrance to the succeeding stage. Rcfcr F ig. 19.37 (velocity diagram for reaction blade). Work done per kg of steam. W= CN (Chi +C, , I = CNIC, cos a+iCr, _ cos o-C, _.)l
  56. 56. .. a “flop. ‘cg Wu 0,, [20, con a-Cu] w . c, = . ;°Eu£': _:_-°_-.2 _ Ea: C12 C‘? 5 = C, ‘ [2p . cos a — p’l . ..uoxn 91 when p - C‘ . 2 The RE. supplied to the fixed blad = 9- . 2g 2 2 The KB. supplied to the moving blade = CALEL . 2 Total energy supplied to the stage, M = Eli + an)? _(’v: L2 2 2 is C, “ = C, for symmetrical triangles. Ah = 9l:2_+Cl2 2 2 2 = 0,2 - 93- . ..u9.2s» Considering the . .MS (Fig. 19.35! Cr‘? .-. C12 +CM2 - 2C, . C, “ . cos (1 Substituting this value at‘ (. ‘,"“’ is cqn. (19.35). we have Total energy supplied to the stage A; . = C3 — (c, * +(: ,,, ’ - 2C1.C~. cos an :40,’ +2C, C~ cosa— CH2 I/ '2 . , H 2 C—'L[l+—‘3‘~. cusu-L-(-‘M-]] 2 C: 2 I = C; [l+2pcoaa-pa] 11.. 51.4. Qfficigncy or the reaction turbine is givtn by-
  57. 57. 858 THERMAL izuaumnuua R flggcoau-E’) = 2p(2ooaa-pl ___ 2(1+2poosn-E‘)-2 (1+2pooau—p’) (l+2poosu-9‘) (1+2pcoaa-p) n 2 - -—l—-5- . ..<19.3o) 1+2pcosa-p 'n: o nu becomes maximum when the value of (1 + 2p cos I: -- 9’) becomes maximum. '15 Fig.19.42‘ The required cquntionis £(l+2pooaa-p’)=0 do 21:05 a-2p-= o p : cos a . ..(19.31) Subatibzting the value of p from cqn. (19.31) into the cqn. (19.30). the value of maximum ¢fl'5€1£'I¢.7 19 Riven 5!. 2 1 2 ’ "‘«-I’--= "$: .5:= "’[ —°3"i"“ 1 + cos’ a 1 + cos a 5’ Hence (n. ),, ,,, = fiafi . ..(19.a2) The variation of 115, with blade speed ratio for tho reaction stone is shown in 1 Fig. 19.42.
  58. 58. STEAM TURl: llNl': S 335 TURBINI-IS El-‘FICIENCIES 1. Blade or diagram efficiency (nA_. ). It is the ratio of nor}: dour on the blade per . ~:: -ound to the one-rgy vnlvrlng (hr blade per .5:-cond. 2. Stage efficiency (n_'“_). The stage efficic-ncy covers all the losses in the no7.zle. ~‘, blades, diaphragms and disc: that are associated with that stage. N = Network done on shaft per stas: _e_per kg of an-um flnwint: .: .». .:- Adiabatic heat drop per stage _ Network done on blade». — Disc friction and win<l. u,: v.- Adiabutic heat drop per stage ' 3. Internal efficiency (ninm_M, ). This la equivalent to the aluf. -{L' elliciency when upplied to the whole turbine. and ia given by ‘ Heat oover_ted_into useful worl-: - _"l'-oFt. nl¥adillbatic heat 4. Overall or turbine efficiency lnonmul. This efficiency covers intemnl and external losses : for example. bearings and atcum friction. leakage, rudiution etc. Work delivered at the turbine coupling in heat units per kg of steam 'l| v‘r*‘. ll 7 _ . "7 ‘ Total adiabatic heat drop 5. Net efficiency or efficiency ratio tum). It is the ratio Brake thermal ¢: lEek_-xaE}- Thermal efficiency on the Rankine cycle Also the actual thcrtnul eflicicncy H‘1“_°_‘? £‘_‘f!3'. ‘.. °1‘_L£“. ‘.? ,!. ’:*_°_'£‘? j9.£’_‘_P£'_ fi9!'_=2%:9¥33_ _ = 'l"_ou-11 heat in steam ut . -stop valve Water heat in cxltuusat - Again. Rankine efficiency "mt: -m. =l = Adiubalic ho. -nt dmp Total heat in steam at stop valve - Wntcr hunt in o. thuu. ~'t A Hr. -at converted into u. ~scl'ul work n, ,,. —— --~~-—-—---~v -———-—-- Total udinbutic heat drop Hence m. .. = n. ,.. ,,. :. It as (hr owmll or net cfficivncy that is meant when the nflicwnqv of a turbine is . spolen of without qualification. TYPES OF POWER IN STEAM TURBINE PRACTICE ln steam turbine performance the following types of power are generally used ' l. Adlabatlc power (A. P.). It is the power based on the total internal steam flow and adiabatic hr; -at drop. 2. Shai. power (S. P.). It is the actual power transmitted by the turbine. 3. Kim power (R. P.). It is the power developed at the rim. It is: also called blade power. Power losses are usually expressed as follows : (i) (P. ,, = Power lost in overcoming disc friction. (ii) (Pym = Power lost in blade uindage lo-. ~e-s.
