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Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 503858 9 pageshttpdxdoiorg1011552013503858
Research ArticleNonlinear Characteristics of Helicopter Rotor Blade AirfoilsAn Analytical Evaluation
Constantin Rotaru
Department of Aviation Integrated Systems Military Technical Academy 39-49 George Cosbuc Avenue 050141 Bucharest Romania
Correspondence should be addressed to Constantin Rotaru rotaruconstantinyahoocom
Received 27 June 2013 Revised 4 September 2013 Accepted 6 September 2013
Academic Editor Tirivanhu Chinyoka
Copyright copy 2013 Constantin Rotaru This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
Some results are presented about the study of airloads of the helicopter rotor blades the aerodynamic characteristics of airfoilsections the physical features and the techniques for modeling the unsteady effects found on airfoil operating under nominallyattached flow conditions away from stall The unsteady problem was approached on the basis of Theodorsenrsquos theory where theaerodynamic response (lift and pitchingmoment) is considered as a sumof noncirculatory and circulatory partsThenoncirculatoryor apparent mass accounts for the pressure forces required to accelerate the fluid in the vicinity of the airfoil The apparent masscontributions to the forces and pitching moments which are proportional to the instantaneous motion are included as part of thequasi-steady result
1 Introduction
Themost important component of the helicopter is the mainrotor for which there is a great deal of activity in developingnew and improved mathematical models that predict theflow physics A high tip speed gives the rotor a high levelof stored rotational kinetic energy and reduces the rotortorque required for a given power but there are two importantfactors that work against the use of a high tip speed com-pressibility effects and noise
The additional effects of compressibility on the over-all rotor profile power requirements when the tip of theadvancing blade approaches and exceeds the drag divergenceMach number were estimated using blade element theorycombined with the airfoil section characteristics [1] A moredetailed analysis of compressibility effects on the rotor mustrepresent the actual nonlinear airfoil characteristics as func-tions of Mach number through stall at each blade elementfollowed by numerical integration
The classical unsteady aerodynamic theories describingthe observed behavior have formed the basis for many typesof rotor analysisThe tools for the analysis of 2D incompress-ible and unsteady aerodynamic problems were extended tocompressible flows being a basis for developing linearized
unsteady aerodynamic models applicable to compressibleflows [2] But while the classical theories assume linearityin the airloads the assumption of linearity can probablybe justified for many of the problems encountered on therotor in practice The advent of nonlinear methods based onCFD solutions to the Euler and Navier-Stokes equations hasprovided new results that justify and define the limits of thelinearmodels andmay give guidance in developing improvedand more practical unsteady aerodynamic models for futureuse in helicopter rotor blade airloads prediction aeroelasticanalysis and rotor design [3] At the blade element level thevarious sources of unsteady effects can be decomposed intoperturbations to the local angle of attack and velocity fieldAt low angle of attack with fully attached flow the varioussources of unsteady effects manifest as moderate amplitudeand phase variations relative to the quasi-steady airloads Athigher angles of attack when time-dependent flow separationfrom the airfoil may be involved a phenomenon character-ized by large overshoots in the values of the lift drag andpitchingmoments relative to the quasi-steady stall valuesmayoccur [4]
The helicopter rotor airfoil must assure a high maximumlift coefficient a high drag divergence Mach number a goodlift-to-drag ratio over a wide range of Mach number and a
2 Mathematical Problems in Engineering
Region ofreverse flow
Disk exceeding drag divergence Mach number
Vinfin
1205952
1205951
Figure 1 Helicopter rotor blade in forward flight
low pitching moment At high angles of attack the adversepressure gradients produced on the upper surface of theairfoil result in a progressive increase in the thickness of theboundary layer and cause some deviation from the linear liftversus angle of attack behavior On many airfoils the onsetof flow separation and stall occurs gradually with increasingangle of attack but on some airfoils (those with sharp leadingedges) the flow separation may occur suddenly
2 Helicopter Rotor Blade Aerodynamics
The region of the rotor disk affected by compressibility effectsis shown in Figure 1 and is defined on the surface wherethe incident Mach number of the flow that is normal to theleading edge of the blade exceeds the drag divergence Machnumber119872119889119889 If119872Ω119877 is the hover tip Mach number then theregion of the disk affected by compressibility effects is definedby
119872119903120595 = 119872Ω119877 (119903 + 120583 sin120595) ge 119872119889119889 (1)
The angular or rotational speed of the rotor is denotedby Ω the rotor radius by 119877 the advanced ratio by 120583 and theradial distance from the rotational axis by 119910 In the aboveequation the nondimensional quantities are 119903 = 119910119877 and120583 = 119881infin cos120572Ω119877 where 120572 is the angle between the forwardvelocity 119881infin and the plane of the rotor [1 2] The azimuthangle for the onset drag divergence 1205951 can be obtained bysetting 119903 = 1 so that
1205951 = arcsin([1120583(119872119889119889
119872Ω119877
minus 1)]) (2)
and 1205952 = 180 minus 1205951The increment in profile power Δ119862119875 associated with this
region on the disk is
Δ119862119875
120590=1
4120587int
1205952
1205951
int
1
119903119889119889
(119903 + 120583 sin120595)3Δ119862119889119903 119889119903 119889120595 (3)
where Δ119862119889 is the extra drag on the blade section when itexceeds the drag divergence Mach number 119872119889119889 and 120590 is
Tip vortexfilament
Ω
Px
120595b
120595w
120595 bminus 120595 w
y
yref
Figure 2 Tip vortex trajectory
rotor solidity coefficient which represents the ratio of theblades area to the rotor disk area For the NACA 0012 airfoilProuty [4] suggests that this can be approximated by
Δ119862119889 (119872) = 125(119872 minus 074)
3 for 119872 ge 074
0 otherwise(4)
The rotor limits may be determined by two conditions onecondition given by advancing blade compressibility effectsand the other condition given by retreating blade stall Ineither case the advancing blade operates at low angle of attack(AoA) but at high subsonic or transonic conditions whereasthe retreating blade operates at low Mach numbers and highlift coefficients
The aerodynamic behavior of airfoils in the high AoAregime is important for predicting the adverse effects pro-duced in the reverse flow regime on the rotor In the reverseflow region the direction of the relative flow vector changesfrom the trailing edge toward the leading edge of the airfoilWhile the fundamental process of the blade wake and tipvortex formation is similar to that found with a fixed wingone difference with helicopter tip vortices is that they arecurved and so they experience a self-induced effect Anothercomplication with helicopter rotors is that the wakes and tipvortices from other blades can lie close to each other and tothe plane of blade rotation and so they have large inducedeffects on the blade lift distribution
If the wake is assumed to be undistorted in the tip pathplane and no wake contraction occurs in the radial direction(Figure 2) then the tip vortex trajectories are described by theequations
119909 = 119877 cos (120595119887 minus 120595119908) + 119877120583120595119908
119910 = 119877 sin (120595119887 minus 120595119908) (5)
where 120595119887 is the position of the blade when the vortex wasformed and 120595119908 is the position of the vortex element relativeto the blade
One important parameter used in the description ofunsteady aerodynamics and unsteady airfoil behavior is thereduced frequency 119896 defined as 119896 = 120596 sdot 119888(2119881) where 120596 is
Mathematical Problems in Engineering 3
Ω middot R + Vinfin
Vinfin
Ω middot R
Ω middot R minus VinfinR
Figure 3 Helicopter main rotor
the angular frequency 119888 is the chord of the airfoil and 119881 isthe flow velocity [3] According to the dimensional analysisthe resultant force 119865 on the airfoil chord 119888 can be written infunctional form as 119865(12058811988121198882) = 119891(Re119872 119896) For 119896 = 0 theflow is steady and for 0 le 119896 le 005 the flow can be consideredquasi-steady that is unsteady effects are generally smallFlows with characteristic reduced frequencies above 005 areconsidered unsteady [5 6] For a helicopter rotor in forwardflight (Figure 3) the local sectional velocity which appearsin the denominator of the reduced frequency expression isconstantly changing
The approach to modeling of unsteady aerodynamiceffects through an extension of steady 2D thin airfoil theorygives a good level of analysis of the problem and providesconsiderable insight into the physics responsible for theunderlying unsteady behavior The Laplacersquos equation forincompressible flow is eliptic therefore the unsteady aero-dynamic theories cannot be obtained in a correspondinganalytical form
3 The Apparent Mass Tensor
The rate of change of the impulse vector in general is not inthe direction of the acceleration of the body The externalforce 119865119890 applied to the body to translate it through thefluid has to be applied in a direction different from that ofthe acceleration of the body through the fluid [7] Physicalconditions that should be satisfied on given boundaries of thefluid (boundary conditions) depend on the assumptionsmade with regard to the nature of the fluid more specificallyon the nature of the differential equations that are assumed togovern the motion of the fluid For a solid-fluid boundary ateach point of the solid-fluid surface at every instant thecomponent normal to the surface of the relative velocitybetween the fluid and the solid must be vanish sdot 119899 = 0where represents the relative velocity and 119899 the normal tothe surface (Figure 4)
If the surface is represented by a scalar function of posi-tion and time 119865( 119903 119905) = 0 then the total time rate of changeis zero
119863 (119865)
119863119905=120597 (119865)
120597119905+ sdot grad (119865) = 0 (6)
on 119865( 119903 119905) = 0
rarrn
rarrVdV
dS
Boundary
Figure 4 Solid-fluid surface
The fluid force acting on a rigid body of arbitrary shapetranslating with a velocity (119905) is given by
= minus∬119878
119901 119899 119889119878 (7)
where 119878 denotes the surface of the body and 119901 is the pressureon the surface of the body In general the body may be trans-lating rotating and deforming consequently the velocity 119880is a function of position on the surface and time If the body isrigid and is in translatorymotion then119880 is a function of timebut uniform over the surface of the body The mathematicalproblem is to determine the external force 119890 applied to thebody to translate it through the fluid According to Newtonrsquossecond law we have
119889
119889119905(119898) = 119890 + (8)
where119898 is the mass of the body The above equation may berewritten as
119890 =119889
119889119905(119898) minus (9)
or
119890 =119889
119889119905(119898 + 119868) (10)
where 119868 is the impulse applied on the fluid and minus119889 119868119889119905 =
The fluid force acting on the body is
=120597
120597119905∯119878
120588120601 119899 119889119878 minus 120588 times∯119878
119899 times grad (120601) 119889119878 (11)
The integral
119868119862 =∯119878
119899 times grad (120601) 119889119878 (12)
is related to the circulation 119862 around the body and 120601 is thevelocity potential
The unit vector 119890 is normal to the cutting planes the unitvector 1198901 is tangent to the curve of intersection between theblade element surface and the cutting plane and the unitvector 1198902 is tangent to the blade element surface
4 Mathematical Problems in Engineering
rarrn
rarre1
rarre2
rarre
dy
dS = dx middot dy
dx
C
Figure 5 Rotor blade element
The component of the vector 119868119862 in the direction 119890 is
119890 sdot 119868119862 =∯119878
119890 sdot 119899 times 119902119889119878 (13)
and the vector 119899 119889119878 corresponding to the surface element119889119878 = 119889119909 sdot 119889119910 may be written as 119899 119889119878 = 119889 times 119889 119910 where119889 = 119889119909 1198901 119889 119910 = 119889119910 1198902 and 119902 = grad(120601)
On the other hand
119899 119889119878 times 119902 = (119889 times 119889 119910) times 119902 = ( 119902 sdot 119889) 119889 119910 minus ( 119902 sdot 119889 119910) 119889
119890 sdot 119899 times 119902 119889119878 = 119890 ( 119902 sdot 119889) 119889 119910 minus 119890 ( 119902 sdot 119889 119910) 119889
(14)
Since the unit vectors 119890 and 1198901are normal it follows that 119890 sdot
119889 = 0 and
119890 sdot ( 119899 times 119902) 119889119878 = ( 119902 sdot 119889) ( 119890 sdot 119889 119910) (15)
The scalar product 119890 sdot 119889 119910 is the normal distance between thecutting planes of the solid body (Figure 5) If we denote 119889ℎ =119890 sdot 119889 119910 it follows that
119890 sdot∯119878
119899 times 119902119889119878 = int
ℎ2
ℎ1
(∮119862
119902 sdot 119889) 119889ℎ = int
ℎ2
ℎ1
Γ119890 (ℎ) 119889ℎ (16)
Here ℎ is the distance measured along the fixed direction 119890and
Γ119890 (ℎ) = ∮119862
119902 sdot 119889 (17)
is the circulation around the curve of intersection betweenthe body surface and the cutting plane The limits ℎ1 andℎ2 denote the extremities of the body measured along thedirection 119890 It follows that for motions without circulation theforce on the body is given by
=120597
120597119905(∯119878
120588120601 119899 119889119878) (18)
where the velocity potential 120601 is the solution of the system
nabla2120601 = 0
grad120601 sdot 119899 =120597120601
120597119899= 119880 (119905) 119899 on 119878
(19)
Since the equation and boundary condition for 120601 are linearthe solution could have the form
120601 = 1206011 + 1206012 + 1206013 (20)
where each of the functions 1206011 1206012 and 1206013 is a solution of theequations
nabla2120601119894 = 0
grad (120601119894) sdot 119899 =120597120601119894
120597119899= 119906119894119899119894 on 119878
(21)
where 119894 = 1 2 or 3In the Cartesian coordinate system the vectors and 119899
have the expressions
= 1199061119894 + 1199062
119895 + 1199063119896
119899 = 1198991119894 + 1198992
119895 + 1198993
(22)
Because time enters through 119906119894 it is convenient to set 120601119894 =119906119894120593119894 so the system (21) takes the form
nabla2120593119894 = 0
grad (120593119894) sdot 119899 =120597120593119894
120597119899= 119899119894 on 119878
(23)
With these considerations the impulse 119868 becomes
minus 119868 = ∯119878
120588120601 119899 119889119878 = ∯119878
120588(
3
sum
119896=1
119906119896120593119896) 119899 119889119878
=
3
sum
119896=1
(∯119878
120588120601119896 119899 119889119878) 119906119896
(24)
The components of the impulse 119868 = 1198681119894 + 1198682
119895 + 1198683 are
1198681 =119895 sdot 119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198991 119889119878) 119906119896
1198682 =119895 sdot 119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198992 119889119878) 119906119896
1198683 = sdot119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198993 119889119878) 119906119896
(25)
The surface integral in above equations may be written asfollows
∯119878
120588120593119896119899119894 119889119878 =∯119878
120588120593119896
120597120593119894
120597119899119889119878 (26)
According to Greenrsquos theorem if 1198911 and 1198912 are two harmonicfunctions then
∯119878
1198911
1205971198912
120597119899119889119878 =∯
119878
1198912
1205971198911
120597119899119889119878 (27)
Introducing the symbol119898119896119894
119898119896119894 = minus∯119878
120588120593119896
120597120593119894
120597119899119889119878 (28)
Mathematical Problems in Engineering 5
with119898119896119894 = 119898119894119896 the components of the impulse 119868 are thereforegiven by
119868119894 =
3
sum
119896=1
119898119894119896119906119896 119894 = 1 2 3 (29)
and the force applied externally to the body is
119890 =119889
119889119905119898 (1199061
119894 + 1199062119895 + 1199063)
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119894
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119895
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
(30)
or
119890 = [(119898 + 11989811)1198891199061
119889119905+ 11989812
1198891199062
119889119905+ 11989813
1198891199063
119889119905] 119894
+ [11989821
1198891199061
119889119905+ (119898 + 11989822)
1198891199062
119889119905+ 11989823
1198891199063
119889119905] 119895
+ [11989831
1198891199061
119889119905+ 11989832
1198891199062
119889119905+ (119898 + 11989833)
1198891199063
119889119905] 119896
(31)
The coefficients119898119894119896 form a set of nine numbers whichmay bedisplayed as an array
(
11989811 11989812 11989813
11989821 11989822 11989823
11989831 11989832 11989833
) (32)
and may be referred to as a virtual mass tensor or virtualmasses that need to be added to the mass of the body in orderto find the force that must be applied to translate it throughthe fluid Introducing the symbol 120575119894119896 defined by 120575119894119896 = 0 if 119894 = 119896
and 120575119894119896 = 1 for 119894 = 119896 (30) may be rewritten as
(119865119890)119894=
3
sum
119896=1
(119898120575119894119896 + 119898119894119896)119889119906119896
119889119905 (33)
For any body there are three perpendicular directions suchthat 119898119894119896 = 0 for 119894 = 119896 so with respect to such axes (33)becomes
(119865119890)119894= (119898 + 119898119894119894)
119889119906119894
119889119905 119894 = 1 2 3 (34)
The sum (119898+119898119894119894) represents the apparentmass for translationin the 119894-direction and the corresponding119898119894119894 is the additionalapparent mass
4 The Airloads on an Oscillating Airfoil
The oscillatory motion of the airfoil can be decomposed intocontributions associated with angle of attack which is equiv-alent to a pure plunging motion (Figure 6) and contributionsassociated with pitching (Figure 7)
Vh
h
x
w(x) = minus
Figure 6 Plunge velocity
V
x
y
w(x) = minus (x minus a)
bb
a
Figure 7 Pitch rate
A plunge velocity ℎproduces a uniformvelocity perturba-tion 119908 that is normal to the chord 119908(119909) = minusℎ and the pitch-rate term produces a linear variation in normal perturbationvelocity
For a pitch rate imposed about an axis at ldquo119886rdquo semichordsfrom the midchord then 119908(119909) = minus(119909 minus 119886) so that theinduced chamber is a parabolic arc
The problem of finding the airloads on an oscillatingairfoil was solved by Theodorsen who gave a solution to theunsteady airloads on a 2D harmonically oscillated airfoil ininviscid incompressible flow with the assumption of smalldisturbances [6] Both the airfoil and its shed wake wererepresented by a vortex sheet with the shedwake extending asa planar surface from the trailing edge downstream to infinityThe assumption of planar wake is justified if the angleof attack disturbances remain relatively small As with thestandard quasi-steady thin airfoil theory the bound vorticity120574119887 can sustain a pressure difference and therefore a liftforceThe wake vorticity 120574119908 must be force-free with zero netpressure jump over the sheet According to the Theodorsenrsquostheory the solution for the loading 120574119887 on the airfoil surfaceunder harmonic forcing conditions is obtained from integralequation
119908 (119909 119905) =1
2120587int
119888
0
120574119887 (119909 119905)
119909 minus 1199090
119889119909 +1
