Helicopter Aerodynamics and Performance Preliminary Remarks
© L. Sankar
1
The problems are many.. Transonic Flow on Advancing Blade Noise
Thrust
Aeroelastic Response
Shock Waves
Unsteady Aerodynamics
0
90
Tip Vortices
Main Rotor / Tail Rotor / Fuselage Flow Interference
Blade-Tip Vortex interactions
V
270
180
Dynamic Stall on Retreating Blade
© L. Sankar
2
The problems are many.. Transonic Flow on Advancing Blade Noise
Thrust
Aeroelastic Response
Shock Waves
Unsteady Aerodynamics
0
90
Tip Vortices
Main Rotor / Tail Rotor / Fuselage Flow Interference
Blade-Tip Vortex interactions
V
270
180
Dynamic Stall on Retreating Blade
© L. Sankar
2
A systematic Approach is necessary • •
A varie variety ty of tools tools are are need needed ed to to under understan stand, d, and and pred predict ict thes these e pheno phenomen mena. a. Tools needed include – Simple back-of-the envelop tools for sizing helicopters, selecting engines, engines, laying out configuration, and predicting performance – Spreadsheets and MATLAB MATLAB scripts for mapping out out the blade loads over the entire rotor disk – High end CFD tools tools for modeling • Airfoi Airfoill and and rotor rotor aerody aerodynam namics ics and design design • Roto Rotor-a r-air irfra frame me int inter erac acti tion ons s • Aero Aeroac acou oust stic ic anal analys yses es
– Elastic and multi-body dynamics modeling modeling tools – Trim analyses, Flight Simulation Simulation software
• • •
In this this work, work, we will will cove coverr most most of the tools tools that that we need, need, exce except pt for for elastic elastic analyses, multi-body dynamics analyses, and flight simulation software. We will will cov cover er bot both h the the basi basics cs,, and and the the appl applic icat atio ions ns.. We will will assum assume e famil familiari iarity ty with with classic classical al low low spee speed d and and high high spee speed d aerodynamics, but nothing more.
© L. Sankar
3
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 © L. Sankar
4
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 0-53406360-8 – Gessow and Myers – Stepniewski & Keys © L. Sankar
5
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.
© L. Sankar
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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]
© L. Sankar
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Earliest Helicopter.. Chinese Top
© L. Sankar
8
Leonardo da Vinci (1480? 1493?)
© L. Sankar
9
Human Powered Flight? Weight 160lbf Rotor Radius ~ 6ft Rotor Area 100 sq.ft Desnity 0.00238 slugs. Ideal Power W
W 2 A
5.33 HP
Actual Power Ideal Power/Figure of Merit
5.33/0.8 6.7 HP © L. Sankar
10
D’AmeCourt (1863) Steam-Propelled Helicopter
© L. Sankar
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Paul Cornu (1907) First man to fly in helicopter mode..
© L. Sankar
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De La Cierva invented Autogyros (1923)
© L. Sankar
13
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 from Coriolis forces © L. Sankar
14
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. © L. Sankar
15
Sikorsky’s R-4
© L. Sankar
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Ways of countering the Reactive Torque
Other possibilities: Tip jets, tip mounted engines © L. Sankar
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Single Rotor Helicopter
© L. Sankar
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Tandem Rotors (Chinook)
© L. Sankar
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Coaxial rotors Kamov KA-52
© L. Sankar
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NOTAR Helicopter
© L. Sankar
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NOTAR Concept
© L. Sankar
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Tilt Rotor Vehicles
© L. Sankar
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Helicopters tend to grow in size..
