#### Transcript Part 2

Steady, Level Forward Flight I . Introductory Remarks © Lakshmi Sankar 2002 1 The Problems are Many.. Transonic Flow on Advancing Blade Noise Thrust Aeroelastic Response Shock Waves Unsteady Aerodynamics 0 90 Main Rotor / Tail Rotor / Fuselage Flow Interference Tip Vortices Blade-Tip Vortex interactions V 180 270 Dynamic Stall on Retreating Blade © Lakshmi Sankar 2002 2 The Dynamic Pressure varies Radially and Azimuthally V 180 Vtip R Retreating Side Advancing Side Vtip R V 90 270 Vtip R V R Reverse Flow Region © Lakshmi Sankar 2002 0 Vtip R 3 Consequences of Forward Flight • The dynamic pressure, and hence the air loads have high harmonic content. Above some speed, vibrations can limit safe operations. • On the advancing side, high dynamic pressure will cause shock waves, and too high a lift (unbalanced). – To counter this, the blade may need to flap up (or pitch down) to reduce the angle of attack. • Low dynamic pressure on the retreating side. • The blade may need to flap down or pitch up to increase angle of attack on the retreating side. This can cause dynamic stall. • Total lift decreases as the forward speed increases as a consequence of these effects, setting a upper limit on forward speed. © Lakshmi Sankar 2002 4 Forward Flight Analysis thus requires • Performance Analysis – How much power is needed? • Blade Dynamics and Control – What is the flapping dynamics? How does the pilot input alters the blade behavior? Is the rotor and the vehicle trimmed? • Airload prediction over the entire rotor disk using blade element theory, which feeds into vibration analysis, aeroelastic studies, and acoustic analyses. • We will look at some of these elements. © Lakshmi Sankar 2002 5 Steady, Level Forward Flight II. Performance © Lakshmi Sankar 2002 6 Inflow Model • To start this effort, we will need a very simple inflow model. • A model proposed by Glauert is used. • This model is phenomenological, not mathematically well founded. • It gives reasonable estimates of inflow velocity at the rotor disk, and is a good starting point. • It also gives the correct results for an elliptically loaded wing. © Lakshmi Sankar 2002 7 Force Balance in Hover Thrust Drag Rotor Disk Drag Weight In hover, T= W That is all! No net drag, or side forces. The drag forces on the individual blades Cancel each other out, when © Lakshmi Sankar 2002 summed up. 8 Force Balance in Forward Flight Thrust, T Vehicle Drag, D Flight Direction Weight, W © Lakshmi Sankar 2002 9 Simplified Picture of Force Balance T aTPP Rotor Disk, referred to As Tip Path Plane (Defined later) c.g. Flight Direction D T sin a TPP D T cos a TPP W W © Lakshmi Sankar 2002 10 Recall the Momentum Model V V+v V+2v © Lakshmi Sankar 2002 11 Glauert’s Conceptual model Freestream, V∞ Freestream, V∞ Induced velocity, v Freestream, V∞ 2v © Lakshmi Sankar 2002 12 Total Velocity at the Rotor Disk V∞cosαTPP V∞sinαTPP Induced Velocity, v Total Velocity V cos a TPP V sin a TPP v 2 © Lakshmi Sankar 2002 2 13 Relationship between Thrust and Velocities In the case of hover and climb, recall T = 2 r A (V+v) v Induced Velocity Total velocity Glauert used the same analogy in forward flight. © Lakshmi Sankar 2002 14 In forward flight.. T 2 rAv V cos aTPP V sin aTPP v 2 2 This is a non-linear equation for induced velocity v, which must be iteratively solved for a given T, A, and tip path plane angle aTPP It is convenient to non-dimensionalize all quantities. © Lakshmi Sankar 2002 15 Non-Dimensional Forms T rAR 2 Edgewise Freestream Component V cos a TPP V Tip Speed R R is called advance ratio, CT Non - dimensinal inflow ratio, i v R Glauert equation in non - dimensiona l form becomes C T 2i tan a TPP i 2 © Lakshmi Sankar 2002 2 16 Approximate Form at High Speed Forward Flight If advance ratio is higher tha n 0.2, and if tip