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CHAPTER Three-Dimensional Flows in Axial Turbomachines 6 It cost much labour and many days before all these things were brought to perfection. Defoe, Robinson Crusoe 6.1 INTRODUCTION In Chapters 4 and 5 the fluid motion through the blade rows of axial turbomachines was assumed to be two-dimensional in the sense that radial (i.e., spanwise) velocities did not exist. This assumption is not unreasonable for axial turbomachines of high hub–tip ratio. However, with hub–tip ratios less than about 4/5, radial velocities through a blade row may become appreciable, the consequent redistribution of mass flow (with respect to radius) seriously affecting the outlet velocity profile (and flow angle distribution). The temporary imbalance between the strong centrifugal forces exerted on the fluid and radial pressures restoring equilibrium is responsible for these radial flows. Thus, to an observer travelling with a fluid particle, radial motion will continue until sufficient fluid is transported (radially) to change the pressure distribution to that necessary for equilibrium. The flow in an annular passage in which there is no radial component of velocity, whose streamlines lie in circular, cylindrical surfaces and which is axisymmetric, is commonly known as radial equilibrium flow. An analysis called the radial equilibrium method, widely used for three-dimensional design calculations in axial compressors and turbines, is based upon the assumption that any radial flow that may occur is completed within a blade row, the flow outside the row then being in radial equilibrium. Figure 6.1 illustrates the nature of this assumption. The other assumption, that the flow is axisymmetric, implies that the effect of the discrete blades is not transmitted to the flow. 6.2 THEORY OF RADIAL EQUILIBRIUM Consider a small element of fluid of mass dm, shown in Figure 6.2, of unit depth and subtending an angle dθ at the axis, rotating about the axis with tangential velocity, cθ at radius r. The element is in radial equilibrium so that the pressure forces balance the centrifugal forces:   1 ð p þ dpÞðr þ dr Þdθ  prdθ  p þ dp drdθ ¼ dmc2θ =r. 2 Writing dm ¼ ρrdθdr and ignoring terms of the second order of smallness this equation reduces to 1 dp c2θ ¼ . r ρ dr © 2010 S. L. Dixon and C. A. Hall. Published by Elsevier Inc. All rights reserved. DOI: 10.1016/B978-1-85617-793-1.00006-7 ð6:1Þ 183 184 CHAPTER 6 Three-Dimensional Flows in Axial Turbomachines Casing Hub Streamlines Axis FIGURE 6.1 Radial Equilibrium Flow Through a Rotor Blade Row Mass/unit depth 5 rddr p1 dp dr p Velocity 5 c  p1 1 2 dp 1 dp 12 p r d FIGURE 6.2 Fluid Element in Radial Equilibrium (cr ¼ 0) If the swirl velocity cθ and density are known functions of the radius, the radial pressure variation along the blade length can be determined: Z tip dr ð6:2aÞ ρc2θ . ptip  proot ¼ r root For an incompressible fluid, Z ptip  proot ¼ ρ tip root c2θ dr . r ð6:2bÞ The stagnation enthalpy is written (with cr ¼ 0) h0 ¼ h þ  1 2 c þ c2θ ; 2 x ð6:3Þ 6.3 The Indirect Problem 185 therefore, dh0 dh dcx dcθ ¼ þ cθ . þ cx dr dr dr dr ð6:4Þ The thermodynamic relation Tds ¼ dh  (1/ρ)dp can be similarly written T ds dh 1 dp ¼  . dr dr ρ dr ð6:5Þ Combining eqns. (6.1), (6.4), and (6.5), eliminating dp/dr and dh/dr, the radial equilibrium equation may be obtained: dh0 ds dcx cθ d  T ¼ cx þ ðrcθ Þ. dr dr r dr dr ð6:6aÞ If the stagnation enthalpy h0 and entropy s remain the same at all radii, dh0/dr ¼ ds/dr ¼ 0, eqn. (6.6a) becomes dcx cθ d cx þ ðrcθ Þ ¼ 0. ð6:6bÞ dr r dr Equation (6.6b) will hold for the flow between the rows of an adiabatic, reversible (ideal) turbomachine in which rotor rows either deliver or receive equal  work at all radii. Now if the flow is incompressible, instead of eqn. (6.3) use p0 ¼ p þ 12 p c2x þ c2θ to obtain 1 dp0 1 dp dcx dcθ ¼ þ cθ . þ cx dr dr ρ dr ρ dr ð6:7Þ 1 dp0 dcx cθ d ¼ cx þ ðrcθ Þ. ρ dr dr r dr ð6:8Þ Combining eqns. (6.1) and (6.7), Equation (6.8) clearly reduces to eqn. (6.6b) in a turbomachine in which equal work is delivered at all radii and the total pressure losses across a row are uniform with radius. Equation (6.6b) may be applied to two sorts of problem: the design (or indirect) problem, in which the tangential velocity distribution is specified and the axial velocity variation is found, or the direct problem, in which the swirl angle distribution is specified, the axial and tangential velocities being determined. 6.3 THE INDIRECT PROBLEM Free-Vortex Flow This is a flow where the product of radius and tangential velocity remains constant (i.e., rcθ ¼ K, a constant). The term vortex free might be more appropriate as the vorticity (to be precise we mean axial vorticity component) is then zero. Consider an element of an ideal inviscid fluid rotating about some fixed axis, as indicated in Figure 6.3. The circulation Γ is defined as the line integral of velocity around a curve enclosing an area A, or Γ ¼ ∮ cds. The vorticity at a point is defined as the limiting value of circulation δΓ divided by area δA, as δA becomes vanishingly small. Thus, vorticity, ω ¼ dΓ/dA. 186 CHAPTER 6 Three-Dimensional Flows in Axial Turbomachines c 1 dc  c r 1dr r d FIGURE 6.3 Circulation About an Element of Fluid For the element shown in Figure 6.3, cr ¼ 0 and   dcθ cθ þ rdθdr, dΓ ¼ ðcθ þ dcθ Þðr þ dr Þdθ  cθ rdθ ¼ dr r ignoring the product of small terms. Thus, ω ¼ dΓ/dA ¼ (1/r)d(rcθ)/dr. If the vorticity is zero, d(rcθ)/dr is also zero and, therefore, rcθ is constant with radius. Putting rcθ ¼ constant in eqn. (6.6b), then dcx/dr ¼ 0 and so cx ¼ a constant. This information can be applied to the incompressible flow through a free-vortex compressor or turbine stage, enabling the radial variation in flow angles, reaction and work to be found. Compressor Stage Consider the case of a compressor stage in which rcθ1 ¼ K1 before the rotor and rcθ2 ¼ K2 after the rotor, where K1 and K2 are constants. The work done by the rotor on unit mass of fluid is ΔW ¼ Uðcθ2  cθ1 Þ ¼ ΩrðK2 =r  Kl =rÞ ¼ constant. Thus, the work done is equal at all radii. The relative flow angles (see Figure 5.2) entering and leaving the rotor are tan β1 ¼ U Ωr  K1 =r  tan α1 ¼ , cx cx tan β2 ¼ U Ωr  K2 =r  tan α2 ¼ . cx cx in which cx1 ¼ cx2 ¼ cx for incompressible flow. 6.3 The Indirect Problem 187 In Chapter 5, reaction in an axial compressor is defined by static enthalpy rise in the rotor . R¼ static enthalpy rise in the stage For a normal stage (α1 ¼ α3) with cx