Helicopter Slides

© L. Sankar Helicopter Aerodynamics

1

Helicopter Aerodynamics and Performance

Preliminary Remarks


2

ThrustAeroelasticResponse

0

270 180

90

Dynamic Stall onRetreating Blade

Blade-Tip Vortexinteractions

UnsteadyAerodynamicsTransonic Flow on

Advancing Blade

Main Rotor / Tail Rotor/ Fuselage

Flow Interference

V

NoiseShock Waves

Tip Vortices

The problems are many..


3

A systematic Approach is necessary

• A variety of tools are needed to understand, and predict these phenomena.• Tools needed include

– Simple back-of-the envelop tools for sizing helicopters, selecting engines, laying out configuration, and predicting performance

– Spreadsheets and MATLAB scripts for mapping out the blade loads over the entire rotor disk

– High end CFD tools for modeling• Airfoil and rotor aerodynamics and design• Rotor-airframe interactions• Aeroacoustic analyses

– Elastic and multi-body dynamics modeling tools– Trim analyses, Flight Simulation software

• In this work, we will cover most of the tools that we need, except for elastic analyses, multi-body dynamics analyses, and flight simulation software.

• We will cover both the basics, and the applications.• We will assume familiarity with classical low speed and high speed

aerodynamics, but nothing more.


4

Plan for the Course• PowerPoint presentations, interspersed with

numerical calculations and spreadsheet applications.

• Part 1: Hover Prediction Methods• Part 2: Forward Flight Prediction Methods• Part 3: Helicopter Performance Prediction

Methods• Part 4: Introduction to Comprehensive Codes

and CFD tools• Part 5: Completion of CFD tools, Discussion of

Advanced Concepts


5

Text Books• Wayne Johnson: Helicopter Theory, Dover

Publications,ISBN-0-486-68230-7• References:

– Gordon Leishman: Principles of Helicopter Aerodynamics, Cambridge Aerospace Series, ISBN 0-521-66060-2

– Prouty: Helicopter Performance, Stability, and Control, Prindle, Weber & Schmidt, ISBN 0-534-06360-8

– Gessow and Myers– Stepniewski & Keys


6

Grading

• 5 Homework Assignments (each worth 5%).• Two quizzes (each worth 25%)• One final examination (worth 25%)• All quizzes and exams will be take-home type.

They will require use of an Excel spreadsheet program, or optionally short computer programs you will write.

• All the material may be submitted electronically.


7

Instructor Info.

• Lakshmi N. Sankar• School of Aerospace Engineering, Georgia

Tech, Atlanta, GA 30332-0150, USA.• Web site: www.ae.gatech.

edu/~lsankar/AE6070.Fall2002• E-mail Address: [email protected]

http://www.ae.gatech.edu/~lsankar/AE6070.Fall2002

mailto:[email protected]


8

Earliest Helicopter..Chinese Top


9

Leonardo da Vinci(1480? 1493?)


10

Human Powered Flight?

HP 6.7 5.33/0.8 Merit of rePower/Figu Ideal Power Actual

33.5A2

W WPower Ideal

slugs. 0.00238Desnitysq.ft 100 AreaRotor

6ft ~RadiusRotor 160lbfWeight

HP


11

D’AmeCourt (1863)Steam-Propelled Helicopter


12

Paul Cornu (1907)First man to fly in helicopter mode..


13

De La Ciervainvented Autogyros (1923)


14

Cierva introduced hinges at the rootthat allowed blades to freely flap

Hinges

Only the lifts were transferred to the fuselage, not unwanted moments.In later models, lead-lag hinges were also used toAlleviate root stresses from Coriolis forces

http://www.infonegocio.com/dagafox/autogiro/herrera.htm


15

Igor Sikorsky Started work in 1907, Patent in 1935

Used tail rotor to counter-act the reactive torque exerted by the rotor on the vehicle.


16

Sikorsky’s R-4


17

Ways of countering the Reactive Torque

Other possibilities: Tip jets, tip mounted engines


18

Single Rotor Helicopter


19

Tandem Rotors (Chinook)


20

Coaxial rotorsKamov KA-52


21

NOTAR Helicopter


22

NOTAR Concept


23

Tilt Rotor Vehicles


24

Helicopters tend to grow in size..