path plane angle is small, far exceeds inflow ratio i so that C T 2i 2 tan a TPP i 2 2 i CT i 2 In practice, advance ratio seldom exceeds 0.4, because of limitations associated with forward speed. © Lakshmi Sankar 2002 17 Variation of Non-Dimensional Inflow with Advance Ratio CT i 2 i CT 2i 2 tan a TPP i2 2 Notice that inflow velocity rapidly decreases with advance ratio. © Lakshmi Sankar 2002 18 Power Consumption in Forward Flight We can compute " Ideal Power" from Glauert' s theory as Thrust times normal velocity at the rotor disk. The actual power will include blade profile power, due to viscous drag on the blade. We will compute it later. P T V sin a TPP v Profile Power Recall Tsin a TPP D Thus, P Tv DV Profile Power Induced Power Parasite Power © Lakshmi Sankar 2002 19 Power Consumption in Level Flight P Tv DV Blade Profile Power The induced power Tv decreases with advance ratio , as discussed earlier. 1 1 Parasite Power DV rV2C D S V rV3C D S 2 2 The parasite power increases as the cube of the velocity (or advance ratio ) and dominates power consumptio n in high speed forward flight. Here C D is vehicle parasite drag coefficien t, and S is the reference area C D is based on. Because there is no agreement on a common reference area, it is customary to supply the product C D S . This product is called f , the equivalent flat plate area. 1 Parasite Power rV3 f 2 © Lakshmi Sankar 2002 20 Power Induced Power Consumption in Forward Flight Induced Power, Tv V∞ © Lakshmi Sankar 2002 21 Power Parasite Power Consumption in Forward Flight V∞ © Lakshmi Sankar 2002 22 Profile Power Consumption in Forward Flight We will later derive : Profile Power rAR Power 3 C d 0 8 1 3 2 where Cd 0 Average Drag Coefficien t of the Airfoil Blade Profile Power V∞ © Lakshmi Sankar 2002 23 Power Power Consumption in Forward Flight Blade Profile Power V∞ © Lakshmi Sankar 2002 24 Non-Dimensional Expressions for Contributions to Power P rAR 3 C P C P ,i C P , parasite C P ,0 Recall : C P Pi Tv Tv T v CT i 3 2 rAR rAR R 1 r V3 f 1 1 f 3 Pparasite rV3 f C P , parasite 2 3 2 rAR 2 A C P,i We will later show from blade element th eory that C P,0 C d , 0 8 1 3 2 1 f 3 C d , 0 Thus, C P CT i 1 3 2 2A 8 © Lakshmi Sankar 2002 25 Empirical Corrections • The performance theory above does not account for – Non-uniform inflow effects – Swirl losses – Tip Losses • It also uses an average drag coefficient. • To account for these, the power coefficient is empirically corrected. © Lakshmi Sankar 2002 26 Empirical Corrections Power Coefficent (Uncorrect ed) 1 f 3 Cd 0 2 CT i 1 3 2A 8 Profile Power Induced power Parasite Power Power Coefficent (Corrected ) 1 f 3 C d 0 2 CT i 1 4.6 2 A © Lakshmi Sankar 8 2002 1.15 27 Excess Power Determines Ability to Climb Rate of Climb= Excess Power/W Excess Power Power Available Power Blade Profile Power © Lakshmi Sankar 2002 V∞ 28 Forward Flight Blade Dynamics © Lakshmi Sankar 2002 29 Background • Helicopter blades are attached to the rotor shaft with a series of hinges: – Flapping hinges (or a soft flex-beam) , that allow blades to freely flap up or down. This ensures that lift is transferred to the shaft, but not the moments. – Lead-Lag Hinges. When the blades rotate, and flap, Coriolis forces are created in the plane of the rotor. In order to avoid unwanted stresses at the blade root, lead-lag hinges are used. – Pitch bearing/pitch-link/swash plate: Used to control the blade pitch. • The blade loads are affected by the motion of the blades about these hinges. • From an aerodynamic perspective, lead-lag motion can be neglected. Pitching and flapping motions must be included. © Lakshmi Sankar 2002 30 © Lakshmi Sankar 2002 31 Articulated Rotor © Lakshmi Sankar 2002 32 Flap Hinge http://www.unicopter.com/0941.html#delta3 http://huizen.dds.nl/ ~w-p/bookaut/scbprts.htm © Lakshmi