AH-64A
AH-64D
Length
58.17 ft (17.73 m)
58.17 ft (17.73 m)
Height
15.24 ft (4.64 m)
13.30 ft (4.05 m)
Wing Span
17.15 ft (5.227 m)
17.15 ft (5.227 m)
Primary Mission Gross Weight
15,075 lb (6838 kg) 11,800 pounds Empty
16,027 lb (7270 kg) Lot 1 Weight
© L. Sankar
24
AH-64A
AH-64D
Length
58.17 ft (17.73 m)
58.17 ft (17.73 m)
Height
15.24 ft (4.64 m)
13.30 ft (4.05 m)
Wing Span
17.15 ft (5.227 m)
17.15 ft (5.227 m)
Primary Mission Gross Weight
15,075 lb (6838 kg) 11,800 pounds Empty
16,027 lb (7270 kg) Lot 1 Weight
Hover In-Ground Effect (MRP)
15,895 ft (4845 m) [Standard Day] 14,845 ft (4525 m) [Hot Day ISA + 15C]
14,650 ft (4465 m) [Standard Day] 13,350 ft (4068 m) [Hot Day ISA + 15 C]
Hover Out-of-Ground Effect (MRP)
12,685 ft (3866 m) [Sea Level Standard Day] 11,215 ft (3418 m) [Hot Day 2000 ft 70 F (21 C)]
10,520 ft (3206 m) [Standard Day] 9,050 ft (2759 m) [Hot Day ISA + 15 C]
Vertical Rate of Climb (MRP)
2,175 fpm (663 mpm) [Sea Level Standard Day] 2,050 fpm (625 mpm) [Hot Day 2000 ft 70 F (21 C)]
1,775 fpm (541 mpm) [Sea Level Standard Day] 1,595 fpm (486 mpm) [Hot Day 2000 ft 70 F (21 C)]
© L. Sankar
2,635 fpm (803 mpm)
25
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.
© L. Sankar
26
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. © L. Sankar
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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.
© L. Sankar
28
Hover Performance Prediction Methods
I. Momentum Theory
© L. Sankar
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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. © L. Sankar
30
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. © L. Sankar
31
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. © L. Sankar
32
Control Volume is a Cylinder V
Station1
2 3
4
Total area S© L. Sankar
Disk area A V+v2 V+v3
V+v4 33
Conservation of Mass Inflow through the top VS Inflow through the side m1 Outflow through the bottom V S - A 4 (V v4 ) A4
© L. Sankar
34
Conservation of Mass through the Rotor Disk Flow through the rotor disk =
m AV v 2 AV v 3
A4 V v 4 Thus v2=v3=v There is no velocity jump across the rotor disk The quantity v is called induced velocity at the rotor disk © L. Sankar
35
Global Conservation of Momentum Momentum inflow through top V 2 S Momentum inflow through the side m1V
A 4 v 4V Momentum outflow through bottom S - A 4 V 2 V v 4 2 A4 Pressure is atmospheric on all the far field boundaries. Thrust , T Momentum rate out Momentum Rate in T A 4 (V v 4 ) v 4
mv 4
Mass flow rate through the rotor disk times Excess velocity between stations 1 and 4 © L. Sankar
36
Conservation of Momentum at the Rotor Disk
p2
V+v
Due to conservation of mass across the Rotor disk, there is no velocity jump. Momentum inflow rate = Momentum outflow rate
p 3
V+v Thus, Thrust T = A(p3-p2)
© L. Sankar
37
Conservation of Energy Consider a particle that traverses from Station 1 to station 4
1
2
V+v
3
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. p2
4
V+v4
p3
1 2 1 2
1
V v 2 p V 2 V v 2 p V v 4 2
p3 p2 © L. Sankar
2 1 2
v 4 V v4 2 38
From the previous slide #38, v 4 p3 p2 V v 4 2 v 4 T A p3 p2 AV v 4 2 From an earlier slide # 36, Thrust equals mass flow rate through the rotor disk times excess velocity between stations 1 and 4
T AV v v 4 Thus, v = v4/2 © L. Sankar
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Induced Velocities V
The excess velocity in the Far wake is twice the induced Velocity at the rotor disk.
V+v
To accommodate this excess Velocity, the stream tube has to contract.