16,027 lb (7270 kg) Lot 1 Weight

15,075 lb (6838 kg)11,800 pounds Empty

Primary Mission Gross Weight

17.15 ft (5.227 m) 17.15 ft (5.227 m) Wing Span

13.30 ft (4.05 m) 15.24 ft (4.64 m) Height

58.17 ft (17.73 m) 58.17 ft (17.73 m) Length

AH-64D AH-64A


25

147 kt (273 kph)[Sea Level Standard Day]149 kt (276 kph)[Hot Day 2000 ft 70 F (21 C)]


Cruise Speed (MCP)



Maximum Level Flight Speed

2,635 fpm (803 mpm)[Sea Level Standard Day]2,600 fpm (793 mpm)[Hot Day 2000 ft 70 F (21 C)]


Maximum Rate of Climb (IRP)



Vertical Rate of Climb (MRP)

10,520 ft (3206 m)[Standard Day]9,050 ft (2759 m)[Hot Day ISA + 15 C]

12,685 ft (3866 m)[Sea Level Standard Day]11,215 ft (3418 m)[Hot Day 2000 ft 70 F (21 C)]

Hover Out-of-Ground Effect (MRP)

14,650 ft (4465 m)[Standard Day]13,350 ft (4068 m)[Hot Day ISA + 15 C]

15,895 ft (4845 m)[Standard Day]14,845 ft (4525 m)[Hot Day ISA + 15C]

Hover In-Ground Effect (MRP)

16,027 lb (7270 kg) Lot 1 Weight

15,075 lb (6838 kg)11,800 pounds Empty

Primary Mission Gross Weight

17.15 ft (5.227 m) 17.15 ft (5.227 m) Wing Span

13.30 ft (4.05 m) 15.24 ft (4.64 m) Height

58.17 ft (17.73 m) 58.17 ft (17.73 m) Length

AH-64D AH-64A


26

Power Plant Limitations

• Helicopters use turbo shaft engines.• Power available is the principal factor.• An adequate power plant is important for

carrying out the missions.• We will look at ways of estimating power

requirements for a variety of operating conditions.


27

High Speed Forward Flight Limitations

• As the forward speed increases, advancing side experiences shock effects, retreating side stalls. This limits thrust available.

• Vibrations go up, because of the increased dynamic pressure, and increased harmonic content.

• Shock Noise goes up.• Fuselage drag increases, and parasite power

consumption goes up as V3.• We need to understand and accurately predict

the air loads in high speed forward flight.


28

Concluding Remarks

• Helicopter aerodynamics is an interesting area.• There are a lot of problems, but there are also

opportunities for innovation.• This course is intended to be a starting point for

engineers and researchers to explore efficient (low power), safer, comfortable (low vibration), environmentally friendly (low noise) helicopters.


29

Hover Performance Prediction Methods

I. Momentum Theory


30

Background

• Developed for marine propellers by Rankine (1865), Froude (1885).

• Extended to include swirl in the slipstream by Betz (1920)

• This theory can predict performance in hover, and climb.

• We will look at the general case of climb, and extract hover as a special situation with zero climb velocity.


31

Assumptions

• Momentum theory concerns itself with the global balance of mass, momentum, and energy.

• It does not concern itself with details of the flow around the blades.

• It gives a good representation of what is happening far away from the rotor.

• This theory makes a number of simplifying assumptions.


32

Assumptions (Continued)

• Rotor is modeled as an actuator disk which adds momentum and energy to the flow.

• Flow is incompressible.• Flow is steady, inviscid, irrotational.• Flow is one-dimensional, and uniform

through the rotor disk, and in the far wake.• There is no swirl in the wake.


33

Control Volume is a CylinderV

Disk area A

Total area S

Station1

2

3

4

V+v2

V+v3

V+v4


34

Conservation of Mass

444

1

)(A-SV bottom he through tOutflowm side he through tInflowVS tophe through tInflow

AvV


35

Conservation of Mass through the Rotor Disk

Flow through the rotor disk =

44

32

vvv

VAVAVAm

Thus v2=v3=v

There is no velocity jump across the rotor diskThe quantity v is called induced velocity at the rotor disk


36

Global Conservation of Momentum

4444

42

42

4

44

1

2

vv)v(A Tin Rate Momentum

-out rate MomentumT,Thrust.boundaries fieldfar the

allon catmospheri is PressurevA-S

bottom through outflow MomentumvA

Vm side he through tinflow MomentumV op through tinflow Momentum

mV

AVV

V

S

Mass flow rate through the rotor disk timesExcess velocity between stations 1 and 4


37

Conservation of Momentum at the Rotor Disk

V+v

V+v

p2

p3

Due to conservation of mass across theRotor disk, there is no velocity jump.