Sankar 2002 33 Pitch Horn http://www.unicopter.com/0941.html#delta3 © Lakshmi Sankar 2002 34 Lead-Lag Motion © Lakshmi Sankar 2002 35 © Lakshmi Sankar 2002 36 © Lakshmi Sankar 2002 37 Coordinate Systems • Before we start defining the blade motion, and the blade angular positions, it is necessary to define what is the coordinate system to use. • Unfortunately, there are many possible coordinate systems. No unique choice. © Lakshmi Sankar 2002 38 Hub Plane z Y-axis is perpendicular to Fuselage symmetry plane And perpendicular to z =180 deg Y Z axis is normal to shaft =270 deg X-axis runs along fuselage symmetry plane And is normal to Z © Lakshmi Sankar 2002 =90 deg =0 deg X 39 Tip Path Plane This plane is defined by two straight lines. The first connects the blade tips at azimuth angle =0 and =180 deg. The second connects the blade tips at X azimuth angle =90 and =270 deg. Z is perpendicular to TPP. In TPP, the blade Z Does not appear to be Flapping.. Blade is at =0 deg. © Lakshmi Sankar 2002 40 Tip Path Plane 2Rb1c b1c b0+b1c Rb0-b1c) Rb0+b1c) X b0-b1c © Lakshmi Sankar 2002 41 No-Feathering Plane (NFP) or Control Plane Blade at =90 deg Blade at =270 deg Pitch link Z Swash Plate Pilot Input X The pilot controls the blade pitch by applying a collective control (all blades pitch up or down by the same amount), or by a cyclic control (which involves tilting the swash-plate). Some of the pitch links move up, while others move down. The airfoils connected © Lakshmi Sankar 2002 42 pitch up or down). Differences between Various Systems • For an observer sitting on the tip path plane, the blade tips appear to be touching the plane all the time. There is no flapping motion in this coordinate system. • For an observer sitting on the swash plate, the pitch links will appear to be stationary. There is no vertical up or down motion of the pitch links, and no pitching motion of the blades either. • In the shaft plane, the blades will appear to pitch and flap, both. • One can use any one of these coordinate systems for blade element theory. Some coordinate systems are easier to work with. For example, in the TPP we can set the flapping motion to zero. © Lakshmi Sankar 2002 43 Forward Flight Blade Flapping Dynamics and Response © Lakshmi Sankar 2002 44 Background • As seen earlier, blades are usually hinged near the root, to alleviate high bending moments at the root. • This allows the blades to flap up and down. • Aerodynamic forces cause the blades to flap up. • Centrifugal forces causes the blades to flap down. • Inertial forces will arise, which oppose the direction of acceleration. • In forward flight, an equilibrium position is achieved, where the net moments at the hinge due to these three types of forces (aerodynamic, centrifugal, inertial) cancel out and go to zero. © Lakshmi Sankar 2002 45 Schematic of Forces and Moments dL r b dCentrifugal Force We assume that the rotor is hinged at the root, for simplicity. This assumption is adequate for most aerodynamic calculations. Effects of hinge offset are discussed in many classical texts. © Lakshmi Sankar 2002 46 Velocity encountered by the Blade Velocity normal to The blade leading edge is r+V∞sin V∞ r © Lakshmi Sankar 2002 X 47 Moment at the Hinge due to Aerodynamic Forces From blade element theory, the lift force dL = 1 2 rc r V sin Cl dr 2 Moment arm = r cosb ~ r 1 2 Counterclockwise moment due to lift = rcr V sin rCl dr 2 Integrating over all such strips, Total counterclockwise moment = r R 1 2 r 0 2 rcr V sin rCl dr © Lakshmi Sankar 2002 48 Moment due to Centrifugal Forces The centrifugal force acting on this strip = r dm 2 rdm 2 Where “dm” is the mass of this strip. This force acts horizontally. The moment arm = r sinb ~ r b Clockwise moment due to centrifugal forces = r r bdm 2 2 Integrating over all such strips, total