2120587int
infin
119888
120574119908 (119909 119905)
119909 minus 1199090
119889119909 (35)
6 Mathematical Problems in Engineering
where119908 is the downwash on the airfoil surface At the trailingedge 120574119887(119888 119905) = 0 and the airfoil circulation Γ(119905) is given by
Γ (119905) = int
119888
0
120574119887 (119909 119905) 119889119909 (36)
So long as the circulation about the airfoil is changing withrespect to time the circulation is continuously shed into thewake and will continuously affect the aerodynamic loads onthe airfoil
For a general motion where an airfoil of chord 119888 = 2119887 isundergoing a combination of pitching (120572 ) and plunging (ℎ)motion in a flow of steady velocity 119881 Theodorsenrsquos solution[8ndash10] for the lift coefficient and pitching moment coefficientcorresponding to midchord11987212 is
119888119897 = 120587119887 [ℎ
1198812+
119881minus119887
1198812119886]
+ 2120587[ℎ
119881+ 120572 +
119887
119881(1
2minus 119886)]119862 (119896)
11988811989812 =120587
2[119887119886ℎ
1198812minus1198872
1198812(1
8+ 1198862) ]
+120587(119886 + 12) [
ℎ
119881+ 120572 + 119887 (
1
2minus 119886)
119881]119862 (119896)
minus120587
2[(1
2minus 119886)
119887
119881]
(37)
where 119886 is the pitch axis location relative to the midchordof the airfoil measured in terms of semichord and 119862(119896) =119865(119896) + 119894119866(119896) is the complex transfer function (known asTheodorsenrsquos function) which accounts for the effects of theshed wake on the unsteady airloads
119862 (119896) =119867(2)
1(119896)
119867(2)
1(119896) + 119894 sdot 119867
(2)
0(119896)
=1198691 (1198691 + 1198690) + 1198841 (1198841 minus 1198690)
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
+ 11989411988411198840 + 11986911198690
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
(38)
with 1198690 1198691 1198840 and 1198841 being Bessel functions of the first andsecond kinds respectively (Figure 8)
The Hankel functions in above expression are
119867(2)
0= 1198690 minus 119894 sdot 1198840
119867(2)
1= 1198691 minus 119894 sdot 1198841
(39)
The real and imaginary parts of 119862(119896) function are plotted inFigure 9
It could be appreciated that 119862(119896) function serves tointroduce an amplitude reduction and phase lag effect on thecirculatory part of the lift response compared to the resultobtained under quasi-steady conditions [11]
1
05
05 10 15
J(0 k)
J(1 k)
Y(0 k)
Y(1 k)
minus1
minus05
Bessel
Bessel
Bessel
Bessel
Figure 8 Bessel functions
Real part of C(k)
Imag
inar
y pa
rt o
f C(k)
0 02 04 06 08 1
0
minus005
minus010
minus015
minus020
minus025
minus030
Figure 9 Theodorsenrsquos function
This effect can be seen if a pure oscillatory variation inangle of attack is considered that is 120572 = 120572119890
119894120596119905 so thecirculatory part of the airfoil lift coefficient is given by
119888119897 = 2120587120572119862 (119896) = 2120587120572 [119865 (119896) + 119894119866 (119896)] (40)
For 119896 = 0 the steady-state lift behavior is obtained that is 119888119897is linearly proportional to 120572 As 119896 is increased the lift plotsdevelop into hysteresis loops and these loops rotate such thatthe amplitude of the lift response (half of the peak-to-peakvalue) decreases with increasing reduced frequency
These loops are circumvented in a counterclockwisedirection such that the lift is lower than the steady value when
Mathematical Problems in Engineering 7
CirculatoryTotal response
Apparent mass
2
15
1
05
0
0 1 2 3 4
k
Nor
mal
ized
lift
ampl
itude
Figure 10 Normalized lift amplitude
120572 is decreasingwith time (ie there is a phase lag) For infinitereduced frequency the circulatory part of the lift amplitude ishalf that at 119896 = 0 and there is no phase lag angle
5 Pure Angle of Attack Oscillations
For a harmonic variation in 120572 that is 120572 = 120572119890119894120596119905 the lift is
119871 = 21205871205881198812119887 [119862 (119896) +
1
2119894120596119887
119881]120572119890119894120596119905 (41)
or in terms of the lift coefficient the result is
119888119897 =119871
1205881198812119887= [2120587 (119865 + 119894119866) + 119894120587119896] 120572119890
119894120596119905 (42)
The term inside the square brackets can be considered the lifttransfer function which accounts for the difference betweenthe unsteady and quasi-steady airloads
The first term inside the brackets is the circulatory termand the second term is the apparentmass contribution whichis proportional to the reduced frequency and leads the forcingby a phase angle of 1205872 The noncirculatory or apparent massterms arise from the nonsteady term of the velocity potentialand account for the pressure forces required to accelerate thefluid in the vicinity of the airfoil
The normalized lift amplitude is
119888119897
2120587120572= (119865 + 119894119866) + 119894
119896
2 (43)
The normalized lift amplitude and phase of lift for pure angleof attack oscillations are presented in Figures 10 and 11 wherethe significance of the apparentmass contribution to both theamplitude and phase can be appreciated
At lower values of reduced frequency the circulatoryterms dominate the solution At higher values of reducedfrequency the apparent mass forces dominate
100
80
60
40
20
0
minus20
Phas
e ang
le [d
egre
es]
CirculatoryTotal response
Apparent mass
0 1 2 3 4
k
Figure 11 Phase angle
6 Pure Plunging Oscillations
For a harmonic plunging motion such as that contributed byblade flapping the forcing is ℎ = ℎ119890119894120596119905 so that ℎ = 119894120596ℎ119890119894120596119905 andℎ = minus120596
2ℎ119890119894120596119905 Substituting into the expression for the lift and
solving for the lift coefficient it gives
119888119897 = [2120587119896 (119894119865 minus 119866) minus 1205871198962]ℎ
119887119890119894120596119905 (44)
The complete term inside the square brackets can be con-sidered as the lift transfer function [12] The circulatory partof the lift response leads the forcing displacement ℎ by aphase angle of 1205872 Also the apparent mass force leads thecirculatory part of the response by a phase angle of 1205872 orthe forcing by a phase angle of 120587 The corresponding pitchingmoment about midchord for this case is
11988811989812 = (120587
4) 1198962 ℎ
119887119890119894120596119905 (45)
7 Pitching Oscillations
For harmonic pitch oscillations additional terms involvingpitch rate appear in the equations for the aerodynamic res-ponse The forcing is given by 120572 = 120572119890119894120596119905 and the pitch rate by = 119894120596120572119890
119894120596119905 In this case the lift coefficient is
119888119897 = 2120587 [119865 (1 + 119894119896) + 119866 (119894 minus 119896)] 120572119890119894120596119905
+ 120587119896(119894 minus119896
2)120572119890119894120596119905
(46)
The lift amplitude initially decreases with increasing 119896 beca-use of the effects of the shed wake and then for 119896 gt 05it begins to increase as the apparent mass forces begin to
8 Mathematical Problems in Engineering
dominate the airloads This is also shown by the phase anglewhich exhibits an increasing lead for 119896 gt 03
Von Karman and Sear analyzed the problem of a thinairfoil moving through a sinusoidal vertical gust field wherethe gust can be considered as an upwash velocity that isuniformly convected by the free streamThe forcing functionin this case is
119908119892 (119909 119905) = sin(120596119892119905 minus120596119892119909
119881) (47)
where 120596119892 is the gust frequency If the gust is referenced tothe airfoil leading edge then 119909 = 0 and 119908119892(119909 119905) becomes119908119892(119909 119905) = sin(120596119892119905) and if the gust is referenced to themidchord then 119909 = 119887 = 1198882 and
120596119892119909
119881=
120596119892 (1198882)
119881=
120596119892119888
2119881= 119896119892
(48)
Therefore119908119892(119905) = sin(120596119892119905)sdotcos(119896119892)minussin(119896119892)sdotcos(120596119892119905) whichis equivalent to a phase shift In this case the lift coefficientcan be written as
119888119897 = 2120587(1199080
119881) 119878 (119896119892) (49)
where 119878(119896119892) is known as Sears function and the gust encoun-ter frequency 119896119892 is given by
119896119892 =2120587119881
120582119892
(50)
and 120582119892 is the wavelength of the gustIn terms of Bessel functions Searsrsquos function is given by
119878 (119896119892) = [1198690 (119896119892) minus 1198941198691 (119896119892)] 119862 (119896119892) + 1198941198691 (119896119892) (51)
The terms of real and imaginary parts are
Real 119878 (119896119892) = 119865 (119896119892) sdot 1198690 (119896119892) + 119866 (119896119892) sdot 1198691 (119896119892)
Im 119878 (119896119892) = 119866 (119896119892) sdot 1198690 (119896119892) minus 119865 (119896119892) sdot 1198691 (119896119892) + 1198691 (119896119892)
(52)
If the gust is referenced to the leading edge of the airfoil thefunction will be called 1198781015840 and can be written as
Real 1198781015840 (119896119892) = Real 119878 (119896119892) sdot cos (119896119892)
+ Im 119878 (119896119892) sdot sin (119896119892)
Im 1198781015840(119896119892) = minusReal 119878 (119896119892) sdot sin (119896119892)
+ Im 119878 (119896119892) sdot cos (119896119892)
(53)
The two results are plotted in Figure 12 The peculiar spiralshape of the 119878 transfer function arises only when the gustfront is referenced to the midchord of the airfoil
The kinematics of the pitching and plunging airfoil ofa typical blade element is the resultant of a combination offorcing from collective and cyclic blade pitch twist angle
03
02
01
00
minus01
minus02
minus03
minus04
minus025 0 025 050 075 1
Imag
inar
y pa
rt
Sears function referenced to the midchordSears function referenced to the leading edge
Real part
Figure 12 Sears function
elastic torsion blade flapping velocity and elastic bendingAt low angles of attack with fully attached flow the varioussources of unsteady effects manifest primarily as moderateamplitude and phase variations relative to the quasi-steadyairloads At higher angles of attack when time-dependentflow separation from the airfoil may be involved the dynamicstall may occurThe amplitude and phase effects produced bythe stalled airloads can lead to various aeroelastic problemson the helicopter rotor that may limit its performance Theneed to control the aerodynamic forces on the rotor requiresthat the pitch of each blade be changed individually as theblades rotate about the shaft
8 Summary
The first flap frequency of a helicopter rigid blade is about105Ω then the reduced frequency at the 75 radius locationis 119896 = 07119888119877 For a blade aspect ratio 119877119888 gt 10 the reducedfrequency is in excess of 007 which is in the unsteadyrange In the case of the first elastic torsion mode which istypically about 3 divide 4Ω at the blade tip the reduced frequencyis in excess of 02 At these reduced frequencies there isa significant amplitude and phasing introduced into theairloads by the effects of the unsteady aerodynamics
When a wingrsquos angle of attack is increased rapidly it canmomentarily generate a higher maximum lift coefficient thanit could if the angle of attack was increased slowly This over-shoot can be related to the change in angle of attack duringthe time required for the air to travel one chord length Thedynamic overshoot is attributed to two effects (for the airfoilsthat stall first at the leading edge) the delay in the separationof the boundary layer and the momentary existence of avortex shed at the leading edge after the boundary layer doesseparate The delay in separation corresponds to the finite
Mathematical Problems in Engineering 9
time required for the aft edge of the separation bubble tomove forward to its bursting positionOn the other hands theairfoil can generate high lift as a result of a vortex that is shedat the leading edge at the instant of stall The vortex travelsback over the top of the airfoil carrying with it a low pressurewave that accounts for the very large lift coefficient Airfoilsthat stall first at the trailing edge also exhibit a dynamicovershoot but considerably less than those airfoils that haveleading edge stall
References
[1] J G Leishman Principles of Helicopter Aerodynamics Cam-bridge University Press 2007
[2] J Seddon and SNewmanBasicHelicopterAerodynamics AIAAEducation Series 2001
[3] M A Ashraf J Young and I C S Lai ldquoOscillation frequencyand amplitude effects on plunging airfoil propulsion and flowperiodicityrdquo The American Institute of Aeronautics and Astro-nautics Journal pp 2308ndash2324 2012
[4] R Prouty Helicopter Performance Stability and Control Kri-eger Gainesville Fla USA 2002
[5] P D Gareth Helicopter Flight Dynamics AIAA EducationSeries 2007
[6] B J Lee and M S Liou ldquoUnsteady adjoint approach for designoptimization of flapping airfoilsrdquo The American Institute ofAeronautics and Astronautics Journal pp 2460ndash2491 2012
[7] K Karamcheti Principles of Ideal-Fluid Aerodynamics KriegerGainesville Fla USA 1966
[8] T Theodorsen ldquoGeneral theory of aerodynamic instability andthe mechanism of flutterrdquo Tech Rep 496 NACA 1934
[9] I EGarrick ldquoOn some reciprocal relations in the theory of non-stationary flowsrdquo NACA Report 629 1938
[10] J Katz and A Plotkin Low Speed Aerodynamics CambridgeUniversity Press 2010
[11] C Rotaru I Circiu and M Boscoianu ldquoComputational meth-ods for the aerodynamic designrdquo Review of the Air Force Acad-emy vol 2 no 17 pp 43ndash48 2010
[12] C Rotaru and V Jaulin ldquoNumerical simulations of Helicopterrotor blade airloadsrdquoMTAReview vol 22 no 2 pp 81ndash90 2012
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2 Mathematical Problems in Engineering
Region ofreverse flow
Disk exceeding drag divergence Mach number
Vinfin
1205952
1205951
Figure 1 Helicopter rotor blade in forward flight
low pitching moment At high angles of attack the adversepressure gradients produced on the upper surface of theairfoil result in a progressive increase in the thickness of theboundary layer and cause some deviation from the linear liftversus angle of attack behavior On many airfoils the onsetof flow separation and stall occurs gradually with increasingangle of attack but on some airfoils (those with sharp leadingedges) the flow separation may occur suddenly
2 Helicopter Rotor Blade Aerodynamics
The region of the rotor disk affected by compressibility effectsis shown in Figure 1 and is defined on the surface wherethe incident Mach number of the flow that is normal to theleading edge of the blade exceeds the drag divergence Machnumber119872119889119889 If119872Ω119877 is the hover tip Mach number then theregion of the disk affected by compressibility effects is definedby
119872119903120595 = 119872Ω119877 (119903 + 120583 sin120595) ge 119872119889119889 (1)
The angular or rotational speed of the rotor is denotedby Ω the rotor radius by 119877 the advanced ratio by 120583 and theradial distance from the rotational axis by 119910 In the aboveequation the nondimensional quantities are 119903 = 119910119877 and120583 = 119881infin cos120572Ω119877 where 120572 is the angle between the forwardvelocity 119881infin and the plane of the rotor [1 2] The azimuthangle for the onset drag divergence 1205951 can be obtained bysetting 119903 = 1 so that
1205951 = arcsin([1120583(119872119889119889
119872Ω119877
minus 1)]) (2)
and 1205952 = 180 minus 1205951The increment in profile power Δ119862119875 associated with this
region on the disk is
Δ119862119875
120590=1
4120587int
1205952
1205951
int
1
119903119889119889
(119903 + 120583 sin120595)3Δ119862119889119903 119889119903 119889120595 (3)
where Δ119862119889 is the extra drag on the blade section when itexceeds the drag divergence Mach number 119872119889119889 and 120590 is
Tip vortexfilament
Ω
Px
120595b
120595w
120595 bminus 120595 w
y
yref
Figure 2 Tip vortex trajectory
rotor solidity coefficient which represents the ratio of theblades area to the rotor disk area For the NACA 0012 airfoilProuty [4] suggests that this can be approximated by
Δ119862119889 (119872) = 125(119872 minus 074)
3 for 119872 ge 074
0 otherwise(4)
The rotor limits may be determined by two conditions onecondition given by advancing blade compressibility effectsand the other condition given by retreating blade stall Ineither case the advancing blade operates at low angle of attack(AoA) but at high subsonic or transonic conditions whereasthe retreating blade operates at low Mach numbers and highlift coefficients
The aerodynamic behavior of airfoils in the high AoAregime is important for predicting the adverse effects pro-duced in the reverse flow regime on the rotor In the reverseflow region the direction of the relative flow vector changesfrom the trailing edge toward the leading edge of the airfoilWhile the fundamental process of the blade wake and tipvortex formation is similar to that found with a fixed wingone difference with helicopter tip vortices is that they arecurved and so they experience a self-induced effect Anothercomplication with helicopter rotors is that the wakes and tipvortices from other blades can lie close to each other and tothe plane of blade rotation and so they have large inducedeffects on the blade lift distribution
If the wake is assumed to be undistorted in the tip pathplane and no wake contraction occurs in the radial direction(Figure 2) then the tip vortex trajectories are described by theequations
119909 = 119877 cos (120595119887 minus 120595119908) + 119877120583120595119908
119910 = 119877 sin (120595119887 minus 120595119908) (5)
where 120595119887 is the position of the blade when the vortex wasformed and 120595119908 is the position of the vortex element relativeto the blade
One important parameter used in the description ofunsteady aerodynamics and unsteady airfoil behavior is thereduced frequency 119896 defined as 119896 = 120596 sdot 119888(2119881) where 120596 is
Mathematical Problems in Engineering 3
Ω middot R + Vinfin
Vinfin
Ω middot R
Ω middot R minus VinfinR
Figure 3 Helicopter main rotor
the angular frequency 119888 is the chord of the airfoil and 119881 isthe flow velocity [3] According to the dimensional analysisthe resultant force 119865 on the airfoil chord 119888 can be written infunctional form as 119865(12058811988121198882) = 119891(Re119872 119896) For 119896 = 0 theflow is steady and for 0 le 119896 le 005 the flow can be consideredquasi-steady that is unsteady effects are generally smallFlows with characteristic reduced frequencies above 005 areconsidered unsteady [5 6] For a helicopter rotor in forwardflight (Figure 3) the local sectional velocity which appearsin the denominator of the reduced frequency expression isconstantly changing
The approach to modeling of unsteady aerodynamiceffects through an extension of steady 2D thin airfoil theorygives a good level of analysis of the problem and providesconsiderable insight into the physics responsible for theunderlying unsteady behavior The Laplacersquos equation forincompressible flow is eliptic therefore the unsteady aero-dynamic theories cannot be obtained in a correspondinganalytical form
3 The Apparent Mass Tensor
The rate of change of the impulse vector in general is not inthe direction of the acceleration of the body The externalforce 119865119890 applied to the body to translate it through thefluid has to be applied in a direction different from that ofthe acceleration of the body through the fluid [7] Physicalconditions that should be satisfied on given boundaries of thefluid (boundary conditions) depend on the assumptionsmade with regard to the nature of the fluid more specificallyon the nature of the differential equations that are assumed togovern the motion of the fluid For a solid-fluid boundary ateach point of the solid-fluid surface at every instant thecomponent normal to the surface of the relative velocitybetween the fluid and the solid must be vanish sdot 119899 = 0where represents the relative velocity and 119899 the normal tothe surface (Figure 4)