V+2v © L. Sankar
40
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
v-
V 2
2
V T 2 2 A
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. © L. Sankar
41
Induced velocity at the rotor disk v-
V 2
2
V T 2 2 A
In Hover, climb velocity V 0 v
T 2 A
© L. Sankar
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Ideal Power Consumed by the Rotor P
Energy flow out - Energy flow in
1
m
V 2v 2
2 2mvV v
1 2
mV
2
T V v V V 2 T T 2 2 2 A In hover, ideal power © L. Sankar
T
T 2 A
43
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 lossesviscous effects, compressibility (shocks), tip losses, swirl, non-uniform flows, etc. © L. Sankar
44
Figure of Merit • Figure of merit is defined as the ratio of ideal power for a rotor FM in hover obtained from momentum theory and the actual power consumed by the rotor.
Ideal Power in Hover Actual Power in Hover T v P
C T
C T 2
C P
• For most rotors, it is between 0.7 and 0.8. © L. Sankar
45
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). © L. Sankar
46
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.
© L. Sankar
47
Example #1 (Continued) Disk Area A 19
2
A 1134.12 square feet Density 0.00238 slugs/cubic feet There are two rotors. T 30250 lbf Induced velocity, v
T 2 A
v 74.86 ft/sec Downwash in the far wake 150 ft/sec ! Ideal Power Tv 30250 x 74.86 lb ft/sec Ideal Power 4117 HP Actual Power ideal Power/FM 4117/0.75 Actual power 5490 HP For the two rotors, total actual power 10980 HP There is 5% transmission loss Power supplied by the engine to the shaft 10980 *1.05 11528 HP © L. Sankar
48
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. © L. Sankar
49
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/m 2). • 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/ft 2). © L. Sankar
50
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. © L. Sankar
51
Non-Dimensional Forms Thrust, Torque, and Power are usually expressed in non - dimensional form. CT CP CQ
T
Thrust Coefficient Power Coefficient
2 A R
P
AR 3
Torque Coefficient
Q
AR R 2
In hover, Power Angualr velocity x Torque P Q CP
C Q © L. Sankar
52
Non-dimensional forms.. Induced inflow i FM
v
1
R R
T 2 A
CT 2
Ideal Power in Hover Actual Power in Hover T v
P
© L. Sankar
C T
C T
2
C P 53
Tip Losses R
BR
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 1
B = Number of blades. © L. Sankar
2C T b 54
Power Consumption in Hover Including Tip Losses.. C P
1 FM
© L. Sankar
1 B
C T
C T
2
55
Hover Performance Prediction Methods II. Blade Element Theory
© L. Sankar
56
Preliminary Remarks • Momentum theory gives rapid, back-ofthe-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.
© L. Sankar
57
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
© L. Sankar
58
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. © L. Sankar
59
Typical Blade Section (Strip) dT r
Tip
T b
dr
dT
Cut Out
R Root Cut-out
Tip
P b
dP
Cut Out © L. Sankar
60
Typical Airfoil Section V v arctan r
Line of Zero Lift
V+v
r
effective = © L. Sankar
61
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.
1
L U U 2 1
2 T
2 P
cC
l
D U T 2 U P 2 cC d
UT=r UP=V+v
2
These forces will be normal to and along the total velocity vector. © L. Sankar
62
Rotation of Forces T
L
Fx
V+v
r
D
dT L cos D sin dr
1
U T 2 U P 2 cC l cos C d sin dr 2
dF x
D cos L sin dr 1
U T 2 U P 2 cC d cos C l sin dr
2 dP U T dF x
rdF X © L. Sankar
63
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.
© L. Sankar
64
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. © L. Sankar
65
Closed Form Expressions r R
V v 2 T cba r dr 2 r r r 0 1
2
r R
V v V v 3 P cba C d r dr 2 r r r r r 0 1
3
© L. Sankar
66
Linearly Twisted Rotor: Thrust Here, we assume that the pitch angle varies as
E Fr 1 3 V v 3 b .75 R / 2 2 2 ca E FR R ca R R 3 2 3 4 2 2 R abc .75 a .75 R C T / 2 / 2 2 R 3 2 3
T
b
where
solidity BladeArea/DiskArea bc / R a Lift Curve slope (~ 2 ) V v Inflow Ratio R © L. Sankar
67
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 Cd0.