Momentum inflow rate = Momentum outflow rate

Thus, Thrust T = A(p3-p2)


38

Conservation of EnergyConsider a particle that traverses fromStation 1 to station 4

We can apply Bernoulli equation betweenStations 1 and 2, and between stations 3 and 4.Recall assumptions that the flow is steady, irrotational, inviscid.

1

2

3

4

V+v

V+v4

44

23

24

23

222

v2v

v21v

21

21v

21

Vpp

VpVp

VpVp


39

44

23

44

23

v2v

v2v

#38, slide previous theFrom

VAppAT

Vpp

From an earlier slide # 36, Thrust equals mass flow rate through the rotor disk times excess velocity between stations 1 and 4

4vv VAT Thus, v = v4/2


40

Induced Velocities

V

V+v

V+2v

The excess velocity in theFar wake is twice the inducedVelocity at the rotor disk.

To accommodate this excessVelocity, the stream tube has to contract.


41

Induced Velocity at the Rotor DiskNow we can compute the induced velocity at the rotor disk in terms of thrust T.

T = Mass flow rate through the rotor disk * (Excess velocity between 1 and 4).

T = 2 A (V+v) v

ATV222

V-v2

There are two solutions. The – sign Corresponds to a wind turbine, where energy Is removed from the flow. v is negative.

The + sign corresponds to a rotor orPropeller where energy is added to the flow.In this case, v is positive.


42

Induced velocity at the rotor disk

AT

ATV

2v

0V velocity climb Hover,In 222

V-v2


43

Ideal Power Consumed by the Rotor

ATVVT

VTVm

mm

P

222

vvv2

V212vV

21

in flowEnergy -out flowEnergy

2

22

In hover, ideal power ATT2


44

Summary• According to momentum theory, the downwash

in the far wake is twice the induced velocity at the rotor disk.

• Momentum theory gives an expression for induced velocity at the rotor disk.

• It also gives an expression for ideal power consumed by a rotor of specified dimensions.

• Actual power will be higher, because momentum theory neglected many sources of losses- viscous effects, compressibility (shocks), tip losses, swirl, non-uniform flows, etc.


45

Figure of Merit• Figure of merit is

defined as the ratio of ideal power for a rotor in hover obtained from momentum theory and the actual power consumed by the rotor.

• For most rotors, it is between 0.7 and 0.8.

P

TT

C

CCT

FM

2Pv

Hoverin Power ActualHoverin Power Ideal


46

Some Observations on Figure of Merit

• Because a helicopter spends considerable portions of time in hover, designers attempt to optimize the rotor for hover (FM~0.8).

• We will discuss how to do this later.• A rotor with a lower figure of merit (FM~0.

6) is not necessarily a bad rotor.• It has simply been optimized for other

conditions (e.g. high speed forward flight).


47

Example #1

• A tilt-rotor aircraft has a gross weight of 60,500 lb. (27500 kg).

• The rotor diameter is 38 feet (11.58 m).• Assume FM=0.75, Transmission

losses=5%• Compute power needed to hover at sea

level on a hot day.


48

Example #1 (Continued)

HP 11528 1.05*10980shaft the toengine by the suppliedPower lossion transmiss5% is There

HP 10980 power actual totalrotors, twoFor theHP 5490 power Actual

4117/0.75Power/FM idealPower ActualHP 4117Power Ideal

ft/sec lb 74.86 x 30250 Tv Power Ideal! ft/sec 150 far wake in theDownwash

ft/sec 86.74vA2

T v velocity,Induced

lbf 30250 T rotors. twoare Therefeet cslugs/cubi 0.00238 Density

feet square 12.113419A AreaDisk 2

A


49

Alternate scenarios

• What happens on a hot day, and/or high altitude?– Induced velocity is higher.– Power consumption is higher

• What happens if the rotor disk area A is smaller?– Induced velocity and power are higher.

• There are practical limits to how large A can be.


50

Disk Loading

• The ratio T/A is called disk loading.• The higher the disk loading, the higher the

induced velocity, and the higher the power.• For helicopters, disk loading is between 5 and

10 lb/ft2 (24 to 48 kg/m2).• Tilt-rotor vehicles tend to have a disk loading of

20 to 40 lbf/ft2. They are less efficient in hover.• VTOL aircraft have very small fans, and have

very high disk loading (500 lb/ft2).


51

Power Loading

• The ratio of thrust to power T/P is called the Power Loading.

• Pure helicopters have a power loading between 6 to 10 lb/HP.

• Tilt-rotors have lower power loading – 2 to 6 lb/HP.