If the surface is represented by a scalar function of posi-tion and time 119865( 119903 119905) = 0 then the total time rate of changeis zero
119863 (119865)
119863119905=120597 (119865)
120597119905+ sdot grad (119865) = 0 (6)
on 119865( 119903 119905) = 0
rarrn
rarrVdV
dS
Boundary
Figure 4 Solid-fluid surface
The fluid force acting on a rigid body of arbitrary shapetranslating with a velocity (119905) is given by
= minus∬119878
119901 119899 119889119878 (7)
where 119878 denotes the surface of the body and 119901 is the pressureon the surface of the body In general the body may be trans-lating rotating and deforming consequently the velocity 119880is a function of position on the surface and time If the body isrigid and is in translatorymotion then119880 is a function of timebut uniform over the surface of the body The mathematicalproblem is to determine the external force 119890 applied to thebody to translate it through the fluid According to Newtonrsquossecond law we have
119889
119889119905(119898) = 119890 + (8)
where119898 is the mass of the body The above equation may berewritten as
119890 =119889
119889119905(119898) minus (9)
or
119890 =119889
119889119905(119898 + 119868) (10)
where 119868 is the impulse applied on the fluid and minus119889 119868119889119905 =
The fluid force acting on the body is
=120597
120597119905∯119878
120588120601 119899 119889119878 minus 120588 times∯119878
119899 times grad (120601) 119889119878 (11)
The integral
119868119862 =∯119878
119899 times grad (120601) 119889119878 (12)
is related to the circulation 119862 around the body and 120601 is thevelocity potential
The unit vector 119890 is normal to the cutting planes the unitvector 1198901 is tangent to the curve of intersection between theblade element surface and the cutting plane and the unitvector 1198902 is tangent to the blade element surface
4 Mathematical Problems in Engineering
rarrn
rarre1
rarre2
rarre
dy
dS = dx middot dy
dx
C
Figure 5 Rotor blade element
The component of the vector 119868119862 in the direction 119890 is
119890 sdot 119868119862 =∯119878
119890 sdot 119899 times 119902119889119878 (13)
and the vector 119899 119889119878 corresponding to the surface element119889119878 = 119889119909 sdot 119889119910 may be written as 119899 119889119878 = 119889 times 119889 119910 where119889 = 119889119909 1198901 119889 119910 = 119889119910 1198902 and 119902 = grad(120601)
On the other hand
119899 119889119878 times 119902 = (119889 times 119889 119910) times 119902 = ( 119902 sdot 119889) 119889 119910 minus ( 119902 sdot 119889 119910) 119889
119890 sdot 119899 times 119902 119889119878 = 119890 ( 119902 sdot 119889) 119889 119910 minus 119890 ( 119902 sdot 119889 119910) 119889
(14)
Since the unit vectors 119890 and 1198901are normal it follows that 119890 sdot
119889 = 0 and
119890 sdot ( 119899 times 119902) 119889119878 = ( 119902 sdot 119889) ( 119890 sdot 119889 119910) (15)
The scalar product 119890 sdot 119889 119910 is the normal distance between thecutting planes of the solid body (Figure 5) If we denote 119889ℎ =119890 sdot 119889 119910 it follows that
119890 sdot∯119878
119899 times 119902119889119878 = int
ℎ2
ℎ1
(∮119862
119902 sdot 119889) 119889ℎ = int
ℎ2
ℎ1
Γ119890 (ℎ) 119889ℎ (16)
Here ℎ is the distance measured along the fixed direction 119890and
Γ119890 (ℎ) = ∮119862
119902 sdot 119889 (17)
is the circulation around the curve of intersection betweenthe body surface and the cutting plane The limits ℎ1 andℎ2 denote the extremities of the body measured along thedirection 119890 It follows that for motions without circulation theforce on the body is given by
=120597
120597119905(∯119878
120588120601 119899 119889119878) (18)
where the velocity potential 120601 is the solution of the system
nabla2120601 = 0
grad120601 sdot 119899 =120597120601
120597119899= 119880 (119905) 119899 on 119878
(19)
Since the equation and boundary condition for 120601 are linearthe solution could have the form
120601 = 1206011 + 1206012 + 1206013 (20)
where each of the functions 1206011 1206012 and 1206013 is a solution of theequations
nabla2120601119894 = 0
grad (120601119894) sdot 119899 =120597120601119894
120597119899= 119906119894119899119894 on 119878
(21)
where 119894 = 1 2 or 3In the Cartesian coordinate system the vectors and 119899
have the expressions
= 1199061119894 + 1199062
119895 + 1199063119896
119899 = 1198991119894 + 1198992
119895 + 1198993
(22)
Because time enters through 119906119894 it is convenient to set 120601119894 =119906119894120593119894 so the system (21) takes the form
nabla2120593119894 = 0
grad (120593119894) sdot 119899 =120597120593119894
120597119899= 119899119894 on 119878
(23)
With these considerations the impulse 119868 becomes
minus 119868 = ∯119878
120588120601 119899 119889119878 = ∯119878
120588(
3
sum
119896=1
119906119896120593119896) 119899 119889119878
=
3
sum
119896=1
(∯119878
120588120601119896 119899 119889119878) 119906119896
(24)
The components of the impulse 119868 = 1198681119894 + 1198682
119895 + 1198683 are
1198681 =119895 sdot 119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198991 119889119878) 119906119896
1198682 =119895 sdot 119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198992 119889119878) 119906119896
1198683 = sdot119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198993 119889119878) 119906119896
(25)
The surface integral in above equations may be written asfollows
∯119878
120588120593119896119899119894 119889119878 =∯119878
120588120593119896
120597120593119894
120597119899119889119878 (26)
According to Greenrsquos theorem if 1198911 and 1198912 are two harmonicfunctions then
∯119878
1198911
1205971198912
120597119899119889119878 =∯
119878
1198912
1205971198911
120597119899119889119878 (27)
Introducing the symbol119898119896119894
119898119896119894 = minus∯119878
120588120593119896
120597120593119894
120597119899119889119878 (28)
Mathematical Problems in Engineering 5
with119898119896119894 = 119898119894119896 the components of the impulse 119868 are thereforegiven by
119868119894 =
3
sum
119896=1
119898119894119896119906119896 119894 = 1 2 3 (29)
and the force applied externally to the body is
119890 =119889
119889119905119898 (1199061
119894 + 1199062119895 + 1199063)
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119894
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119895
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
(30)
or
119890 = [(119898 + 11989811)1198891199061
119889119905+ 11989812
1198891199062
119889119905+ 11989813
1198891199063
119889119905] 119894
+ [11989821
1198891199061
119889119905+ (119898 + 11989822)
1198891199062
119889119905+ 11989823
1198891199063
119889119905] 119895
+ [11989831
1198891199061
119889119905+ 11989832
1198891199062
119889119905+ (119898 + 11989833)
1198891199063
119889119905] 119896
(31)
The coefficients119898119894119896 form a set of nine numbers whichmay bedisplayed as an array
(
11989811 11989812 11989813
11989821 11989822 11989823
11989831 11989832 11989833
) (32)
and may be referred to as a virtual mass tensor or virtualmasses that need to be added to the mass of the body in orderto find the force that must be applied to translate it throughthe fluid Introducing the symbol 120575119894119896 defined by 120575119894119896 = 0 if 119894 = 119896
and 120575119894119896 = 1 for 119894 = 119896 (30) may be rewritten as
(119865119890)119894=
3
sum
119896=1
(119898120575119894119896 + 119898119894119896)119889119906119896
119889119905 (33)
For any body there are three perpendicular directions suchthat 119898119894119896 = 0 for 119894 = 119896 so with respect to such axes (33)becomes
(119865119890)119894= (119898 + 119898119894119894)
119889119906119894
119889119905 119894 = 1 2 3 (34)
The sum (119898+119898119894119894) represents the apparentmass for translationin the 119894-direction and the corresponding119898119894119894 is the additionalapparent mass
4 The Airloads on an Oscillating Airfoil
The oscillatory motion of the airfoil can be decomposed intocontributions associated with angle of attack which is equiv-alent to a pure plunging motion (Figure 6) and contributionsassociated with pitching (Figure 7)
Vh
h
x
w(x) = minus
Figure 6 Plunge velocity
V
x
y
w(x) = minus (x minus a)
bb
a
Figure 7 Pitch rate
A plunge velocity ℎproduces a uniformvelocity perturba-tion 119908 that is normal to the chord 119908(119909) = minusℎ and the pitch-rate term produces a linear variation in normal perturbationvelocity
For a pitch rate imposed about an axis at ldquo119886rdquo semichordsfrom the midchord then 119908(119909) = minus(119909 minus 119886) so that theinduced chamber is a parabolic arc
The problem of finding the airloads on an oscillatingairfoil was solved by Theodorsen who gave a solution to theunsteady airloads on a 2D harmonically oscillated airfoil ininviscid incompressible flow with the assumption of smalldisturbances [6] Both the airfoil and its shed wake wererepresented by a vortex sheet with the shedwake extending asa planar surface from the trailing edge downstream to infinityThe assumption of planar wake is justified if the angleof attack disturbances remain relatively small As with thestandard quasi-steady thin airfoil theory the bound vorticity120574119887 can sustain a pressure difference and therefore a liftforceThe wake vorticity 120574119908 must be force-free with zero netpressure jump over the sheet According to the Theodorsenrsquostheory the solution for the loading 120574119887 on the airfoil surfaceunder harmonic forcing conditions is obtained from integralequation
119908 (119909 119905) =1
2120587int
119888
0
120574119887 (119909 119905)
119909 minus 1199090
119889119909 +1
2120587int
infin
119888
120574119908 (119909 119905)
119909 minus 1199090
119889119909 (35)
6 Mathematical Problems in Engineering
where119908 is the downwash on the airfoil surface At the trailingedge 120574119887(119888 119905) = 0 and the airfoil circulation Γ(119905) is given by
Γ (119905) = int
119888
0
120574119887 (119909 119905) 119889119909 (36)
So long as the circulation about the airfoil is changing withrespect to time the circulation is continuously shed into thewake and will continuously affect the aerodynamic loads onthe airfoil
For a general motion where an airfoil of chord 119888 = 2119887 isundergoing a combination of pitching (120572 ) and plunging (ℎ)motion in a flow of steady velocity 119881 Theodorsenrsquos solution[8ndash10] for the lift coefficient and pitching moment coefficientcorresponding to midchord11987212 is
119888119897 = 120587119887 [ℎ
1198812+
119881minus119887
1198812119886]
+ 2120587[ℎ
119881+ 120572 +
119887
119881(1
2minus 119886)]119862 (119896)
11988811989812 =120587
2[119887119886ℎ
1198812minus1198872
1198812(1
8+ 1198862) ]
+120587(119886 + 12) [
ℎ
119881+ 120572 + 119887 (
1
2minus 119886)
119881]119862 (119896)
minus120587
2[(1
2minus 119886)
119887
119881]
(37)
where 119886 is the pitch axis location relative to the midchordof the airfoil measured in terms of semichord and 119862(119896) =119865(119896) + 119894119866(119896) is the complex transfer function (known asTheodorsenrsquos function) which accounts for the effects of theshed wake on the unsteady airloads
119862 (119896) =119867(2)
1(119896)
119867(2)
1(119896) + 119894 sdot 119867
(2)
0(119896)
=1198691 (1198691 + 1198690) + 1198841 (1198841 minus 1198690)
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
+ 11989411988411198840 + 11986911198690
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
(38)
with 1198690 1198691 1198840 and 1198841 being Bessel functions of the first andsecond kinds respectively (Figure 8)
The Hankel functions in above expression are
119867(2)
0= 1198690 minus 119894 sdot 1198840
119867(2)
1= 1198691 minus 119894 sdot 1198841
(39)
The real and imaginary parts of 119862(119896) function are plotted inFigure 9
It could be appreciated that 119862(119896) function serves tointroduce an amplitude reduction and phase lag effect on thecirculatory part of the lift response compared to the resultobtained under quasi-steady conditions [11]
1
05
05 10 15
J(0 k)
J(1 k)
Y(0 k)
Y(1 k)
minus1
minus05
Bessel
Bessel
Bessel
Bessel
Figure 8 Bessel functions
Real part of C(k)
Imag
inar
y pa
rt o
f C(k)
0 02 04 06 08 1
0
minus005
minus010
minus015
minus020
minus025
minus030
Figure 9 Theodorsenrsquos function
This effect can be seen if a pure oscillatory variation inangle of attack is considered that is 120572 = 120572119890
119894120596119905 so thecirculatory part of the airfoil lift coefficient is given by
119888119897 = 2120587120572119862 (119896) = 2120587120572 [119865 (119896) + 119894119866 (119896)] (40)
For 119896 = 0 the steady-state lift behavior is obtained that is 119888119897is linearly proportional to 120572 As 119896 is increased the lift plotsdevelop into hysteresis loops and these loops rotate such thatthe amplitude of the lift response (half of the peak-to-peakvalue) decreases with increasing reduced frequency
These loops are circumvented in a counterclockwisedirection such that the lift is lower than the steady value when
Mathematical Problems in Engineering 7
CirculatoryTotal response
Apparent mass
2
15
1
05
0
0 1 2 3 4
k
Nor
mal
ized
lift
ampl
itude
Figure 10 Normalized lift amplitude
120572 is decreasingwith time (ie there is a phase lag) For infinitereduced frequency the circulatory part of the lift amplitude ishalf that at 119896 = 0 and there is no phase lag angle
5 Pure Angle of Attack Oscillations
For a harmonic variation in 120572 that is 120572 = 120572119890119894120596119905 the lift is
119871 = 21205871205881198812119887 [119862 (119896) +
1
2119894120596119887
119881]120572119890119894120596119905 (41)
or in terms of the lift coefficient the result is
119888119897 =119871
1205881198812119887= [2120587 (119865 + 119894119866) + 119894120587119896] 120572119890
119894120596119905 (42)
The term inside the square brackets can be considered the lifttransfer function which accounts for the difference betweenthe unsteady and quasi-steady airloads
The first term inside the brackets is the circulatory termand the second term is the apparentmass contribution whichis proportional to the reduced frequency and leads the forcingby a phase angle of 1205872 The noncirculatory or apparent massterms arise from the nonsteady term of the velocity potentialand account for the pressure forces required to accelerate thefluid in the vicinity of the airfoil
The normalized lift amplitude is
119888119897
2120587120572= (119865 + 119894119866) + 119894
119896
2 (43)
The normalized lift amplitude and phase of lift for pure angleof attack oscillations are presented in Figures 10 and 11 wherethe significance of the apparentmass contribution to both theamplitude and phase can be appreciated
At lower values of reduced frequency the circulatoryterms dominate the solution At higher values of reducedfrequency the apparent mass forces dominate
100
80
60
40
20
0
minus20
Phas
e ang
le [d
egre
es]
CirculatoryTotal response
Apparent mass
0 1 2 3 4
k
Figure 11 Phase angle
6 Pure Plunging Oscillations
For a harmonic plunging motion such as that contributed byblade flapping the forcing is ℎ = ℎ119890119894120596119905 so that ℎ = 119894120596ℎ119890119894120596119905 andℎ = minus120596
2ℎ119890119894120596119905 Substituting into the expression for the lift and
solving for the lift coefficient it gives
119888119897 = [2120587119896 (119894119865 minus 119866) minus 1205871198962]ℎ
119887119890119894120596119905 (44)
The complete term inside the square brackets can be con-sidered as the lift transfer function [12] The circulatory partof the lift response leads the forcing displacement ℎ by aphase angle of 1205872 Also the apparent mass force leads thecirculatory part of the response by a phase angle of 1205872 orthe forcing by a phase angle of 120587 The corresponding pitchingmoment about midchord for this case is
11988811989812 = (120587
4) 1198962 ℎ
119887119890119894120596119905 (45)
7 Pitching Oscillations
For harmonic pitch oscillations additional terms involvingpitch rate appear in the equations for the aerodynamic res-ponse The forcing is given by 120572 = 120572119890119894120596119905 and the pitch rate by = 119894120596120572119890
119894120596119905 In this case the lift coefficient is
119888119897 = 2120587 [119865 (1 + 119894119896) + 119866 (119894 minus 119896)] 120572119890119894120596119905
+ 120587119896(119894 minus119896
2)120572119890119894120596119905
(46)
The lift amplitude initially decreases with increasing 119896 beca-use of the effects of the shed wake and then for 119896 gt 05it begins to increase as the apparent mass forces begin to
8 Mathematical Problems in Engineering
dominate the airloads This is also shown by the phase anglewhich exhibits an increasing lead for 119896 gt 03
Von Karman and Sear analyzed the problem of a thinairfoil moving through a sinusoidal vertical gust field wherethe gust can be considered as an upwash velocity that isuniformly convected by the free streamThe forcing functionin this case is
119908119892 (119909 119905) = sin(120596119892119905 minus120596119892119909
119881) (47)
where 120596119892 is the gust frequency If the gust is referenced tothe airfoil leading edge then 119909 = 0 and 119908119892(119909 119905) becomes119908119892(119909 119905) = sin(120596119892119905) and if the gust is referenced to themidchord then 119909 = 119887 = 1198882 and
120596119892119909
119881=
120596119892 (1198882)
119881=
120596119892119888
2119881= 119896119892
(48)
Therefore119908119892(119905) = sin(120596119892119905)sdotcos(119896119892)minussin(119896119892)sdotcos(120596119892119905) whichis equivalent to a phase shift In this case the lift coefficientcan be written as
119888119897 = 2120587(1199080
119881) 119878 (119896119892) (49)
where 119878(119896119892) is known as Sears function and the gust encoun-ter frequency 119896119892 is given by
119896119892 =2120587119881
120582119892
(50)
and 120582119892 is the wavelength of the gustIn terms of Bessel functions Searsrsquos function is given by
119878 (119896119892) = [1198690 (119896119892) minus 1198941198691 (119896119892)] 119862 (119896119892) + 1198941198691 (119896119892) (51)
The terms of real and imaginary parts are
Real 119878 (119896119892) = 119865 (119896119892) sdot 1198690 (119896119892) + 119866 (119896119892) sdot 1198691 (119896119892)
Im 119878 (119896119892) = 119866 (119896119892) sdot 1198690 (119896119892) minus 119865 (119896119892) sdot 1198691 (119896119892) + 1198691 (119896119892)
(52)
If the gust is referenced to the leading edge of the airfoil thefunction will be called 1198781015840 and can be written as
Real 1198781015840 (119896119892) = Real 119878 (119896119892) sdot cos (119896119892)
+ Im 119878 (119896119892) sdot sin (119896119892)
Im 1198781015840(119896119892) = minusReal 119878 (119896119892) sdot sin (119896119892)
+ Im 119878 (119896119892) sdot cos (119896119892)
(53)
The two results are plotted in Figure 12 The peculiar spiralshape of the 119878 transfer function arises only when the gustfront is referenced to the midchord of the airfoil
The kinematics of the pitching and plunging airfoil ofa typical blade element is the resultant of a combination offorcing from collective and cyclic blade pitch twist angle
03
02
01
00
minus01
minus02
minus03
minus04
minus025 0 025 050 075 1
Imag
inar
y pa
rt
Sears function referenced to the midchordSears function referenced to the leading edge
Real part
Figure 12 Sears function
elastic torsion blade flapping velocity and elastic bendingAt low angles of attack with fully attached flow the varioussources of unsteady effects manifest primarily as moderateamplitude and phase variations relative to the quasi-steadyairloads At higher angles of attack when time-dependentflow separation from the airfoil may be involved the dynamicstall may occurThe amplitude and phase effects produced bythe stalled airloads can lead to various aeroelastic problemson the helicopter rotor that may limit its performance Theneed to control the aerodynamic forces on the rotor requiresthat the pitch of each blade be changed individually as theblades rotate about the shaft