C P C T Induced Power © L. Sankar
C d 0 8 Profile Power
68
Closed Form Expressions for Ideally Twisted Rotor
C T C P
tip R
a 4
r
C T © L. Sankar
tip
C d 0 8
Same as linearly Twisted rotor! 69
Figure of Merit according to Blade Element Theory C T FM C T C d 0 / 8 where,
Inflow Ratio (V v)/R Solidity Blade Area/Disk Area 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. © L. Sankar
70
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.
Average Lift Coefficient Cl R
T b
1
2 cr C dr 2
bc Cl 2 R 3
l
6
0
C T Cl
T
R R
6
2
2
bc Cl
R
6
Cl 6
C T
Rotor will stall if average lift coefficient exceeds 1.2, or so. Thus, in practice, CT/ is limited to 0.2 or so. © L. Sankar
71
Optimum Lift Coefficient in Hover C T FM C d 0 C T 8
In hover,
CT
2
C T 3 / 2 FM
2
3/ 2 T
C
2 If C T
C d 0 8
C l / 6
FM is maximized if C d0 / C l 3 / 2 is minimized. © L. Sankar
72
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.
C P C T
1.15 © L. Sankar
C d 0 8
73
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)
© L. Sankar
74
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 [Cd0/8] • Find an empirical fit for variation of with altitude: 4.2553
0.00198h 1 sea level 288.16 © L. Sankar
75
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 , 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. © L. Sankar 76 A 10 500 ft
Hover Performance Prediction Methods III. Combined Blade Element-Momentum (BEM) Theory
© L. Sankar
77
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. © L. Sankar
78
Consider an Annulus of the rotor Disk Area = 2rdr dr
r
Mass flow rate =2r V+vdr dT = (Mass flow rate) * (twice the induced velocity at the annulus) = 4r(V+v)vdr
© L. Sankar
79
Blade Elements Captured by the Annulus
dr
Thrust generated by these blade elements: dT b
1 2
r 2 c C l dr
V v abc r dr r 2
r
1
© L. Sankar
2
80
Equate the Thrust for the Elements from the Momentum and Blade Element Approaches a a r c 0 8 R 8 2
where,
c
V
2
a c a r a c 8 R 16 2 16 2
R V v R
Total Inflow Velocity from Combined Blade Element-Momentum Theory © L. Sankar
81
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: 2
a c a r a c 8 R 16 2 16 2 Note that inflow is uniform if = CR/r . This twist is therefore 82 called the ideal twist. © L. Sankar
Effect of Inflow on Power in Hover P induced R
R
R
0
0
vdT 4 r v3dr R
T dT 4 r v 2 dr 0
constraint
0
We wish to minimize induced power, for a specified value of T. Therefore, we minimize P - T where is a Lagrangean multiplier .
R 3 2 4 r v 4r v dr 0 0 R
P - T 0
Variation of a functional
4 r 3v 2 2 vvdr 0
0
The only way the integral will vanish for all possible variations v is if 3v 2
2 v 0
Since is a contant (Lagrangean multiplier ), it follows that v must be a constant. Uniform inflow produces least induced power, for a specified level of thrust! © L. Sankar
83
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. © L. Sankar
84
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 3 maximized if C l 2 is maximized. C d
• 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 C l, optimum . © L. Sankar
85
Optimum rotor (continued..) 3
All radial stations will operate at an optimum a at which
Cl 2 Cd
is maximum.
Once angle of attack is selected, we find from v v - arctan and R r
C T 2
This determines how the blade must be twisted.
© L. Sankar
86
Variation of Chord for the Optimum Rotor dT b
1 2
r c C l dr 2
dT = (Mass flow rate) * (twice the induced velocity at the annulus) = 4r(v)vdr Compare these two. Note that C l is a constant (the optimum value). It follows that
8v 2 1 Const r 2 R RC l r r bc
Local solidity © L. Sankar
87
Planform of Optimum Rotor Root Cut out
Chord is proportional to 1/r
Tip
r=R
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. © L. Sankar
88
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. © L. Sankar
89
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.
