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Flight Dynamics and Aircraft Performance
Lecture 9:Helicopters
G. DimitriadisUniversity of Liege 1
Textbooks
• Bramwell’s Helicopter Dynamics, A. R. S. Bramwell, G. Done, D. Balmford, Butterworth-Heinemann, 2001
• Basic Helicopter Aerodynamics, J. Seddon, BSP (Blackwell Scientific Publications) Professional, 1990
• Principles of Helicopter Aerodynamics, J. G. Leishman, Cambridge University Press, 2000 2
Introduction• Helicopters can do all this:
Hovering flight, reverse Rolls, backflips etc
3
Not helicopters• Autogyros, gyrogliders etc are not
helicopters.
Focke-Wulf Fw 61, 1936
Cierva C.4, 1923
Fa 330, Gyroglider (or rotorkite),1943
Cierva C.19, 1932
4
Helicopters
Breguet-Dorand, Gyroplane Laboratoire,1935
Gyroplane de Breguet, 1907
Helicoptère de Paul Cornu, 1907
Pescara Helicopter No 3, 1924
5
Belgian First
• The first ever tandem rotor helicopter was built by Nicolas Florine.
• It first flew in 1933 at the Laboratoire Aérotechnique de Belgique (now Von Karman Institute).
6
Modern helicoptersBell 204/205, 1955 Aerospatiale Alouette II, 1955 CH-47, Chinook, 1957
Kamov 50, 1982 Eurocopter Tiger, 1991Mil Mi-26, Hind, 1977 7
Largest helicopter ever built• Mil-V12
8
How does a helicopter fly?
• By accelerating downwards a column of air through the rotor.
The rotor creates a pressure difference Δpwhich accelerates flow through it. The velocity far upstream is 0, at the rotor vi and far downstream v∞.
Disk pi<p∞
p∞
p∞
pi+Δp>p∞
v=0
vi
v∞Flow field Pressure
fieldVelocity field
9
Pressure change
• Using Bernoulli’s equation on the upstream flow (assuming incompressibility) we have:
• On the downstream flow we have:
• So that
p∞ = pi +12ρvi
2
p∞ +12ρv∞
2 = Δp + pi +12ρvi
2
Δp =12ρv∞
2 (1)10
Rotor thrust
• Mass flow through the rotor:
• Far downstream: the momentum flow, i.e. the momentum of the mass that flowed through the rotor is equal to:
• The thrust is the difference in momentum flow, i.e.
m = ρAvi
Jdownstream = m v∞ = ρAviv∞
T = Jdownstream − Jupstream = ρAviv∞ (2) 11
Airspeed at infinity• Noting that the pressure change across
the rotor is a measure of the thrust,
• We can combine with equation (1) to show that
• And that• Define: vi=induced velocity• Define w=T/A=disc loading• Define P=Tvi=induced power of the rotor
Δp =TA
= ρviv∞
v∞ = 2vi
T = 2ρAvi2
12
Thrust for vertical climb
• If the helicopter is climbing at speed Vc
The airspeed upstream is equal to Vc+vi. Donwstream it is equal to Vc+v∞.Bernoulli upstream:
Bernoulli downstream:
p∞ +12ρ Vc + v∞( )2 = Δp + pi +
12ρ Vc + vi( )2
p∞ +12ρ Vc( )2 = pi +
12ρ Vc + vi( )2
Vc
Vc+vi
Vc+v∞13
Thrust for vertical climb (2)
• The pressure change is therefore
• The thrust is given by:
• Combining with (3) gives v∞=2vi, i.e.
• The induced power is
T = ρA Vc + vi( ) Vc + v∞( ) − ρA Vc + vi( )Vc = ρA Vc + vi( )v∞
Pi = T Vc + vi( )
Δp =12ρv∞ 2Vc + v∞( ) (3)
T = 2ρA Vc + vi( )vi
14
Induced velocity and climb velocity
• Consider a hover case where the thrust is equal to Th, the power to Ph and the induced airspeed to vh.
• Consider climbing flight at the same thrust, Th. The rotor climbs but also induces a velocity vi≠vh.
