Unit V - TURBINES > Principle of operation of axial flow turbines, > Work done and pressure rise, > Degree of Reaction, > Types of design of turbines, > Turbine blade cooling, > Velocity diagrams, > Limitations of radial flow turbines, compressor; > Turbine matching — materials for turbine blades. MIT-ADT University, Pune 2020Impulse and Reaction Turbines * Work can be extracted from a gas at a higher inlet pressure to the lower back pressure by allowing it to flow through a turbine. In a turbine as the gas passes through, it expands. The work done by the gas is equivalent to the change of its enthalpy, It is a well knoivn fact that the turbines operate on the momentum principle. Part of the energy of the gas during expansion is converted into kinetic energy in the flow nozzles. The gas leaves these stationary nozzles at a relatively higher velocity. Then it is made to impinge of the blades over the turbine rotor or wheel. Momentum imparted to the blades turns the wheel. Thus, the two primary parts of the turbine are the stator nozzles, and the turbine rotor blades. Most turbines possess more than one stage with their respective wheels mounted on a common shaft. The stage consists of a ring of fixed nozzle blades followed by the rotor blade ring, However, a nozzleless stage with only the rotor is also possible as in the case of an inward flow radial turbine, In certain requirements, the flow in a turbine stage may be partly radial and partly axial which is called a mixed stage. It combines the advantage of both axial and radial types. Normally, a turbine stage is classified as . “+ (i) an impulse stage, and (ii) a reaction stage.ion Turbines Impulse and Rea + An impulse stage is characterized by the expansion of the gas which occurs only in the stator nozzles. The rotor blades act as directional vanes to deflect the direction of the flow. % Further, they convert the kinetic energy of the gas into work by changing the monientum of the gas more or less at constant-pressure. “+ A reaction stage is one in which expansion of the gas takes place both in the stator and in the rotor, The function of the stator is the same as that in the impulse stage, but the function in the rotor is two fold. (i) the rotor converts the kinetic energy of the gas into work, and (ii) contributes a reaction force on the rotor blades. The reaction force is due to the increase in the velocity of the gas relative to the blades. This results from the expansion of the gas during its passage through the rotor. Inlet . Jade row i ete KE CC wo} bl Axial clearance “outA Single Impulse Stage - Impulse machines are those in which there is no change of static or pressure head of the Muid in the rotor. The rotor blades cause only energy transfer and there is no energy transformation. * The energy transformation from pressure or static head to kinetic energy or vice versa takes place in fixed blades only, To give an example, the transfer of kinetic energy to the rotor in an impulse turbine from a high velocity fluid occurs only due to the impulsive action of the fluid on the rotor. Figure shows an impulse turbine stage. As can be scen from the figure that in the rotor blade passages of an impulse turbine there is no acceleration of the fluid, i.c., there is ho energy transformation. Hence, the chances are greater for separation due to boundary layer growth on the blades surface. Due to this, the rotor blade passages of the impulse machine suffer greater losses giving lower stage efficiencies. The paddle wheel, Pelton wheel and Curtis steam turbine are some of the examples of impuls?|A Single Impulse Stage Nozzle Rotor 1 Energy transfer Energy transformation ~ - Velocity PressureA Single Reaction Stage those, in which, changes in static or pressure h the energy transformation occurs both in fixed formation. “> The reaction machines are head occur both in the rotor and stator blade passages. Here, as well as moving blades. ‘The rotor experiences both energy’ transfer as well as energy trans considered to be more efficient. This is mainly due to continuous sh Therefore, reaction turbines are Je stage reaction turbine along with pressure acceleration of flow with lower losses. Figure shows @ sing! and velocity changes when the fluid passes through a turbine stage. 