• VTOL vehicles have the lowest power loading – less than 2 lb/HP.


52

Non-Dimensional Forms

QC

P

2Q

3P

2T

CQP

Torquelocity x Angualr ve Power hover,In RAR

QtCoefficien TorqueC

RAPtCoefficienPower C

RATtCoefficienThrust C

form. ldimensiona-nonin expressedusually arePower and Torque, Thrust,


53

Non-dimensional forms..

P

TT

i

C

CCT

FM

2Pv

Hoverin Power ActualHoverin Power Ideal

2

CA2

TR

1R

v inflow Induced T


54

Tip Losses

R A portion of the rotor near theTip does not produce much liftDue to leakage of air fromThe bottom of the disk to the top.

One can crudely account for it byUsing a smaller, modified radius

BR, where

bC

B T21

BR

B = Number of blades.


55

Power Consumption in HoverIncluding Tip Losses..

211 T

TPCC

BFMC


56

Hover PerformancePrediction Methods

II. Blade Element Theory


57

Preliminary Remarks

• Momentum theory gives rapid, back-of-the-envelope estimates of Power.

• This approach is sufficient to size a rotor (i.e. select the disk area) for a given power plant (engine), and a given gross weight.

• This approach is not adequate for designing the rotor.


58

Drawbacks of Momentum Theory

• It does not take into account– Number of blades– Airfoil characteristics (lift, drag, angle of zero

lift)– Blade planform (taper, sweep, root cut-out)– Blade twist distribution– Compressibility effects


59

Blade Element Theory• Blade Element Theory rectifies many of these

drawbacks. First proposed by Drzwiecki in 1892.• It is a “strip” theory. The blade is divided into a

number of strips, of width r.• The lift generated by that strip, and the power

consumed by that strip, are computed using 2-D airfoil aerodynamics.

• The contributions from all the strips from all the blades are summed up to get total thrust, and total power.


60

Typical Blade Section (Strip)

R

dr

r

Tip

OutCut

Tip

OutCut

dPbP

dTbT

dT

Root Cut-out


61

Typical Airfoil Section

r

V varctan

r

V+v

Line of Zero Lift

effective =


62

Sectional ForcesOnce the effective angle of attack is known, we can look-up the lift and drag coefficients for the airfoil section at that strip.

We can subsequently compute sectional lift and drag forces per foot (or meter) of span.

dPT

lPT

cCUUD

cCUUL

21

21

22

22

These forces will be normal to and along the total velocity vector.

UT=r

UP=V+v


63

Rotation of Forces

rV+v

L

D

T

Fx

XxT

ldPT

x

dlPT

rdFdFUdP

drCCcUU

drLDdF

drCCcUU

drDLdT

sincos21

sincos

sincos21

sincos

22

22


64

Approximate Expressions

• The integration (or summation of forces) can only be done numerically.

• A spreadsheet may be designed. A sample spreadsheet is being provided as part of the course notes.

• In some simple cases, analytical expressions may be obtained.


65

Closed Form Integrations• The chord c is constant. Simple linear twist.• The inflow velocity v and climb velocity V are small.

Thus, << 1.

• We can approximate cos( ) by unity, and approximate sin( ) by ( ).

• The lift coefficient is a linear function of the effective angle of attack, that is, Cl=a() where a is the lift curve slope.

• For low speeds, a may be set equal to 5.7 per radian.

• Cd is small. So, Cd sin() may be neglected.

• The in-plane velocity r is much larger than the normal component V+v over most of the rotor.


66

Closed Form Expressions

drrCrr

Vrr

VcbaP

drrrr

VcbaT

Rr

rd

Rr

r

3

0

3

2

0

2

vv21

v21


67

Linearly Twisted Rotor: ThrustHere, we assume that the pitch angle varies as

E Fr

RvV

aRbc

where

aR

abcC

RRcabRRvVFREcabT

RT

R

Ratio Inflow

)2(~ slope CurveLift /DiskAreaBladeArea/solidity

2/32

2/32

2/3224

331

2

75.75.

75.232


68

Linearly Twisted RotorNotice that the thrust coefficient is linearly proportional to the pitch angle at the 75% Radius.

This is why the pitch angle is usually defined at 75% R in industry.The expression for power may be integrated in a similar manner, if the drag coefficient Cd is assumed to be a constant, equal to Cd0.

80d

TPCCC

Induced Power Profile Power


69

Closed Form Expressions forIdeally Twisted Rotor

rRtip

tipTaC

4

C CC

P Td

0

8Same as linearlyTwisted rotor!