clockwise moment = r R 2 2 2 r b dm I b r 0 © Lakshmi Sankar 2002 49 Moment at the hinge due to Inertial forces Small segment of mass dm With acceleration rb r b Flap Hinge b is positive if blade is flapping up Associated moment at the hinge = r rb dm Integrate over all such segments: Resulting clockwise moment at the hinge= © Lakshmi Sankar 2002 Ib 50 At equilibrium.. Ib I 2 b Tip 2 1 rcCl r V sin rdr 2 Root Note that the left hand side of this ODE resembles a spring-mass system, with a natural frequency of . We will later see that the right hand side forcing term has first harmonic (terms containing t), second, and higher Harmonic content. The system is thus in resonance. Fortunately, there is Lakshmi Sankar 2002 Adequate aerodynamic©damping. 51 How does the blade dynamics behave when there is a forcing function component on the right hand side of the form Asint, and a damping term on the left hand side of form c db/dt ? To find out let us solve the equation: 2 Ib cb I b A sin t To solve this equation, we will assume a solution of form: b B sin t C cos t b A A cos t sin t 2 c c In other words, the blade response will be proportional to the amplitude A of the resonance force, but will lag the force by 90 degrees. © Lakshmi Sankar 2002 52 What happens when the pilot tilts the swash plate back? • The blade, when it reaches 90 degrees azimuth, pitches up. • Lift goes up instantly. • The blade response occurs 90 degrees later (recall the phase lag). • The front part of the rotor disk tilts up. • The exact opposite happens with the blade at • 270 degree azimuth. Blade at =90 deg © Lakshmi Sankar 2002 Blade at =270 deg Z X 53 What happens when the pilot tilts the swash plate back? TPP T Blade reaches its Highest position At =180 deg Blade reaches its lowest position At =0 deg NFP SwashPlate The tip path plane tilts back. The thrust points backwards. The vehicle will tend to decelerate. © Lakshmi Sankar 2002 54 What happens when the pilot tilts the swash plate to his/her right? • Blade at = 180 deg pitches up. Lift goes up. • Blade responds by flapping up, and reaches its maximum response 90 degrees later, at = 270 deg. • The opposite occurs with the blade at = 0 deg. • TPP tilts towards the pilot’s right. • The vehicle will sideslip. © Lakshmi Sankar 2002 T TPP NFP Helicopter viewed from Aft of the pilot 55 Blade Flapping Motion The blade flapping dynamics equation Ib I 2 b 2 Tip 1 rcCl r V sin rdr 2 Root has the general solution of the form b b 0 b1c cos b1s sin b 2c cos 2 b 2 s sin 2 First and higher harmonics Coning Angle =t © Lakshmi Sankar 2002 56 b1c determines the fore-aft tilt of TPP b1c TPP X This angle is measured in one of the three coordinate systems (Shaft plane, TPP, or NFP) that we have chosen to work with. Some companies like to use TPP, others like NFP or shaft plane. If TPP is our reference coordinate system, what will b1c be? Ans: Zero. © Lakshmi Sankar 2002 57 b1s determines the lateral tilt of TPP b1s Y Helicopter viewed from the back of the pilot © Lakshmi Sankar 2002 58 Pilot Input 75% R 0 1c cos 1s sin Collective Lateral control Longitudinal control The pitch is always, by convention Specified at 75%R. The pitch at all the other radial locations may be found If we know the linear twist distribution. The pilot applies a collective pitch by vertically raising the swash plate up or down. All the blades collectively, and equally pitch up or down. The pilot applies longitudinal control (i.e. tilts the TPP fore and aft) by Tilting the swash plate fore or aft as discussed earlier. Lateral control means tilt the swash plate (and the TPP) laterally. © Lakshmi Sankar 2002 59 Longitudinal control Blade at 90 deg azimuth 1s b1c X 1s NFP Note that 1s+ b1c is independent of the © Lakshmi Sankar 2002 are measured. coordinate system in which these angles 60 