4} The degree of reaction of a turbomachines stage may be defined as the ratio of the static or pressure head change occurring in the rotor to the total change across the stage. + A 50% or half degree reaction machine has some special characteristics. & Axial-flow turbines with fifty per cent reaction have symmetrical blades in their rotors and stators. It may be noted that the velocity triangles at the entry and exit of a 50% reaction stage are also symmetrical. ‘} Hero's turbine, the lawn sprinkler and Parson's steam turbine are some of the examples of reactio?. machines.Nozzle Rotor | Energy transfer Energy transformation Fig. 11.3 A reaction turbine stage j— Fived bile Rotor - ¢ ? “Nelociy * Pressure: Fig. Hal Vieiation of pressure ane velocity thromgh a two-stage reaction tnsbiveVelocity Triangle for Turbine Staion © Nozzle blades Staton @ Station @ Rotor blades Station @ Fig. Ha Velocity triaughes for a turbine stageThe following tigonometrical relations ean be deduced from the velocity Ariangles shown in F Lh = rpeosag = meos/e (11.2) ~? ea = esinaz = wet = wysin gy ba (11.3) — a 7 sin (az — 13) sin (90 au __ sin (ag ~ () hid o cos dy Gas = wycosihy = egeosng * — (1L5) cis cysinay = ta—u = wysingy—w (116)2 $e = eysinag+ cysinay = (wysin gy + u) + (wy sin = we sin det is sin Bs It is often assmmed that the axial velocity component remains constant through the stage, For sich a condition § fe = ta = tae = ta Co = creosay = eacosag = eycosny Ca = wreosfy = wrcosfs (11.8) Equation 11.7 for constant axial velocity yietds a useful relation, ( fanny +tanas = tanjy + tan (119) ( |11.5 EXPRESSION FOR WORK OUTPUT " Though force and torque are exerted on both stationary and moving, blades: alike, work can only be done on the moving rotor blad Dlades transfer encrgy from the fluid to the shaft. ‘The axial turbine (ey = uy = 2) can be written as ‘Thus the rotor tage work in an Wo = uper = unces 7 = ule — (—e3)} (11.10) = ulers + ers) This equation can nlso be expressed in another useful form! 2) cin)" oe . (14.11) w The first ter 42) in the bracket depends on the nozzle or fixed blade angle (a2) and the ratio “>. The contribution of the second term ( Lo the work is generally sinall’ [tf is also observed that the kinerie energy of the fluid leaving the stage is greater for larger valucs of e73. The leaving loss from the stage is minimum when es, — 0, ie. when the discharge from the stage is axial (ey; = Cas). However, this condition gives lesser an be scen from Eqs. 1110 and 1.1L. age worl: asIf the swirl at the exit of the stage is zeros, i.e., cr3 = 0, Eq. LL-LL yields Wo= (S2)ut = (282) "bae (1.32) J lasso Thus, when the exit swirl is zero, the work output is a funetion of speed ratio, fixed blade angle and peripheral speed11.6 BLADE LOADING AND FLOW COEFFICIENTS a Performance of turbomachines are charac parameters. For example, the loading coc (@) have been defined as ized by various dimensi ont (¢) and the flow ¢oeflici Vv aA (11.13) (11.14) u Since the work, (WW), in Eq. 11.13 is frequently referred to as the blade or stage work, the coefficient (#4) would also be known as the blade or stage loading, coefficient. From Eqs. 11.10, 11-13 and 11.14, for constant axial velocity (ca), it can be shown that wy = (tanaz+tanas) = (tan G2 + tan Js) (11.15) The $— w plots are us of diffe ful in comparing the performances of various stages ont sizes and geometries11.7. BLADE AND STAGE EFFICIENCIES Even though the blade and stage work (outputs) are the same, the blade and stage efficiencies need not be equal. This is because the energy inputs to the rotor blades and the stage (fixed blade ring plus the rotor) are different. The blade efficiency is also known as the utilization factor (¢) which is an index of the energy utilizing capability of the rotor blades. ‘Thus, ce = ™ Rotor blade work V = = — 11.16 Energy supplied to the rotor blades = (1.16) The blade work can be written as the sum of changes in the various kinetic cnergies: Wo = wee + 1 Lee ee : zs 3 +5 U3 — uh) + 5 (wh — we) (11.17)in y change within the rotor blades. The energy supplicd to the rotor blades is the absolute kinetic energ: the jet at the entry plus the kinet: Ey = 53+ § (wh — 08) + $ (08 ~ 08) (1.18) Substituting