70

Figure of Merit according to Blade Element Theory

AreaArea/Disk Blade Solidity Rv)/(V Ratio Inflow

,8/0

whereCCCFM

dT

T

High solidity (lot of blades, wide-chord, large blade area) leads to higherPower consumption, and lower figure of merit.

Figure of merit can be improved with the use of low drag airfoils.


71

Average Lift Coefficient• Let us assume that

every section of the entire rotor is operating at an optimum lift coefficient.

• Let us assume the rotor is untapered.

T

T

R

CRbc

RRTC

RbcdrrcbT

6C

6C

6C

6CC

21

CtCoefficienLift Average

l

ll22

32l

l0

2

l

Rotor will stall if average lift coefficient exceeds 1.2, or so.

Thus, in practice, CT/ is limited to 0.2 or so.


72

Optimum Lift Coefficient in Hover

minimized. is/C if maximized is FM

6/C If82

2

2C hover,In

8

2/3d0

T

02/3

2/3

T

0

l

l

dT

T

dT

T

CC

CC

C

FM

CC

CFM


73

Drawbacks of Blade Element Theory• It does not handle tip losses.

– Solution: Numerically integrate thrust from the cutout to BR, where B is the tip loss factor. Integrate torque from cut-out all the way to the tip.

• It assumes that the induced velocity v is uniform.• It does not account for swirl losses.• The Predicted power is sometimes empirically

corrected for these losses.

15.18

0

d

TPCCC


74

Example(From Leishman)

• Gross Weight = 16,000lb• Main rotor radius = 27 ft• Tail rotor radius 5.5 ft• Chord=1.7 ft (main), Tail rotor chord=0.8 ft• No. of blades =4 (Main rotor), 4 (tail rotor)• Tip speed= 725 ft/s (main), 685 ft/s (tail)• K=1.15, Cd0=0.008• Available HP =3000Transmission losses=10%• Estimate hover ceiling (as density altitude)


75

Step I• Multiply 3000 HP by 550 ft.lb/sec.• Divide this by 1.10 to account for available

power to the two rotors (10% transmission loss).

• We will use non-dimensional form of power into dimensional forms, as shown below:

• P=Tv+(R)3A [Cd0/8]• Find an empirical fit for variation of with

altitude: 2553.4

16.28800198.01

h

levelsea


76

Step 2• Assume an altitude, h. Compute density, .• Do the following for main rotor:

– Find main rotor area A– Find v as [T/(2A)]1/2 Note T= Vehicle weight in lbf.– Insert supplied values of , Cd0, W to find main rotor P.– Divide this power by angular velocity W to get main rotor

torque.– Divide this by the distance between the two rotor shafts

to get tail rotor thrust.• Now that the tail rotor thrust is known, find tail rotor

power in the same way as the main rotor.• Add main rotor and tail rotor powers. Compare with

available power from step 1.• Increase altitude, until required power = available

power.• Answer = 10,500 ft


77

Hover PerformancePrediction Methods

III. Combined Blade Element-Momentum (BEM) Theory


78

Background

• Blade Element Theory has a number of assumptions.

• The biggest (and worst) assumption is that the inflow is uniform.

• In reality, the inflow is non-uniform.• It may be shown from variational calculus

that uniform inflow yields the lowest induced power consumption.


79

Consider an Annulus of the rotor Disk

r

dr

Area = 2rdr

Mass flow rate =2rV+vdr

dT = (Mass flow rate) * (twice the induced velocity at the annulus) = 4r(V+v)vdr


80

Blade Elements Captured by the Annulus

r

dr

Thrust generated by these blade elements:

drrvVrabc

drCcrbdT l

2

2

21

21


81

Equate the Thrust for the Elementsfrom the

Momentum and Blade Element Approaches

Rv

,

088

2

VRV

whereRraa

c

c

2168216

2cc a

Rraa

Total Inflow Velocity from CombinedBlade Element-Momentum Theory


82

Numerical Implementation of Combined BEM Theory

• The numerical implementation is identical to classical blade element theory.

• The only difference is the inflow is no longer uniform. It is computed using the formula given earlier, reproduced below:

2168216

2cc a

Rraa

Note that inflow is uniform if = CR/r . This twist is therefore called the ideal twist.


