Helicopter-Slides aerodynamics and performance .pdf

Heli

1. © L. Sankar
Helicopter Aerodynamics
1
Helicopter Aerodynamics and
Performance
Preliminary Remarks

2. © L. Sankar
Helicopter Aerodynamics
2
Thrust
Aeroelastic
Response
0
270
180
90
Dynamic Stall on
Retreating Blade
Blade-Tip Vortex
interactions
Unsteady
Aerodynamics
Transonic Flow on
Advancing Blade
Main Rotor / Tail Rotor
/ Fuselage
Flow Interference

V
Noise
Shock
Waves
Tip Vortices
The problems are many..

3. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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]

8. © L. Sankar
Helicopter Aerodynamics
8
Earliest Helicopter..
Chinese Top

9. © L. Sankar
Helicopter Aerodynamics
9
Leonardo da Vinci
(1480? 1493?)

10. © L. Sankar
Helicopter Aerodynamics
10
Human Powered Flight?
HP 6.7 5.33/0.8
Merit of rePower/Figu Ideal Power Actual
33.5
A2
W
WPower Ideal
slugs. 0.00238Desnity
sq.ft 100 AreaRotor
6ft ~RadiusRotor
160lbfWeight






HP


11. © L. Sankar
Helicopter Aerodynamics
11
D’AmeCourt (1863)
Steam-Propelled Helicopter

12. © L. Sankar
Helicopter Aerodynamics
12
Paul Cornu (1907)
First man to fly in helicopter mode..

13. © L. Sankar
Helicopter Aerodynamics
13
De La Cierva
invented Autogyros (1923)

14. © L. Sankar
Helicopter Aerodynamics
14
Cierva introduced hinges at the root
that 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 to
Alleviate root stresses from Coriolis forces

15. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
16
Sikorsky’s R-4

17. © L. Sankar
Helicopter Aerodynamics
17
Ways of countering the
Reactive Torque
Other possibilities: Tip jets, tip mounted engines

18. © L. Sankar
Helicopter Aerodynamics
18
Single Rotor Helicopter

19. © L. Sankar
Helicopter Aerodynamics
19
Tandem Rotors (Chinook)

20. © L. Sankar
Helicopter Aerodynamics
20
Coaxial rotors
Kamov KA-52

21. © L. Sankar
Helicopter Aerodynamics
21
NOTAR Helicopter

22. © L. Sankar
Helicopter Aerodynamics
22
NOTAR Concept

23. © L. Sankar
Helicopter Aerodynamics
23
Tilt Rotor Vehicles

24. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
25
147 kt (273 kph)
[Sea Level Standard Day]
149 kt (276 kph)
[Hot Day 2000 ft 70 F (21
C)]
150 kt (279 kph)
[Sea Level Standard Day]
153 kt (284 kph)
[Hot Day 2000 ft 70 F (21 C)]
Cruise Speed (MCP)
147 kt (273 kph)
[Sea Level Standard Day]
149 kt (276 kph)
[Hot Day 2000 ft 70 F (21
C)]
150 kt (279 kph)
[Sea Level Standard Day]
153 kt (284 kph)
[Hot Day 2000 ft 70 F (21 C)]
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)]
2,915 fpm (889 mpm)
[Sea Level Standard Day]
2,890 fpm (881 mpm)
[Hot Day 2000 ft 70 F (21 C)]
Maximum Rate of Climb
(IRP)
1,775 fpm (541 mpm)
[Sea Level Standard Day]
1,595 fpm (486 mpm)
[Hot Day 2000 ft 70 F (21
C)]
2,175 fpm (663 mpm)
[Sea Level Standard Day]
2,050 fpm (625 mpm)
[Hot Day 2000 ft 70 F (21 C)]
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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 V
3
.
•We need to understand and accurately predict
the air loads in high speed forward flight.

28. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
29
Hover Performance
Prediction Methods
I. Momentum Theory

30. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
33
Control Volume is a Cylinder
V
Disk area A
Total area S
Station1
2
3
4
V+v
2
V+v
3
V+v
4

34. © L. Sankar
Helicopter Aerodynamics
34
Conservation of Mass
 
444
1
)(A-SV bottom he through tOutflow
m side he through tInflow
VS tophe through tInflow
AvV





35. © L. Sankar
Helicopter Aerodynamics
35
Conservation of Mass through the
Rotor Disk
Flow through the rotor disk =
  
 
44
32
v
vv


VA
VAVAm


Thus v
2
=v
3
=v
There is no velocity jump across the rotor disk
The quantity v is called induced velocity at the rotor disk

36. © L. Sankar
Helicopter Aerodynamics
36
Global Conservation of Momentum
   
4444
4
2
4
2
4
44
1
2
vv)v(A T
in Rate Momentum
-out rate MomentumT,Thrust
.boundaries fieldfar the
allon catmospheri is Pressure
vA-S
bottom through outflow Momentum
vA
Vm side he through tinflow Momentum
V op through tinflow Momentum
mV
AVV
V
S











Mass flow rate through the rotor disk times
Excess velocity between stations 1 and 4

37. © L. Sankar
Helicopter Aerodynamics
37
Conservation of Momentum at the
Rotor Disk
V+v
V+v
p
2
p
3
Due to conservation of mass across the
Rotor disk, there is no velocity jump.
Momentum inflow rate = Momentum outflow rate
Thus, Thrust T = A(p
3
-p
2
)

38. © L. Sankar
Helicopter Aerodynamics
38
Conservation of Energy
Consider a particle that traverses from
Station 1 to station 4
We can apply Bernoulli equation between
Stations 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+v
4

  
4
4
23
2
4
2
3
22
2
v
2
v
v
2
1
v
2
1
2
1
v
2
1











Vpp
VpVp
VpVp




39. © L. Sankar
Helicopter Aerodynamics
39
 
4
4
23
4
4
23
v
2
v
v
2
v
#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

4
vvVAT
Thus, v = v
4
/2

40. © L. Sankar
Helicopter Aerodynamics
40
Induced Velocities
V
V+v
V+2v
The excess velocity in the
Far wake is twice the induced
Velocity at the rotor disk.
To accommodate this excess
Velocity, the stream tube
has to contract.

41. © L. Sankar
Helicopter Aerodynamics
41
Induced Velocity at the Rotor Disk
Now 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
A
TV
222
V
-v
2







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 or
Propeller where energy is added to the flow.
In this case, v is positive.

42. © L. Sankar
Helicopter Aerodynamics
42
Induced velocity at the rotor disk
A
T
A
TV


2
v
0V velocity climb Hover,In
222
V
-v
2










43. © L. Sankar
Helicopter Aerodynamics
43
Ideal Power Consumed by the Rotor
 





















A
TVV
T
VT
Vm
mm
P
222
v
vv2
V
2
1
2vV
2
1
in flowEnergy -out flowEnergy
2
22
In hover, ideal power
A
T
T
2


44. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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
T
T
C
C
C
T
FM
2
P
v
Hoverin Power Actual
Hoverin Power Ideal



46. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
48
Example #1 (Continued)

HP 11528 1.05*10980shaft the toengine by the suppliedPower
lossion transmiss5% is There
HP 10980 power actual totalrotors, twoFor the
HP 5490 power Actual
4117/0.75Power/FM idealPower Actual
HP 4117Power Ideal
ft/sec lb 74.86 x 30250 Tv Power Ideal
! ft/sec 150 far wake in theDownwash
ft/sec 86.74v
A2
T
v velocity,Induced
lbf 30250 T rotors. twoare There
feet cslugs/cubi 0.00238 Density
feet square 12.1134
19A AreaDisk
2















A

49. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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/ft
2
(24 to 48 kg/m
2
).
•Tilt-rotor vehicles tend to have a disk loading of
20 to 40 lbf/ft
2
. They are less efficient in hover.
•VTOL aircraft have very small fans, and have
very high disk loading (500 lb/ft
2
).

51. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
52
Non-Dimensional Forms



Q
C








P
2Q
3P
2T
C
QP
Torquelocity x Angualr ve Power hover,In
RAR
Q
tCoefficien TorqueC
RA
P
tCoefficienPower C
RA
T
tCoefficienThrust C
form. ldimensiona-nonin
expressedusually arePower and Torque, Thrust,




53. © L. Sankar
Helicopter Aerodynamics
53
Non-dimensional forms..
P
T
T
i
C
C
C
T
FM
2
P
v
Hoverin Power Actual
Hoverin Power Ideal

2
C
A2
T
R
1
R
v
inflow Induced
T










54. © L. Sankar
Helicopter Aerodynamics
54
Tip Losses
R
A portion of the rotor near the
Tip does not produce much lift
Due to leakage of air from
The bottom of the disk to the top.
One can crudely account for it by
Using a smaller, modified radius
BR, where
b
C
B
T
2
1
BR
B = Number of blades.

55. © L. Sankar
Helicopter Aerodynamics
55
Power Consumption in Hover
Including Tip Losses..
2
11
T
TP
C
C
BFM
C 

56. © L. Sankar
Helicopter Aerodynamics
56
Hover Performance
Prediction Methods
II. Blade Element Theory

57. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
60
Typical Blade Section (Strip)
R
dr
r






Tip
OutCut
Tip
OutCut
dPbP
dTbT
dT
Root Cut-out

61. © L. Sankar
Helicopter Aerodynamics
61
Typical Airfoil Section








r
Vv
arctan
r
V+v
Line of Zero Lift



effective
= 

62. © L. Sankar
Helicopter Aerodynamics
62
Sectional Forces
Once 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

2
1

2
1
22
22




These forces will be normal to and along
the total velocity vector.
U
T
=r
U
P
=V+v

63. © L. Sankar
Helicopter Aerodynamics
63
Rotation of Forces
r
V+v
L
D
T
F
x
  
   
  
   
XxT
ldPT
x
dlPT
rdFdFUdP
drCCcUU
drLDdF
drCCcUU
drDLdT









sincos
2
1
sincos
sincos
2
1
sincos
22
22

64. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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.
•C
d
is small. So, C
d
sin() may be neglected.
•The in-plane velocity r is much larger than the normal
component V+v over most of the rotor.

66. © L. Sankar
Helicopter Aerodynamics
66
Closed Form Expressions
drrC
rr
V
rr
V
cbaP
drr
rr
V
cbaT
Rr
r
d
Rr
r
3
0
3
2
0
2
vv
2
1
v
2
1












































67. © L. Sankar
Helicopter Aerodynamics
67
Linearly Twisted Rotor: Thrust
Here, we assume that the pitch angle varies as
EFr

R
vV
a
Rbc
where
a
R
abc
C
RRca
b
R
R
vV
FREca
b
T
R
T
R









































Ratio Inflow
)2(~ slope CurveLift
/DiskAreaBladeArea/solidity
2/
32
2/
32
2/
3224
3
3
1
2
75.75.
75.232












68. © L. Sankar
Helicopter Aerodynamics
68
Linearly Twisted Rotor
Notice 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 C
d
is assumed to be a
constant, equal to C
d0
.
8
0d
TP
C
CC


Induced Power Profile Power

69. © L. Sankar
Helicopter Aerodynamics
69
Closed Form Expressions for
Ideally Twisted Rotor
r
R
tip

  


tipT
a
C
4
C C
C
P T
d
 

0
8
Same as linearly
Twisted rotor!

70. © L. Sankar
Helicopter Aerodynamics
70
Figure of Merit according to Blade
Element Theory
AreaArea/Disk Blade Solidity
Rv)/(V Ratio Inflow
,
8/
0








where
CC
C
FM
dT
T
High solidity (lot of blades, wide-chord, large blade area) leads to higher
Power consumption, and lower figure of merit.
Figure of merit can be improved with the use of low drag airfoils.

71. © L. Sankar
Helicopter Aerodynamics
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
C
R
bc
RR
T
C
Rbc
drrcbT
6C
6
C
6
C
6
C
C
2
1
CtCoefficienLift Average
l
ll
22
32
l
l
0
2
l








Rotor will stall if average lift coefficient exceeds 1.2, or so.
Thus, in practice, C
T
/ is limited to 0.2 or so.