8 Summary
The first flap frequency of a helicopter rigid blade is about105Ω then the reduced frequency at the 75 radius locationis 119896 = 07119888119877 For a blade aspect ratio 119877119888 gt 10 the reducedfrequency is in excess of 007 which is in the unsteadyrange In the case of the first elastic torsion mode which istypically about 3 divide 4Ω at the blade tip the reduced frequencyis in excess of 02 At these reduced frequencies there isa significant amplitude and phasing introduced into theairloads by the effects of the unsteady aerodynamics
When a wingrsquos angle of attack is increased rapidly it canmomentarily generate a higher maximum lift coefficient thanit could if the angle of attack was increased slowly This over-shoot can be related to the change in angle of attack duringthe time required for the air to travel one chord length Thedynamic overshoot is attributed to two effects (for the airfoilsthat stall first at the leading edge) the delay in the separationof the boundary layer and the momentary existence of avortex shed at the leading edge after the boundary layer doesseparate The delay in separation corresponds to the finite
Mathematical Problems in Engineering 9
time required for the aft edge of the separation bubble tomove forward to its bursting positionOn the other hands theairfoil can generate high lift as a result of a vortex that is shedat the leading edge at the instant of stall The vortex travelsback over the top of the airfoil carrying with it a low pressurewave that accounts for the very large lift coefficient Airfoilsthat stall first at the trailing edge also exhibit a dynamicovershoot but considerably less than those airfoils that haveleading edge stall
References
[1] J G Leishman Principles of Helicopter Aerodynamics Cam-bridge University Press 2007
[2] J Seddon and SNewmanBasicHelicopterAerodynamics AIAAEducation Series 2001
[3] M A Ashraf J Young and I C S Lai ldquoOscillation frequencyand amplitude effects on plunging airfoil propulsion and flowperiodicityrdquo The American Institute of Aeronautics and Astro-nautics Journal pp 2308ndash2324 2012
[4] R Prouty Helicopter Performance Stability and Control Kri-eger Gainesville Fla USA 2002
[5] P D Gareth Helicopter Flight Dynamics AIAA EducationSeries 2007
[6] B J Lee and M S Liou ldquoUnsteady adjoint approach for designoptimization of flapping airfoilsrdquo The American Institute ofAeronautics and Astronautics Journal pp 2460ndash2491 2012
[7] K Karamcheti Principles of Ideal-Fluid Aerodynamics KriegerGainesville Fla USA 1966
[8] T Theodorsen ldquoGeneral theory of aerodynamic instability andthe mechanism of flutterrdquo Tech Rep 496 NACA 1934
[9] I EGarrick ldquoOn some reciprocal relations in the theory of non-stationary flowsrdquo NACA Report 629 1938
[10] J Katz and A Plotkin Low Speed Aerodynamics CambridgeUniversity Press 2010
[11] C Rotaru I Circiu and M Boscoianu ldquoComputational meth-ods for the aerodynamic designrdquo Review of the Air Force Acad-emy vol 2 no 17 pp 43ndash48 2010
[12] C Rotaru and V Jaulin ldquoNumerical simulations of Helicopterrotor blade airloadsrdquoMTAReview vol 22 no 2 pp 81ndash90 2012
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 3
Ω middot R + Vinfin
Vinfin
Ω middot R
Ω middot R minus VinfinR
Figure 3 Helicopter main rotor
the angular frequency 119888 is the chord of the airfoil and 119881 isthe flow velocity [3] According to the dimensional analysisthe resultant force 119865 on the airfoil chord 119888 can be written infunctional form as 119865(12058811988121198882) = 119891(Re119872 119896) For 119896 = 0 theflow is steady and for 0 le 119896 le 005 the flow can be consideredquasi-steady that is unsteady effects are generally smallFlows with characteristic reduced frequencies above 005 areconsidered unsteady [5 6] For a helicopter rotor in forwardflight (Figure 3) the local sectional velocity which appearsin the denominator of the reduced frequency expression isconstantly changing
The approach to modeling of unsteady aerodynamiceffects through an extension of steady 2D thin airfoil theorygives a good level of analysis of the problem and providesconsiderable insight into the physics responsible for theunderlying unsteady behavior The Laplacersquos equation forincompressible flow is eliptic therefore the unsteady aero-dynamic theories cannot be obtained in a correspondinganalytical form
3 The Apparent Mass Tensor
The rate of change of the impulse vector in general is not inthe direction of the acceleration of the body The externalforce 119865119890 applied to the body to translate it through thefluid has to be applied in a direction different from that ofthe acceleration of the body through the fluid [7] Physicalconditions that should be satisfied on given boundaries of thefluid (boundary conditions) depend on the assumptionsmade with regard to the nature of the fluid more specificallyon the nature of the differential equations that are assumed togovern the motion of the fluid For a solid-fluid boundary ateach point of the solid-fluid surface at every instant thecomponent normal to the surface of the relative velocitybetween the fluid and the solid must be vanish sdot 119899 = 0where represents the relative velocity and 119899 the normal tothe surface (Figure 4)
If the surface is represented by a scalar function of posi-tion and time 119865( 119903 119905) = 0 then the total time rate of changeis zero
119863 (119865)
119863119905=120597 (119865)
120597119905+ sdot grad (119865) = 0 (6)
on 119865( 119903 119905) = 0
rarrn
rarrVdV
dS
Boundary
Figure 4 Solid-fluid surface
The fluid force acting on a rigid body of arbitrary shapetranslating with a velocity (119905) is given by
= minus∬119878
119901 119899 119889119878 (7)
where 119878 denotes the surface of the body and 119901 is the pressureon the surface of the body In general the body may be trans-lating rotating and deforming consequently the velocity 119880is a function of position on the surface and time If the body isrigid and is in translatorymotion then119880 is a function of timebut uniform over the surface of the body The mathematicalproblem is to determine the external force 119890 applied to thebody to translate it through the fluid According to Newtonrsquossecond law we have
119889
119889119905(119898) = 119890 + (8)
where119898 is the mass of the body The above equation may berewritten as
119890 =119889
119889119905(119898) minus (9)
or
119890 =119889
119889119905(119898 + 119868) (10)
where 119868 is the impulse applied on the fluid and minus119889 119868119889119905 =
The fluid force acting on the body is
=120597
120597119905∯119878
120588120601 119899 119889119878 minus 120588 times∯119878
119899 times grad (120601) 119889119878 (11)
The integral
119868119862 =∯119878
119899 times grad (120601) 119889119878 (12)
is related to the circulation 119862 around the body and 120601 is thevelocity potential
The unit vector 119890 is normal to the cutting planes the unitvector 1198901 is tangent to the curve of intersection between theblade element surface and the cutting plane and the unitvector 1198902 is tangent to the blade element surface
4 Mathematical Problems in Engineering
rarrn
rarre1
rarre2
rarre
dy
dS = dx middot dy
dx
C
Figure 5 Rotor blade element
The component of the vector 119868119862 in the direction 119890 is
119890 sdot 119868119862 =∯119878
119890 sdot 119899 times 119902119889119878 (13)
and the vector 119899 119889119878 corresponding to the surface element119889119878 = 119889119909 sdot 119889119910 may be written as 119899 119889119878 = 119889 times 119889 119910 where119889 = 119889119909 1198901 119889 119910 = 119889119910 1198902 and 119902 = grad(120601)
On the other hand
119899 119889119878 times 119902 = (119889 times 119889 119910) times 119902 = ( 119902 sdot 119889) 119889 119910 minus ( 119902 sdot 119889 119910) 119889
119890 sdot 119899 times 119902 119889119878 = 119890 ( 119902 sdot 119889) 119889 119910 minus 119890 ( 119902 sdot 119889 119910) 119889
(14)
Since the unit vectors 119890 and 1198901are normal it follows that 119890 sdot
119889 = 0 and
119890 sdot ( 119899 times 119902) 119889119878 = ( 119902 sdot 119889) ( 119890 sdot 119889 119910) (15)
The scalar product 119890 sdot 119889 119910 is the normal distance between thecutting planes of the solid body (Figure 5) If we denote 119889ℎ =119890 sdot 119889 119910 it follows that
119890 sdot∯119878
119899 times 119902119889119878 = int
ℎ2
ℎ1
(∮119862
119902 sdot 119889) 119889ℎ = int
ℎ2
ℎ1
Γ119890 (ℎ) 119889ℎ (16)
Here ℎ is the distance measured along the fixed direction 119890and
Γ119890 (ℎ) = ∮119862
119902 sdot 119889 (17)
is the circulation around the curve of intersection betweenthe body surface and the cutting plane The limits ℎ1 andℎ2 denote the extremities of the body measured along thedirection 119890 It follows that for motions without circulation theforce on the body is given by
=120597
120597119905(∯119878
120588120601 119899 119889119878) (18)
where the velocity potential 120601 is the solution of the system
nabla2120601 = 0
grad120601 sdot 119899 =120597120601
120597119899= 119880 (119905) 119899 on 119878
(19)
Since the equation and boundary condition for 120601 are linearthe solution could have the form
120601 = 1206011 + 1206012 + 1206013 (20)
where each of the functions 1206011 1206012 and 1206013 is a solution of theequations
nabla2120601119894 = 0
grad (120601119894) sdot 119899 =120597120601119894
120597119899= 119906119894119899119894 on 119878
(21)
where 119894 = 1 2 or 3In the Cartesian coordinate system the vectors and 119899
have the expressions
= 1199061119894 + 1199062
119895 + 1199063119896
119899 = 1198991119894 + 1198992
119895 + 1198993
(22)
Because time enters through 119906119894 it is convenient to set 120601119894 =119906119894120593119894 so the system (21) takes the form
nabla2120593119894 = 0
grad (120593119894) sdot 119899 =120597120593119894
120597119899= 119899119894 on 119878
(23)
With these considerations the impulse 119868 becomes
minus 119868 = ∯119878
120588120601 119899 119889119878 = ∯119878
120588(
3
sum
119896=1
119906119896120593119896) 119899 119889119878
=
3
sum
119896=1
(∯119878
120588120601119896 119899 119889119878) 119906119896
(24)
The components of the impulse 119868 = 1198681119894 + 1198682
119895 + 1198683 are
1198681 =119895 sdot 119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198991 119889119878) 119906119896
1198682 =119895 sdot 119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198992 119889119878) 119906119896
1198683 = sdot119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198993 119889119878) 119906119896
(25)
The surface integral in above equations may be written asfollows
∯119878
120588120593119896119899119894 119889119878 =∯119878
120588120593119896
120597120593119894
120597119899119889119878 (26)
According to Greenrsquos theorem if 1198911 and 1198912 are two harmonicfunctions then
∯119878
1198911
1205971198912
120597119899119889119878 =∯
119878
1198912
1205971198911
120597119899119889119878 (27)
Introducing the symbol119898119896119894
119898119896119894 = minus∯119878
120588120593119896
120597120593119894
120597119899119889119878 (28)
Mathematical Problems in Engineering 5
with119898119896119894 = 119898119894119896 the components of the impulse 119868 are thereforegiven by
119868119894 =
3
sum
119896=1
119898119894119896119906119896 119894 = 1 2 3 (29)
and the force applied externally to the body is
119890 =119889
119889119905119898 (1199061
119894 + 1199062119895 + 1199063)
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119894
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119895
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
(30)
or
119890 = [(119898 + 11989811)1198891199061
119889119905+ 11989812
1198891199062
119889119905+ 11989813
1198891199063
119889119905] 119894
+ [11989821
1198891199061
119889119905+ (119898 + 11989822)
1198891199062
119889119905+ 11989823
1198891199063
119889119905] 119895
+ [11989831
1198891199061
119889119905+ 11989832
1198891199062
119889119905+ (119898 + 11989833)
1198891199063
119889119905] 119896
(31)
The coefficients119898119894119896 form a set of nine numbers whichmay bedisplayed as an array
(
11989811 11989812 11989813
11989821 11989822 11989823
11989831 11989832 11989833
) (32)
and may be referred to as a virtual mass tensor or virtualmasses that need to be added to the mass of the body in orderto find the force that must be applied to translate it throughthe fluid Introducing the symbol 120575119894119896 defined by 120575119894119896 = 0 if 119894 = 119896
and 120575119894119896 = 1 for 119894 = 119896 (30) may be rewritten as
(119865119890)119894=
3
sum
119896=1
(119898120575119894119896 + 119898119894119896)119889119906119896
119889119905 (33)
For any body there are three perpendicular directions suchthat 119898119894119896 = 0 for 119894 = 119896 so with respect to such axes (33)becomes
(119865119890)119894= (119898 + 119898119894119894)
119889119906119894
119889119905 119894 = 1 2 3 (34)
The sum (119898+119898119894119894) represents the apparentmass for translationin the 119894-direction and the corresponding119898119894119894 is the additionalapparent mass
4 The Airloads on an Oscillating Airfoil
The oscillatory motion of the airfoil can be decomposed intocontributions associated with angle of attack which is equiv-alent to a pure plunging motion (Figure 6) and contributionsassociated with pitching (Figure 7)
Vh
h
x
w(x) = minus
Figure 6 Plunge velocity
V
x
y
w(x) = minus (x minus a)
bb
a
Figure 7 Pitch rate
A plunge velocity ℎproduces a uniformvelocity perturba-tion 119908 that is normal to the chord 119908(119909) = minusℎ and the pitch-rate term produces a linear variation in normal perturbationvelocity
For a pitch rate imposed about an axis at ldquo119886rdquo semichordsfrom the midchord then 119908(119909) = minus(119909 minus 119886) so that theinduced chamber is a parabolic arc
The problem of finding the airloads on an oscillatingairfoil was solved by Theodorsen who gave a solution to theunsteady airloads on a 2D harmonically oscillated airfoil ininviscid incompressible flow with the assumption of smalldisturbances [6] Both the airfoil and its shed wake wererepresented by a vortex sheet with the shedwake extending asa planar surface from the trailing edge downstream to infinityThe assumption of planar wake is justified if the angleof attack disturbances remain relatively small As with thestandard quasi-steady thin airfoil theory the bound vorticity120574119887 can sustain a pressure difference and therefore a liftforceThe wake vorticity 120574119908 must be force-free with zero netpressure jump over the sheet According to the Theodorsenrsquostheory the solution for the loading 120574119887 on the airfoil surfaceunder harmonic forcing conditions is obtained from integralequation
119908 (119909 119905) =1
2120587int
119888
0
120574119887 (119909 119905)
119909 minus 1199090
119889119909 +1
2120587int
infin
119888
120574119908 (119909 119905)
119909 minus 1199090
119889119909 (35)
6 Mathematical Problems in Engineering
where119908 is the downwash on the airfoil surface At the trailingedge 120574119887(119888 119905) = 0 and the airfoil circulation Γ(119905) is given by
Γ (119905) = int
119888
0
120574119887 (119909 119905) 119889119909 (36)
So long as the circulation about the airfoil is changing withrespect to time the circulation is continuously shed into thewake and will continuously affect the aerodynamic loads onthe airfoil
For a general motion where an airfoil of chord 119888 = 2119887 isundergoing a combination of pitching (120572 ) and plunging (ℎ)motion in a flow of steady velocity 119881 Theodorsenrsquos solution[8ndash10] for the lift coefficient and pitching moment coefficientcorresponding to midchord11987212 is
119888119897 = 120587119887 [ℎ
1198812+
119881minus119887
1198812119886]
+ 2120587[ℎ
119881+ 120572 +
119887
119881(1
2minus 119886)]119862 (119896)
11988811989812 =120587
2[119887119886ℎ
1198812minus1198872
1198812(1
8+ 1198862) ]
+120587(119886 + 12) [
ℎ
119881+ 120572 + 119887 (
1
2minus 119886)
119881]119862 (119896)
minus120587
2[(1
2minus 119886)
119887
119881]
(37)
where 119886 is the pitch axis location relative to the midchordof the airfoil measured in terms of semichord and 119862(119896) =119865(119896) + 119894119866(119896) is the complex transfer function (known asTheodorsenrsquos function) which accounts for the effects of theshed wake on the unsteady airloads
119862 (119896) =119867(2)
1(119896)
119867(2)
1(119896) + 119894 sdot 119867
(2)
0(119896)
=1198691 (1198691 + 1198690) + 1198841 (1198841 minus 1198690)
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
+ 11989411988411198840 + 11986911198690
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
(38)
with 1198690 1198691 1198840 and 1198841 being Bessel functions of the first andsecond kinds respectively (Figure 8)
The Hankel functions in above expression are
119867(2)
0= 1198690 minus 119894 sdot 1198840
119867(2)
1= 1198691 minus 119894 sdot 1198841
(39)
The real and imaginary parts of 119862(119896) function are plotted inFigure 9
It could be appreciated that 119862(119896) function serves tointroduce an amplitude reduction and phase lag effect on thecirculatory part of the lift response compared to the resultobtained under quasi-steady conditions [11]
1
05
05 10 15
J(0 k)
J(1 k)
Y(0 k)
Y(1 k)
minus1
minus05
Bessel
Bessel
Bessel
Bessel
Figure 8 Bessel functions
Real part of C(k)
Imag
inar
y pa
rt o
f C(k)
0 02 04 06 08 1
0
minus005
minus010
minus015
minus020
minus025
minus030
Figure 9 Theodorsenrsquos function
This effect can be seen if a pure oscillatory variation inangle of attack is considered that is 120572 = 120572119890
119894120596119905 so thecirculatory part of the airfoil lift coefficient is given by
119888119897 = 2120587120572119862 (119896) = 2120587120572 [119865 (119896) + 119894119866 (119896)] (40)
For 119896 = 0 the steady-state lift behavior is obtained that is 119888119897is linearly proportional to 120572 As 119896 is increased the lift plotsdevelop into hysteresis loops and these loops rotate such thatthe amplitude of the lift response (half of the peak-to-peakvalue) decreases with increasing reduced frequency
These loops are circumvented in a counterclockwisedirection such that the lift is lower than the steady value when
Mathematical Problems in Engineering 7
CirculatoryTotal response
Apparent mass
2
15
1
05
0
0 1 2 3 4
k
Nor
mal
ized
lift
ampl
itude
Figure 10 Normalized lift amplitude
120572 is decreasingwith time (ie there is a phase lag) For infinitereduced frequency the circulatory part of the lift amplitude ishalf that at 119896 = 0 and there is no phase lag angle
5 Pure Angle of Attack Oscillations
For a harmonic variation in 120572 that is 120572 = 120572119890119894120596119905 the lift is
119871 = 21205871205881198812119887 [119862 (119896) +
1
2119894120596119887
119881]120572119890119894120596119905 (41)
or in terms of the lift coefficient the result is
119888119897 =119871
1205881198812119887= [2120587 (119865 + 119894119866) + 119894120587119896] 120572119890
119894120596119905 (42)
The term inside the square brackets can be considered the lifttransfer function which accounts for the difference betweenthe unsteady and quasi-steady airloads
The first term inside the brackets is the circulatory termand the second term is the apparentmass contribution whichis proportional to the reduced frequency and leads the forcingby a phase angle of 1205872 The noncirculatory or apparent massterms arise from the nonsteady term of the velocity potentialand account for the pressure forces required to accelerate thefluid in the vicinity of the airfoil
The normalized lift amplitude is
119888119897
2120587120572= (119865 + 119894119866) + 119894
119896
2 (43)
The normalized lift amplitude and phase of lift for pure angleof attack oscillations are presented in Figures 10 and 11 wherethe significance of the apparentmass contribution to both theamplitude and phase can be appreciated