  59. 59. 360 THERMAL BKIINEERINO Let us consider the case of an impulse turbine. Let "'1. be the total internal steam new in kgls. Refer Fig. 19.48. The line (1-2) represents the adiabatic or isentropic expansion of steam in the nozzle from pressure p top, But the actual path oftho stage point during expansion in nozzles is shown by (1-3) which tulcca into account the effect of ‘noules losses’ Then, LP. - nu, 0:, -11,) kw . ..u9.3s) Alter expansion In the nozzle the steam enters the blades where the R. P. is developed. Due to blade friction the steam is somewhat reheated and this reheating is shown by (34) along the constant pressure p, line just for convenience. But in actual practice though the pressure at outlet of the blade is equnl to thnt at the inlet. the pressure in the blade channels is not constant. However, with this simplification ; l'I (Enlhalpy) P2 <——-——-——-—? + 8 (Enitopy) Fig. 19.-13 RP. = tit, 01,- It. ) kW . ..(19.34) 4-5 shows the further reheating due to friction and blade windaze and these losses are given mm! = na, (as-1:. ) kW . ..(19.35) New points 1 and 5 are the initial and fine] etngo points respectively for l! single stage Impulse turbine. It. therefore. follows that sr. = m, u:, — 1:, » kW. .. .u9.s6) REACTION TURBINES Example 19.29. The fiallowing data refer to a particular stage of a Parson‘. -1 reaction tur- bine : Speed of the turbine = 1500 r. p.m. Mean diametcruftlu: rotor = lnutrc Stage efficiency : 80 per cent
  60. 60. STEAM TURBINES “I Blade outlet angle = 20' Speed ratio = 0.7 Determine the available ismtropic enthalpy drop in the stage. Solution. Mean diameter of the rotor, D = 1 In Turbine speed. N = 1600 r. p.m. Biado outlet angle, 9 = 20' Speed ratio, p = 95- = 0.7 0: Stage cmciency, n_, _‘, = Isentr-epic enthalpy drop : SEDN : : x 1 x 1500 Blade speed, C” = -6—o- = T = 78.54 in/ s But p = (7:31 = 0.7 (given) 1 ‘ g ‘7 . _. 7&4 . = 112,2 3.}; 0.7 0.7 In Parson's turbine a = Q. With the above data known. the velocity diagram for the turbine can be drawn to a suitable scale as ahown in Fig. 19.44. .9 on ---------<--------- n-o¢oo+Q-o--¢- .5’ Fig. 19.4-8 By measurement (from the diagram) C. ., = 106.5 In/ s; c, , : 27 nxla C, ,;(C, , +C, , ) _ um‘, = ? -‘——'-ha . where in‘, = uscntropic enthalpy drop. _ 78.64(108.25 + 27) ' . . o "' 3 h, ,x1ooo
  61. 61. 888 ‘I1-«II-IRMAL ENGINEERING 78.5I(103.25 4 27) hi, -71~§: :-1-6-MT-— —' 13.08 kn’ Hence. isentropic enthalpy drop - 13.08 kJ/ kg. (A. ns. ) Example 19.30. In a reaction turbine. the blade tips are inclined at 35“ and 20" in the direction of motion. The guide blades are of the same shape as the moving blades. but reversed in direction. A! a certain place in the turbine, the drum diameter is I metre and the blades are )0 cm high. At this place. the . -'!4"arn ham :3 pr-e. ~:~: un- of 1.75 but and dryness 0.935. If the speed of this turbine is 250 r. p.m. and the steam fin. -L'('5 through the bladex without shock. find the mass of stc. -am Row and power developed in the ring of moving blades. Fm. V3.45 Solution. Refer I-‘lg. 19.45. Angle5,u= ¢=20". and9=B=35 Mum drum diameter. D" = 1 + 0.1 = 1.1 In Area of flow V x. Duh, where h is Lhc height of blade : : x 1.1 x 0.1 0.3156 In’ Steam pres: -‘um = 1.75 bar Dryness fraction of stream. 