72. © L. Sankar
Helicopter Aerodynamics
72
Optimum Lift Coefficient in Hover
minimized. is
/C if maximized is FM
6/C If
82
2
2
C
hover,In
8
2/3
d0
T
0
2/3
2/3
T
0
l
l
dT
T
d
T
T
C
C
CC
C
FM
C
C
C
FM













73. © L. Sankar
Helicopter Aerodynamics
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.
15.1
8
0





d
TP
C
CC

74. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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)
3
A [C
d0
/8]
•Find an empirical fit for variation of  with
altitude:
2553.4
16.288
00198.0
1








h
levelsea



76. © L. Sankar
Helicopter Aerodynamics
76
Step 2
•Assume an altitude, h. Compute density, .
•Do the following for main rotor:
–Find main rotor area A
–Find v as [T/(2A)]
1/2
Note T= Vehicle weight in lbf.
–Insert supplied values of , C
d0
, 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. © L. Sankar
Helicopter Aerodynamics
77
Hover Performance
Prediction Methods
III. Combined Blade Element-Momentum
(BEM) Theory

78. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
79
Consider an Annulus of the rotor
Disk
r
dr
Area = 2rdr
Mass flow rate =2rV+vdr
dT = (Mass flow rate) * (twice
the induced velocity at the
annulus)
= 4r(V+v)vdr

80. © L. Sankar
Helicopter Aerodynamics
80
Blade Elements Captured by the
Annulus
r
dr
Thrust generated by these
blade elements:

 dr
r
vV
rabc
drCcrbdT
l












2
2
2
1
2
1

81. © L. Sankar
Helicopter Aerodynamics
81
Equate the Thrust for the Elements
from the
Momentum and Blade Element
Approaches
R
v
,
0
88
2












V
R
V
where
R
raa
c
c




















2168216
2
cc a
R
raa 



Total Inflow Velocity from Combined
Blade Element-Momentum Theory

82. © L. Sankar
Helicopter Aerodynamics
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
2
cc a
R
raa 



Note that inflow is uniform if = CR/r . This twist is therefore
called the ideal twist.

83. © L. Sankar
Helicopter Aerodynamics
83
Effect of Inflow on Power in Hover
 
 
 
 
thrust!of level specified afor power, inducedleast produces inflow Uniform
constant. 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
2
0
2
0
23
0
2
0
0
3
0























R
R
RR
RR
induced
r
drrr
drr
drrP
Variation of a functional
constraint

84. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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 C
l
and C
d
are produced.
•We will call this C
l
the optimum lift coefficient C
l,
optimum
.
d
l
C
C
2
3

86. © L. Sankar
Helicopter Aerodynamics
86
Optimum rotor (continued..)
twisted.bemust blade thehow determines This
2R
v
and
r
v
arctan-
from find weselected, is attack of angle Once
maximum. is
C
C
at which a optimuman at operate willstations radial All
d
2
3
l
T
C












87. © L. Sankar
Helicopter Aerodynamics
87
Variation of Chord for the Optimum
Rotor
drCcrbdT
l

2
2
1

dT = (Mass flow rate) * (twice the induced velocity at the annulus)
= 4r(v)vdr
Compare these two. Note that C
l
is a constant (the optimum value).
It follows that

r
Const
rRCR
bc
r
l











18v
2
2


Local solidity

88. © L. Sankar
Helicopter Aerodynamics
88
Planform of Optimum Rotor
Root
Cut out
Tip
Chord is proportional to 1/r
Such planforms and twist distributions are hard to manufacture, and are optimum
only 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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
90
Prandtl’s Tip Loss Model
Prandtl suggests that we multiply the sectional inflow by
a function F, which goes to zero at the tip, and unity in the interior.




rb
f
where
earcCosF
f




1
2
,
2
When there are infinite number of blades,
F approaches unity, there is no tip loss.