At lower values of reduced frequency the circulatoryterms dominate the solution At higher values of reducedfrequency the apparent mass forces dominate
100
80
60
40
20
0
minus20
Phas
e ang
le [d
egre
es]
CirculatoryTotal response
Apparent mass
0 1 2 3 4
k
Figure 11 Phase angle
6 Pure Plunging Oscillations
For a harmonic plunging motion such as that contributed byblade flapping the forcing is ℎ = ℎ119890119894120596119905 so that ℎ = 119894120596ℎ119890119894120596119905 andℎ = minus120596
2ℎ119890119894120596119905 Substituting into the expression for the lift and
solving for the lift coefficient it gives
119888119897 = [2120587119896 (119894119865 minus 119866) minus 1205871198962]ℎ
119887119890119894120596119905 (44)
The complete term inside the square brackets can be con-sidered as the lift transfer function [12] The circulatory partof the lift response leads the forcing displacement ℎ by aphase angle of 1205872 Also the apparent mass force leads thecirculatory part of the response by a phase angle of 1205872 orthe forcing by a phase angle of 120587 The corresponding pitchingmoment about midchord for this case is
11988811989812 = (120587
4) 1198962 ℎ
119887119890119894120596119905 (45)
7 Pitching Oscillations
For harmonic pitch oscillations additional terms involvingpitch rate appear in the equations for the aerodynamic res-ponse The forcing is given by 120572 = 120572119890119894120596119905 and the pitch rate by = 119894120596120572119890
119894120596119905 In this case the lift coefficient is
119888119897 = 2120587 [119865 (1 + 119894119896) + 119866 (119894 minus 119896)] 120572119890119894120596119905
+ 120587119896(119894 minus119896
2)120572119890119894120596119905
(46)
The lift amplitude initially decreases with increasing 119896 beca-use of the effects of the shed wake and then for 119896 gt 05it begins to increase as the apparent mass forces begin to
8 Mathematical Problems in Engineering
dominate the airloads This is also shown by the phase anglewhich exhibits an increasing lead for 119896 gt 03
Von Karman and Sear analyzed the problem of a thinairfoil moving through a sinusoidal vertical gust field wherethe gust can be considered as an upwash velocity that isuniformly convected by the free streamThe forcing functionin this case is
119908119892 (119909 119905) = sin(120596119892119905 minus120596119892119909
119881) (47)
where 120596119892 is the gust frequency If the gust is referenced tothe airfoil leading edge then 119909 = 0 and 119908119892(119909 119905) becomes119908119892(119909 119905) = sin(120596119892119905) and if the gust is referenced to themidchord then 119909 = 119887 = 1198882 and
120596119892119909
119881=
120596119892 (1198882)
119881=
120596119892119888
2119881= 119896119892
(48)
Therefore119908119892(119905) = sin(120596119892119905)sdotcos(119896119892)minussin(119896119892)sdotcos(120596119892119905) whichis equivalent to a phase shift In this case the lift coefficientcan be written as
119888119897 = 2120587(1199080
119881) 119878 (119896119892) (49)
where 119878(119896119892) is known as Sears function and the gust encoun-ter frequency 119896119892 is given by
119896119892 =2120587119881
120582119892
(50)
and 120582119892 is the wavelength of the gustIn terms of Bessel functions Searsrsquos function is given by
119878 (119896119892) = [1198690 (119896119892) minus 1198941198691 (119896119892)] 119862 (119896119892) + 1198941198691 (119896119892) (51)
The terms of real and imaginary parts are
Real 119878 (119896119892) = 119865 (119896119892) sdot 1198690 (119896119892) + 119866 (119896119892) sdot 1198691 (119896119892)
Im 119878 (119896119892) = 119866 (119896119892) sdot 1198690 (119896119892) minus 119865 (119896119892) sdot 1198691 (119896119892) + 1198691 (119896119892)
(52)
If the gust is referenced to the leading edge of the airfoil thefunction will be called 1198781015840 and can be written as
Real 1198781015840 (119896119892) = Real 119878 (119896119892) sdot cos (119896119892)
+ Im 119878 (119896119892) sdot sin (119896119892)
Im 1198781015840(119896119892) = minusReal 119878 (119896119892) sdot sin (119896119892)
+ Im 119878 (119896119892) sdot cos (119896119892)
(53)
The two results are plotted in Figure 12 The peculiar spiralshape of the 119878 transfer function arises only when the gustfront is referenced to the midchord of the airfoil
The kinematics of the pitching and plunging airfoil ofa typical blade element is the resultant of a combination offorcing from collective and cyclic blade pitch twist angle
03
02
01
00
minus01
minus02
minus03
minus04
minus025 0 025 050 075 1
Imag
inar
y pa
rt
Sears function referenced to the midchordSears function referenced to the leading edge
Real part
Figure 12 Sears function
elastic torsion blade flapping velocity and elastic bendingAt low angles of attack with fully attached flow the varioussources of unsteady effects manifest primarily as moderateamplitude and phase variations relative to the quasi-steadyairloads At higher angles of attack when time-dependentflow separation from the airfoil may be involved the dynamicstall may occurThe amplitude and phase effects produced bythe stalled airloads can lead to various aeroelastic problemson the helicopter rotor that may limit its performance Theneed to control the aerodynamic forces on the rotor requiresthat the pitch of each blade be changed individually as theblades rotate about the shaft
8 Summary
The first flap frequency of a helicopter rigid blade is about105Ω then the reduced frequency at the 75 radius locationis 119896 = 07119888119877 For a blade aspect ratio 119877119888 gt 10 the reducedfrequency is in excess of 007 which is in the unsteadyrange In the case of the first elastic torsion mode which istypically about 3 divide 4Ω at the blade tip the reduced frequencyis in excess of 02 At these reduced frequencies there isa significant amplitude and phasing introduced into theairloads by the effects of the unsteady aerodynamics
When a wingrsquos angle of attack is increased rapidly it canmomentarily generate a higher maximum lift coefficient thanit could if the angle of attack was increased slowly This over-shoot can be related to the change in angle of attack duringthe time required for the air to travel one chord length Thedynamic overshoot is attributed to two effects (for the airfoilsthat stall first at the leading edge) the delay in the separationof the boundary layer and the momentary existence of avortex shed at the leading edge after the boundary layer doesseparate The delay in separation corresponds to the finite
Mathematical Problems in Engineering 9
time required for the aft edge of the separation bubble tomove forward to its bursting positionOn the other hands theairfoil can generate high lift as a result of a vortex that is shedat the leading edge at the instant of stall The vortex travelsback over the top of the airfoil carrying with it a low pressurewave that accounts for the very large lift coefficient Airfoilsthat stall first at the trailing edge also exhibit a dynamicovershoot but considerably less than those airfoils that haveleading edge stall
References
[1] J G Leishman Principles of Helicopter Aerodynamics Cam-bridge University Press 2007
[2] J Seddon and SNewmanBasicHelicopterAerodynamics AIAAEducation Series 2001
[3] M A Ashraf J Young and I C S Lai ldquoOscillation frequencyand amplitude effects on plunging airfoil propulsion and flowperiodicityrdquo The American Institute of Aeronautics and Astro-nautics Journal pp 2308ndash2324 2012
[4] R Prouty Helicopter Performance Stability and Control Kri-eger Gainesville Fla USA 2002
[5] P D Gareth Helicopter Flight Dynamics AIAA EducationSeries 2007
[6] B J Lee and M S Liou ldquoUnsteady adjoint approach for designoptimization of flapping airfoilsrdquo The American Institute ofAeronautics and Astronautics Journal pp 2460ndash2491 2012
[7] K Karamcheti Principles of Ideal-Fluid Aerodynamics KriegerGainesville Fla USA 1966
[8] T Theodorsen ldquoGeneral theory of aerodynamic instability andthe mechanism of flutterrdquo Tech Rep 496 NACA 1934
[9] I EGarrick ldquoOn some reciprocal relations in the theory of non-stationary flowsrdquo NACA Report 629 1938
[10] J Katz and A Plotkin Low Speed Aerodynamics CambridgeUniversity Press 2010
[11] C Rotaru I Circiu and M Boscoianu ldquoComputational meth-ods for the aerodynamic designrdquo Review of the Air Force Acad-emy vol 2 no 17 pp 43ndash48 2010
[12] C Rotaru and V Jaulin ldquoNumerical simulations of Helicopterrotor blade airloadsrdquoMTAReview vol 22 no 2 pp 81ndash90 2012
Submit your manuscripts athttpwwwhindawicom
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Stochastic AnalysisInternational Journal of
4 Mathematical Problems in Engineering
rarrn
rarre1
rarre2
rarre
dy
dS = dx middot dy
dx
C
Figure 5 Rotor blade element
The component of the vector 119868119862 in the direction 119890 is
119890 sdot 119868119862 =∯119878
119890 sdot 119899 times 119902119889119878 (13)
and the vector 119899 119889119878 corresponding to the surface element119889119878 = 119889119909 sdot 119889119910 may be written as 119899 119889119878 = 119889 times 119889 119910 where119889 = 119889119909 1198901 119889 119910 = 119889119910 1198902 and 119902 = grad(120601)
On the other hand
119899 119889119878 times 119902 = (119889 times 119889 119910) times 119902 = ( 119902 sdot 119889) 119889 119910 minus ( 119902 sdot 119889 119910) 119889
119890 sdot 119899 times 119902 119889119878 = 119890 ( 119902 sdot 119889) 119889 119910 minus 119890 ( 119902 sdot 119889 119910) 119889
(14)
Since the unit vectors 119890 and 1198901are normal it follows that 119890 sdot
119889 = 0 and
119890 sdot ( 119899 times 119902) 119889119878 = ( 119902 sdot 119889) ( 119890 sdot 119889 119910) (15)
The scalar product 119890 sdot 119889 119910 is the normal distance between thecutting planes of the solid body (Figure 5) If we denote 119889ℎ =119890 sdot 119889 119910 it follows that
119890 sdot∯119878
119899 times 119902119889119878 = int
ℎ2
ℎ1
(∮119862
119902 sdot 119889) 119889ℎ = int
ℎ2
ℎ1
Γ119890 (ℎ) 119889ℎ (16)
Here ℎ is the distance measured along the fixed direction 119890and
Γ119890 (ℎ) = ∮119862
119902 sdot 119889 (17)
is the circulation around the curve of intersection betweenthe body surface and the cutting plane The limits ℎ1 andℎ2 denote the extremities of the body measured along thedirection 119890 It follows that for motions without circulation theforce on the body is given by
=120597
120597119905(∯119878
120588120601 119899 119889119878) (18)
where the velocity potential 120601 is the solution of the system
nabla2120601 = 0
grad120601 sdot 119899 =120597120601
120597119899= 119880 (119905) 119899 on 119878
(19)
Since the equation and boundary condition for 120601 are linearthe solution could have the form
120601 = 1206011 + 1206012 + 1206013 (20)
where each of the functions 1206011 1206012 and 1206013 is a solution of theequations
nabla2120601119894 = 0
grad (120601119894) sdot 119899 =120597120601119894
120597119899= 119906119894119899119894 on 119878
(21)
where 119894 = 1 2 or 3In the Cartesian coordinate system the vectors and 119899
have the expressions
= 1199061119894 + 1199062
119895 + 1199063119896
119899 = 1198991119894 + 1198992
119895 + 1198993
(22)
Because time enters through 119906119894 it is convenient to set 120601119894 =119906119894120593119894 so the system (21) takes the form
nabla2120593119894 = 0
grad (120593119894) sdot 119899 =120597120593119894
120597119899= 119899119894 on 119878
(23)
With these considerations the impulse 119868 becomes
minus 119868 = ∯119878
120588120601 119899 119889119878 = ∯119878
120588(
3
sum
119896=1
119906119896120593119896) 119899 119889119878
=
3
sum
119896=1
(∯119878
120588120601119896 119899 119889119878) 119906119896
(24)
The components of the impulse 119868 = 1198681119894 + 1198682
119895 + 1198683 are
1198681 =119895 sdot 119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198991 119889119878) 119906119896
1198682 =119895 sdot 119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198992 119889119878) 119906119896
1198683 = sdot119868 =
3
sum
119896=1
(minus∯119878
1205881205931198961198993 119889119878) 119906119896
(25)
The surface integral in above equations may be written asfollows
∯119878
120588120593119896119899119894 119889119878 =∯119878
120588120593119896
120597120593119894
120597119899119889119878 (26)
According to Greenrsquos theorem if 1198911 and 1198912 are two harmonicfunctions then
∯119878
1198911
1205971198912
120597119899119889119878 =∯
119878
1198912
1205971198911
120597119899119889119878 (27)
Introducing the symbol119898119896119894
119898119896119894 = minus∯119878
120588120593119896
120597120593119894
120597119899119889119878 (28)
Mathematical Problems in Engineering 5
with119898119896119894 = 119898119894119896 the components of the impulse 119868 are thereforegiven by
119868119894 =
3
sum
119896=1
119898119894119896119906119896 119894 = 1 2 3 (29)
and the force applied externally to the body is
119890 =119889
119889119905119898 (1199061
119894 + 1199062119895 + 1199063)
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119894
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119895
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
(30)
or
119890 = [(119898 + 11989811)1198891199061
119889119905+ 11989812
1198891199062
119889119905+ 11989813
1198891199063
119889119905] 119894
+ [11989821
1198891199061
119889119905+ (119898 + 11989822)
1198891199062
119889119905+ 11989823
1198891199063
119889119905] 119895
+ [11989831
1198891199061
119889119905+ 11989832
1198891199062
119889119905+ (119898 + 11989833)
1198891199063
119889119905] 119896
(31)
The coefficients119898119894119896 form a set of nine numbers whichmay bedisplayed as an array
(
11989811 11989812 11989813
11989821 11989822 11989823
11989831 11989832 11989833
) (32)
and may be referred to as a virtual mass tensor or virtualmasses that need to be added to the mass of the body in orderto find the force that must be applied to translate it throughthe fluid Introducing the symbol 120575119894119896 defined by 120575119894119896 = 0 if 119894 = 119896
and 120575119894119896 = 1 for 119894 = 119896 (30) may be rewritten as
(119865119890)119894=
3
sum
119896=1
(119898120575119894119896 + 119898119894119896)119889119906119896
119889119905 (33)
For any body there are three perpendicular directions suchthat 119898119894119896 = 0 for 119894 = 119896 so with respect to such axes (33)becomes
(119865119890)119894= (119898 + 119898119894119894)
119889119906119894
119889119905 119894 = 1 2 3 (34)
The sum (119898+119898119894119894) represents the apparentmass for translationin the 119894-direction and the corresponding119898119894119894 is the additionalapparent mass
4 The Airloads on an Oscillating Airfoil
The oscillatory motion of the airfoil can be decomposed intocontributions associated with angle of attack which is equiv-alent to a pure plunging motion (Figure 6) and contributionsassociated with pitching (Figure 7)
Vh
h
x
w(x) = minus
Figure 6 Plunge velocity
V
x
y
w(x) = minus (x minus a)
bb
a
Figure 7 Pitch rate
A plunge velocity ℎproduces a uniformvelocity perturba-tion 119908 that is normal to the chord 119908(119909) = minusℎ and the pitch-rate term produces a linear variation in normal perturbationvelocity
For a pitch rate imposed about an axis at ldquo119886rdquo semichordsfrom the midchord then 119908(119909) = minus(119909 minus 119886) so that theinduced chamber is a parabolic arc
The problem of finding the airloads on an oscillatingairfoil was solved by Theodorsen who gave a solution to theunsteady airloads on a 2D harmonically oscillated airfoil ininviscid incompressible flow with the assumption of smalldisturbances [6] Both the airfoil and its shed wake wererepresented by a vortex sheet with the shedwake extending asa planar surface from the trailing edge downstream to infinityThe assumption of planar wake is justified if the angleof attack disturbances remain relatively small As with thestandard quasi-steady thin airfoil theory the bound vorticity120574119887 can sustain a pressure difference and therefore a liftforceThe wake vorticity 120574119908 must be force-free with zero netpressure jump over the sheet According to the Theodorsenrsquostheory the solution for the loading 120574119887 on the airfoil surfaceunder harmonic forcing conditions is obtained from integralequation
119908 (119909 119905) =1
2120587int
119888
0
120574119887 (119909 119905)
119909 minus 1199090
119889119909 +1
2120587int
infin
119888
120574119908 (119909 119905)
119909 minus 1199090
119889119909 (35)
6 Mathematical Problems in Engineering
where119908 is the downwash on the airfoil surface At the trailingedge 120574119887(119888 119905) = 0 and the airfoil circulation Γ(119905) is given by
Γ (119905) = int
119888
0
120574119887 (119909 119905) 119889119909 (36)
So long as the circulation about the airfoil is changing withrespect to time the circulation is continuously shed into thewake and will continuously affect the aerodynamic loads onthe airfoil
For a general motion where an airfoil of chord 119888 = 2119887 isundergoing a combination of pitching (120572 ) and plunging (ℎ)motion in a flow of steady velocity 119881 Theodorsenrsquos solution[8ndash10] for the lift coefficient and pitching moment coefficientcorresponding to midchord11987212 is
119888119897 = 120587119887 [ℎ
1198812+
119881minus119887
1198812119886]
+ 2120587[ℎ
119881+ 120572 +
119887
119881(1
2minus 119886)]119862 (119896)
11988811989812 =120587
2[119887119886ℎ
1198812minus1198872
1198812(1
8+ 1198862) ]
+120587(119886 + 12) [
ℎ
119881+ 120572 + 119887 (
1
2minus 119886)
119881]119862 (119896)
minus120587
2[(1
2minus 119886)
119887
119881]
(37)
where 119886 is the pitch axis location relative to the midchordof the airfoil measured in terms of semichord and 119862(119896) =119865(119896) + 119894119866(119896) is the complex transfer function (known asTheodorsenrsquos function) which accounts for the effects of theshed wake on the unsteady airloads
119862 (119896) =119867(2)
1(119896)
119867(2)
1(119896) + 119894 sdot 119867
(2)
0(119896)
=1198691 (1198691 + 1198690) + 1198841 (1198841 minus 1198690)
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
+ 11989411988411198840 + 11986911198690
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
(38)
with 1198690 1198691 1198840 and 1198841 being Bessel functions of the first andsecond kinds respectively (Figure 8)
The Hankel functions in above expression are
119867(2)
0= 1198690 minus 119894 sdot 1198840
119867(2)
1= 1198691 minus 119894 sdot 1198841
(39)
The real and imaginary parts of 119862(119896) function are plotted inFigure 9
It could be appreciated that 119862(119896) function serves tointroduce an amplitude reduction and phase lag effect on thecirculatory part of the lift response compared to the resultobtained under quasi-steady conditions [11]
1
05
05 10 15
J(0 k)
J(1 k)
Y(0 k)
Y(1 k)
minus1
minus05
Bessel
Bessel
Bessel
Bessel
Figure 8 Bessel functions
Real part of C(k)
Imag
inar
y pa
rt o
f C(k)
0 02 04 06 08 1
0
minus005
minus010
minus015
minus020
minus025
minus030
Figure 9 Theodorsenrsquos function
This effect can be seen if a pure oscillatory variation inangle of attack is considered that is 120572 = 120572119890
119894120596119905 so thecirculatory part of the airfoil lift coefficient is given by
119888119897 = 2120587120572119862 (119896) = 2120587120572 [119865 (119896) + 119894119866 (119896)] (40)
For 119896 = 0 the steady-state lift behavior is obtained that is 119888119897is linearly proportional to 120572 As 119896 is increased the lift plotsdevelop into hysteresis loops and these loops rotate such thatthe amplitude of the lift response (half of the peak-to-peakvalue) decreases with increasing reduced frequency
These loops are circumvented in a counterclockwisedirection such that the lift is lower than the steady value when
Mathematical Problems in Engineering 7
CirculatoryTotal response
Apparent mass
2
15
1
05
0