• It is easy to see that• So that
vh2 = Vc + vi( )vi
vivh
= −Vc
2vh+
Vc
2vh
#
$ %
&
' (
2
+1 15
Induced power and climb power
• Therefore, we can write that• Leading to:
Pi = Th Vc + vi( )
PiPh
=Th Vc + vi( )
Thvh=
Vc
2vh+
Vc
2vh
"
# $
%
& '
2
+1
16
Realistic climbing rotor wake
• The results shown before assume that the wake is a column with a smooth and continuous vertical velocity distribution
• A real wake is much more complex
17
Realistic climbing rotor wake (2)
• Depending on the rotation speed, climb speed, blade span and blade twist, the blade can produce:– Lift near the tip
(the wake curls upwards)
– Downforcenear the root (the wake curls downwards)
Blade
Tip vortex
Inner vortexsheet
18
Realistic descending rotor wake
• When the helicopter is descending, the rotor descends into its own wake.
19
Descent• Climb is an easy case. The rotor wake lies
under the rotor and the rotor itself climbs into a smooth airflow.
• On the contrary, when the helicopter is descending, the rotor descends into its own wake.
• There are three different possibilities:– Vortex ring flow: The rotor tips are caught inside
their own vortex rings.– Turbulent wake state: The rate of descent is so
high that the rotor wake develops upwards but is quite turbulent.
– Windmill brake state: The rate is even higher. The rotor wake develops upwards but is well defined.
20
Descent cases
Vortex ring flow, slow descent
Vortex ring flow, faster descent
Turbulent wake state
Windmill brake state
21
Vortex ring state• Denote by Vd the descent speed.• If Vd=O(vh), i.e the induced velocity in hover,
then some of the air recirculates around the rotor.
• Effectively, the rotor wake is squashed onto the rotor.
• The phenomenon leads to very high descent speeds and loss of stability.
• Recovery can be accomplished by moving the helicopter forward so that the rotor encounters clean air as its wake lies behind it.
22
Windmill brake state
• At much higher descent rates, i.e. Vd>>vh, the rotor wake develops upwards.
• The wake is well defined.• The airflow decelerates on passing
through the rotor.• The turbulent wake state lies between
the vortex ring and windmill brake states. The rotor acts as a bluff body.
23
Safe descent• So how can a helicopter achieve a safe
descent?• There are two methods:
– Descend very slowly so that Vd<<vh and the rotor wake effectively descends with the rotor.
– Descend with a forward velocity component so that the rotor wake lies behind the rotor.
• It must be said that near the ground the descent speed will be necessarily low.
• Additionally, near the ground the helicopter can benefit from the ground effect. 24
Ground effect
• A helicopter hovering near the ground benefits from a large improvement in efficiency.
• The vertical velocity of the wake on the ground must be equal to zero.
• Therefore, the induced velocity of the rotor is very low. As P=Tvi, the power required to produce the same amount of thrust is much lower near the ground.
25
Induced velocity in ground effect
• Induced velocity in ground effect divided by induced velocity in free air.
r
h=height above groundr=distance from centre of rotor divided by R.
26
Blade Element Method• Blade Element
Method (BEM), also known as strip theory in aeroelasticity.
• It consists of estimating the aerodynamic forces on a small element of a blade, dy.
R
y
dy
c
ΩR
ψ
ψ=azimuth angle 27
Blade Element• The blade can have a pitch angle of θ. It
also features an inflow angle φ=tan-1[(Vc+vi)/Ωy].
• Its true angle of attack is given by α=θ-φ.
28
Blade element lift and drag
• The blade element lift and drag are given by:
• Where cl and cd come from the sectional characteristics of the blade element.
• The thrust is given by:
• The in-plane torque is given by
dL =12ρU 2ccldy
dD =12ρU 2ccddy
dT = dLcosφ − dDsinφ
dQ = dLsinφ + dDcosφ( )y 29
Approximations
• The inflow angle is assumed to be small.
• The drag coefficient is assumed to be much smaller than the lift coefficient.