91. © L. Sankar
Helicopter Aerodynamics
91
Incorporation of Tip Loss Model in
BEM
All we need to do is multiply the lift due to inflow by F.
r
dr
Thrust generated by the annulus:
dT =
= 4rF(V+v)vdr

92. © L. Sankar
Helicopter Aerodynamics
92
Resulting Inflow (Hover)




















1
32
1
16
16816
2
R
r
a
F
F
a
F
a
R
r
F
a
F
a








93. © L. Sankar
Helicopter Aerodynamics
93
Hover Performance
Prediction Methods
IV. Vortex Theory

94. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
95
Background (Continued)
•Vortex theory addresses some of the drawbacks
of combined blade element-momentum theory
methods, at high thrust settings (high C
T
/).
•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. © L. Sankar
Helicopter Aerodynamics
96
Kutta-Joukowsky Theorem
r
V+v
T
F
x
T  (r) 
F
x
= (V+v) 
 : Bound Circulation surrounding
the airfoil section.
This circulation is physically stored
As vorticity in the boundary Layer
over the airfoil

97. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
99
Landgrebe’s Curve Fit for the
Tip Vortex Contraction

R
v
v
2v
R
R
R 707.0
2
v


100. © L. Sankar
Helicopter Aerodynamics
100
Radial Contraction
blade thefrom measuredFilament
vortex theofPosition Azimuthal
AgeVortex
27145.0
78.0
)1(
R
R
: vortex tip theofposition Radial
v
vortex v









T
C
A
eAA

101. © L. Sankar
Helicopter Aerodynamics
101
Vertical Descent Rate

v
Z
v
Initial descent is slow
Descent is faster
After the first blade
Passes over the vortex

102. © L. Sankar
Helicopter Aerodynamics
102
Landgrebe’s Curve Fit for
Tip Vortex Descent Rate
degrees twist,2
degrees twist,1
21
1
01.0
001.025.0
2

2
k
2

2
0










TT
T
VV
V
VV
V
CCk
C
k
bbb
k
R
z
b
k
R
z

















twist,degrees
: Blade twist=Tip Pitch angle – Root Pitch Angle
This quantity is usually negative.

103. © L. Sankar
Helicopter Aerodynamics
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 C
T
/.

104. © L. Sankar
Helicopter Aerodynamics
104
Tip Vortex Representation in
Computational 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. © L. Sankar
Helicopter Aerodynamics
105
Tip Vortex Representation
Control Points on the Lifting Line where induced flow is calculated
15
degrees
The x,y,z positions of the
End points of each segment
Are computed using
Landgrebe’s
Prescribed Wake Model
Inner Wake
(Optional)
Lifting Line

106. © L. Sankar
Helicopter Aerodynamics
106
Biot-Savart Law
1
r
Segment
Control Point
2
r

107. © L. Sankar
Helicopter Aerodynamics
107
Biot-Savart Law (Continued) 
  
21
2
2
2
1
22
21
2
21
21
21
21
21
2
1
4 rrrrrrrrr
rr
rr
rr
rrV
c
induced












Core radius used to keep
Denominator from going to zero.

108. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
110
Free Wake Trajectories
(Calculations by Leishman)

111. © L. Sankar
Helicopter Aerodynamics
111
Vertical Descent of Rotors

112. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
113
Phase I: Power Needed in
Climb and Hover
Climb Velocity, V
Power

















A
TVV
T
VTP
222
v
2
Descent

114. © L. Sankar
Helicopter Aerodynamics
114
Non-Dimensional Form
It 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
A2
T
v velocity inflow
hoverby lizeddimensiona-non
islocity descent veor Climb




115. © L. Sankar
Helicopter Aerodynamics
115
Momentum Theory gives incorrect
Estimates of Power in Descent
V/v
h
(V+v)/v
h
ClimbDescent
No matter how fast we descend, positive power is
still required if we use the above formula.
This is incorrect!