Eqs. 11.17 and 11.18 into Eq. 11.16, we get = 13) + (w} = w3) = 11.19 c ™ 2 — wh) +(e — wa) (21.19) For axial machines, 1 = tg = ta, =m = (11.20) B+ (we = w' To avoid any confusion between the stage and blade efficiencies, the term utilization factor (¢) will be used in place of blade efficiency in this book.The energy supplicd to the rotor blades is the absolute kinctic energy in the jet at the entry plus the kinetic energy change within the rotor blades. + 5 (wi — 08) + 5 (u8 - 08) (11.18) 1 Ey = We Substituting Eqs. 11.17 and 11.18 into Eq. 11.16, we get coe m = GEA + d= 18) + (WR) yy ° B+ (ws — w3) + (uz — U5) a For axial machines, u = u2 = us, . " « = m = (BS) (wi — ws) (11.20) B+ (wz — w: To avoid any confusion between the stage and blade efficiencies, the term utilization factor (¢) will be used in place of blade efficiency in this book.11.8 MAXIMUM UTILIZATION FACTOR FOR A SINGLE NWPULSE STAGE s, ther variations of the Figure 11.5 shows a single is no static stage impulse turbine. In thi pressure change in the rotor. ‘The pressure and velocity fluid passing through the stage is also shown in Fig. 11.5. Because of the pressure drop in the nozzle blade row, the absolute velocity of the fluid increases correspondingly and transformation of energy occurs in the nozzle. However, the tran: of energy occurs across the rotor blade row. Therefore the absolute Muid velocity decreases through the rotor blade row, shown in the figure. The velocity triangles for a single impulse stage are as shown in Fig. 11.6. For frictionless flow in the absence of any pressure drop through the rotor blades the relative velocities at their entry and exit are very much the same (ez © we). To obtain this condition the rotor blade angles must be equal. Therefore the utilization factor is given by ‘. as 2B Substituting from Eq. 11.7 and noting ws = ws and 2 = 34, we getFrom the velocity trinnale an oe © fey Cin. 11-6). . Ler Cons Sh ey ~» mm ACen ee D} c1nzay Rhows that phe wei finetion of ein Lae) correspon vadnic obo Velitewenie ate . _ Boo nines 22 = 0 1 on eon — gxines cieey urine = — an : crt.ey Banations 11.3 and 11.23 give 5 te sin G2 wy sin Bs ue (1.24) Substituting this in Eq. 11.6 yields cs = 0. This means that the exit from the stage should be axial, ie. ca = eas. This result can also be obtained by mere physical interpretation of the mechanics of flow. Combining Eq. 11 and 11.12,Fig. 11.6 Velocity triangles for a single impulse stage with negative swirl at exitWesa0 = Wop = 20 (11.25) Velocity triangles to satisfy these conditions combination of Eqs. 11.21 and 11.22 gives are shown in Fig: 11.7. ‘The Emax = sin? a2 (11.26) The rotor blade angles can also be determined for these conditions u tans = tani, = (11.27) Substituting from Eq. 11.22, ' . tanZ,; = tanfg = § tanas (11.28) Thus, for maximum utilization factor, the rotor blade angles are fixed by the nozzle air angle at exit.Multistage Impulse TurbineIL. Velocity triangles for a two-st turbine. © pressure-compounded impulseReaction Turbine Nozzle Rotor Energy transfer q Energy transformation ; ~ Velocity i {— Pressure Fig. 11.3 A reaction turbine stage Ry R,) — F= Fixed blade R= Rotor STL TIN: Velocity ; 7 Pressureho2 rai "08 vet Enthalpy Entropy rbine stage un for flow through a t + dings 3 Enthalpy-entror Pig. 11.‘The reversible adiabatic (isentropic) wing rows of blades Is represented ~. Stagnation points Oy and Oy expan: hy processes 1 dofined by the following ve hor = (aio) hove = (nan ‘The ideal work in the stage is give We = hor hhose = Cy(Tin — Tv) (aay ‘The actual expansi sible adiabatic) is represented by ¢ 1-2-3, and the actual states of the gas at the exits of the fixed woving blades are represented by: points 2 and 3 respectiv nee, nergy transfer (work) between const The loss duc to irreversibi loss coeflicient (As = ay = sur) is ate = 2Qr,— Ty) (cre)Ht commenpeondings to Spansion process (2 AuMformation nel trniinfer of Ano dita sta he stage sv In the moying binde rews represents both yensey Nowefore, tlie eliffeweree bythe Watlon Onthnlpion nt nutes 2nd H gives the netinl vale ey ae hon — Me = hor rotor the relative Hew apr SY Obnerver, Pleret Malps: bn tlie aacw ins tae ule syntenn) thee wenggintion, OF COORdIALON POH Ln cones