F
2
arcCos e
When there are infinite number of blades, F approaches unity, there is no tip loss.
where, f
f
b 1 r 2
© L. Sankar
90
Incorporation of Tip Loss Model in BEM All we need to do is multiply the lift due to inflow by F.
dr
Thrust generated by the annulus:
dT = = 4r F(V+v)vdr
r
© L. Sankar
91
Resulting Inflow (Hover) 2
a a r a 16 F 8 F R 16 F 32 F r a 1 1 16 F a R
© L. Sankar
92
Hover Performance Prediction Methods IV. Vortex Theory
© L. Sankar
93
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. © L. Sankar
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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. © L. Sankar
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Kutta-Joukowsky Theorem : Bound Circulation surrounding
T
r
the airfoil section.
Fx V+v
This circulation is physically stored As vorticity in the boundary Layer over the airfoil
T (r) Fx= (V+v) © L. Sankar
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Representation of Bound and Trailing Vorticies
Since vorticity can not abruptly increase in space, trailing vortices develop. Some have clockwise rotation, © L. Sankar 97 others have counterclockwise rotation.
Robin Gray’s Conceptual Model
Inner wake descends at a near constant velocity. It descends faster near the tip than at the © L. Sankar root.
Tip Vortex has a Contraction that can be fitted with an exponential curve fit. 98
Landgrebe’s Curve Fit for the Tip Vortex Contraction Rv
Rv v
R 2
0.707 R
2v
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Radial Contraction Radial position of the tip vortex : R vortex
A (1 A)e
v
R A 0.78
0.145 27C T v Vortex Age Azimuthal Position of the vortex Filament measured from the blade
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Vertical Descent Rate r e t a s l a d e r x e f t i s t b v o t i r s t e n f h e c s h e e r e v D e r t o t e s f A s s P a
Zv
s l o w s i t n e s c e d l a i t i I n
v
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Landgrebe’s Curve Fit for Tip Vortex Descent Rate z V R z V R
k 1 V k 1
2 b
0 V
2 b
2 k 2 V b
V
2 b
C T 0.001 twist, degrees k 1 0.25 k 2 C T 0.01 C T twist, degrees
twist,degrees: Blade twist=Tip Pitch angle – Root Pitch Angle This quantity is usually negative. © L. Sankar
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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/. © L. Sankar
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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. © L. Sankar
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Tip Vortex Representation Control Points on the Lifting Line where induced flow is calculated Lifting Line 15 degrees
e k a l ) W a r o n e n i I n O p t ( The x,y,z positions of the End points of each segment Are computed using Landgrebe’s Prescribed Wake Model © L. Sankar
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Biot-Savart Law Control Point
r 2
r 1 Segment
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Biot-Savart Law (Continued) r 1 r 2 r 1 r 2 1 r 1r 2 V induced r 1 r 2 2 2 4 r 1r 2 r 1 r 2 r c2 r 12 r 22 2r 1 r 2
Core radius used to keep Denominator from going to zero.
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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. © L. Sankar
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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 selfinduced velocity at these end points. • Their positions are updated at the end of each time step. © L. Sankar
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Free Wake Trajectories (Calculations by Leishman)
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Vertical Descent of Rotors
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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. © L. Sankar
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Phase I: Power Needed in Climb and Hover Power
Descent
Climb Velocity, V
P T V v
V V 2 T T 2 2 2 A © L. Sankar
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Non-Dimensional Form It is convenient to non-dimensionalize these graphs, so that universal behavior of a variety of rotors can be studied.
Climb or descent velocity is non - dimensionalized by hover inflow velocity v h
T 2 A
Power T(V v) is non - dimensionalized by Tv h © L. Sankar
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Momentum Theory gives incorrect Estimates of Power in Descent (V+v)/vh
P T V v
V V 2 T 0 T 2 2 2 A Descent
Climb
V/vh
No matter how fast we descend, positive power is still required if we use the above formula. This is incorrect! © L. Sankar
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The reason.. V is down
V is up
V+v is down V+v is down
V is down V+2v is down
V is down
Climb or hover Physically acceptable Flow © L. Sankar
V is up
V+2v is down
V is up
Descent: Everything inside Slipstream is down 116 Outside flow is up
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
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Vortex Ring State (V is up, V+v is down, V+2v is down) The rotor pushes tip vortices down. V+v is down
Oncoming air at the bottom pushes them up Vortices get trapped in a donut-shaped ring. The ring periodically grows and bursts.