Lateral Control Blade at =180 Line parallel to NFP 1c TPP b1s NFP 1c Y Note that 1c-b1s is independent of the coordinate system in which 1c And b1s were measured. © Lakshmi Sankar 2002 61 As a result.. b1c 1s b1c b1s 1c b1s NFP 1s TPP NFP 1c TPP © Lakshmi Sankar 2002 62 In the future.. • We will always see b1c+1s appear in pair. • We will always see b1s-1c appear in pair. • As far as the blade sections are concerned, to them it does not matter if the aerodynamic loads on them are caused by one degree of pitch that the pilot inputs in the form of 1c or 1s, or by one degree of flapping (b1c or b1s). • One degree of pitch= One degree of flap. © Lakshmi Sankar 2002 63 Forward Flight Angle of Attack of the Airfoil And Sectional Load Calculations © Lakshmi Sankar 2002 64 The angle of attack of an airfoil depends on • Pilot input: collective and cyclic pitch • How the blade is twisted • Inflow due to freestream component, and induced inflow • Velocity of the air normal to blade chord, caused by the blade flapping • Anhedral and dihedral effects due to coning of the blades. • The first two bullets are self-evident. Let us look at the other contributors to the angle of attack. © Lakshmi Sankar 2002 65 Blade Flapping Effect r db/dt The blade section flaps up at a velocity rb . The airfoil thinks it is experienci ng a downwash of equal magnitude. © Lakshmi Sankar 2002 66 Angle of Attack caused by the Inflow and Freestream Rotor Disk V∞sina+v a V∞ r+V∞cosa © Lakshmi Sankar 2002 67 Anhedral/Dihedral Effect V∞ V∞sinb V∞sinb V∞cosaV∞ Blade at =180 will See an upwash equal to V∞sinb Blade at =0 will See an upwash equal to V∞sinb Blade at any will see an upwash equal to - V∞sinb cos © Lakshmi Sankar 2002 68 Summing them all up.. a UP UT= r+V∞cosa U P UP a tan UT U T where, U P V sin a v rb V cos sin b 1 © Lakshmi Sankar 2002 69 Small Angle of Attack Assumptions • The angle of attack a (which is the angle between the freestream and the rotor disk) is small. • The cyclic and collective pitch angles are all small. • The coning and flapping angles are all small. • Cos(a) = Cos(b)= Cos() ~ 1 • sin(a) ~ a, sin(b) ~ b, sin() ~ © Lakshmi Sankar 2002 70 Angle of Attack a effecive UP U P U T U P arctan UT UT UT 0 tw r 1c cos 1s sin R U T r V cos a s sin r V sin U P R rb V cos a s b cos R rb Vb cos s Subscript s: All angles are in the shaft plane © Lakshmi Sankar 2002 71 Angle of attack (continued) U T U P r r V sin 0 tw 1c cos 1s sin R Rs rb Vb cos db db d db b dt d dt d b1s cos b1c sin © Lakshmi Sankar 2002 72 Angle of Attack (Continued) After some minor algebra, U T U P r 0 1c b1s cos 1s b1c sin V 0 sin V 1c b1s cos sin V 1s b1c sin 2 Vb 0 cos Va TPP v Notice b1c+1s appears in pair, as pointed out earlier. Also 1c-b1s appears in pairs. One degree of pitching is equivalent to one degree of Flapping. © Lakshmi Sankar 2002 73 Relationship between aTPP and as as b1c aTPP=as+b1c © Lakshmi Sankar 2002 74 Angle of Attack (Concluded) r 0 1c b1s cos 1s b1c sin V sin V b cos sin U U P 1 0 1c 1s a effective T UT U T V 1s b1c sin 2 Vb 0 cos Va TPP v © Lakshmi Sankar 2002 75 Calculation of Sectional Loads • Once the angle of attack at a blade section is computed as shown in the previous slide, one can compute lift, drag, and pitching moment coefficients. • This can be done in a number of ways. Modern rotorcraft performance codes (e.g. CAMRAD) give the user numerous choices on the way the force coefficients are computed. © Lakshmi Sankar 2002 76 Some choices for Computing Sectional Loads as a function of a • In analytical work, it is customary to use Cl=aa, Cd=Cd0 = constant, and Cm= Cmo, a constant. Here “a” is the lift curve slope, close to 2. • In simple computer based simulations using Excel of a program (that we will show later), these loads are corrected for compressibility using Prandtl-Glauert