0 1 2 3 4
k
Nor
mal
ized
lift
ampl
itude
Figure 10 Normalized lift amplitude
120572 is decreasingwith time (ie there is a phase lag) For infinitereduced frequency the circulatory part of the lift amplitude ishalf that at 119896 = 0 and there is no phase lag angle
5 Pure Angle of Attack Oscillations
For a harmonic variation in 120572 that is 120572 = 120572119890119894120596119905 the lift is
119871 = 21205871205881198812119887 [119862 (119896) +
1
2119894120596119887
119881]120572119890119894120596119905 (41)
or in terms of the lift coefficient the result is
119888119897 =119871
1205881198812119887= [2120587 (119865 + 119894119866) + 119894120587119896] 120572119890
119894120596119905 (42)
The term inside the square brackets can be considered the lifttransfer function which accounts for the difference betweenthe unsteady and quasi-steady airloads
The first term inside the brackets is the circulatory termand the second term is the apparentmass contribution whichis proportional to the reduced frequency and leads the forcingby a phase angle of 1205872 The noncirculatory or apparent massterms arise from the nonsteady term of the velocity potentialand account for the pressure forces required to accelerate thefluid in the vicinity of the airfoil
The normalized lift amplitude is
119888119897
2120587120572= (119865 + 119894119866) + 119894
119896
2 (43)
The normalized lift amplitude and phase of lift for pure angleof attack oscillations are presented in Figures 10 and 11 wherethe significance of the apparentmass contribution to both theamplitude and phase can be appreciated
At lower values of reduced frequency the circulatoryterms dominate the solution At higher values of reducedfrequency the apparent mass forces dominate
100
80
60
40
20
0
minus20
Phas
e ang
le [d
egre
es]
CirculatoryTotal response
Apparent mass
0 1 2 3 4
k
Figure 11 Phase angle
6 Pure Plunging Oscillations
For a harmonic plunging motion such as that contributed byblade flapping the forcing is ℎ = ℎ119890119894120596119905 so that ℎ = 119894120596ℎ119890119894120596119905 andℎ = minus120596
2ℎ119890119894120596119905 Substituting into the expression for the lift and
solving for the lift coefficient it gives
119888119897 = [2120587119896 (119894119865 minus 119866) minus 1205871198962]ℎ
119887119890119894120596119905 (44)
The complete term inside the square brackets can be con-sidered as the lift transfer function [12] The circulatory partof the lift response leads the forcing displacement ℎ by aphase angle of 1205872 Also the apparent mass force leads thecirculatory part of the response by a phase angle of 1205872 orthe forcing by a phase angle of 120587 The corresponding pitchingmoment about midchord for this case is
11988811989812 = (120587
4) 1198962 ℎ
119887119890119894120596119905 (45)
7 Pitching Oscillations
For harmonic pitch oscillations additional terms involvingpitch rate appear in the equations for the aerodynamic res-ponse The forcing is given by 120572 = 120572119890119894120596119905 and the pitch rate by = 119894120596120572119890
119894120596119905 In this case the lift coefficient is
119888119897 = 2120587 [119865 (1 + 119894119896) + 119866 (119894 minus 119896)] 120572119890119894120596119905
+ 120587119896(119894 minus119896
2)120572119890119894120596119905
(46)
The lift amplitude initially decreases with increasing 119896 beca-use of the effects of the shed wake and then for 119896 gt 05it begins to increase as the apparent mass forces begin to
8 Mathematical Problems in Engineering
dominate the airloads This is also shown by the phase anglewhich exhibits an increasing lead for 119896 gt 03
Von Karman and Sear analyzed the problem of a thinairfoil moving through a sinusoidal vertical gust field wherethe gust can be considered as an upwash velocity that isuniformly convected by the free streamThe forcing functionin this case is
119908119892 (119909 119905) = sin(120596119892119905 minus120596119892119909
119881) (47)
where 120596119892 is the gust frequency If the gust is referenced tothe airfoil leading edge then 119909 = 0 and 119908119892(119909 119905) becomes119908119892(119909 119905) = sin(120596119892119905) and if the gust is referenced to themidchord then 119909 = 119887 = 1198882 and
120596119892119909
119881=
120596119892 (1198882)
119881=
120596119892119888
2119881= 119896119892
(48)
Therefore119908119892(119905) = sin(120596119892119905)sdotcos(119896119892)minussin(119896119892)sdotcos(120596119892119905) whichis equivalent to a phase shift In this case the lift coefficientcan be written as
119888119897 = 2120587(1199080
119881) 119878 (119896119892) (49)
where 119878(119896119892) is known as Sears function and the gust encoun-ter frequency 119896119892 is given by
119896119892 =2120587119881
120582119892
(50)
and 120582119892 is the wavelength of the gustIn terms of Bessel functions Searsrsquos function is given by
119878 (119896119892) = [1198690 (119896119892) minus 1198941198691 (119896119892)] 119862 (119896119892) + 1198941198691 (119896119892) (51)
The terms of real and imaginary parts are
Real 119878 (119896119892) = 119865 (119896119892) sdot 1198690 (119896119892) + 119866 (119896119892) sdot 1198691 (119896119892)
Im 119878 (119896119892) = 119866 (119896119892) sdot 1198690 (119896119892) minus 119865 (119896119892) sdot 1198691 (119896119892) + 1198691 (119896119892)
(52)
If the gust is referenced to the leading edge of the airfoil thefunction will be called 1198781015840 and can be written as
Real 1198781015840 (119896119892) = Real 119878 (119896119892) sdot cos (119896119892)
+ Im 119878 (119896119892) sdot sin (119896119892)
Im 1198781015840(119896119892) = minusReal 119878 (119896119892) sdot sin (119896119892)
+ Im 119878 (119896119892) sdot cos (119896119892)
(53)
The two results are plotted in Figure 12 The peculiar spiralshape of the 119878 transfer function arises only when the gustfront is referenced to the midchord of the airfoil
The kinematics of the pitching and plunging airfoil ofa typical blade element is the resultant of a combination offorcing from collective and cyclic blade pitch twist angle
03
02
01
00
minus01
minus02
minus03
minus04
minus025 0 025 050 075 1
Imag
inar
y pa
rt
Sears function referenced to the midchordSears function referenced to the leading edge
Real part
Figure 12 Sears function
elastic torsion blade flapping velocity and elastic bendingAt low angles of attack with fully attached flow the varioussources of unsteady effects manifest primarily as moderateamplitude and phase variations relative to the quasi-steadyairloads At higher angles of attack when time-dependentflow separation from the airfoil may be involved the dynamicstall may occurThe amplitude and phase effects produced bythe stalled airloads can lead to various aeroelastic problemson the helicopter rotor that may limit its performance Theneed to control the aerodynamic forces on the rotor requiresthat the pitch of each blade be changed individually as theblades rotate about the shaft
8 Summary
The first flap frequency of a helicopter rigid blade is about105Ω then the reduced frequency at the 75 radius locationis 119896 = 07119888119877 For a blade aspect ratio 119877119888 gt 10 the reducedfrequency is in excess of 007 which is in the unsteadyrange In the case of the first elastic torsion mode which istypically about 3 divide 4Ω at the blade tip the reduced frequencyis in excess of 02 At these reduced frequencies there isa significant amplitude and phasing introduced into theairloads by the effects of the unsteady aerodynamics
When a wingrsquos angle of attack is increased rapidly it canmomentarily generate a higher maximum lift coefficient thanit could if the angle of attack was increased slowly This over-shoot can be related to the change in angle of attack duringthe time required for the air to travel one chord length Thedynamic overshoot is attributed to two effects (for the airfoilsthat stall first at the leading edge) the delay in the separationof the boundary layer and the momentary existence of avortex shed at the leading edge after the boundary layer doesseparate The delay in separation corresponds to the finite
Mathematical Problems in Engineering 9
time required for the aft edge of the separation bubble tomove forward to its bursting positionOn the other hands theairfoil can generate high lift as a result of a vortex that is shedat the leading edge at the instant of stall The vortex travelsback over the top of the airfoil carrying with it a low pressurewave that accounts for the very large lift coefficient Airfoilsthat stall first at the trailing edge also exhibit a dynamicovershoot but considerably less than those airfoils that haveleading edge stall
References
[1] J G Leishman Principles of Helicopter Aerodynamics Cam-bridge University Press 2007
[2] J Seddon and SNewmanBasicHelicopterAerodynamics AIAAEducation Series 2001
[3] M A Ashraf J Young and I C S Lai ldquoOscillation frequencyand amplitude effects on plunging airfoil propulsion and flowperiodicityrdquo The American Institute of Aeronautics and Astro-nautics Journal pp 2308ndash2324 2012
[4] R Prouty Helicopter Performance Stability and Control Kri-eger Gainesville Fla USA 2002
[5] P D Gareth Helicopter Flight Dynamics AIAA EducationSeries 2007
[6] B J Lee and M S Liou ldquoUnsteady adjoint approach for designoptimization of flapping airfoilsrdquo The American Institute ofAeronautics and Astronautics Journal pp 2460ndash2491 2012
[7] K Karamcheti Principles of Ideal-Fluid Aerodynamics KriegerGainesville Fla USA 1966
[8] T Theodorsen ldquoGeneral theory of aerodynamic instability andthe mechanism of flutterrdquo Tech Rep 496 NACA 1934
[9] I EGarrick ldquoOn some reciprocal relations in the theory of non-stationary flowsrdquo NACA Report 629 1938
[10] J Katz and A Plotkin Low Speed Aerodynamics CambridgeUniversity Press 2010
[11] C Rotaru I Circiu and M Boscoianu ldquoComputational meth-ods for the aerodynamic designrdquo Review of the Air Force Acad-emy vol 2 no 17 pp 43ndash48 2010
[12] C Rotaru and V Jaulin ldquoNumerical simulations of Helicopterrotor blade airloadsrdquoMTAReview vol 22 no 2 pp 81ndash90 2012
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Mathematical Problems in Engineering 5
with119898119896119894 = 119898119894119896 the components of the impulse 119868 are thereforegiven by
119868119894 =
3
sum
119896=1
119898119894119896119906119896 119894 = 1 2 3 (29)
and the force applied externally to the body is
119890 =119889
119889119905119898 (1199061
119894 + 1199062119895 + 1199063)
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119894
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
119895
+119889
119889119905(119898111199061 + 119898121199062 + 119898131199063)
(30)
or
119890 = [(119898 + 11989811)1198891199061
119889119905+ 11989812
1198891199062
119889119905+ 11989813
1198891199063
119889119905] 119894
+ [11989821
1198891199061
119889119905+ (119898 + 11989822)
1198891199062
119889119905+ 11989823
1198891199063
119889119905] 119895
+ [11989831
1198891199061
119889119905+ 11989832
1198891199062
119889119905+ (119898 + 11989833)
1198891199063
119889119905] 119896
(31)
The coefficients119898119894119896 form a set of nine numbers whichmay bedisplayed as an array
(
11989811 11989812 11989813
11989821 11989822 11989823
11989831 11989832 11989833
) (32)
and may be referred to as a virtual mass tensor or virtualmasses that need to be added to the mass of the body in orderto find the force that must be applied to translate it throughthe fluid Introducing the symbol 120575119894119896 defined by 120575119894119896 = 0 if 119894 = 119896
and 120575119894119896 = 1 for 119894 = 119896 (30) may be rewritten as
(119865119890)119894=
3
sum
119896=1
(119898120575119894119896 + 119898119894119896)119889119906119896
119889119905 (33)
For any body there are three perpendicular directions suchthat 119898119894119896 = 0 for 119894 = 119896 so with respect to such axes (33)becomes
(119865119890)119894= (119898 + 119898119894119894)
119889119906119894
119889119905 119894 = 1 2 3 (34)
The sum (119898+119898119894119894) represents the apparentmass for translationin the 119894-direction and the corresponding119898119894119894 is the additionalapparent mass
4 The Airloads on an Oscillating Airfoil
The oscillatory motion of the airfoil can be decomposed intocontributions associated with angle of attack which is equiv-alent to a pure plunging motion (Figure 6) and contributionsassociated with pitching (Figure 7)
Vh
h
x
w(x) = minus
Figure 6 Plunge velocity
V
x
y
w(x) = minus (x minus a)
bb
a
Figure 7 Pitch rate
A plunge velocity ℎproduces a uniformvelocity perturba-tion 119908 that is normal to the chord 119908(119909) = minusℎ and the pitch-rate term produces a linear variation in normal perturbationvelocity
For a pitch rate imposed about an axis at ldquo119886rdquo semichordsfrom the midchord then 119908(119909) = minus(119909 minus 119886) so that theinduced chamber is a parabolic arc
The problem of finding the airloads on an oscillatingairfoil was solved by Theodorsen who gave a solution to theunsteady airloads on a 2D harmonically oscillated airfoil ininviscid incompressible flow with the assumption of smalldisturbances [6] Both the airfoil and its shed wake wererepresented by a vortex sheet with the shedwake extending asa planar surface from the trailing edge downstream to infinityThe assumption of planar wake is justified if the angleof attack disturbances remain relatively small As with thestandard quasi-steady thin airfoil theory the bound vorticity120574119887 can sustain a pressure difference and therefore a liftforceThe wake vorticity 120574119908 must be force-free with zero netpressure jump over the sheet According to the Theodorsenrsquostheory the solution for the loading 120574119887 on the airfoil surfaceunder harmonic forcing conditions is obtained from integralequation
119908 (119909 119905) =1
2120587int
119888
0
120574119887 (119909 119905)
119909 minus 1199090
119889119909 +1
2120587int
infin
119888
120574119908 (119909 119905)
119909 minus 1199090
119889119909 (35)
6 Mathematical Problems in Engineering
where119908 is the downwash on the airfoil surface At the trailingedge 120574119887(119888 119905) = 0 and the airfoil circulation Γ(119905) is given by
Γ (119905) = int
119888
0
120574119887 (119909 119905) 119889119909 (36)
So long as the circulation about the airfoil is changing withrespect to time the circulation is continuously shed into thewake and will continuously affect the aerodynamic loads onthe airfoil
For a general motion where an airfoil of chord 119888 = 2119887 isundergoing a combination of pitching (120572 ) and plunging (ℎ)motion in a flow of steady velocity 119881 Theodorsenrsquos solution[8ndash10] for the lift coefficient and pitching moment coefficientcorresponding to midchord11987212 is
119888119897 = 120587119887 [ℎ
1198812+
119881minus119887
1198812119886]
+ 2120587[ℎ
119881+ 120572 +
119887
119881(1
2minus 119886)]119862 (119896)
11988811989812 =120587
2[119887119886ℎ
1198812minus1198872
1198812(1
8+ 1198862) ]
+120587(119886 + 12) [
ℎ
119881+ 120572 + 119887 (
1
2minus 119886)
119881]119862 (119896)
minus120587
2[(1
2minus 119886)
119887
119881]
(37)
where 119886 is the pitch axis location relative to the midchordof the airfoil measured in terms of semichord and 119862(119896) =119865(119896) + 119894119866(119896) is the complex transfer function (known asTheodorsenrsquos function) which accounts for the effects of theshed wake on the unsteady airloads
119862 (119896) =119867(2)
1(119896)
119867(2)
1(119896) + 119894 sdot 119867
(2)
0(119896)
=1198691 (1198691 + 1198690) + 1198841 (1198841 minus 1198690)
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
+ 11989411988411198840 + 11986911198690
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
(38)
with 1198690 1198691 1198840 and 1198841 being Bessel functions of the first andsecond kinds respectively (Figure 8)
The Hankel functions in above expression are
119867(2)
0= 1198690 minus 119894 sdot 1198840
119867(2)
1= 1198691 minus 119894 sdot 1198841
(39)
The real and imaginary parts of 119862(119896) function are plotted inFigure 9
It could be appreciated that 119862(119896) function serves tointroduce an amplitude reduction and phase lag effect on thecirculatory part of the lift response compared to the resultobtained under quasi-steady conditions [11]
1
05
05 10 15
J(0 k)
J(1 k)
Y(0 k)
Y(1 k)
minus1
minus05
Bessel
Bessel
Bessel
Bessel
Figure 8 Bessel functions
Real part of C(k)
Imag
inar
y pa
rt o
f C(k)
0 02 04 06 08 1
0
minus005
minus010
minus015
minus020
minus025
minus030
Figure 9 Theodorsenrsquos function
This effect can be seen if a pure oscillatory variation inangle of attack is considered that is 120572 = 120572119890
119894120596119905 so thecirculatory part of the airfoil lift coefficient is given by
119888119897 = 2120587120572119862 (119896) = 2120587120572 [119865 (119896) + 119894119866 (119896)] (40)
For 119896 = 0 the steady-state lift behavior is obtained that is 119888119897is linearly proportional to 120572 As 119896 is increased the lift plotsdevelop into hysteresis loops and these loops rotate such thatthe amplitude of the lift response (half of the peak-to-peakvalue) decreases with increasing reduced frequency
These loops are circumvented in a counterclockwisedirection such that the lift is lower than the steady value when
Mathematical Problems in Engineering 7
CirculatoryTotal response
Apparent mass
2
15
1
05
0
0 1 2 3 4
k
Nor
mal
ized
lift
ampl
itude
Figure 10 Normalized lift amplitude
120572 is decreasingwith time (ie there is a phase lag) For infinitereduced frequency the circulatory part of the lift amplitude ishalf that at 119896 = 0 and there is no phase lag angle
5 Pure Angle of Attack Oscillations
For a harmonic variation in 120572 that is 120572 = 120572119890119894120596119905 the lift is
119871 = 21205871205881198812119887 [119862 (119896) +
1
2119894120596119887
119881]120572119890119894120596119905 (41)
or in terms of the lift coefficient the result is
119888119897 =119871
1205881198812119887= [2120587 (119865 + 119894119866) + 119894120587119896] 120572119890
119894120596119905 (42)
The term inside the square brackets can be considered the lifttransfer function which accounts for the difference betweenthe unsteady and quasi-steady airloads
The first term inside the brackets is the circulatory termand the second term is the apparentmass contribution whichis proportional to the reduced frequency and leads the forcingby a phase angle of 1205872 The noncirculatory or apparent massterms arise from the nonsteady term of the velocity potentialand account for the pressure forces required to accelerate thefluid in the vicinity of the airfoil
The normalized lift amplitude is
119888119897
2120587120572= (119865 + 119894119866) + 119894
119896
2 (43)
The normalized lift amplitude and phase of lift for pure angleof attack oscillations are presented in Figures 10 and 11 wherethe significance of the apparentmass contribution to both theamplitude and phase can be appreciated
At lower values of reduced frequency the circulatoryterms dominate the solution At higher values of reducedfrequency the apparent mass forces dominate
100
80
60
40
20
0
minus20
Phas
e ang
le [d
egre
es]
CirculatoryTotal response
Apparent mass
0 1 2 3 4
k
Figure 11 Phase angle
6 Pure Plunging Oscillations
For a harmonic plunging motion such as that contributed byblade flapping the forcing is ℎ = ℎ119890119894120596119905 so that ℎ = 119894120596ℎ119890119894120596119905 andℎ = minus120596
2ℎ119890119894120596119905 Substituting into the expression for the lift and
solving for the lift coefficient it gives
119888119897 = [2120587119896 (119894119865 minus 119866) minus 1205871198962]ℎ
119887119890119894120596119905 (44)
The complete term inside the square brackets can be con-sidered as the lift transfer function [12] The circulatory partof the lift response leads the forcing displacement ℎ by aphase angle of 1205872 Also the apparent mass force leads thecirculatory part of the response by a phase angle of 1205872 orthe forcing by a phase angle of 120587 The corresponding pitchingmoment about midchord for this case is