Pose = : fav bed om : Crea) She enehatpy ane press by oho Pole wings coemete nevings Few Of binde: kre Stee 2S or: (11.49) (Ape. ee (14.50) ws CAT <1). Chere in Henle: Us based On enthalpy nel11.12.2. Degree of Reaction ree of reaction of a turbine stage can be defined ina mumber of ways; = it can be expressed in terms of pressures or velocities or enthalpies or the flow geometry in the stage. (a) A definition on the basis of the isentropic change given by the following relation: the stage is Isentropic change of enthalpy in the rotor Isentropic change of enthalpy in the stage RAh, hy - a = Ll fy Ah, Tay = hey Gees ‘wuming p= constant, Eq. 11.51 can be transformed in terms of pressures. ap ot Jin Sp pap 5 R ie 11.52) uap maa ao (b) Definitions in Eqs. 11.51 and 11.52 are not. wi ince the actual enthalpy changes md work are more important or turbine stages, a definition of the degree of reaction based on these quantities is more logical and useful. It may be noted that the change of enthalpy through the rotor and the total change in the stage are hyg—hg = (11.53) thor = hos = ho = hos (users = users) (11.54)the ratio of th of the degree of rene= er sen Equations 1 yield (157) 4. Therofore, from Fig. Hl, a 2uler + cay . eae (c) Be How through the stage ation 115 alo be expressed i terms of the ge For constant axinl velocity,The Turbine: @ MIT-ADT Ke > The high pressure and temperature gas that leaves the combustor is’ directed into ‘ turbine. The turbine may be thought of as a valve because on one side it has a high pressure gas and on the other side it has a low-pressure gas of the exhaust nozzle or the tailpipe. > Areal flow process in an un cooled turbine involves ir-reversibilities such as frictiona losses in the boundary layer, tip clearance flows, and shock losses in transonic turbine stages. : > The viscous dominated losses, that is, boundary layer separation, reattachment, and tit vortex flows, are concentrated near the end walls, thus a special attention in turbine flow optimization is made on the end-wall regions. = © MIT-ADT University, Pune 2020Practuat = MM Cg — Ns) = HAM sctwat Prideat = By (My — Msg) = MAN isentropic ‘Actual power ile Ideal power ‘© MIT-ADT University, Pune 2020Pracruat = M4 (Oy — Ns) = Aly geraat Prideat = (Mg — Massy = 2A, isentropic Ng lis _ Ah ctual Tag Mss Alii jcontropie Actual power Ideal power r= 1=Ts/Ta og ae wh. —T55/T lon? > We may also define a small-stage efficiency for a turbine, as we did in a comggssor, and . ‘ er ay dm . . call it the turbine polytropic efficiency et. «= j= sp Using the equation o: opic efficiency, we get oda Y I 1. or = n= T5/Ts = Ps/Pes ‘© MIT-ADT University, Pune 2020> The temperature ratio parameter zt across the turbine is established via a power balance between the turbine, compressor, and other shaft power extraction (e.g,, eleétric generator) on the gas generator. > Let us first consider the power balance between the turbine and compressor in its simplest form and then try to build on the added parameters. P= Be Ang + SDigg — bys) = ig lng — hg) which simplifies to the following nondimensional form (1+), = 2) = Ale — D> In a simple cycle analysis, it is customary to lump all power dissipation and power extraction terms into a single mechanical efficiency parameter mm that is multiplied by the turbine shaft power, to derive the compressor shaft power, that is, c= Nm where 17m is the mechanical efficiency parameter that needs to be specified a’priori, for example, tm = 0.95. ‘Therefore, an uncooled turbine is assumed to be adiabatic. Q=0 P= Malt AUN) mormm 73 (Prs! Pu em nS Tes! Tes . = @ O— i ‘© MIT-ADT University, Pune 2020Q1) Consider an un cooled gas turbine with its inlet condition the same as the exit condition of the combustor described in Previous Example. The turbine adiabatic efficiéncy is 88%. The turbine produces a shaft power to drive the compressor and other accessories at got = 45 MW. Assuming that the gas properties in the turbine are the same as the bumer exit in Previous Example, calculate (a) Turbine exit total temperature tt5 in K, (b) Turbine polytropic efficiency, et, (c) Turbine exit total pressure pt5 in kpa & (d) Turbine shaft power got based on turbine expansion ATt © MIT-ADT University, Pune 2020