constant across the stage, the reaction was shown to be cx ðtan β1 þ tan β2 Þ. ð5:21Þ R¼ 2U Substituting values of tan β1 and tan β2 into eqn. (5.21), the reaction becomes k R ¼ 1 2, r ð6:9Þ where k ¼ ðK1 þ K2 Þ=ð2ΩÞ. It will be clear that, as k is positive, the reaction increases from root to tip. Likewise, from eqn. (6.1) we observe that as c2θ /r is always positive (excepting cθ ¼ 0), so static pressure increases from root to tip. For the free-vortex flow rcθ ¼ K, the static pressure variation can be shown to be p/ρ ¼ constant  K2/(2r2) upon integrating eqn. (6.1). Example 6.1 An axial flow compressor stage is designed to give free-vortex tangential velocity distributions for all radii before and after the rotor blade row. The tip diameter is constant and 1.0 m; the hub diameter is 0.9 m and constant for the stage. At the rotor tip the flow angles are as follows: Absolute inlet angle, α1 ¼ 30°; Relative inlet angle, β1 ¼ 60°; Absolute outlet angle, α21 ¼ 60°; Relative outlet angle, β2 ¼ 30°. Determine (i) the axial velocity; (ii) the mass flow rate; (iii) the power absorbed by the stage; (iv) the flow angles at the hub; (v) the reaction ratio of the stage at the hub; given that the rotational speed of the rotor is 6000 rev/min and the gas density is 1.5 kg/m3, which can be assumed constant for the stage. It can be further assumed that stagnation enthalpy and entropy are constant before and after the rotor row for the purpose of simplifying the calculations. Solution (i) The rotational speed, Ω ¼ 2πN/60 ¼ 628.4 rad/s. Therefore, blade tip speed, Ut ¼ Ωrt ¼ 314.2 m/s, and blade speed at hub, Uh ¼ Ωrh ¼ 282.5 m/s. From the velocity diagram for the stage (e.g., Figure 5.2), the blade tip speed is pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi Ut ¼ cx ðtan 60° þ tan 60°Þ ¼ cx 3 þ 1= 3 . 188 CHAPTER 6 Three-Dimensional Flows in Axial Turbomachines Therefore, cx ¼ 136 m/s, constant at all radii by eqn. (6.6b). _ ¼ πðrt2  rh2 Þρcx ¼ π(0.52  0.452)1.5  136 ¼ 30.4 kg/s. (ii) The rate of mass flow, m (iii) The power absorbed by the stage, _ c ¼ mU _ t ðcθ2t  cθ1t Þ W _ t cx ðtan α2t  tan α1t Þ ¼ mU pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ¼ 30.4  314.2  136ð 3  1= 3Þ ¼ 1.5 MW. (iv) At inlet to the rotor tip, pffiffiffi cθ1t ¼ cx tan α1 ¼ 136= 3 ¼ 78.6 m=s. The absolute flow is a free vortex, rcθ ¼ constant. Therefore, cθ1h ¼ cθ1t(rt/rh) ¼ 78.6  0.5/0.45 ¼ 87.3 m/s. At outlet to the rotor tip, pffiffiffi cθ2t ¼ cx tan α2 ¼ 136  3 ¼ 235.6 m=s. Therefore, cθ2h ¼ cθ2t(rt/rh) ¼ 235.6  0.5/0.45 ¼ 262 m/s. The flow angles at the hub are tan α1 ¼ cθ1h =cx ¼ 87:3=136 ¼ 0:642, tan β1 ¼ Uh =cx  tan α1 ¼ 1:436, tan α2 ¼ cθ2h =cx ¼ 262=136 ¼ 1:928, tan β2 ¼ Uh =cx  tan α2 ¼ 0:152. Thus, α1 ¼ 32.75°, β1 ¼ 55.15°, α2 ¼ 62.6°, β2 ¼ 8.64° at the hub. (v) The reaction at the hub can be found by several methods. With eqn. (6.9), R ¼ 1  k=r 2 , and noticing that, from symmetry of the velocity triangles, R ¼ 0:5 at r ¼ rt , then k ¼ 0:5rt2 . Therefore, Rh ¼ 1  0.5ð0.5=0.45Þ2 ¼ 0.382. The velocity triangles will be asymmetric and similar to those in Figure 5.5(b). The simplicity of the flow under free-vortex conditions is, superficially, very attractive to the designer and many compressors have been designed to conform