0
222
v
2

















A
TVV
T
VTP


116. © L. Sankar
Helicopter Aerodynamics
116
The reason..
Climb or hover
Physically acceptable Flow
V is down
V+v is down
V+2v is down
V is down
V is down
V is up
V+v is down
V+2v is down
V is up V is up
Descent: Everything inside
Slipstream is down
Outside flow is up

117. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
118
Vortex Ring State
(V is up, V+v is down, V+2v is down)
V is up
V 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. © L. Sankar
Helicopter Aerodynamics
119
Performance in Vortex Ring State
V/v
h
Climb
Descent
Momentum Theory
Vortex Ring State
Experimental data
Has scatter
Cross-over
At V=-1.71v
h
Power/TV
h

120. © L. Sankar
Helicopter Aerodynamics
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 is
down
Rotor looks and behaves like a bluff
Body (or disk). The vortices look
Like wake behind the bluff body.
Again, the flow is unsteady,
Can not analyze using momentum
theory
Need empirical data.

121. © L. Sankar
Helicopter Aerodynamics
121
Performance in Turbulent Wake
State
V/v
h
Climb
Descent
Momentum Theory
Cross-over
At V=-1.71v
h
Vortex Ring State
Turbulent
Wake State
Notice power is –ve
Engine need not supply power
Power/TV
h

122. © L. Sankar
Helicopter Aerodynamics
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 = - 2Av(V+v)
Notice the minus sign. This is because
v (down) and V+v (up) have opposite signs.
The product must be positive..

123. © L. Sankar
Helicopter Aerodynamics
123
Power is Extracted in
Wind Mill Brake State
mill. windain as ,freestream thefrom
extracted ispower case, In this
extracted. ispower means 0 P
consumed ispower means 0 P
descent is 0 V climb, is 0 V
:conventionSign
)v(
222
V
v
get to
v)v(V-2T
:equation thesolvecan We
2












VTP
A
TV
A



124. © L. Sankar
Helicopter Aerodynamics
124
Physical Mechanism for Wind Mill
Power Extraction
r
V+v
Total 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. © L. Sankar
Helicopter Aerodynamics
125
Complete Performance Map
V/v
h
Climb
Descent
Momentum Theory
Cross-over
At V=-1.71v
h
Vortex Ring State
Power/TV
h
Turbulent Wake
State
Wind Mill Brake State

126. © L. Sankar
Helicopter Aerodynamics
126
Consider the cross-over Point
h
-1.7vV
extracted.nor
supplied,neither
ispower speed, at this
descents vehicle theIf

 
!!parachute! a as good As
A. area equivalent with parachute a as
tcoefficien drag same thehasrotor The
4.1
2
vUse
v7.1
2
1
T
:follows asrotor theoft coefficien
drag theestimatecan We
h
2
h



D
D
C
A
T
AC



127. © L. Sankar
Helicopter Aerodynamics
127
Hover Performance
Coning Angle Calculations

128. © L. Sankar
Helicopter Aerodynamics
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. © L. Sankar
Helicopter Aerodynamics
129
Schematic of Forces and Moments

0
dL
dCentrifugal
Force
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. © L. Sankar
Helicopter Aerodynamics
130
Moment at the Hinge due to
Aerodynamic Forces
From blade element theory, the lift force dL =
 drCrcdrCvrc
ll
222
2
1
2
1
 
Moment arm = r cos
0
~ r
Counterclockwise moment due to lift =
drrCrc
l
2
2
1

Integrating over all such strips,
Total counterclockwise moment =




Rr
r
l
drrCrc
0
2
2
1


131. © L. Sankar
Helicopter Aerodynamics
131
Moment due to Centrifugal Forces
The centrifugal force acting on this strip =

rdm
r
dmr
2
2


Where “dm” is the mass of this strip.
This force acts horizontally.
The moment arm = r sin
0
~ r
0
Clockwise moment due to centrifugal forces =
dmr
0
22

Integrating over all such strips, total clockwise moment =
0
2
0
0
22
 


Idmr
Rr
r

132. © L. Sankar
Helicopter Aerodynamics
132
At equilibrium..



Rr
r
l
drrCrcI
0
2
0
2
2
1


 










Rr
r
effective
Rr
r
l
R
r
d
R
r
I
acR
I
drCcr
0
3
4
0
3
0
2
1




Lock Number, 

133. © L. Sankar
Helicopter Aerodynamics
133
Lock Number, 
•The quantity =acR
4
/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.