• Therefore:
dT ≈ dLU ≈ ΩydQ ≈ φdL + dD( )y
30
Non-dimensionalizations• Define the following non-dimensional
quantities:
• Also, for a rotor with N blades define the solidity factor as:
r = y /R =ΩyΩR
=UΩR
λ =Vc + viΩR
= rφ = inflow factor
dCT =dT
ρA ΩR( )2
dCQ =dQ
ρA ΩR( )2R
σ =blade areadisc area
=NcRπR2 =
NcπR 31
Total thrust and torque
• After non-dimensionalization, the blade element forces can be integrated over the blade span to yield:
• The rotor power requirement is given by P=ΩQ. Non-dimensionalising:
CT = dCT0
1
∫ dr =σ2
CLr2
0
1
∫ dr
CQ = dCQ0
1
∫ dr =σ2
φCL + CD( )r30
1
∫ dr
CP =P
ρA ΩR( )3= CQ 32
Thrust Approximation
• For attached flow, the lift coefficient of a blade element is given by
• where a is the lift curve slope. The thrust coefficient becomes
• So that, finally,
cl = aα = a θ − φ( )
CT =σ2
a θ − φ( )r20
1
∫ dr =σa2
θr2 − λr( )0
1
∫ dr =σa2
θ3−λ2
( )
* +
CT =σa2
θ3−λ2
& '
( )
(4) 33
Thrust in hover
• If the rotor is in hover, and
• Then, from (4),
• Which is a nonlinear equation relating pitch angle θ to thrust. It can be solved inversely as:
T = 2ρAvi2
CT = 2λ2
CT =σa2
θ3−12
CT
2%
& '
(
) *
θ =6σa
CT +32
CT
234
About twist
• As shown earlier, helicopter blades produce little lift near the centre of the rotor because of the low linear speed.
• Define the sectional lift as
• For the case where a=2π,
• Define
l =dLdy
=12ρU 2ccl =
12ρU 2ca θ − φ( )
lρ ΩR( )2c
= πr2 θ − φ( )
cl =l
ρ ΩR( )2c= πr2 θ − φ( ) 35
Effect of twist
• Adding geometric twist to the blade can increase the sectional lift coefficient near the centre of the rotor.
• This generally means increasing the twist towards the centre.
• Consider two cases:– Case θ=θ0. The pitch is constant over the
blade. – Case θ=θ1+θ2r. The pitch varies over the
blade, i.e. there is geometric twist. For the pitch to be higher near centre of the rotor, θ2<0 and θ1>θ0. 36
Twist example
Keep in mind that this result was obtained using BEM. 3D effects near the wingtip have been ignored
37
Ideal twist• The ideal twist distribution is obtained when θr is
constant, i.e. θr=θ0.• This is a nonlinear twist that cannot be implemented at
the blade root but it is ideal because it corresponds to the minimum induced power.
38
Forward Flight• Forward flight
is different to vertical climb and hover!
• It creates a total thrust that is not centered on the rotor.
• This thrust causes a significant rolling moment on the rotor, making the helicopter impossible to fly.
Ω
Forward velocity V
V+ΩR
V-ΩR
V
Reversedvelocity
39
Avoiding the rolling moment
• The way to cancel the rolling moment is to allow the blade to flap.
• The additional lift of the advancing blade causes an upward flapping motion.
• Similarly, the lower lift of the retreating blade causes a downward flapping motion.
• Therefore, the rolling moment is not transmitted to the helicopter.
40
Flapping
• Flapping is a stable motion because flapping up causes the lift to drop and flapping down to increase
V+Ωr
β r
Advancing bladeflaps upwards
α<θ
θ θ
Retreating bladeflaps downwards
Ωr-V
β rα>θ
41
Corioli’s moments
• The flapping motion causes Corioli’s moments on the blades:
ΩR
(1-e)ΩRcosβ+ΩeR
The Corioli’s moment is due to the inequality of the tip speeds of the flapped and unflapped blades.It can cause a yawing moment on the helicopter 42
Lagging motion
• The way to avoid the yaw moment due to flapping is to allow the blade to lag:
43
Pitching (feathering)
• The rotor is not only the lifting surface but also the propulsion and main control system.
• The main means of control of the rotor is the changing of the pitch of the blades (also known as feathering).