to this flow. Constant (1945; 1953) may be consulted for an account of early British compressor design methods. Figure 6.4 illustrates the variation of fluid angles and Mach numbers of a typical compressor stage designed for free-vortex flow. Characteristic of this flow are the large fluid deflections near the inner wall and high Mach numbers near the outer wall, both effects being deleterious to efficient performance. A further serious disadvantage is the large amount of rotor twist from root to tip, which adds to the expense of blade manufacture. 6.3 The Indirect Problem Fluid angle, deg 60 40 189 2 1 1 20 0.9 0 0.8 or t Ro 0.7 Stat or Inlet mach number 2 0.6 0.6 0.7 0.8 0.9 Radius ratio, r/r t 0.5 1.0 FIGURE 6.4 Variation of Fluid Angles and Mach Numbers of a Free-Vortex Compressor Stage with Radius (Adapted from Howell, 1945) Many types of vortex design have been proposed to overcome some of the disadvantages set by free-vortex design and several of these are compared by Horlock (1958). Radial equilibrium solutions for the work and axial velocity distributions of some of these vortex flows in an axial compressor stage follow. Forced Vortex This is sometimes called solid-body rotation because cθ varies directly with r. At entry to the rotor assume h01 is constant and cθ1 ¼ K1r. With eqn. (6.6b),  2    d cx1 d 2 ¼ K1 r ¼ K1 dr dr 2 and, after integrating, c2x1 ¼ constant  2k12 r2. ð6:10Þ 190 CHAPTER 6 Three-Dimensional Flows in Axial Turbomachines After the rotor cθ2 ¼ K2r and h02  h01 ¼ U(cθ2  cθ1) ¼ Ω(K2  K1)r2. Thus, as the work distribution is non-uniform, the radial equilibrium equation in the form eqn. (6.6a) is required for the flow after the rotor:     dh02 d c2x2 d K2 r 2 . ¼ 2ΩðK2  K1 Þr ¼ þ K2 dr 2 dr dr After re-arranging and integrating, c2x2 ¼ constant  2½K22  ΩðK2  K1 Þr 2 . ð6:11Þ The constants of integration in eqns. (6.10) and (6.11) can be found from the continuity of mass flow, i.e., Z rt Z rt m_ ¼ cx1 rdr ¼ cx2 rdr, ð6:12Þ 2πρ rh rh which applies to the assumed incompressible flow. General Whirl Distribution The tangential velocity distribution is given by cθ1 ¼ ar n  b=r ðbefore rotorÞ, ð6:13aÞ cθ2 ¼ ar n þ b=r ðafter rotorÞ. ð6:13bÞ The distribution of work for all values of the index n is constant with radius so that, if h01 is uniform, h02 is also uniform with radius. From eqns. (6.13a and b), ΔW ¼ h02  h01 ¼ Uðcθ2  cθ1 Þ ¼ 2bΩ. ð6:14Þ Selecting different values of n gives several of the tangential velocity distributions commonly used in compressor design. With n ¼ 0, or zero power blading, it leads to the so-called exponential type of stage design (included as an exercise at the end of this chapter). With n ¼ 1, or first power blading, the stage design is called (incorrectly, as it transpires later) constant reaction. First Power Stage Design For a given stage temperature rise the discussion in Chapter 5 would suggest the choice of 50% reaction at all radii for the highest stage efficiency. With swirl velocity distributions, cθ1 ¼ ar  b=r, cθ2 ¼ ar þ b=r ð6:15Þ before and after the rotor, respectively; and rewriting the expression for reaction, eqn. (5.21), as cx R ¼ 1 ðtan α1 þ tan α2 Þ, ð6:16Þ 2U then, using eqn. (6.15), R ¼ 1  a=Ω ¼ constant. ð6:17Þ 6.3 The Indirect Problem 