V is up
V is up
Flow is highly unsteady. © L. Sankar
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Can only be empirically analyzed.
Performance in Vortex Ring State Descent
Power/TVh
Climb
Vortex Ring State r y o e h T m u t e n m o M
Experimental data Has scatter
V/vh
Cross-over At V=-1.71vh © L. Sankar
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Turbulent Wake State (V is up, V+v is up, V+2v is down) V+v is up
Rotor looks and behaves like a bluff Body (or disk). The vortices look Like wake behind the bluff body.
V is up
V+2v is down V is up
Again, the flow is unsteady, Can not analyze using momentum theory Need empirical data.
© L. Sankar
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Performance in Turbulent Wake State Descent
Power/TVh
Vortex Ring State
Turbulent Wake State
Cross-over At V=-1.71vh Notice power is –ve © L. Sankar Engine need not supply power
Climb
y r o e h T m u t n e o m M
V/vh
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Wind Mill Brake State (V is up, V+v is up, V+2v is up) V+v is up
Flow is well behaved. No trapped vortices, no wake. Momentum theory can be used. V is up
V is up
V+2v up
T = - 2 Av(V+v)
Notice the minus sign. This is because v (down) and V+v (up) have opposite signs. © L. Sankar 122 The product must be positive..
Power is Extracted in Wind Mill Brake State
We can solve the equation : T -2 Av(V v) to get v
V
V
2
T
2 2 2 A P T (V v) Sign convention : V 0 is climb, V 0 is descent P 0 means power is consumed P 0 means power is extracted. In this case, power is extracted from the freestream, as in a wind mill. © L. Sankar
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Physical Mechanism for Wind Mill Power Extraction Lift
t o r c e V c i ty o l e V T o t a l
V+v
r 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. © L. Sankar
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Complete Performance Map Descent
Power/TVh
Vortex Ring State
Turbulent Wake State
Climb
y r o e h T m u t n e o m M
V/vh
Cross-over At V=-1.71vh
Wind Mill Brake State © L. Sankar
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Consider the cross-over Point We can estimate the drag coefficient of the rotor as follows :
If the vehicle descents at this speed, power is
T
1 2
AC D 1.7 v h 2
neither supplied, nor extracted.
Use v h
V -1.7v h
C D
T 2 A
1 .4
The rotor has the same drag coefficient as a parachute with equivalent area A. As good as a parachute!!! © L. Sankar
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Hover Performance Coning Angle Calculations
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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. © L. Sankar
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Schematic of Forces and Moments dL r
0
dCentrifugal Force
We assume that the rotor is hinged at the root, for simplicity. This assumption is adequate for most aerodynamic calculations. Effects of hinge offset are discussed in many classical texts. © L. Sankar
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Moment at the Hinge due to Aerodynamic Forces From blade element theory, the lift force dL =
1 2
cr v 2
2
C dr l
1 2
cr 2 C l dr
Moment arm = r cos 0 ~ r 1 2 c r rC l dr Counterclockwise moment due to lift =
2
Integrating over all such strips, Total counterclockwise moment = r R
1
2 cr rC dr 2
l
r 0
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Moment due to Centrifugal Forces The centrifugal force acting on this strip =
r
2
dm
r 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 =
2
2
r
2 rdm
0 dm
Integrating over all such strips, total clockwise moment = r R
2 2 2 0 r dm I 0
r 0
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At equilibrium.. r R
I
2
0
1
2
cr
2
rC l dr
r 0 r R
0
1 cr 3C l dr 2 r 0
I
acR I
4 r R
3
r r r 0 R effective d R
Lock Number, © L. Sankar
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