83

Effect of Inflow on Power in Hover

thrust!of level specified afor power, inducedleast produces inflow Uniformconstant. a bemust that vfollowsit ),multipliern (Lagrangeacontant a is Since

0v2v3 if is v s variationpossible allfor vanish willintegral heonly way t The

0vdrv2v34

0v4v4

0T-P .multiplier Lagrangean a is whereT-P minimize weTherefore,T. of valuespecified afor power, induced minimize wish toWe

v4dTT

v4vdT

20

2

0

23

0

2

0

0

3

0

R

R

RR

RR

induced

r

drrr

drr

drrP

Variation of a functional

constraint


84

Ideal Rotor vs. Optimum Rotor• Ideal rotor has a non-linear twist: = CR/r• This rotor will, according to the BEM theory, have a

uniform inflow, and the lowest induced power possible.• The rotor blade will have very high local pitch angles

near the root, which may cause the rotor to stall.• Ideally Twisted rotor is also hard to manufacture. • For these reasons, helicopter designers strive for

optimum rotors that minimize total power, and maximize figure of merit.

• This is done by a combination of twist, and taper, and the use of low drag airfoil sections.


85

Optimum Rotor• We try to minimize total power (Induced power +

Profile Power) for a given T.• In other words, an optimum rotor has the maximum

figure of merit.• From earlier work (see slide 72), figure of merit is

maximized if is maximized.

• All the sections of the rotor will operate at the angle of attack where this value of Cl and Cd are produced.

• We will call this Cl the optimum lift coefficient Cl,optimum .

d

lC

C 23


86

Optimum rotor (continued..)

twisted.bemust blade thehow determines This2R

v and r

varctan-

from find weselected, is attack of angle Once

maximum. is CCat which a optimuman at operate willstations radial All

d

23

l

TC


87

Variation of Chord for the Optimum Rotor

drCcrbdT l 2

21

dT = (Mass flow rate) * (twice the induced velocity at the annulus) = 4r(v)vdr

Compare these two. Note that Cl is a constant (the optimum value).

It follows that

r

ConstrRCR

bcrl

18v

2

2

Local solidity


88

Planform of Optimum RotorRootCut out

Tip

Chord is proportional to 1/r

Such planforms and twist distributions are hard to manufacture, and are optimumonly at one thrust setting.

Manufacturers therefore use a combination of linear twist, and linear variation in chord (constant taper ratio) to achieve optimum performance.

r=R r


89

Accounting for Tip Losses• We have already accounted for two sources of

performance loss-non-uniform inflow, and blade viscous drag.

• We can account for compressibility wave drag effects and associated losses, during the table look-up of drag coefficient.

• Two more sources of loss in performance are tip losses, and swirl.

• An elegant theory is available for tip losses from Prandtl.


90

Prandtl’s Tip Loss ModelPrandtl suggests that we multiply the sectional inflow by a function F, which goes to zero at the tip, and unity in the interior.

rbf

where

earcCosF f

12

,

2

When there are infinite number of blades, F approaches unity, there is no tip loss.


91

Incorporation of Tip Loss Model in BEM

All we need to do is multiply the lift due to inflow by F.

r

drThrust generated by the annulus:

dT = = 4rF(V+v)vdr


92

Resulting Inflow (Hover)

132116

16816

2

Rr

aF

Fa

Fa

Rr

Fa

Fa


93

Hover Performance Prediction Methods

IV. Vortex Theory


94

BACKGROUND• Extension of Prandtl’s Lifting Line Theory• Uses a combination of

– Kutta-Joukowski Theorem– Biot-Savart Law– Empirical Prescribed Wake or Free Wake Representation of Tip

Vortices and Inner Wake• Robin Gray proposed the prescribed wake model in

1952.• Landgrebe generalzied Gray’s model with extensive

experimental data.• Vortex theory was the extensively used in the 1970s and

1980s for rotor performance calculations, and is slowly giving way to CFD methods.


95

Background (Continued)

• Vortex theory addresses some of the drawbacks of combined blade element-momentum theory methods, at high thrust settings (high CT/).

• At these settings, the inflow velocity is affected by the contraction of the wake.

• Near the tip, there can be an upward directed inflow (rather than downward directed) due to this contraction, which increases the tip loading, and alters the tip power consumption.


96

Kutta-Joukowsky Theorem

r

V+v

T

Fx

T (r)

Fx= (V+v)

: Bound Circulation surroundingthe airfoil section.

This circulation is physically stored As vorticity in the boundary Layerover the airfoil


97

Representation of Bound and Trailing Vorticies

Since vorticity can not abruptly increase in space, trailing vortices develop. Some have clockwise rotation, others have counterclockwise rotation.


98

Robin Gray’s Conceptual Model

Tip Vortex has a Contraction that can be fitted with an exponential curve fit.