• Pitch control can be either collective (all blades change pitch at the same time) or cyclic (the pitch change depends on whether the blade is advancing or retreating). 44
Westland Wessex hub
Flap hinge
Lag hinge
Pitch control
Pitchbearing
45
Westland LynxHingeless rotor: the blades are not hinged, they are solidly connected to the rotor hub. However, they have flexible elements near the root which allow flap and lag degrees of freedom, restrained by the stiffness of these elements.
Pitch bearings
Lag dampers
Flexible elements
46
Helicopter control• Control of the helicopter is handled almost
exclusively by the rotor. There are two parameters of importance:– Magnitude of rotor thrust– Line of action of rotor thrust
• Both of these parameters are controlled by rotor pitch. – Collective pitch increases the magnitude of the
thrust.– Cyclic pitch can change the line of action of
the thrust47
Collective vs cyclic pitchThe swashplate mechanism:-Lifting or lowering the swashplate increases or decreases collective pitch.-Tilting the swashplate introduces cyclic pitch.-In this case cyclic pitch is used to increase the angle of attack of the retreating blade. 48
Cyclic pitch• Cyclic pitch changes the pitch angle θ with
azimuth angle ψ.• This change is usually expressed as a first
order Fourier series:
A1, the lateral cyclic coefficient, applies maximum/minimum pitch when the blades are at ψ=0o/ψ=180o. The blade response is phased by 90o, hence the lateral effect.B1, the longitudinal cyclic coefficient, applies maximum/minimum pitch when the blades are at ψ=90o/ψ=270o. Again, the blade response if phased by 90o.
θ ψ( ) = θ0 − A1 cosψ − B1 sinψ
49
Tip Path Plane• Using cyclic pitch it is possible to incline
the rotor without inclining the rotor shaft.• The line of action of the thrust is
perpendicular to the blade Tip Path Plane: T
50
Forward flight, Forward C.G
D
mg
C.G
TCase where the Centre of Gravity lies in front of the rotor shaft.
In this case, the resultant of the weight and drag on the fuselage lies on the same line of action as the thrust.
Tip Path Plane
51
Forward flight, Aft C.G
D
mg
C.G
TTip Path Plane Case where the
Centre of Gravity lies aft of the rotor shaft.
Again, the resultant of the weight and drag on the fuselage lies on the same line of action as the thrust.The pitch angle of the fuselage is much smaller than in the forward C.G. case.
52
Direct Head Moment
D
mg
C.G
TTip Path Plane
MfIn a more general case, the drag on the fuselage will also cause a fuselage pitching moment, Mf. This moment will be counteracted by the fact that the thrust and resultant of fuselage weight and drag are not colinear.
53
How to start going forward
• A hovering helicopter has no forward velocity.• The pilot uses cyclic pitch to tip the Tip Path
Plane forward and tilt the thrust vector forward.• The helicopter picks up forward speed. • The fuselage develops drag and pitches nose
down.• Now the rotor shaft is also pitched nose down;
there is no more need to apply cyclic pitch to the rotor.
T T T
W W WD D
54
Drag• There are two main sources of drag:
– Fuselage drag– Rotor drag
• Fuselage drag is usually calculated in terms of the so-called equivalent flat plate area
• Rotor drag is subdivided into – profile drag – induced drag
55
Fuselage Drag• There are two source of fuselage drag:
– Parasite drag– Interference drag
• Parasite drag has many sources:
• Interference drag is caused by the interaction of flow coming from these different components. 56
Parasite drag examples
Define D=1/2ρV2SFP, SFP being the equivalent flat plate area,i.e. the area of a flat plate that has the same drag as the fuselage.
57
Rotor drag• There are two main contributions to rotor
drag:– Profile drag– Induced drag
• The profile drag is evaluated with respect to the drag of the chosen airfoil section and the angle of attack of the blade using blade element theory.
• The induced drag can be assumed to be small for forward steady flight.
58
Power required for forward flight
µ =V cosαD
ΩR
There is an optimum advance ratio, μ, requiring minimum power.
Maximum forward speed
Maximum climb rate
59