191 Implicit in eqn. (6.16) is the assumption that the axial velocity across the rotor remains constant which, of course, is tantamount to ignoring radial equilibrium. The axial velocity must change in crossing the rotor row so that eqn. (6.17) is only a crude approximation at the best. Just how crude is this approximation will be indicated later. Assuming constant stagnation enthalpy at entry to the stage, integrating eqn. (6.6b), the axial velocity distributions before and after the rotor are   1 2 2 ð6:18aÞ cx1 ¼ constant  4a ar  b ln r , 2 c2x2   1 2 ¼ constant  4a ar þ b ln r . 2 More conveniently, these expressions can be written non-dimensionally as  2 "  2  2  # cx1 2a 1 r b r , ¼ A1   2 ln Ω 2 rt art rt Ut    #  2 "  2 cx2 2 2a 1 r b r , ¼ A2  þ 2 ln Ω 2 rt art rt Ut ð6:18bÞ ð6:19aÞ ð6:19bÞ in which Ut ¼ Ωrt is the tip blade speed. The constants A1, A2 are not entirely arbitrary as the continuity equation, eqn. (6.12), must be satisfied. Example 6.2 As an illustration consider a single stage of an axial-flow air compressor of hub–tip ratio 0.4 with a nominally constant reaction (i.e., according to eqn. 6.17) of 50%. Assuming incompressible, inviscid flow, a blade tip speed of 300 m/s, a blade tip diameter of 0.6 m, and a stagnation temperature rise of 16.1°C, determine the radial equilibrium values of axial velocity before and after the rotor. The axial velocity far upstream of the rotor at the casing is 120 m/s. Take Cp for air as 1.005 kJ/(kg°C). Solution The constants in eqn. (6.19a) can be easily determined. From eqn. (6.17), 2a=Ω ¼ 2ð1  RÞ ¼ 1:0. Combining eqns. (6.14) and (6.17), b ΔW Cp  ΔT0 1005  16:1 ¼ ¼ ¼ 0:18. ¼ art2 2Ω2 ð1  RÞrt2 2Ut2 ð1  RÞ 3002 The inlet axial velocity distribution is completely specified and the constant A1 solved. From eqn. (6.19a),    2 cx1 1 ¼ A1  ðr=rt Þ2  0:18 ln ðr=rtÞ . Ut 2 At r ¼ rt, cx1/ Ut ¼ 0.4 and, hence, A1 ¼ 0.66. 192 CHAPTER 6 Three-Dimensional Flows in Axial Turbomachines Although an explicit solution for A2 can be worked out from eqn. (6.19b) and eqn. (6.12), it is far speedier to use a semigraphical procedure. For an arbitrarily selected value of A2, the distribution of cx2/Ut is known. Values of (r/rt)  (cx2/Ut) and (r/rt)  (cx1/Ut) are plotted against r/rt and the areas under these curves compared. New values of A2 are then chosen until eqn. (6.12) is satisfied. This procedure is quite rapid and normally requires only two or three attempts to give a satisfactory solution. Figure 6.5 shows the final solution of cx2/Ut obtained after three attempts. The solution is "      # cx2 1 r 2 r . þ 0:18 ln ¼ 0:56  Ut 2 rt rt It is illuminating to calculate the actual variation in reaction taking account of the change in axial velocity. From eqn. (5.20) the true reaction across a normal stage is 0 R¼ w21  w22 . 2Uðcθ2  cθ1 Þ From the velocity triangles, Figure 5.2, w21  w22 ¼ ðwθ1 þ wθ2 Þðwθ1  wθ2 Þ þ ðc2x1  c2x2 Þ. As wθ1 þ wθ2 ¼ 2U  (cθ1 þ cθ2) and wθ1  wθ2 ¼ cθ2  cθ1, 0 R ¼ 1 cθ1 þ cθ2 c2x1  c2x2 þ . 2U 2Uðcθ2  cθ1 Þ c x2 Ut 0.8 Final curve cx 1 Ut 0.6 1.0 0.8 cx 0.6 Nominal reaction 0.4 0.2 n ctio e Tru rea 0.2 0 0 0.4 0.5 0.6 0.7 Radius ratio, r/r t FIGURE 6.5 Solution of Exit Axial-Velocity Profile for a First Power Stage 0.8 0.9 20.2 1.0 Reaction U t 0.4