Inner wake descends at a near constant velocity. It descends faster near the tip than at the root.


99

Landgrebe’s Curve Fit for theTip Vortex Contraction

Rv

v 2v

RRR 707.02v


100

Radial Contraction

blade thefrom measuredFilament vortex theofPosition Azimuthal

AgeVortex 27145.0

78.0

)1(R

R: vortex tip theofposition Radial

v

vortex v

TCA

eAA


101

Vertical Descent Rate

v

Zv

Initial descent is slow

Descent is faster

After the firs

t blade

Passes over the vortex


102

Landgrebe’s Curve Fit forTip Vortex Descent Rate

degrees twist,2

degrees twist,1

21

1

01.0

001.025.0

2 2k2

20

TT

T

VVV

VVV

CCk

Ck

bbbk

Rz

bk

Rz

twist,degrees: Blade twist=Tip Pitch angle – Root Pitch AngleThis quantity is usually negative.


103

Circulation Coupled Wake Model

• Landgrebe’s earlier curve fits (1972) were based on the thrust coefficient, blade twist (change in the pitch angle between tip and root, usually negative).

• He subsequently found (1977) that better curve fits are obtained if the tip vortex trajectory is fitted on the basis of peak bound circulation, rather than CT/.


104

Tip Vortex Representation inComputational Analyses

• The tip vortex is a continuous helical structure.• This continuous structure is broken into

piecewise straight line segments, each representing 15 degrees to 30 degrees of vortex age.

• The tip vortex strength is assumed to be the maximum bound circulation. Some calculations assume it to be 80% of the peak circulation.

• The vortex is assumed to have a small core of an empirically prescribed radius, to keep induced velocities finite.


105

Tip Vortex RepresentationControl Points on the Lifting Line where induced flow is calculated

15 degrees

The x,y,z positions of theEnd points of each segmentAre computed usingLandgrebe’s Prescribed Wake Model

Inner

Wak

e

(Opti

onal)

Lifting Line


106

Biot-Savart Law

1rSegment

Control Point

2r


107

Biot-Savart Law (Continued)

212

22

122

212

21

21

2121

21 2

1

4 rrrrrrrrrrr

rrrrrrV

cinduced

Core radius used to keepDenominator from going to zero.


108

Overview of Vortex Theory Based Computations (Code supplied)

• Compute inflow using BEM first, using Biot-Savart law during subsequent iterations.

• Compute radial distribution of Loads.• Convert these loads into circulation strengths. Compute

the peak circulation strength. This is the strength of the tip vortex.

• Assume a prescribed vortex trajectory. • Discard the induced velocities from BEM, use induced

velocities from Biot-Savart law.• Repeat until everything converges. During each iteration,

adjust the blade pitch angle (trim it) if CT computed is too small or too large, compared to the supplied value.


109

Free Wake Models• These models remove the need for empirical

prescription of the tip vortex structure.• We march in time, starting with an initial guess

for the wake.• The end points of the segments are allowed to

freely move in space, convected the self-induced velocity at these end points.

• Their positions are updated at the end of each time step.


110

Free Wake Trajectories(Calculations by Leishman)


111

Vertical Descent of Rotors


112

Background

• We now discuss vertical descent operations, with and without power.

• Accurate prediction of performance is not done. (The engine selection is done for hover or climb considerations. Descent requires less power than these more demanding conditions).

• Discussions are qualitative.• We may use momentum theory to guide the

analysis.


113

Phase I: Power Needed in Climb and Hover

Climb Velocity, V

Power

ATVVT

VTP

222

v2

Descent


114

Non-Dimensional FormIt is convenient to non-dimensionalize these graphs, so that universal behavior of a variety of rotors can be studied.

h

h

Tvby lizeddimensiona-non is v)T(VPower

A2Tv velocity inflow

hoverby lizeddimensiona-nonislocity descent veor Climb


115

Momentum Theory gives incorrect Estimates of Power in Descent

V/vh

(V+v)/vh

ClimbDescent

No matter how fast we descend, positive power is still required if we use the above formula.This is incorrect!

0222

v2

ATVVT

VTP


116

The reason..

Climb or hoverPhysically acceptable Flow

V is down

V+v is down

V+2v is downV is down

V is down

V is up

V+v is down

V+2v is downV is up V is up

Descent: Everything insideSlipstream is downOutside flow is up


117

In reality..

• The rotor in descent operates in a number of stages, depending on how fast the vertical descent is in comparison to hover induced velocity.– Vortex Ring State– Turbulent Wake State– Windmill Brake State


118

Vortex Ring State(V is up, V+v is down, V+2v is down)

V is upV is up

V+v is down

The rotor pushes tip vortices down.