Rule. • They are also corrected for sweep or yaw effects, in some instances. © Lakshmi Sankar 2002 77 Prandtl-Glauert Rule V a UP UT= r+V∞cosa ψ X Compute Mach number=M= UT/a∞ Cl ,corrected © Lakshmi Sankar 2002 Cl ,incompressible 1 M 2 78 Corrections for sweep or yaw angle of the velocity • The angle of attack is modified. amodified=a cosL L is the local sweep angle of the blade ¼ L chord line. Vnormal to blade leading edge Y X © Lakshmi Sankar 2002 79 More accurate ways of computing loads • In more sophisticated computer programs (e.g. industry codes), a table look up is used to compute lift, drag, and pitching moments from stored airfoil database. • These tables only contain steady airfoil loads. Thus, the analysis is quasi-steady. • In some cases, corrections are made for unsteady aerodynamic effects, and dynamic stall. • It is important to correct for dynamic stall effects in high speed forward flight to get the vibration levels of the vehicle correct. © Lakshmi Sankar 2002 80 Calculation of Sectional Forces • After Cl, Cd, Cm are found, one can find the lift, drag, and pitching moments per unit span. • These loads are normal to, and along the total velocity, and must be rotated appropriately. L’=Lift per foot of span a D’ UT= r+V∞cosa 1 rU 2 cCl 2 1 rc(U T2 U P2 )aa eff 2 1 © Lakshmi Sankar 2002 rcaU T U T U P 2 L 81 Forward Flight Integration of Sectional Loads To get Total loads © Lakshmi Sankar 2002 82 Background • In the previous sections, we discussed how to compute the angle of attack of a typical blade element. • We also discussed how to compute lift, drag, and pitching moment coefficients. • We also discussed how to compute sectional lift and drag forces per unit span. • We mentioned that these loads must be rotated to get components normal to, and along reference plane. • In this section, we discuss how to integrate these loads. • In computer codes, these integrations are done numerically. • Analytical integration under simplifying assumptions will be given here to illustrate the process. © Lakshmi Sankar 2002 83 Assumptions for Analytical Integration • • • • • c= constant (untapered rotor) v = constant (uniform inflow) Cd = constant Linearly twisted rotor No cut out, no tip losses. © Lakshmi Sankar 2002 84 Blade Section a UP UT= r+V∞cosa © Lakshmi Sankar 2002 85 Effective Angle of Attack As discussed earlier, a eff UP UT U T r V sin U P V sin a TPP v V b cos RTPP V b cos L 1 1 1 rcClU 2 rcaU T2a eff rca U T2 U T U P 2 2 2 © Lakshmi Sankar 2002 86 Some algebra first.. r 2 r V sin 0 tw 1c cos 1s sin 1 L rca R 2 r V sin R V b cos 3 r2 2 2 2 2 2 r r 2 V r sin V sin 2 V sin 0 0 tw tw 0 R R r V 2 sin 2 2 r 2 cos 2rV sin cos 1c 1c tw R 2 1 2 rca V 1c sin cos 2 2 r 2 sin 2rV sin 2 V 2 sin 3 1s 1s 1s 2 RTPP r V RTPP sin rV b cos V2 b cos sin Notice that we have first, second, and third harmonics present! These fluctuations will be felt by the passengers/pilots as © Lakshmi Sankar 2002 vibratory loads. 87 Thrust 1 T b 2 R L dr d 0 0 2 Thrust is computed by integrating the lift radially to get instantaneous thrust force at the hub, then averaging the thrust force over the entire rotor disk, and multiplying the force per blade by the number of blades. Computer codes will do the integrations numerically, without Sankar 88 any of the assumptions© Lakshmi we had to2002 make. Anlaytical Integration of Thrust We can interchange the order of integration. Integrate with respect to ψ first. Use the formulas such as 2 2 2 0 0 0 2 sin d cos d sin cosd 0 2 sin cosd 0 0 2 2 0 0 2 2 sin d cos d © Lakshmi Sankar 2002 89 Result of Azimuthal Integration 0 2 2 2 0 r 2 V 2 3 1 1 r 2 r 2 Ld rca tw V tw 2 0 2 R R 2 rV1s Rr TPP © Lakshmi Sankar 2002 90 Next perform radial integration and Normalize 3 3 tw 2 R R 2 2 2 0 0 V R tw V R 1 3 2 4 4 T rabc 2 2 2 R R 2 V TPP 1s TPP 2 2 1s