11988811989812 = (120587
4) 1198962 ℎ
119887119890119894120596119905 (45)
7 Pitching Oscillations
For harmonic pitch oscillations additional terms involvingpitch rate appear in the equations for the aerodynamic res-ponse The forcing is given by 120572 = 120572119890119894120596119905 and the pitch rate by = 119894120596120572119890
119894120596119905 In this case the lift coefficient is
119888119897 = 2120587 [119865 (1 + 119894119896) + 119866 (119894 minus 119896)] 120572119890119894120596119905
+ 120587119896(119894 minus119896
2)120572119890119894120596119905
(46)
The lift amplitude initially decreases with increasing 119896 beca-use of the effects of the shed wake and then for 119896 gt 05it begins to increase as the apparent mass forces begin to
8 Mathematical Problems in Engineering
dominate the airloads This is also shown by the phase anglewhich exhibits an increasing lead for 119896 gt 03
Von Karman and Sear analyzed the problem of a thinairfoil moving through a sinusoidal vertical gust field wherethe gust can be considered as an upwash velocity that isuniformly convected by the free streamThe forcing functionin this case is
119908119892 (119909 119905) = sin(120596119892119905 minus120596119892119909
119881) (47)
where 120596119892 is the gust frequency If the gust is referenced tothe airfoil leading edge then 119909 = 0 and 119908119892(119909 119905) becomes119908119892(119909 119905) = sin(120596119892119905) and if the gust is referenced to themidchord then 119909 = 119887 = 1198882 and
120596119892119909
119881=
120596119892 (1198882)
119881=
120596119892119888
2119881= 119896119892
(48)
Therefore119908119892(119905) = sin(120596119892119905)sdotcos(119896119892)minussin(119896119892)sdotcos(120596119892119905) whichis equivalent to a phase shift In this case the lift coefficientcan be written as
119888119897 = 2120587(1199080
119881) 119878 (119896119892) (49)
where 119878(119896119892) is known as Sears function and the gust encoun-ter frequency 119896119892 is given by
119896119892 =2120587119881
120582119892
(50)
and 120582119892 is the wavelength of the gustIn terms of Bessel functions Searsrsquos function is given by
119878 (119896119892) = [1198690 (119896119892) minus 1198941198691 (119896119892)] 119862 (119896119892) + 1198941198691 (119896119892) (51)
The terms of real and imaginary parts are
Real 119878 (119896119892) = 119865 (119896119892) sdot 1198690 (119896119892) + 119866 (119896119892) sdot 1198691 (119896119892)
Im 119878 (119896119892) = 119866 (119896119892) sdot 1198690 (119896119892) minus 119865 (119896119892) sdot 1198691 (119896119892) + 1198691 (119896119892)
(52)
If the gust is referenced to the leading edge of the airfoil thefunction will be called 1198781015840 and can be written as
Real 1198781015840 (119896119892) = Real 119878 (119896119892) sdot cos (119896119892)
+ Im 119878 (119896119892) sdot sin (119896119892)
Im 1198781015840(119896119892) = minusReal 119878 (119896119892) sdot sin (119896119892)
+ Im 119878 (119896119892) sdot cos (119896119892)
(53)
The two results are plotted in Figure 12 The peculiar spiralshape of the 119878 transfer function arises only when the gustfront is referenced to the midchord of the airfoil
The kinematics of the pitching and plunging airfoil ofa typical blade element is the resultant of a combination offorcing from collective and cyclic blade pitch twist angle
03
02
01
00
minus01
minus02
minus03
minus04
minus025 0 025 050 075 1
Imag
inar
y pa
rt
Sears function referenced to the midchordSears function referenced to the leading edge
Real part
Figure 12 Sears function
elastic torsion blade flapping velocity and elastic bendingAt low angles of attack with fully attached flow the varioussources of unsteady effects manifest primarily as moderateamplitude and phase variations relative to the quasi-steadyairloads At higher angles of attack when time-dependentflow separation from the airfoil may be involved the dynamicstall may occurThe amplitude and phase effects produced bythe stalled airloads can lead to various aeroelastic problemson the helicopter rotor that may limit its performance Theneed to control the aerodynamic forces on the rotor requiresthat the pitch of each blade be changed individually as theblades rotate about the shaft
8 Summary
The first flap frequency of a helicopter rigid blade is about105Ω then the reduced frequency at the 75 radius locationis 119896 = 07119888119877 For a blade aspect ratio 119877119888 gt 10 the reducedfrequency is in excess of 007 which is in the unsteadyrange In the case of the first elastic torsion mode which istypically about 3 divide 4Ω at the blade tip the reduced frequencyis in excess of 02 At these reduced frequencies there isa significant amplitude and phasing introduced into theairloads by the effects of the unsteady aerodynamics
When a wingrsquos angle of attack is increased rapidly it canmomentarily generate a higher maximum lift coefficient thanit could if the angle of attack was increased slowly This over-shoot can be related to the change in angle of attack duringthe time required for the air to travel one chord length Thedynamic overshoot is attributed to two effects (for the airfoilsthat stall first at the leading edge) the delay in the separationof the boundary layer and the momentary existence of avortex shed at the leading edge after the boundary layer doesseparate The delay in separation corresponds to the finite
Mathematical Problems in Engineering 9
time required for the aft edge of the separation bubble tomove forward to its bursting positionOn the other hands theairfoil can generate high lift as a result of a vortex that is shedat the leading edge at the instant of stall The vortex travelsback over the top of the airfoil carrying with it a low pressurewave that accounts for the very large lift coefficient Airfoilsthat stall first at the trailing edge also exhibit a dynamicovershoot but considerably less than those airfoils that haveleading edge stall
References
[1] J G Leishman Principles of Helicopter Aerodynamics Cam-bridge University Press 2007
[2] J Seddon and SNewmanBasicHelicopterAerodynamics AIAAEducation Series 2001
[3] M A Ashraf J Young and I C S Lai ldquoOscillation frequencyand amplitude effects on plunging airfoil propulsion and flowperiodicityrdquo The American Institute of Aeronautics and Astro-nautics Journal pp 2308ndash2324 2012
[4] R Prouty Helicopter Performance Stability and Control Kri-eger Gainesville Fla USA 2002
[5] P D Gareth Helicopter Flight Dynamics AIAA EducationSeries 2007
[6] B J Lee and M S Liou ldquoUnsteady adjoint approach for designoptimization of flapping airfoilsrdquo The American Institute ofAeronautics and Astronautics Journal pp 2460ndash2491 2012
[7] K Karamcheti Principles of Ideal-Fluid Aerodynamics KriegerGainesville Fla USA 1966
[8] T Theodorsen ldquoGeneral theory of aerodynamic instability andthe mechanism of flutterrdquo Tech Rep 496 NACA 1934
[9] I EGarrick ldquoOn some reciprocal relations in the theory of non-stationary flowsrdquo NACA Report 629 1938
[10] J Katz and A Plotkin Low Speed Aerodynamics CambridgeUniversity Press 2010
[11] C Rotaru I Circiu and M Boscoianu ldquoComputational meth-ods for the aerodynamic designrdquo Review of the Air Force Acad-emy vol 2 no 17 pp 43ndash48 2010
[12] C Rotaru and V Jaulin ldquoNumerical simulations of Helicopterrotor blade airloadsrdquoMTAReview vol 22 no 2 pp 81ndash90 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Mathematical Problems in Engineering
where119908 is the downwash on the airfoil surface At the trailingedge 120574119887(119888 119905) = 0 and the airfoil circulation Γ(119905) is given by
Γ (119905) = int
119888
0
120574119887 (119909 119905) 119889119909 (36)
So long as the circulation about the airfoil is changing withrespect to time the circulation is continuously shed into thewake and will continuously affect the aerodynamic loads onthe airfoil
For a general motion where an airfoil of chord 119888 = 2119887 isundergoing a combination of pitching (120572 ) and plunging (ℎ)motion in a flow of steady velocity 119881 Theodorsenrsquos solution[8ndash10] for the lift coefficient and pitching moment coefficientcorresponding to midchord11987212 is
119888119897 = 120587119887 [ℎ
1198812+
119881minus119887
1198812119886]
+ 2120587[ℎ
119881+ 120572 +
119887
119881(1
2minus 119886)]119862 (119896)
11988811989812 =120587
2[119887119886ℎ
1198812minus1198872
1198812(1
8+ 1198862) ]
+120587(119886 + 12) [
ℎ
119881+ 120572 + 119887 (
1
2minus 119886)
119881]119862 (119896)
minus120587
2[(1
2minus 119886)
119887
119881]
(37)
where 119886 is the pitch axis location relative to the midchordof the airfoil measured in terms of semichord and 119862(119896) =119865(119896) + 119894119866(119896) is the complex transfer function (known asTheodorsenrsquos function) which accounts for the effects of theshed wake on the unsteady airloads
119862 (119896) =119867(2)
1(119896)
119867(2)
1(119896) + 119894 sdot 119867
(2)
0(119896)
=1198691 (1198691 + 1198690) + 1198841 (1198841 minus 1198690)
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
+ 11989411988411198840 + 11986911198690
(1198691 + 1198840)2+ (1198840 minus 1198691)
2
(38)
with 1198690 1198691 1198840 and 1198841 being Bessel functions of the first andsecond kinds respectively (Figure 8)
The Hankel functions in above expression are
119867(2)
0= 1198690 minus 119894 sdot 1198840
119867(2)
1= 1198691 minus 119894 sdot 1198841
(39)
The real and imaginary parts of 119862(119896) function are plotted inFigure 9
It could be appreciated that 119862(119896) function serves tointroduce an amplitude reduction and phase lag effect on thecirculatory part of the lift response compared to the resultobtained under quasi-steady conditions [11]
1
05
05 10 15
J(0 k)
J(1 k)
Y(0 k)
Y(1 k)
minus1
minus05
Bessel
Bessel
Bessel
Bessel
Figure 8 Bessel functions
Real part of C(k)
Imag
inar
y pa
rt o
f C(k)
0 02 04 06 08 1
0
minus005
minus010
minus015
minus020
minus025
minus030
Figure 9 Theodorsenrsquos function
This effect can be seen if a pure oscillatory variation inangle of attack is considered that is 120572 = 120572119890
119894120596119905 so thecirculatory part of the airfoil lift coefficient is given by
119888119897 = 2120587120572119862 (119896) = 2120587120572 [119865 (119896) + 119894119866 (119896)] (40)
For 119896 = 0 the steady-state lift behavior is obtained that is 119888119897is linearly proportional to 120572 As 119896 is increased the lift plotsdevelop into hysteresis loops and these loops rotate such thatthe amplitude of the lift response (half of the peak-to-peakvalue) decreases with increasing reduced frequency
These loops are circumvented in a counterclockwisedirection such that the lift is lower than the steady value when
Mathematical Problems in Engineering 7
CirculatoryTotal response
Apparent mass
2
15
1
05
0
0 1 2 3 4
k
Nor
mal
ized
lift
ampl
itude
Figure 10 Normalized lift amplitude
120572 is decreasingwith time (ie there is a phase lag) For infinitereduced frequency the circulatory part of the lift amplitude ishalf that at 119896 = 0 and there is no phase lag angle
5 Pure Angle of Attack Oscillations
For a harmonic variation in 120572 that is 120572 = 120572119890119894120596119905 the lift is
119871 = 21205871205881198812119887 [119862 (119896) +
1
2119894120596119887
119881]120572119890119894120596119905 (41)
or in terms of the lift coefficient the result is
119888119897 =119871
1205881198812119887= [2120587 (119865 + 119894119866) + 119894120587119896] 120572119890
119894120596119905 (42)
The term inside the square brackets can be considered the lifttransfer function which accounts for the difference betweenthe unsteady and quasi-steady airloads
The first term inside the brackets is the circulatory termand the second term is the apparentmass contribution whichis proportional to the reduced frequency and leads the forcingby a phase angle of 1205872 The noncirculatory or apparent massterms arise from the nonsteady term of the velocity potentialand account for the pressure forces required to accelerate thefluid in the vicinity of the airfoil
The normalized lift amplitude is
119888119897
2120587120572= (119865 + 119894119866) + 119894
119896
2 (43)
The normalized lift amplitude and phase of lift for pure angleof attack oscillations are presented in Figures 10 and 11 wherethe significance of the apparentmass contribution to both theamplitude and phase can be appreciated
At lower values of reduced frequency the circulatoryterms dominate the solution At higher values of reducedfrequency the apparent mass forces dominate
100
80
60
40
20
0
minus20
Phas
e ang
le [d
egre
es]
CirculatoryTotal response
Apparent mass
0 1 2 3 4
k
Figure 11 Phase angle
6 Pure Plunging Oscillations
For a harmonic plunging motion such as that contributed byblade flapping the forcing is ℎ = ℎ119890119894120596119905 so that ℎ = 119894120596ℎ119890119894120596119905 andℎ = minus120596
2ℎ119890119894120596119905 Substituting into the expression for the lift and
solving for the lift coefficient it gives
119888119897 = [2120587119896 (119894119865 minus 119866) minus 1205871198962]ℎ
119887119890119894120596119905 (44)
The complete term inside the square brackets can be con-sidered as the lift transfer function [12] The circulatory partof the lift response leads the forcing displacement ℎ by aphase angle of 1205872 Also the apparent mass force leads thecirculatory part of the response by a phase angle of 1205872 orthe forcing by a phase angle of 120587 The corresponding pitchingmoment about midchord for this case is
11988811989812 = (120587
4) 1198962 ℎ
119887119890119894120596119905 (45)
7 Pitching Oscillations
For harmonic pitch oscillations additional terms involvingpitch rate appear in the equations for the aerodynamic res-ponse The forcing is given by 120572 = 120572119890119894120596119905 and the pitch rate by = 119894120596120572119890
119894120596119905 In this case the lift coefficient is
119888119897 = 2120587 [119865 (1 + 119894119896) + 119866 (119894 minus 119896)] 120572119890119894120596119905
+ 120587119896(119894 minus119896
2)120572119890119894120596119905
(46)
The lift amplitude initially decreases with increasing 119896 beca-use of the effects of the shed wake and then for 119896 gt 05it begins to increase as the apparent mass forces begin to
8 Mathematical Problems in Engineering
dominate the airloads This is also shown by the phase anglewhich exhibits an increasing lead for 119896 gt 03
Von Karman and Sear analyzed the problem of a thinairfoil moving through a sinusoidal vertical gust field wherethe gust can be considered as an upwash velocity that isuniformly convected by the free streamThe forcing functionin this case is
119908119892 (119909 119905) = sin(120596119892119905 minus120596119892119909
119881) (47)
where 120596119892 is the gust frequency If the gust is referenced tothe airfoil leading edge then 119909 = 0 and 119908119892(119909 119905) becomes119908119892(119909 119905) = sin(120596119892119905) and if the gust is referenced to themidchord then 119909 = 119887 = 1198882 and
120596119892119909
119881=
120596119892 (1198882)
119881=
120596119892119888
2119881= 119896119892
(48)
Therefore119908119892(119905) = sin(120596119892119905)sdotcos(119896119892)minussin(119896119892)sdotcos(120596119892119905) whichis equivalent to a phase shift In this case the lift coefficientcan be written as
119888119897 = 2120587(1199080
119881) 119878 (119896119892) (49)
where 119878(119896119892) is known as Sears function and the gust encoun-ter frequency 119896119892 is given by
119896119892 =2120587119881
120582119892
(50)
and 120582119892 is the wavelength of the gustIn terms of Bessel functions Searsrsquos function is given by
119878 (119896119892) = [1198690 (119896119892) minus 1198941198691 (119896119892)] 119862 (119896119892) + 1198941198691 (119896119892) (51)
The terms of real and imaginary parts are
Real 119878 (119896119892) = 119865 (119896119892) sdot 1198690 (119896119892) + 119866 (119896119892) sdot 1198691 (119896119892)
Im 119878 (119896119892) = 119866 (119896119892) sdot 1198690 (119896119892) minus 119865 (119896119892) sdot 1198691 (119896119892) + 1198691 (119896119892)
(52)
If the gust is referenced to the leading edge of the airfoil thefunction will be called 1198781015840 and can be written as
Real 1198781015840 (119896119892) = Real 119878 (119896119892) sdot cos (119896119892)
+ Im 119878 (119896119892) sdot sin (119896119892)
Im 1198781015840(119896119892) = minusReal 119878 (119896119892) sdot sin (119896119892)
+ Im 119878 (119896119892) sdot cos (119896119892)
(53)
The two results are plotted in Figure 12 The peculiar spiralshape of the 119878 transfer function arises only when the gustfront is referenced to the midchord of the airfoil
The kinematics of the pitching and plunging airfoil ofa typical blade element is the resultant of a combination offorcing from collective and cyclic blade pitch twist angle
03
02
01
00
minus01
minus02
minus03
minus04
minus025 0 025 050 075 1
Imag
inar
y pa
rt
Sears function referenced to the midchordSears function referenced to the leading edge
Real part
Figure 12 Sears function
elastic torsion blade flapping velocity and elastic bendingAt low angles of attack with fully attached flow the varioussources of unsteady effects manifest primarily as moderateamplitude and phase variations relative to the quasi-steadyairloads At higher angles of attack when time-dependentflow separation from the airfoil may be involved the dynamicstall may occurThe amplitude and phase effects produced bythe stalled airloads can lead to various aeroelastic problemson the helicopter rotor that may limit its performance Theneed to control the aerodynamic forces on the rotor requiresthat the pitch of each blade be changed individually as theblades rotate about the shaft
8 Summary
The first flap frequency of a helicopter rigid blade is about105Ω then the reduced frequency at the 75 radius locationis 119896 = 07119888119877 For a blade aspect ratio 119877119888 gt 10 the reducedfrequency is in excess of 007 which is in the unsteadyrange In the case of the first elastic torsion mode which istypically about 3 divide 4Ω at the blade tip the reduced frequencyis in excess of 02 At these reduced frequencies there isa significant amplitude and phasing introduced into theairloads by the effects of the unsteady aerodynamics
When a wingrsquos angle of attack is increased rapidly it canmomentarily generate a higher maximum lift coefficient thanit could if the angle of attack was increased slowly This over-shoot can be related to the change in angle of attack duringthe time required for the air to travel one chord length Thedynamic overshoot is attributed to two effects (for the airfoilsthat stall first at the leading edge) the delay in the separationof the boundary layer and the momentary existence of avortex shed at the leading edge after the boundary layer doesseparate The delay in separation corresponds to the finite
Mathematical Problems in Engineering 9
time required for the aft edge of the separation bubble tomove forward to its bursting positionOn the other hands theairfoil can generate high lift as a result of a vortex that is shedat the leading edge at the instant of stall The vortex travelsback over the top of the airfoil carrying with it a low pressurewave that accounts for the very large lift coefficient Airfoilsthat stall first at the trailing edge also exhibit a dynamicovershoot but considerably less than those airfoils that haveleading edge stall
References
[1] J G Leishman Principles of Helicopter Aerodynamics Cam-bridge University Press 2007
[2] J Seddon and SNewmanBasicHelicopterAerodynamics AIAAEducation Series 2001
[3] M A Ashraf J Young and I C S Lai ldquoOscillation frequencyand amplitude effects on plunging airfoil propulsion and flowperiodicityrdquo The American Institute of Aeronautics and Astro-nautics Journal pp 2308ndash2324 2012