Oncoming air at the bottom pushes them up

Vortices get trapped in a donut-shaped ring.

The ring periodically grows and bursts.

Flow is highly unsteady.

Can only be empirically analyzed.


119

Performance in Vortex Ring State

V/vh

ClimbDescent

Momentum TheoryVortex Ring State

Experimental data Has scatter

Cross-overAt V=-1.71vh

Power/TVh


120

Turbulent Wake State(V is up, V+v is up, V+2v is down)

V is up

V is up

V+v is up

V+2v isdown

Rotor looks and behaves like a bluffBody (or disk). The vortices lookLike wake behind the bluff body.

Again, the flow is unsteady,Can not analyze using momentum theory

Need empirical data.


121

Performance in Turbulent Wake State

V/vh

ClimbDescent

Momen

tum T

heor

y


Vortex Ring State

TurbulentWake State

Notice power is –veEngine need not supply power

Power/TVh


122

Wind Mill Brake State(V is up, V+v is up, V+2v is up)

V is up

V is up

V+v is up

V+2v up

Flow is well behaved.

No trapped vortices, no wake.

Momentum theory can be used.

T = - 2Av(V+v)

Notice the minus sign. This is becausev (down) and V+v (up) have opposite signs. The product must be positive..


123

Power is Extracted in Wind Mill Brake State

mill. windain as ,freestream thefromextracted ispower case, In thisextracted. ispower means 0 Pconsumed ispower means 0 P

descent is 0 V climb, is 0 V:conventionSign

)v(222

Vv

get tov)v(V-2T

:equation thesolvecan We

2

VTPATV

A


124

Physical Mechanism for Wind Mill Power Extraction

r

V+vTotal Velocity Vector

Lift

The airfoil experiences an induced thrust, rather than induced drag!This causes the rotor to rotate without any need for supplying power or torque. This is called autorotation.Pilots can take advantage of this if power is lost.


125

Complete Performance Map

V/vh

ClimbDescent

Momen

tum T

heor

y


Vortex Ring State

Power/TVh

Turbulent WakeState

Wind Mill Brake State


126

Consider the cross-over Point

h-1.7vV extracted.nor

supplied,neither ispower speed, at this descents vehicle theIf

!!parachute! a as good AsA. area equivalent with parachute a as

tcoefficien drag same thehasrotor The4.1

2 vUse

v7.121T

:follows asrotor theoft coefficiendrag theestimatecan We

h

2h

D

D

CAT

AC


127

Hover Performance

Coning Angle Calculations


128

Background• Blades are usually hinged near the root, to

alleviate high bending moments at the root.• This allows the blades t flap up and down.• Aerodynamic forces cause the blades to flap up.• Centrifugal forces causes the blades to flap

down.• In hover, an equilibrium position is achieved,

where the net moments at the hinge due to the opposing forces (aerodynamic and centrifugal) cancel out and go to zero.


129

Schematic of Forces and Moments

0

dL

dCentrifugalForce

r

We assume that the rotor is hinged at the root, for simplicity.This assumption is adequate for most aerodynamic calculations.Effects of hinge offset are discussed in many classical texts.


130

Moment at the Hinge due toAerodynamic Forces

From blade element theory, the lift force dL =

drCrcdrCvrc ll222

21

21

Moment arm = r cos0 ~ rCounterclockwise moment due to lift = drrCrc l

2

21

Integrating over all such strips,Total counterclockwise moment =

Rr

rldrrCrc

0

2

21


131

Moment due to Centrifugal ForcesThe centrifugal force acting on this strip = rdm

rdmr 2

2

Where “dm” is the mass of this strip.This force acts horizontally. The moment arm = r sin0 ~ r0Clockwise moment due to centrifugal forces = dmr 0

22

Integrating over all such strips, total clockwise moment =

02

00

22

IdmrRr

r


132

At equilibrium..

Rr

rldrrCrcI

0

20

2

21

Rr

reffective

Rr

rl

Rrd

Rr

IacR

I

drCcr

0

340

3

021

Lock Number,


133

Lock Number, • The quantity =acR4/I is called the Lock number. • It is a measure of the balance between the aerodynamic

forces and inertial forces on the rotor.• In general has a value between 8 and 10 for

articulated rotors (i.e. rotors with flapping and lead-lag hinges).

• It has a value between 5 and 7 for hingeless rotors. • We will later discuss optimum values of Lock number.

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