TPP TPP 3 2 tw 2 CT 1 1 2 3 2 4 2 2 a 0 Note that we will get the hover expressions back if advance ratio is set to zero. © Lakshmi Sankar 2002 91 Torque and Power • We next look at how to compute the instantaneous torque and power on a blade. • These are azimuthally-averaged to get total torque and total power. • It is simpler to look at profile and induced components of torque are power separately. © Lakshmi Sankar 2002 92 Profile Drag r D’ 1 2 D rU T cCd , 0 2 where U T r V sin We will assume chord c and drag coefficient Cd0 are constant. © Lakshmi Sankar 2002 93 Integration of Profile Torque 1 Q0 b 2 Non-dimensionalize: Final result: R 0 0 rDdr 2 CQ , 0 Q 2 r R AR CQ,0 © Lakshmi Sankar 2002 Cd ,0 8 1 2 94 Profile Power 1 Q0 b 2 R 0 0 rDdr 2 Non-dimensionalize: Final result: 1 P0 b 2 CP ,0 R 0 0 U T Ddr 2 P0 3 r R A CP , 0 © Lakshmi Sankar 2002 Cd ,0 8 1 3 2 95 Induced Drag L’ a Di UP UT= r+V∞cosa UP 1 2 UP Di L rcClU T UT 2 UT 1 rcClU TU P 2 © Lakshmi Sankar 2002 96 Induced Torque and Power 2 R 0 0 rDi dr d 2 R 1 Pi U T Di dr d 2 0 0 1 Qi 2 Performing the analytical integration, CPi i CT This is a familiar result. Induced Power = Thrust times Induced Velocity! © Lakshmi Sankar 2002 97 In-Plane Forces • In addition to thrust, that act normal to the rotor disk (or along the z-axis in the coordinate selected by the user), the blade sections generate in-plane forces. • These forces must be integrated to get net force along the x- axis. This is called the H-force. • These forces must be integrated to get net forces along the Y- axis. This is called the Yforce. • These forces will have inviscid components, and viscous components. © Lakshmi Sankar 2002 98 Origin of In-Plane Forces L’sinb=L’b L’ L’b Y b X One source of in-plane forces is the tilting of the Thrust due to the blade coning angle. © Lakshmi Sankar 2002 99 Origin of In-Plane Forces-II V∞ Radial flow causes radial Skin friction forces A component of the free-stream flows along the blade, in the radial direction. This causes radial skin friction forces. This is hard to quantify,©and isSankar usually Lakshmi 2002 neglected. 100 Origin of In-Plane Forces III D=Di+D0 Sectional drag (which is made of inviscid induced drag, and viscous drag) Can give rise to components along the X- direction (H-force), and Y- direction (Y-Force). Engineers are interested in both the instantaneous values (which determine © Lakshmi Sankar 2002 101 Vibration levels), as well as azimuthal averages (which determine force balance). Closed Form Expressions for CH and CY • Under our assumptions of constant chord, linear twist, linear aerodynamics, and uniform inflow, these forces may be integrated radially, and averaged azimuthally. • The H- forces and the Y- forces are nondimensionalized the same way thrust is nondimensionalized. • Many text books (e.g. Leishman, Prouty) give exact expressions for these coefficients. © Lakshmi Sankar 2002 102 Closed Form Expressions C H TPP TPP tw 2 0 2 Cd 0 a 1c b 0 TPP 1s 4 2 6 4 b 02 4 3 2 1c 4 b 0 0 3 tw 4 TPP C a YTPP 3 2 1s b 0 2 6 1 3 2 b 0 TPP © Lakshmi Sankar 2002 103 Forward Flight Calculation of Blade Flapping Dynamics © Lakshmi Sankar 2002 104 Background • In the previous sections, we developed expressions for sectional angle of attack, sectional loads, total thrust, torque, power, H-force and Y-force. • These equations assumed that the blade flapping dynamics a priori. • The blade flapping coefficients are determined by solving the ODE that covers flapping. © Lakshmi Sankar 2002 105 Flapping Dynamics Ib I 2 b Tip 2 1 rcCl r V sin rdr 2 Root r 0 1c b1s cos 1s b1c sin V sin V b cos sin U U P a 0 1c 1s Cl aa effective a T UT U T V 1s b1c sin 2 Vb 0 cos Va TPP v © Lakshmi Sankar 2002 106 Solution Process Assume the blade flapping dynamics equation Ib I 2 b Tip 2 1 rcCl r V sin rdr 2 Root has the general solution of the form b b 0 b1c cos b1s sin b 2 c cos 2 b 2 s sin 2 Plug in the solution on both the left and right sides. The right side Can be integrated analytically, subject to usual assumptions. Equate coefficients on the left side and right, term by term. For example coefficient with sinψ on the left with the © Lakshmi Sankar 2002 similar term on right. 