[4] R Prouty Helicopter Performance Stability and Control Kri-eger Gainesville Fla USA 2002
[5] P D Gareth Helicopter Flight Dynamics AIAA EducationSeries 2007
[6] B J Lee and M S Liou ldquoUnsteady adjoint approach for designoptimization of flapping airfoilsrdquo The American Institute ofAeronautics and Astronautics Journal pp 2460ndash2491 2012
[7] K Karamcheti Principles of Ideal-Fluid Aerodynamics KriegerGainesville Fla USA 1966
[8] T Theodorsen ldquoGeneral theory of aerodynamic instability andthe mechanism of flutterrdquo Tech Rep 496 NACA 1934
[9] I EGarrick ldquoOn some reciprocal relations in the theory of non-stationary flowsrdquo NACA Report 629 1938
[10] J Katz and A Plotkin Low Speed Aerodynamics CambridgeUniversity Press 2010
[11] C Rotaru I Circiu and M Boscoianu ldquoComputational meth-ods for the aerodynamic designrdquo Review of the Air Force Acad-emy vol 2 no 17 pp 43ndash48 2010
[12] C Rotaru and V Jaulin ldquoNumerical simulations of Helicopterrotor blade airloadsrdquoMTAReview vol 22 no 2 pp 81ndash90 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 7
CirculatoryTotal response
Apparent mass
2
15
1
05
0
0 1 2 3 4
k
Nor
mal
ized
lift
ampl
itude
Figure 10 Normalized lift amplitude
120572 is decreasingwith time (ie there is a phase lag) For infinitereduced frequency the circulatory part of the lift amplitude ishalf that at 119896 = 0 and there is no phase lag angle
5 Pure Angle of Attack Oscillations
For a harmonic variation in 120572 that is 120572 = 120572119890119894120596119905 the lift is
119871 = 21205871205881198812119887 [119862 (119896) +
1
2119894120596119887
119881]120572119890119894120596119905 (41)
or in terms of the lift coefficient the result is
119888119897 =119871
1205881198812119887= [2120587 (119865 + 119894119866) + 119894120587119896] 120572119890
119894120596119905 (42)
The term inside the square brackets can be considered the lifttransfer function which accounts for the difference betweenthe unsteady and quasi-steady airloads
The first term inside the brackets is the circulatory termand the second term is the apparentmass contribution whichis proportional to the reduced frequency and leads the forcingby a phase angle of 1205872 The noncirculatory or apparent massterms arise from the nonsteady term of the velocity potentialand account for the pressure forces required to accelerate thefluid in the vicinity of the airfoil
The normalized lift amplitude is
119888119897
2120587120572= (119865 + 119894119866) + 119894
119896
2 (43)
The normalized lift amplitude and phase of lift for pure angleof attack oscillations are presented in Figures 10 and 11 wherethe significance of the apparentmass contribution to both theamplitude and phase can be appreciated
At lower values of reduced frequency the circulatoryterms dominate the solution At higher values of reducedfrequency the apparent mass forces dominate
100
80
60
40
20
0
minus20
Phas
e ang
le [d
egre
es]
CirculatoryTotal response
Apparent mass
0 1 2 3 4
k
Figure 11 Phase angle
6 Pure Plunging Oscillations
For a harmonic plunging motion such as that contributed byblade flapping the forcing is ℎ = ℎ119890119894120596119905 so that ℎ = 119894120596ℎ119890119894120596119905 andℎ = minus120596
2ℎ119890119894120596119905 Substituting into the expression for the lift and
solving for the lift coefficient it gives
119888119897 = [2120587119896 (119894119865 minus 119866) minus 1205871198962]ℎ
119887119890119894120596119905 (44)
The complete term inside the square brackets can be con-sidered as the lift transfer function [12] The circulatory partof the lift response leads the forcing displacement ℎ by aphase angle of 1205872 Also the apparent mass force leads thecirculatory part of the response by a phase angle of 1205872 orthe forcing by a phase angle of 120587 The corresponding pitchingmoment about midchord for this case is
11988811989812 = (120587
4) 1198962 ℎ
119887119890119894120596119905 (45)
7 Pitching Oscillations
For harmonic pitch oscillations additional terms involvingpitch rate appear in the equations for the aerodynamic res-ponse The forcing is given by 120572 = 120572119890119894120596119905 and the pitch rate by = 119894120596120572119890
119894120596119905 In this case the lift coefficient is
119888119897 = 2120587 [119865 (1 + 119894119896) + 119866 (119894 minus 119896)] 120572119890119894120596119905
+ 120587119896(119894 minus119896
2)120572119890119894120596119905
(46)
The lift amplitude initially decreases with increasing 119896 beca-use of the effects of the shed wake and then for 119896 gt 05it begins to increase as the apparent mass forces begin to
8 Mathematical Problems in Engineering
dominate the airloads This is also shown by the phase anglewhich exhibits an increasing lead for 119896 gt 03
Von Karman and Sear analyzed the problem of a thinairfoil moving through a sinusoidal vertical gust field wherethe gust can be considered as an upwash velocity that isuniformly convected by the free streamThe forcing functionin this case is
119908119892 (119909 119905) = sin(120596119892119905 minus120596119892119909
119881) (47)
where 120596119892 is the gust frequency If the gust is referenced tothe airfoil leading edge then 119909 = 0 and 119908119892(119909 119905) becomes119908119892(119909 119905) = sin(120596119892119905) and if the gust is referenced to themidchord then 119909 = 119887 = 1198882 and
120596119892119909
119881=
120596119892 (1198882)
119881=
120596119892119888
2119881= 119896119892
(48)
Therefore119908119892(119905) = sin(120596119892119905)sdotcos(119896119892)minussin(119896119892)sdotcos(120596119892119905) whichis equivalent to a phase shift In this case the lift coefficientcan be written as
119888119897 = 2120587(1199080
119881) 119878 (119896119892) (49)
where 119878(119896119892) is known as Sears function and the gust encoun-ter frequency 119896119892 is given by
119896119892 =2120587119881
120582119892
(50)
and 120582119892 is the wavelength of the gustIn terms of Bessel functions Searsrsquos function is given by
119878 (119896119892) = [1198690 (119896119892) minus 1198941198691 (119896119892)] 119862 (119896119892) + 1198941198691 (119896119892) (51)
The terms of real and imaginary parts are
Real 119878 (119896119892) = 119865 (119896119892) sdot 1198690 (119896119892) + 119866 (119896119892) sdot 1198691 (119896119892)
Im 119878 (119896119892) = 119866 (119896119892) sdot 1198690 (119896119892) minus 119865 (119896119892) sdot 1198691 (119896119892) + 1198691 (119896119892)
(52)
If the gust is referenced to the leading edge of the airfoil thefunction will be called 1198781015840 and can be written as
Real 1198781015840 (119896119892) = Real 119878 (119896119892) sdot cos (119896119892)
+ Im 119878 (119896119892) sdot sin (119896119892)
Im 1198781015840(119896119892) = minusReal 119878 (119896119892) sdot sin (119896119892)
+ Im 119878 (119896119892) sdot cos (119896119892)
(53)
The two results are plotted in Figure 12 The peculiar spiralshape of the 119878 transfer function arises only when the gustfront is referenced to the midchord of the airfoil
The kinematics of the pitching and plunging airfoil ofa typical blade element is the resultant of a combination offorcing from collective and cyclic blade pitch twist angle
03
02
01
00
minus01
minus02
minus03
minus04
minus025 0 025 050 075 1
Imag
inar
y pa
rt
Sears function referenced to the midchordSears function referenced to the leading edge
Real part
Figure 12 Sears function
elastic torsion blade flapping velocity and elastic bendingAt low angles of attack with fully attached flow the varioussources of unsteady effects manifest primarily as moderateamplitude and phase variations relative to the quasi-steadyairloads At higher angles of attack when time-dependentflow separation from the airfoil may be involved the dynamicstall may occurThe amplitude and phase effects produced bythe stalled airloads can lead to various aeroelastic problemson the helicopter rotor that may limit its performance Theneed to control the aerodynamic forces on the rotor requiresthat the pitch of each blade be changed individually as theblades rotate about the shaft
8 Summary
The first flap frequency of a helicopter rigid blade is about105Ω then the reduced frequency at the 75 radius locationis 119896 = 07119888119877 For a blade aspect ratio 119877119888 gt 10 the reducedfrequency is in excess of 007 which is in the unsteadyrange In the case of the first elastic torsion mode which istypically about 3 divide 4Ω at the blade tip the reduced frequencyis in excess of 02 At these reduced frequencies there isa significant amplitude and phasing introduced into theairloads by the effects of the unsteady aerodynamics
When a wingrsquos angle of attack is increased rapidly it canmomentarily generate a higher maximum lift coefficient thanit could if the angle of attack was increased slowly This over-shoot can be related to the change in angle of attack duringthe time required for the air to travel one chord length Thedynamic overshoot is attributed to two effects (for the airfoilsthat stall first at the leading edge) the delay in the separationof the boundary layer and the momentary existence of avortex shed at the leading edge after the boundary layer doesseparate The delay in separation corresponds to the finite
Mathematical Problems in Engineering 9
time required for the aft edge of the separation bubble tomove forward to its bursting positionOn the other hands theairfoil can generate high lift as a result of a vortex that is shedat the leading edge at the instant of stall The vortex travelsback over the top of the airfoil carrying with it a low pressurewave that accounts for the very large lift coefficient Airfoilsthat stall first at the trailing edge also exhibit a dynamicovershoot but considerably less than those airfoils that haveleading edge stall
References
[1] J G Leishman Principles of Helicopter Aerodynamics Cam-bridge University Press 2007
[2] J Seddon and SNewmanBasicHelicopterAerodynamics AIAAEducation Series 2001
[3] M A Ashraf J Young and I C S Lai ldquoOscillation frequencyand amplitude effects on plunging airfoil propulsion and flowperiodicityrdquo The American Institute of Aeronautics and Astro-nautics Journal pp 2308ndash2324 2012
[4] R Prouty Helicopter Performance Stability and Control Kri-eger Gainesville Fla USA 2002
[5] P D Gareth Helicopter Flight Dynamics AIAA EducationSeries 2007
[6] B J Lee and M S Liou ldquoUnsteady adjoint approach for designoptimization of flapping airfoilsrdquo The American Institute ofAeronautics and Astronautics Journal pp 2460ndash2491 2012
[7] K Karamcheti Principles of Ideal-Fluid Aerodynamics KriegerGainesville Fla USA 1966
[8] T Theodorsen ldquoGeneral theory of aerodynamic instability andthe mechanism of flutterrdquo Tech Rep 496 NACA 1934
[9] I EGarrick ldquoOn some reciprocal relations in the theory of non-stationary flowsrdquo NACA Report 629 1938
[10] J Katz and A Plotkin Low Speed Aerodynamics CambridgeUniversity Press 2010
[11] C Rotaru I Circiu and M Boscoianu ldquoComputational meth-ods for the aerodynamic designrdquo Review of the Air Force Acad-emy vol 2 no 17 pp 43ndash48 2010
[12] C Rotaru and V Jaulin ldquoNumerical simulations of Helicopterrotor blade airloadsrdquoMTAReview vol 22 no 2 pp 81ndash90 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Mathematical Problems in Engineering
dominate the airloads This is also shown by the phase anglewhich exhibits an increasing lead for 119896 gt 03
Von Karman and Sear analyzed the problem of a thinairfoil moving through a sinusoidal vertical gust field wherethe gust can be considered as an upwash velocity that isuniformly convected by the free streamThe forcing functionin this case is
119908119892 (119909 119905) = sin(120596119892119905 minus120596119892119909
119881) (47)
where 120596119892 is the gust frequency If the gust is referenced tothe airfoil leading edge then 119909 = 0 and 119908119892(119909 119905) becomes119908119892(119909 119905) = sin(120596119892119905) and if the gust is referenced to themidchord then 119909 = 119887 = 1198882 and
120596119892119909
119881=
120596119892 (1198882)
119881=
120596119892119888
2119881= 119896119892
(48)
Therefore119908119892(119905) = sin(120596119892119905)sdotcos(119896119892)minussin(119896119892)sdotcos(120596119892119905) whichis equivalent to a phase shift In this case the lift coefficientcan be written as
119888119897 = 2120587(1199080
119881) 119878 (119896119892) (49)
where 119878(119896119892) is known as Sears function and the gust encoun-ter frequency 119896119892 is given by
119896119892 =2120587119881
120582119892
(50)
and 120582119892 is the wavelength of the gustIn terms of Bessel functions Searsrsquos function is given by
119878 (119896119892) = [1198690 (119896119892) minus 1198941198691 (119896119892)] 119862 (119896119892) + 1198941198691 (119896119892) (51)
The terms of real and imaginary parts are
Real 119878 (119896119892) = 119865 (119896119892) sdot 1198690 (119896119892) + 119866 (119896119892) sdot 1198691 (119896119892)
Im 119878 (119896119892) = 119866 (119896119892) sdot 1198690 (119896119892) minus 119865 (119896119892) sdot 1198691 (119896119892) + 1198691 (119896119892)
(52)
If the gust is referenced to the leading edge of the airfoil thefunction will be called 1198781015840 and can be written as
Real 1198781015840 (119896119892) = Real 119878 (119896119892) sdot cos (119896119892)
+ Im 119878 (119896119892) sdot sin (119896119892)
Im 1198781015840(119896119892) = minusReal 119878 (119896119892) sdot sin (119896119892)
+ Im 119878 (119896119892) sdot cos (119896119892)
(53)
The two results are plotted in Figure 12 The peculiar spiralshape of the 119878 transfer function arises only when the gustfront is referenced to the midchord of the airfoil
The kinematics of the pitching and plunging airfoil ofa typical blade element is the resultant of a combination offorcing from collective and cyclic blade pitch twist angle
03
02
01
00
minus01
minus02
minus03
minus04
minus025 0 025 050 075 1
Imag
inar
y pa
rt
Sears function referenced to the midchordSears function referenced to the leading edge
Real part
Figure 12 Sears function
elastic torsion blade flapping velocity and elastic bendingAt low angles of attack with fully attached flow the varioussources of unsteady effects manifest primarily as moderateamplitude and phase variations relative to the quasi-steadyairloads At higher angles of attack when time-dependentflow separation from the airfoil may be involved the dynamicstall may occurThe amplitude and phase effects produced bythe stalled airloads can lead to various aeroelastic problemson the helicopter rotor that may limit its performance Theneed to control the aerodynamic forces on the rotor requiresthat the pitch of each blade be changed individually as theblades rotate about the shaft
8 Summary
The first flap frequency of a helicopter rigid blade is about105Ω then the reduced frequency at the 75 radius locationis 119896 = 07119888119877 For a blade aspect ratio 119877119888 gt 10 the reducedfrequency is in excess of 007 which is in the unsteadyrange In the case of the first elastic torsion mode which istypically about 3 divide 4Ω at the blade tip the reduced frequencyis in excess of 02 At these reduced frequencies there isa significant amplitude and phasing introduced into theairloads by the effects of the unsteady aerodynamics
When a wingrsquos angle of attack is increased rapidly it canmomentarily generate a higher maximum lift coefficient thanit could if the angle of attack was increased slowly This over-shoot can be related to the change in angle of attack duringthe time required for the air to travel one chord length Thedynamic overshoot is attributed to two effects (for the airfoilsthat stall first at the leading edge) the delay in the separationof the boundary layer and the momentary existence of avortex shed at the leading edge after the boundary layer doesseparate The delay in separation corresponds to the finite
Mathematical Problems in Engineering 9
time required for the aft edge of the separation bubble tomove forward to its bursting positionOn the other hands theairfoil can generate high lift as a result of a vortex that is shedat the leading edge at the instant of stall The vortex travelsback over the top of the airfoil carrying with it a low pressurewave that accounts for the very large lift coefficient Airfoilsthat stall first at the trailing edge also exhibit a dynamicovershoot but considerably less than those airfoils that haveleading edge stall
References
[1] J G Leishman Principles of Helicopter Aerodynamics Cam-bridge University Press 2007
[2] J Seddon and SNewmanBasicHelicopterAerodynamics AIAAEducation Series 2001
[3] M A Ashraf J Young and I C S Lai ldquoOscillation frequencyand amplitude effects on plunging airfoil propulsion and flowperiodicityrdquo The American Institute of Aeronautics and Astro-nautics Journal pp 2308ndash2324 2012
[4] R Prouty Helicopter Performance Stability and Control Kri-eger Gainesville Fla USA 2002
[5] P D Gareth Helicopter Flight Dynamics AIAA EducationSeries 2007
[6] B J Lee and M S Liou ldquoUnsteady adjoint approach for designoptimization of flapping airfoilsrdquo The American Institute ofAeronautics and Astronautics Journal pp 2460ndash2491 2012
[7] K Karamcheti Principles of Ideal-Fluid Aerodynamics KriegerGainesville Fla USA 1966
[8] T Theodorsen ldquoGeneral theory of aerodynamic instability andthe mechanism of flutterrdquo Tech Rep 496 NACA 1934
[9] I EGarrick ldquoOn some reciprocal relations in the theory of non-stationary flowsrdquo NACA Report 629 1938
[10] J Katz and A Plotkin Low Speed Aerodynamics CambridgeUniversity Press 2010
[11] C Rotaru I Circiu and M Boscoianu ldquoComputational meth-ods for the aerodynamic designrdquo Review of the Air Force Acad-emy vol 2 no 17 pp 43ndash48 2010
[12] C Rotaru and V Jaulin ldquoNumerical simulations of Helicopterrotor blade airloadsrdquoMTAReview vol 22 no 2 pp 81ndash90 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 9
time required for the aft edge of the separation bubble tomove forward to its bursting positionOn the other hands theairfoil can generate high lift as a result of a vortex that is shedat the leading edge at the instant of stall The vortex travelsback over the top of the airfoil carrying with it a low pressurewave that accounts for the very large lift coefficient Airfoilsthat stall first at the trailing edge also exhibit a dynamicovershoot but considerably less than those airfoils that haveleading edge stall
References
[1] J G Leishman Principles of Helicopter Aerodynamics Cam-bridge University Press 2007
[2] J Seddon and SNewmanBasicHelicopterAerodynamics AIAAEducation Series 2001
[3] M A Ashraf J Young and I C S Lai ldquoOscillation frequencyand amplitude effects on plunging airfoil propulsion and flowperiodicityrdquo The American Institute of Aeronautics and Astro-nautics Journal pp 2308ndash2324 2012
[4] R Prouty Helicopter Performance Stability and Control Kri-eger Gainesville Fla USA 2002
[5] P D Gareth Helicopter Flight Dynamics AIAA EducationSeries 2007
[6] B J Lee and M S Liou ldquoUnsteady adjoint approach for designoptimization of flapping airfoilsrdquo The American Institute ofAeronautics and Astronautics Journal pp 2460ndash2491 2012
[7] K Karamcheti Principles of Ideal-Fluid Aerodynamics KriegerGainesville Fla USA 1966
[8] T Theodorsen ldquoGeneral theory of aerodynamic instability andthe mechanism of flutterrdquo Tech Rep 496 NACA 1934
[9] I EGarrick ldquoOn some reciprocal relations in the theory of non-stationary flowsrdquo NACA Report 629 1938
[10] J Katz and A Plotkin Low Speed Aerodynamics CambridgeUniversity Press 2010
[11] C Rotaru I Circiu and M Boscoianu ldquoComputational meth-ods for the aerodynamic designrdquo Review of the Air Force Acad-emy vol 2 no 17 pp 43ndash48 2010
[12] C Rotaru and V Jaulin ldquoNumerical simulations of Helicopterrotor blade airloadsrdquoMTAReview vol 22 no 2 pp 81ndash90 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of