lecture_1

16.333:Lecture#1
Equilibrium States
Aircraft performance Introduction to basic terms
Fall200416.3331–1
Aircraft Performance •Accelerated horizontalflight­balance of forces
–Engine thrust T
–Lift L(⊥to V)
–Drag D(�to V)
–Weight W
dV
T−D=m=0for steadyflight
dt
and
L−W=0
Define L=
•1
2
ρV2SC L where
–ρ–air density(standard tables)
–S–gross wing area=¯c×b,
–c¯=mean chord
–b=wing span
–A R–wing aspect ratio=b/c
¯
–Q=1
2
ρV2dynamic pressure –V=speed relative to the air
� �
Fall 2004 16.333 1–2
– C L lift coefficient – for low Mach number, C L = C L α (α − α0) 3 α angle of incidence of wind to the wing 3 α0 is the angle associated with zero lift
• Back to the performance:
1 L = ρV 2
SC L and L = mg
2 2mg
which implies that V = ρSC L
so that
V ∝ C −1/2
L
and we can relate the effect of speed to wing lift
• A key number is stall speed, which is the lowest speed that an aircraft can fly steadily 2mg
V s =
ρSC L max
where typically get C L max at αmax = 10◦
Fall 2004
16.333 1–3
Steady Gliding Flight
• Aircraft at a steady glide angle of γ
• Assume forces are in equilibrium
L − mg cos γ = 0 (1) D + mg sin γ = 0
(2)
Gives that
D C D
tan γ = L ≡
C L
⇒ Minimum gliding angle obtained when C D /C L is a minimum – High L/D gives a low gliding angle
• Note: typically C D = C D min +
2 L
C πA Re
where
is the zero lift (friction/parasitic) drag
– C D min – C 2
lift induced drag
gives the L – e is Oswald’s efficiency factor ≈ 0.7 − 0.85
Fall 2004 16.333 1–4
• Total drag then given by
11 �� D = ρV 2SC D = ρV 2S L
(3)C D min + kC 2
22 1(mg )2 =2 ρV 2SC D min + k 1 ρV 2
S
(4)
2
• So that the speed for minimum drag is
�� �1/4
2mg
k V min drag =
ρS C D min
•Equations:
T−D−W sinγ=0(5)
L−W cosγ=0(6)�which gives
L
cosγ=0
T−D−
sinγ
so that
T−D
tanγ=
L
•Consistent with1–3if T=0since thenγas defined above is negative •Note that for smallγ,tanγ≈γ≈sinγ
R/C=V sinγ≈Vγ≈(T−D)V
L
so that the rate of climb is approximately equal to the excess power available(above that needed to maintain levelflight)
•Equations:
L sinφ=centrifugal force(7)
mV2
=
R
(8)
L cosφ=W=mg(9)
⇒tanφ=V2
Rg
V=Rω
=
Vω
g
(10)
•Note:obtain R min at C L
max
R min(1
2
ρV2SC L
max
)sinφ=
W V2
g
⇒R min=
W/S
1/2ρgC L
max
sinφmax
where W/S is the wing loading andφmax<30◦
Fall 2004 16.333 1–7
• Define load factor N = L/mg . i.e. ratio of lift in turn to weight
N = sec φ = (1 + tan 2 φ)
1/2
�
(11) tan φ = N 2 − 1
(12)
so that
V 2 V 2
R = g tan φ = g √
N 2 − 1
• For a given load factor (wing strength)
R ∝ V 2
• Compare straight level with turning flight
– If same light coefficient
L mg 1 C L = 1 = ρV 2S ≡ N mg ρV t 2
S
ρV 2S 1 222 so that V t = √
N V gives the speed increase (more lift)
• Note that C L constant C D constant ⇒ D ∝ V 2C D
⇒ ⇒ T t ∝ D t ∝ V t 2C D ∼ N
D so that must increase throttle or will descend in the turn。