107 Final Form 2 80% R b 1c 1s TPP 2 b0 1 tw 8 60 6 6 8 3 75% R TPP 3 4 b 1c 1s 3 2 1 2 4 b 0 b 1s 1c 3 1 2 1 2 © Lakshmi Sankar 2002 108 Level Flight Calculation of Trim Conditions Including Fuselage Aerodynamics © Lakshmi Sankar 2002 109 Background • By trim conditions we mean the operating conditions of the entire vehicle, including the main rotor, tail rotor, and the fuselage, needed to maintain steady level flight. • The equations are all non-linear, algebraic, and coupled. • An iterative procedure is therefore needed. © Lakshmi Sankar 2002 110 Horizontal Force Balance HM HT DF V∞ Total Drag= Fuselage Drag (DF) + H-force on main rotor (HM) + H-force on the tail rotor (HT) © Lakshmi Sankar 2002 111 Vertical Force Balance LF GW Vertical Force = GW- Lift generated by the fuselage, LF © Lakshmi Sankar 2002 112 Tip Path Plane Angle T aTPP V∞ T cos a TPP GW LF T sin a TPP DF H M H T a TPP DF H M H T tan GW LF 1 © Lakshmi Sankar 2002 113 Fuselage Lift and Drag • These are functions of the fuselage geometry, and its attitude (or angle of attack). • This information is currently obtained from wind tunnel studies, and stored as a database in computer codes. © Lakshmi Sankar 2002 114 Fuselage Angle of Attack • Extracted from – Tip path angle – Blade flapping dynamics – Downwash felt by the fuselage from the main rotor – Shaft inclination angle. © Lakshmi Sankar 2002 115 Shaft Inclination Angle is Fuselage Reference Axis © Lakshmi Sankar 2002 116 Relationship between Tip Path Plane Angle of Attack and Shaft Angle of Attack V ∞ as TPP b1c Shaft Plane aTPP aTPP a s b1c © Lakshmi Sankar 2002 117 Angle of Attack of the Fuselage • Start with tip path plane angle of attack. • Subtract b1c to get shaft angle of attack • Subtract the inclination of the shaft • Subtract angle of attack reduction associated with the downwash from the rotor v a F a TPP b1c is V © Lakshmi Sankar 2002 118 Iterative process • Assume angle of attack for fuselage (zero deg). • Find LF and DF from wind tunnel tables. • Compute needed T. during the first iteration, T is approximately GW-LF. Use this info. to find main rotor torque, main rotor H-force, tail rotor thrust needed to counteract main rotor torque, and tail rotor H- force. • From blade trim equations, find b1c. • Find tip path plane angle of attack. • Recompute fuselage angle of attack. • When iterations have converged, find main and tail rotor power. Add them up. Add transmission losses to get total power needed. © Lakshmi Sankar 2002 119 Autorotation in Forward Flight • The calculations described for steady level flight can be modified to handle autorotative descent in forward flight. • Power needed is supplied by the time rate of loss in potential energy. © Lakshmi Sankar 2002 120 Descent HM HT LF Wsinχ χ W cosχ Rate of descent V∞ DF descent ve locity tan V 1 © Lakshmi Sankar 2002 121 Tip Path Plane Angle in Descent T cos a TPP GW cos LF T sin a TPP DF H M H T GW sin a TPP DF H M H T GW sin tan GW cos L F 1 © Lakshmi Sankar 2002 122 Iterative Procedure • The iterative procedure involves – assume a rate of descent – Iterate on fuselage angle of attack to achieve forces to balance, as done previously in steady level flight. – Compute the power needed to operate= main rotor+ tail rotor+ transmission losses. – Equate this power needed with the power available from loss of potential energy= GW * Rate of descent. – Iterate until power needed = power available © Lakshmi Sankar 2002 123