1 : 0.935 Speed of the turbine. N : 250 r. p.m. Rate ofstexun flow. In, : Blade speed. C“ = -——: €0N = -M 1'61: 250 = 14.4 m/ s With the above given data the velocity diagram can be drawn to :1 suitable mule ma shown in Fig. 19.45. By measurement (from diagram) : C. .., = so m/5 ; C. _., = 15.45 mm ; C, , 2 C‘, = 10.8 m/ —.-« From steam tables corresponding to 1.75 bar pmaeum. L‘; ~. Specific volume of drysatumt»: -d steam r 1.004 m“/ kg x = 0.935 (given) Specific volume of wet steam : xv‘, : 0.935 x 1.004 : 0.938 In“/ kg
  62. 62. Mean flow rate is prim-n by : m. = Arm of flow - Vt-lucity of How Q3456 110.8 398 Hpueific vulumcenf utomrn 0.938 . ' kw, ’ Power dovelnpod. I‘ : rh_,1(. ‘"‘ + (‘ | C,, , : V u', V,v7 * W 1000 k 3.93610 4 i5.45I >r I-| .v“| = ‘mm Z = 2.b kw. (Anm) Example 19.31. In a rearlinn turhimn the fixed blmles and moving bltules are of the same shape‘ 3'-ut n'n‘rs¢'d in direvlinn. The nn; ,vla-s of the reeeiring lips are 35" and 0/ 50‘? ‘1i3Ch“’Ei"I| ? tips 90“. Find the pnu-vr (i¢‘u'Iup1'd per pair ofbladt. -s for u stmni c. -nnsulrtptiun of 2.5 kg/3, when the blade spun‘! is :30 In If I/ w heat «Imp per pair is 10.04 la-lb"/ e;. _:_. find the :3 tr, -iosncy of the pair. Solution. Angles of roroiving tips, 0 = |$ : 35" Anglos of di. <eharg: in; : tips. at = 0 = 20’ Steam ennsumptimi, Iii‘, : 2.? ) kg . ~' Blade speed. ('_, = 50 m. .~: Heat drop per pair, I: V, = 10.04 kvl kg Power developed per pair of blades : Refer Fig. ISL-16. NS - l’(. t = I52 n1"a Work done pm‘ pm! per M: an! nlrauu -' “hr, ."u, h ‘. m_N ‘u ' ‘I _. N M '_'. ."I ~ 7600 ‘ : ' ; ' - I nun / [um mm. mm: Eficicncy of the pair: - 19 kW. QAHI. ) . Work dune par pair pet kg of steam Effiueney = — - -’ ‘ ' hd* 75°” . 0.757 . 75.-ms. (An) A L“ ‘ moa x1000
  63. 63. '3' Ixulple 18.88. A «huge of a IurIu‘m- with I’arnmn'x blading tlvlivern dry MW steam at 2.7 bar from lhv fixml hlmlru at 90 m/ u. 'I'Iu- "It'll" hlmlr Iwiuh! is 40 mm. and (I3. moving blade‘ en‘! angvlr is 20". Thv «mu! m-lm-cry n/ ‘strum in . '!/ -I n/ ' Ilw hlmlr v-rlority at the mam radius. Stmm is supplml to (M . -zlngr u! Ihv rnlv n/ ' . ‘N)00 lag/ In. 'l'Iu- a-/ Ii-r-I uf lhr hlmlv tip lhickneu on the annulus own run he Inugla-vlml. (. 'nlruIahr : u Thr win-1'1 s}wu*ui in r. p.m. ; iii! Tlw ¢! a'u; ,-rum [mu'o'r ; um The‘ oiiagrmn r ? civIu'}' . ' Him The 4-nlhn/ py «Imp of Mr . -do-nm in this uog¢_ Solution. The velocity diagram is shown in Fig. 19.47 ml and thv bludo whnc-I annulug 1, mproscntvd in Fig. 19.47 (In). Pressure = 2.7 bar. I = 1. (‘l = 90 unis. I: = 40 mm = 0.04 m a = o = 20?. C, -, = Q L :54 (', __, Rate of steam supply = 9000 kgm. u‘) Wheel speed, N : C‘, = 3.4 (75; = (71 sin 20 : 9!) . -sin '30" : .'¥().78 m. -"S (‘M = 30.78 x 4'3 = -v1I. l|-I nys h Avaa Am: (M Ma. 19.47

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